问:
Cambridge Admissions Exam Statistics 1999

My work:

Since $f$ is the pdf we must have $\int_{0}^{1} Ax\,\mathrm dx=1 \implies A=2$.  Let $Y$ be the number of currants in my portion. We have $Y\sim B(4,x)$. For the expectation
$$E(Y)=4x,$$
however I don't know how to continue. I had a thought of using the expectation for $x$ in this, however, I can't statistically justify it.

For the second part, we require
$$P\left(X\geq\frac{1}{2}\mid Y=4\right)=\frac{P\left(X\geq\frac{1}{2},Y=4\right)}{P(Y=4)}.$$

I am completely stuck on how to approach this. I checked the student room for solutions but they seem to disagree with themselves and with other solutions on other websites. If anybody could help me in understanding this problem, and the technique required for it I would be really thankful.

答:
The probability that a currant is in proportion x is x.  Probability all 4 are in that portion is x^4.  Integral of this is x^5/5.

So the integral with limits 0.5 to 1 is the probability of getting all 4 if the proportion is greater than 0.5 which is 1/5 - 1/160 = 31/160.  The probability of getting all 4 no matter what proportion was taken is the same integral from 0 to 1 = 1/5.  It's a conditional probability question (not using the Ax distribution).  The answer is then 31/160 divided by 1/5 = 31/32.


问:
Smoothing a collar twist.
I am currently reading an article https://arxiv.org/pdf/1802.02609.pdf, and a question arose.

The setting is at the beginning of section 2. Let $M$ be a closed, oriented, smooth $d$-manifold, and fix an embedded disck $D^d\subset M$ and a colloar neighborhood $[0,1]\times S^{d-1}\subset M\backslash \text{int}(D^d)$ such that $\{1\}\times S^{d-1}$ is identified with $\partial D^d$. Let $\gamma\in \Omega SO(d)$ and define a path $\lambda:I \to \text{Homeo}(M)$ by
$$ \lambda(t)(m)=\begin{cases} \gamma(t)m & m\in D^d \\ (s,\gamma(s)x) & m=(s,x)\in [0,t]\times S^{d-1} \\ (s,\gamma(t)x) & m=(s,x)\in(t,0]\times S^{d+1} \\ m & \text{otherwise}\end{cases}
$$
The autor says, that one can then use a suitable family of bump functions to make it into a path which takes values in $\text{Diff}^+(M)$ the topological group of orientation-preserving diffeomorphisms. Furthermore, this should be done such that the endpoints of $\lambda$ are preserved.

The idea i had was to use a standard partition of unity argument for a fixed covering of $[0,1]\times S^{d-1}$ as in Lee's book in the section for Whitney approximation theorem. Namely, one embeds $M\subset \mathbb{R}^n$ and gets a tubular neighborhood $U$. We can then use the POU, to make that path into a path $I\to C^\infty(M,U)$ and postcomposing with the projection $U\to M$ yields pointwise a smooth map $M\to U \to M$. However, doesn't this only yield a path $I\to C^\infty(M,M)$? Futhermore, this doesn't fix the endpoints. Any help or ideas would be appreciated.

Edit: As mentioned in the comments of the accepted answer, one can assume that $\gamma$ is a smooth loop for the purpose in the article. Moreover, i don't think it is possible to give a nice solution to the question as asked without this assumption if there even exists one.

答:
It seems that a more straightforward approach is available; the interpolation doesn't require anything as involved as a partition of unity.

Choose a smooth function $\psi:\mathbb{R}\to\mathbb{R}$ satisfying $\psi(x)=0$ for $x<1/4$, $\psi(x)=1$ for $x>3/4$. We can then define the path $\lambda$ by
$$
\lambda(t)(m)=\begin{cases}
\gamma(t)m & m\in D^d \\
(s,\gamma(t\psi(s))x) & m=(s,x)\in[0,1]\times S^{d-1} \\
m & \text{otherwise}
\end{cases}
$$
This path only agrees with the original definition outside of the collar (necessarily, since the endpoints may fail to be smooth on the collar as originally defined), but it should still have all of the desired properties.

ContentLicense: CC BY-SA 4.0

Title: Smoothing a collar twist.


问:
demonstrate the following argument
argument to be demonstrated:

$A ∨ (B ∧ C)$

$B → D$

$C → E$

$D ∧ E → A ∨ C$

$¬A$

$∴ C$

My attempt at this mathematical demonstration was as follows:

$A ∨ (B ∧ C)$

$≡ (A ∨ B) ∧ (A ∨ C)$

$≡ (A ∨ A) ∧ (B ∨ C)$

$≡ A ∧ (B ∨ C)$

$≡ (A ∧ B) ∨ (A ∧ C)$

答: Note the first and last givens,  you are given "A or" something, and then "Not A"  That immediately gives you that the other part of the not must be true, so you get B and C.   Those then lead to give you D and E from the next two givens.   That then leads to A or C.  But once again, we know A is false, so...</p>

Title: demonstrate the following argument


问:
How can I prove that the l.h.s equals the r.h.s?
I can't move forward.  Can anyone help?

$$\frac{k (2k) + 2 }{ k (k + 1) }= \frac{2k + 2 }{ k + 2}$$

I'm trying to prove that the left side equals the right side.

It started like this

$$\frac{2k }{ k + 1} + \frac{1}{1+2+3+...+(k+1)}= \frac{2(k + 1) }{ (k + 1) + 1}$$

答: Perhaps you made a typo, or it might be a trick question: for $k=1$ we get $\frac {1}{2} = \frac 43$, which is false.


答:
Why would you WANT to "make both sides equal".  That's a very unusual thing to do.

More often people want to "solve for k" or "prove an identity" as Joe suggested.

I would start by trying to "solve for k", If it turns out that k can be any number, then it is an identity.

You have $\frac{k(2k+ 1)}{k(k+1)}= \frac{2k+ 2}{k+2}$.
The first thing you can do is cancel the "k" in both numerator and denominator in the left. That leaves $\frac{2k+ 1}{k+ 1}= \frac{2k+ 2}{k+ 2}$.

Eliminate the fractions (I really don't like fractions!) by multiplying both sides by $(k+ 1)(k+ 2)$
$(2k+ 1)(k+ 2)= (2k+ 2)(k+ 1)$
$2k^2+ 5k+ 2= 2k^2+ 4k+ 2$
Subtract $2k^2+ 4k+ 2$ from both sides to get
k^2= 0.

NO, this is not an identity.  It is only true for k= 0.

Title: How can I prove that the l.h.s equals the r.h.s?

Body:
I can't move forward.  Can anyone help?

$$\frac{k (2k) + 2 }{ k (k + 1) }= \frac{2k + 2 }{ k + 2}$$

I'm trying to prove that the left side equals the right side.

It started like this

$$\frac{2k }{ k + 1} + \frac{1}{1+2+3+...+(k+1)}= \frac{2(k + 1) }{ (k + 1) + 1}$$


答: $1+2+\cdot+(k+1)={(k+1)(k+2)\over2}$, not $k(k+1)\over2$. If you make that change in the steps you carried out (in the image attached to the original version of the posted question), you should see that you get an identity.


问:
Does my proof of the irrationality of $\ln(2)$ hold up?
At first, I thought that proving the irrationality of $\ln(2)$ was so easy that it was trivial. However, someone in the comments of the $115$ vote answer at Can an irrational number raised to an irrational power be rational? said that "OTOH, proving that e and log2 are irrational is not trivial." (Ihf said this). Thus, I decided to give it a so to see how difficult it is.

First, we assume that $\ln(2)$ is rational and can thus be written as the fraction $\frac{m}{n}$ with integers $m,n$ and $m$ and $n$ in lowest terms, meaning that $\gcd(m,n)$ $=1$. So we have that $$\ln(2) = \frac{m}{n}  \implies$$
$$2 = e^\frac{m}{n}  \implies$$ $$2^n = e^m $$

Since $n$ is an integer $2^n$ is an integer as well. For $\frac{m}{n}$ to be a valid fraction, we need $m$ to be an integer as well. However, since $m$ must be an integer, $e^m$ cannot be an integer because it is only an integer when we have $e^{\ln(a)}$ for some natural number $a$, but we know that $\ln(a)$ can't be an integer, with exception of $\ln(1) = 0$, but $m$ isn't $0$ because that would make our fraction $0$. Thus $2^n$ is an integer, while $e^m$ isn't, which is a contradiction. Thus our initial assumption that $\ln(2)$ is rational and can be written in lowest terms as $\frac{m}{n}$ is false, meaning that $\ln(2)$ is irrational. This proof seemed too easy, what did I do wrong?

答:
The proof can be streamlined in the following way. Suppose that $\log 2=m/n$ for two coprime integers $m$ and $n$. Then,
$$
e^{m/n}=2 \implies e^m=2^n \, .
$$
Hence, $e$ would be a root of the polynomial
$$
x^m-2^n=0 \, .
$$
But this would imply that $e$ is algebraic, which is untrue. Hence, $\log 2$ is irrational.


问:
Using the chain rule to differentiate composite functions (exponential of an exponential)
How would I differentiate the following two functions using the chain rule.

$t=(e^{2y})^{e^{y^2}+1}$

$t=e^{{(y^3-\ln(y)+1)}^5}$.

Body:
hint

For the first:

$$t=e^{e^{y^2+1}\ln(e^{2y})}$$

$$=e^{2ye^{y^2+1}}=e^{f(y)}$$

the derivative is

$$f'(y)e^{f(y)}$$

with
$$f(y)=2ye^{y^2+1}=2g(y)e^{h(y)}$$
and
$$f'(y)=2\Bigl(g'(y)+g(y)h'(y)\Bigr)e^{h(y)}$$


Title: Using the chain rule to differentiate composite functions (exponential of an exponential)

Body:
How would I differentiate the following two functions using the chain rule.

$t=(e^{2y})^{e^{y^2}+1}$

$t=e^{{(y^3-\ln(y)+1)}^5}$.

Body:
$t=e^{(y^3−log(y)+1)^5}$

Here $t=e^{f(y)}$ and so $t' = e^{f(y)}\cdot f'(y)$.

Above $f(y) = (y^3−log(y)+1)^5$ and so $f'(y) = 5(y^3−log(y)+1)^4\cdot(3y^2 -\frac{1}{y})$.


答:
I don't know what book you are using, but the rule to differentiate $k^m$ is just a combination of the exponential rule and the power rule:

$$ d(k^m) = mk^{m-1}\,dk + \ln(k)\,k^m\,dm $$

So, for the first one, just set $k = e^{2y}$ and $m = e^{y^2+1}$.
This means that $dk = 2e^{2y}\,dy$ and $dm = 2y\,e^{y^2 + 1}\,dy$.

Putting those in gives:

$$ (e^{y^2+1})^{e^{2y} - 1} e^{2y}\frac{dy}{dt} + \ln(e^{2y}) (e^{2y})^{e^{y^2+1}}(2y\,e^{y^2 + 1})\frac{dy}{dt} $$

You can use the same process for the second one as well.


问:
What does the word ''root'' really mean?
So I have been asked this question:

Suppose $$\theta = 0.5\pi + \sin(\theta).$$
Verify by calculation that $\theta$ lies between $2.2$ and $2.4.$

What my book has done is put all the terms on the left-hand site and replace the $0$ on the right-hand side with $f(\theta)$.

My question is, why am I not asked to find the root of $\theta$? Why am I asked to find $\theta$? And from the function what happens with the $0$?

答:
"Roots" refer to the values that make a polynomial equal to zero.  For example, if $p(x)=x^2-2x+1$ is your polynomial, then to find the roots you set up the equation $x^2-2x+1=0$ and find values of $x$ that solve it.

If you have a pre-existing equation, you don't call solutions to it "roots."

Roots of polynomials are important because they have to do with factoring.  In the above polynomial, $x=1$ is a root, and with some more work we can factor it as $(x-1)^2$.

(Note: it's reasonably common to talk about roots of a function as well, but at least in my experience it's more common for non-polynomial functions to call them "zeros.")


问:
Show that if $\lim f(x)=L$ then $\lim|f(x)|=|L|$. And, consequently if$ f(x)$ is continuous, then $|f(x)|$ is also continuous.
Since $\lim f(x)=L$ then I know that I can use the straight definition of continuity with $|x-c|$, but I'm confused as to how the absolute value can be added in to be shown. </p>

答:
Let $A\subset \mathbb R$, let $f:A\rightarrow \mathbb R$ and let $c\in \mathbb R$ be a cluster point of $A$. If $\lim_{x\to c}f$ exists, and if $|f|$ denotes the function defined for $x\in A$ by $|f|(x):=|f(x)|$, then

$$\forall \varepsilon > 0 \exists \delta\left(\varepsilon \right)>0,0<|x-c|<\delta\left(\varepsilon\right)\implies |f(x)-L|<\varepsilon$$

But if $$w,z\in \mathbb R\implies ||w|-|z||\leq|w-z|$$
Then $f(x),L\in \mathbb R$
$$||f(x)|-|L||\leq |f(x)-L|<\varepsilon$$
$$\therefore \lim_{x\to c}|f|=|L|=\left|\lim_{x\to c}f\right|$$
If $|L|=|f(c)|$ then $|f(x)|$ is continuous in point $c$.

ContentLicense: CC BY-SA 3.0


问:
Find the $3 \times3$ matrix (all non-zero entries) that has eigenvalues $\lambda = 1, \lambda = 3, \lambda = 5.$

Find the $3 \times3$ matrix (all non-zero entries) that has eigenvalues $\lambda = 1, \lambda = 3, \lambda = 5.$

I'm having trouble figuring out how to approach this question. I was thinking of using the characteristic polynomial, but not entirely sure.

答:
Copying an answer from the comments to take this off the unanswered queue.

Start with $D = \left[ \matrix{1 & 0 & 0 \\0 & 3 & 0 \\0 & 0 & 5} \right]$ and experiment with different invertible $3 \times 3$ matrices $P$, producing matrices $P D P^{-1}$ which will all have eigenvalues $1$, $3$, $5$. In all likelihood you will stumble upon a suitable matrix $P$ in a few tries if not faster.


问:
Evaluate $\frac{4}{\sin^2 20^\circ} - \frac{4}{\sin^2 40^\circ} + 64\sin^2 20^\circ$
Evaluate the following expression:$$\frac{4}{\sin^2 20^\circ} - \frac{4}{\sin^2 40^\circ} + 64\sin^2 20^\circ$$
I tried combining the whole into a single fraction and using double-angle identity, product/sum to sum/product, but it didn't work.

Body: Formulae for $\sin4x,\,\sin5x$ provide another solution. Let $S:=\sin^220^\circ\ne\sin^260^\circ=\tfrac34$ so$$\begin{align}0&=\frac{\sin^2100^\circ-\sin^280^\circ}{S}\\&=(5-20S+16S^2)^2-16(1-S)(1-2S)^2\\&=256S^4-576S^3+432S^2-120S+9\\&=(4S-3)(64S^3-96S^2+36S-3)\\\implies64S^3-96S^2+36S-3&=0\\\implies\frac{4}{S}-\frac{1}{S(1-S)}+64S&=\frac{3-4S+64S^2-64S^3}{S(1-S)}\\&=32.\end{align}$$


问:
Combinatorial interpretation of factorization identity
For $a,b,c\in\mathbb{N}$, it is the case that $a^{bc}-1=(a^b-1)(1+a^b+a^{2b}+\cdots+a^{b(c-1)})=(a^c-1)(1+a^c+a^{2c}+\cdots a^{c(b-1)})$. It is perhaps slightly more suggestive to write this as $a^{bc}-1=(a^b-1)\displaystyle{\sum_{n<c}a^{bn}}=(a^c-1)\displaystyle{\sum_{n<b}}a^{cn}$

On its own, this identity gives factorization information about natural numbers of this specific form. For example, $63=2^6-1=(2^2-1)(1+2^2+2^4)=3\times 21=(2^3-1)(1+2^3)=7\times 9$.

My question is: Is there a way to understand this factorization in terms of functions? The total number of functions $f:b\times c\to a$ equals $a^{bc}$, so the number of non-zero functions is $a^{bc}-1$. As a special case, we can take $a=2$ to interpret things as subsets so that the number of non-empty subsets $\varnothing\subsetneq X\subseteq p\cdot q\in\mathbb{N}$ is equal to $2^{pq}-1=(2^{p}-1)\displaystyle{\sum_{n<q}2^{p\cdot n}}$.

Evidently, the non-empty subsets of $p\cdot q$ correspond to first picking a non-empty subset of $p$ and then picking $n<q$ along with an arbitrary subset of $p\cdot n$. And moreover, this is symmetric in $p$ and $q$. However, I cannot figure out how this actually works out. I have tried writing out a $p\times q$ rectangle for small $p,q$ but the non-empty subsets don't seem to have any obvious properties that split into the necessary factorization (that I can find). In particular, I am having a hard time understanding how the quantity $\displaystyle{\sum_{n<q}a^{n\cdot q}}$ might be given a combinatorial interpretation in the context of functions into $a$.

Thanks!

Edit: Thanks to Mike for the idea to consider which is the rightmost column not all zeros. I realized it's the same thing to consider $\displaystyle{\sum_{k<n}2^k}=2^n-1$ by the correspondence $(k\in n,X\subseteq k)\mapsto (\varnothing\subsetneq  X\cup\{k\}\subseteq n)$

答:
$a^{bc}-1$ is the number of $b\times c$ grids ($b$ rows, $c$ columns) of numbers, each of between $0$ and $a-1$, such that not all the entries are $0$.

$(a^b-1)\cdot a^{bn}$ is the number of binary grids where rightmost nonzero column is column number $n+1$. There are $a^{bn}$ ways to fill the leftmost $n$ columns arbitrarily, then $a^b-1$ ways to fill column number $n+1$ so it is not entirely zero, and then the remaining columns are forced to be zero.

Summing the last quantity over $n=0$ to $c-1$, you get all possible nonzero grids. Of course, these can be thought of as functions from a set of size $bc$ to a set of size $a$.


问:
Are continuous convex functions subharmonic?
We say that a continuous function $u:\mathbb{R}^d\to \mathbb{R}$ is subharmonic if it satisfies the mean value property
$$u(x)\leq \frac{1}{|\partial
 B_r(x)|}\int_{\partial B_r(x)}u(y)\,\mathrm{d}y \qquad (\star)$$  for
any ball $B_r(x)\subset \mathbb{R}^d$.

Let $u:\mathbb{R}^d\to \mathbb{R}$ be a convex function (hence,
continuous). Is $u$ subharmonic?

If $u\in C^2(\mathbb{R}^d)$, this is  true. Using a second-order Taylor expansion we have
\begin{align*}\int_{\partial B_r(x)}(u(y)-u(x))\,\mathrm{d}y&=\int_{\partial B_r(x)}\left(\nabla u(x)\cdot(y-x)+\frac{1}{2}(y-x)D^2u(\xi)(y-x)^t\right)\,\mathrm{d}y.\end{align*}
The first term in the above integral vanishes by symmetry, the second is non-negative because $D^2u(\xi)$ is a positive semi-definite matrix. Therefore, ($\star$) is proven.

If $d=1$, the statement is true when $u$ is continuous, in general. Indeed since balls reduce to intervals, ($\star$) is easily shown to be equivalent to $u$ being midpoint-convex.

I'm not sure how to attack the problem in higher dimensions. Of course $(\star)$ is true for affine functions in any dimension, and I'd like to use the fact that the graph of a convex function lies below that of an affine function, loosely speaking. However, to close the estimate I would need $u$ to be equal to the affine function at the boundary of the ball, and this is not necessarily possible.

答:
Martin gives a very reasonable answer extending the question's second argument; here's how to extend the first one.

If $u: \mathbb{R}^{d} \to \mathbb{R}$ is continuous and convex, then mollification of $u$ gives a family of smooth functions $(u^{\epsilon})_{\epsilon > 0}$ such that $u^{\epsilon} \to u$ locally uniformly in $\mathbb{R}^{d}$.  (Recall that mollification means that we define $u^{\epsilon} = \rho^{\epsilon} * u$, where $\rho^{\epsilon}(x) = \epsilon^{-d} \rho(\epsilon^{-1} x)$ and $\rho \in C^{\infty}_{c}(\mathbb{R}^{d})$ is a non-negative, even function with $\int_{\mathbb{R}^{d}} \rho(x) \, dx = 1$.)

It is not hard to show that, for each $\epsilon > 0$, $u^{\epsilon}$ is convex.  Indeed, if $x, y \in \mathbb{R}^{d}$ and $\lambda \in [0,1]$, then
\begin{align*}
u^{\epsilon}((1- \lambda)x + \lambda y) &= \int_{\mathbb{R}^{d}} u((1 - \lambda)x + \lambda y - \xi) \rho^{\epsilon}(\xi) \, d \xi \\
&\leq \int_{\mathbb{R}^{d}} ((1 - \lambda) u(x - \xi) + \lambda u(y - \xi)) \rho^{\epsilon}(\xi) \, d \xi \\
&= (1 - \lambda) u^{\epsilon}(x) + \lambda u^{\epsilon}(y).
\end{align*}
Thus, $u^{\epsilon}$ is subharmonic.

Now if $x \in \mathbb{R}^{d}$ and $r > 0$, then
\begin{equation*}
u(x) = \lim_{\epsilon \to 0^{+}} u^{\epsilon}(x) \leq \lim_{\epsilon \to 0^{+}} \frac{1}{|\partial B_{r}(x)|} \int_{\partial B_{r}(x)} u^{\epsilon}(y) \, dy = \frac{1}{|\partial B_{r}(x)|} \int_{\partial B_{r}(x)} u(y) \, dy.
\end{equation*}
Therefore, $u$ is also subharmonic.


答:
I do not agree with the previous solution.

The conditional probability is a binomial

$$(Y|X=x)\sim B(4;x)$$

with $E(Y|X=x)=4x$ thus the expectation of Y is

$$E(Y)=E(E(Y|X))=E(4X)=4E(X)=4\cdot\frac{2}{3}=\frac{8}{3}$$

With standard reasoning the joint distribution of $(X,Y)$ is

$$f_{XY}(x,y)=f_X(x)f_{Y|X}(y|x)=2\binom{4}{y}x^{y+1}(1-x)^{4-y}$$

Integrating (easy, using Beta distribution) you get the marginal Y

$$f_Y(y)=\int_0^1 f(x,y)dx=\frac{y+1}{15}$$

where $y=0,1,2,3,4,$

you can check that $E(Y)=\frac{8}{3}$ as calculated before in a different way.

Now having $f(x,y)$ and $f(y)$ you can calculate the conditional probability they request. It results to me

$$P(X\geq 0.5|Y=4)=\frac{\int_{1/2}^{1}2x^5dx}{\frac{5}{15}}=\frac{63}{64}$$

Title: Cambridge Admissions Exam Statistics 1999

Body:
My work:

Since $f$ is the pdf we must have $\int_{0}^{1} Ax\,\mathrm dx=1 \implies A=2$.  Let $Y$ be the number of currants in my portion. We have $Y\sim B(4,x)$. For the expectation
$$E(Y)=4x,$$
however I don't know how to continue. I had a thought of using the expectation for $x$ in this, however, I can't statistically justify it.

For the second part, we require
$$P\left(X\geq\frac{1}{2}\mid Y=4\right)=\frac{P\left(X\geq\frac{1}{2},Y=4\right)}{P(Y=4)}.$$

I am completely stuck on how to approach this. I checked the student room for solutions but they seem to disagree with themselves and with other solutions on other websites. If anybody could help me in understanding this problem, and the technique required for it I would be really thankful.


问:
Finding tangent planes using differential geometry "language"
I'm studying differential geometry and I have a couple of questions (which are related).

I'm supposed to find the tangent planes to the surface given by $x^2+y^2-z^2=1$ at points $(x,y,0)$.

Is it enough to consider only the "upper" part of this hyperboloid, by using $\varphi(x,y)=(x,y,\sqrt{x^2+y^2-1})$ and work it as I would with the graph of functions?

If so, why? My reasoning was that even if I had to "complete" the surface with the lower part of the hyperboloid,  $\tilde{\varphi}(x,y)=(x,y,-\sqrt{x^2+y^2-1})$, the tangent planes would have to agree at $(x,y,0)$ so I can choose either one. I guess this would imply that it is sufficient to investigate tangent planes in any local parametrization that contains the points of interest.

Supose I can proceed as stated. Then I know how to find the tangent plane as I used to do in multivariable calculus, but I was also given the following deffinition:

Given a local parametrization $\varphi: U \subset \mathbb{R}^2 \to S$ where $S$ is a regular surface and a point $p \in \varphi(U)$. The tangent plane to this point is :

$$T_pS=D\varphi(\varphi^{-1}(p))(\mathbb{R}^2)$$

Which leads me to my next problem:

I'm having a hard time understanding the differential notation.

I tried to apply this to $\varphi(x,y)=(x,y,\sqrt{x^2+y^2-1})$ at points $p=(x_0,y_0,0)$.

As far as I'm concerned, $\varphi^{-1}(p)=(x_0,y_0)$ and

$$
D\varphi((x_0,y_0))=\begin{bmatrix}
1 &0 \\ 0& 1 \\ \dfrac{x_0}{\sqrt{x_0^2+y_0^2-1}} &\dfrac{y_0}{\sqrt{x_0^2+y_0^2-1}}\\
\end{bmatrix}
$$

I don't know what to do next. What does the $\mathbb{R^2}$ has to do in the definition of the tangent plane, and how do I get to the tangent plane expression from here?

This was a long question but I would really appreciate anyone taking the time to help me here :´)

答:
I agree with Trandus's answer, but here is a slightly different version.

Question 1: Your $\varphi$ is differentiable only when $x^2 + y^2 > 1$; this shows up when you examine the denominators in your calculation of $D\varphi$. It would be fine to use it, were the question not asking for the tangent plane at points $(x,y,0)$ which exactly have $x^2 + y^2 =1$. In other words, the question is asking for tangent planes at points where that plane is vertical, so it won't be okay to use a function of $(x,y)$ to describe them.

You could find another parametrization by describing your surface as a graph over, say, the $(y,z)$ plane: solving the given equation for $x$ gives $x = \pm\sqrt{1 + z^2 - y^2}$, and then defining $\psi(y,z) = (\sqrt{1 + z^2 -y^2}, y, z)$ is a parametrization you can use to find tangent planes at points $(x,y,0)$ with $x >0$. For $x < 0$ choose the other sign for the square root, and if $x= 0$ do a similar construction solving for $y$ instead.

Question 2. As Trandus says, the columns in the differential of $\psi$ will give a basis for the tangent plane. If you prefer to find an equation of the plane rather than a basis, another method is to remember that the gradient of a function is always normal to the level sets. Looking at the function $f(x,y,z) = x^2 + y^2 - z^2$, your surface is the level set $f(x,y,z) = 1$. So compute that the gradient is $\nabla f = \langle 2x, 2y, -2z\rangle$, which gives you a normal vector to the desired plane. At the point $(a,b,0)$ this becomes $\langle 2a, 2b, 0\rangle$, and the plane with this normal vector is given by the equation
$$
2a(x-a) + 2b(y-b) = 0.
$$
Note that in ``differential geometry language'', the fact that $\nabla f$ is a normal vector for the tangent to the level set is expressed by the fact that the tangent space of the level set of $f$ at a point $p$ is equal to the kernel of the differential of $f$ at $p$.


问:
Show $(a_{1_n}, a_{2_n}, \ldots, a_{k_n}) \to (a_1, a_2, \ldots a_k) $ if $\lim_{n \to \infty} a_{j_n} = a_j$ for $1 \le j \le k$
For any $j$ and $\epsilon > 0$, there is some $N_j$ s.t. $n \ge N_j \implies |a_{j_n} - a_j| < \epsilon.$ If we let $\epsilon = \sqrt k$, then $k \ge 1 \iff \sqrt k \ge 1 \iff |a_{j_n} - a_j| < 1$. Then if $N = \max\{N_i\}_{i=1}^k$, we have for any $n \ge N$, $$|(a_{1_n}, a_{2_n}, \ldots, a_{k_n}) - (a_1, a_2, \ldots a_k)|  \\ = \sqrt{\sum_{j=1}^k|a_{j_n} - a_j|^2} \\ < \sqrt{\sum_{j=1}^k(1)^2} \\ = \sqrt k = \epsilon$$

To me this looks like it works. But we want the inequality above to hold for arbitrary $\epsilon > 0$, yet I defined $\epsilon:= \sqrt k.$ If the proof above is not correct, how can I fix it? Thanks.

EDIT:

Following @Alan's answer below, the proof above is fixed as follows:

For any $j$ and $\epsilon > 0$, there is some $N_j$ s.t. $n \ge N_j \implies |a_{j_n} - a_j| < \epsilon.$ Then if $N = \max\{N_i\}_{i=1}^k$, we have for any $n \ge N$, $$|(a_{1_n}, a_{2_n}, \ldots, a_{k_n}) - (a_1, a_2, \ldots a_k)|  \\ = \sqrt{\sum_{j=1}^k|a_{j_n} - a_j|^2} \\ < \sqrt{\sum_{j=1}^k\left(\frac{\epsilon}{\sqrt k}\right)^2} \\ = \sqrt{\sum_{j=1}^k\frac{\epsilon^2}{k}}  = \epsilon$$

edit:

What's up with the downvote, though? We are not allowed to make mistakes on this site? If I knew the proof, I wouldn't have come here. LOL.

答:
Given $\varepsilon >0$, there exist naturals $N_1, N_2\cdots, N_n$ such that for each $j\in\{1,2,\cdots,n\}$,
$$|a_{j_n} - a_j| < \frac{\varepsilon}{\sqrt{k}}\: \: \forall\: n\geqslant N_{j}.$$
Now, it immediately follows that
$$\sqrt{\sum_{j=1}^k|a_{j_n} - a_j|^2}<\varepsilon\:\: \forall \:n\geqslant \underset{j}{\max}N_j.$$
The proof is complete.


问:
Finding the area of this figure inside a circle?
Let's say I have a unit circle and I draw a triangular sector with an angle $\pi/2$. Next to it, I draw a sector with an angle $\pi/8$, next $\pi/18$, so that the $n^{\text{th}}$ sector has an angle of $\pi/(2n^2)$. Now I am only looking at the area of the triangle inside each sector. What is the area of the triangles if I were to draw infinitely many of them, as shown in the figure below?

I was only able to find an approximation for this. I got $f(1)/2 + \int_1^\infty f(x)dx$, where $f(x)=(1/2)\sin(\pi/(2x^2)) \approx 1.95$. How would you find the exact area of this?

答:
Area of a sector with angle $\theta$ is $\frac{\theta}{2}$ (Can you see this?).

Now, in your case, the area is given by

$ A = \sum_{x=1}^{x=\infty}\frac{\pi}{4x^{2}} = \frac{\pi}{4}\sum_{x=1}^{\infty}\frac{1}{x^{2}}$

Do you know the value of the last sum? It is $\frac{\pi^{2}}{6}$.


问:
Show that every group of order 15 is cyclic using class equation.
I am willing to show that every group of order 15 is cyclic, using class equation.

Let $G$ be a group of order 15. If $G$ is abelian, then $G=Z(G)$ and so for each $a\in G=Z(G)$ we have $cl(a)=\{a\}$. Hence only possible class equation is
$$15=1+1+1+\cdots+1+1(15~\text{times})$$
In this case, $G$ is isomorphic to either $\mathbb{Z}_{15}$ or the external direct product $\mathbb{Z}_3\times \mathbb{Z}_5$. But since $\mathbb{Z}_{15}\simeq \mathbb{Z}_3\times \mathbb{Z}_5$, it follows that $G$ is cyclic.

We now show if $G$ is non-abelian then contradiction will appear.

When $G$ is non-abelian, then $G\neq Z(G)$. Here $|Z(G)|\in \{1,3,5\}$. But if $|Z(G)|=3$ then $|G/Z(G)|=5$ a prime, so $G/Z(G)$ is cyclic and hence $G$ becomes Abelian, contradiction.

Similalry $|Z(G)|\neq 5$ as well and so only possibility is $|Z(G)|=1$. Then the class equation reads
$$15=|G|=|Z(G)|+\sum |cl(a)|=1+\sum |cl(a)|$$
where the sum is taken over the orders of all non-singleton conjugacy classes $cl(a)$ in $G$.

Let there be $x_3$ and $x_5$ number of conjugacy classes of order 3 and 5 respectively in $G$. Then we must have
$$15=1+3x_3+5x_5\Rightarrow 14=3x_3+5x_5$$
which is satisfied by $x_3=3, x_5=1$ so that ultimately the class equation  becomes
$$15=1+(3+3+3)+5$$

I do not know how to bring contradiction here. Any help?

Thanks in well advance.

答:
Why $15 =1 + 3 + 3 + 3 + 5$ cannot be the class equation of group of order 15.

We know that number of elements of order $5$ can be $4,8,12$,..

And

Number of elements of order $3$ can be $2,4,6$,..

So the only way, the given class equation would make sense is by having $8$ elements of order $5$ which fit in class of sizes ($3,5$) and $6$ elements of order 3 which fit in class of sizes ($3,3$) ..

BUT ,if that's the case , since there are $8$ elements of order $5$, there are $2$ subgroups of order $5$ (H and K ) .. then HK order will be $25$, which is not possible (because HK is subset of G, G contains $15$ elements)


答:
The union of the sectors should be a sector of angle
$$\sum_{n=1}^\infty\frac{\pi}{2n^2}=\frac{\pi}{2}\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi}{2}\cdot\frac{\pi^2}{6}=\frac{\pi^3}{12}$$
and the area of a sector of angle $\theta$ in a circle of radius $r$ is $\frac{\theta}{2\pi}\cdot\pi r^2$, so the area of the union of the sectors is
$$\frac{\pi^3/12}{2\pi}\cdot\pi=\frac{\pi^3}{24}$$</p>

Title: Finding the area of this figure inside a circle?

Body:
Let's say I have a unit circle and I draw a triangular sector with an angle $\pi/2$. Next to it, I draw a sector with an angle $\pi/8$, next $\pi/18$, so that the $n^{\text{th}}$ sector has an angle of $\pi/(2n^2)$. Now I am only looking at the area of the triangle inside each sector. What is the area of the triangles if I were to draw infinitely many of them, as shown in the figure below?

I was only able to find an approximation for this. I got $f(1)/2 + \int_1^\infty f(x)dx$, where $f(x)=(1/2)\sin(\pi/(2x^2)) \approx 1.95$. How would you find the exact area of this?


问:
Is there any bijective mapping between a non-convex polygon and a convex polygon?
Given an arbitrary non-convex polygon (2d) with $n$ vertices, is there any procedure by which to map it to a single convex polygon, and then back again? Maybe operating somehow on the angles between edges to map the image to the right (below) to the image to the left? If so, is there any generalization to polytopes (>2d)? Any guidance is appreciated.

答:
If both polygons have a similar triangulation, i.e. can be decomposed to the same number of nondegenerate triangles with the same connectivity graph, then yes.

(And such mapping is trivial, as it is a set of mappings between corresponding triangles.)

Whether this is a necessary (and not just sufficient) condition, I do not know.

This extends trivially to higher dimensions. In 2D, each cell in the grid is a 2-simplex, i.e. triangle; in 3D, a 3-simplex, i.e. a tetrahedron; in 4D, a 4-simplex, i.e. a 5-cell, and so on.

In the example case, the two have a similar triangulation using three triangles, giving a trivial bijective mapping between the two:

When morphing between images, the shapes are covered by a (triangular) grid where grid vertices are at similar details, and the mapping itself is usually bicubic (using coordinates of more than one grid cell, so that grid cells are less visible in the morphed output).

Edited to describe the affine (linear) map between triangles:

An affine map between two triangles is trivial, by using barycentric coordinates $(u, v, w)$: If $\vec{p}_u$, $\vec{p}_v$, and $\vec{p}_w$ are the vertices of the triangle, then barycentric coordinates $(u, v, w)$ correspond to point $u \vec{p}_u + v \vec{p}_v + w \vec{p}_w$.

By definition, $u + v + w = 1$.  Point $(u, v, w)$ is within the triangle if and only if $0 \le u \le 1$, $0 \le v \le 1$, and $0 \le w \le 1$.

Because $u + v + w = 1$ by definition, one of them (usually $w$, via $w = 1 - u - v$) is left out for simplicity.

If we have a triangle with vertex $\vec{p}_0 = (x_0 , y_0)$, and the other two vertices at $\vec{p}_u = (x_0 + x_u , y_0 + y_u)$ and $\vec{p}_v = (x_0 + x_v , y_0 + y_v)$, then
$$\left\lbrace\begin{aligned}
x(u, v) &= x_0 + u x_u + v x_v \\
y(u, v) &= y_0 + u y_u + v y_v \\
\end{aligned}\right. \quad \iff \quad \left\lbrace\begin{aligned}
u(x, y) &= \displaystyle \frac{(x - x_0)v_y - (y - y_0)v_x}{u_x v_y - u_y v_x} \\
v(x, y) &= \displaystyle \frac{u_x (y - y_0) - u_y (x - x_0)}{u_x v_y - u_y v_x} \\
\end{aligned}\right.$$
In a computer program, you'll want to use form
$$\left\lbrace\begin{aligned}
u_i &= (x - X_i) A_i + (y - Y_i) B_i \\
v_i &= (x - X_i) C_i + (y - Y_i) D_i \\
\end{aligned}\right.$$
for each source triangle $i$, the correct one being the one for which $0 \le u_i \le 1$, $0 \le v_i \le 1$, $0 \le u_i + v_i \le 1$.  The six constants ($X_i$, $Y_i$, $A_i$, $B_i$, $C_i$, $D_i$) are calculated from the three vertices of triangle $i$, $(x_0, y_0)$, $(x_1, y_1)$, and $(x_2, y_2)$, via
$$\begin{aligned}
X_i &= x_0 \\
Y_i &= y_0 \\
A_i &= \displaystyle\frac{y_2 - y_0}{(x_1 - x_0)(y_2 - y_0) - (y_1 - y_0)(x_2 - x_0)} \\
B_i &= \displaystyle\frac{x_0 - x_2}{(x_1 - x_0)(y_2 - y_0) - (y_1 - y_0)(x_2 - x_0)} \\
C_i &= \displaystyle\frac{y_0 - y_1}{(x_1 - x_0)(y_2 - y_0) - (y_1 - y_0)(x_2 - x_0)} \\
D_i &= \displaystyle\frac{x_1 - x_0}{(x_1 - x_0)(y_2 - y_0) - (y_1 - y_0)(x_2 - x_0)} \\
\end{aligned}$$
This is numerically stable and accurate with floating-point numbers (noting that it will fail for degenerate triangles, because the denominator will be zero), and fast enough to find the correct source triangle when given an arbitrary point $(x, y)$ that may or may not be inside the polygon.

When the source triangle $i$ is known, the inverse is trivial. The point corresponding to $(u, v)$ in the target triangle with vertices at $(x_0, y_0)$, $(x_1, y_1)$, $(x_2, y_2)$, is
$$\left\lbrace\begin{aligned}
x &= (1 - u - v) x_0 + u x_1 + v x_2 \\
y &= (1 - u - v) y_0 + u y_1 + v y_2 \\
\end{aligned}\right.$$
which is also numerically stable for points within the triangle.

Therefore, for each triangle, you'll need to store the vertex coordinates and the four helper constants, for a total of ten floating-point numbers (scalars) per triangle.

Whenever you have two corresponding triangles (no matter which is source and which target), choose the base vertex so that it is closest to a right angle in both triangles, to maximize numerical stability.

While there probably is an algorithm for obtaining a similar triangulation for two arbitrary polygons, I personally do not know it; hopefully others will pipe up and suggest some.


问:
Tricky trigonometric equation
I stumbled upon this trigonometric equation:
$$2\cos^2(x)-5x+2=0$$
for $0<x<2\pi$, and I don't know how to solve it analytically (the task explicitly says not to use a calculator).

I know that you can rearrange the question like this:
$$2\cos^2(x)-5x+2=0$$
$$2\cos^2(x)-1=5x-3$$
$$\cos(2x)=5x-3$$

but I don't know how this should be useful.

My textbook says that the solutions are $\frac \pi 3$ and $\frac {5\pi} 3$. However, these are the solutions to the equation:
$$2\cos^2(x)-5\cos(x)+2=0$$
so I'm sure my textbook made a little mistake there (and I know how to solve such an equation).
Nonetheless I'm curious how to solve the "wrong" equation. Is there even a way to solve it without graphs?

答: Question has been answered, this is simply a comment that I can accept to close this question.</p>

Title: Tricky trigonometric equation


问:
Cambridge Admissions Exam Statistics 1999

My work:

Since $f$ is the pdf we must have $\int_{0}^{1} Ax\,\mathrm dx=1 \implies A=2$.  Let $Y$ be the number of currants in my portion. We have $Y\sim B(4,x)$. For the expectation
$$E(Y)=4x,$$
however I don't know how to continue. I had a thought of using the expectation for $x$ in this, however, I can't statistically justify it.

For the second part, we require
$$P\left(X\geq\frac{1}{2}\mid Y=4\right)=\frac{P\left(X\geq\frac{1}{2},Y=4\right)}{P(Y=4)}.$$

I am completely stuck on how to approach this. I checked the student room for solutions but they seem to disagree with themselves and with other solutions on other websites. If anybody could help me in understanding this problem, and the technique required for it I would be really thankful.

答:
I agree with @tommik.

Since $E(X)=\int_0^12x^2dx=\tfrac23$, the answer to (i) is $\tfrac234=\tfrac83$.

For (ii) let $Y$ denote the number of currants in the portion so $Y|X=x\sim B(4,\,x)$, and $P(Y=4|X=x)=x^4$. Since $P(X\le\tfrac12)=\int_0^{1/2}2xdx=\tfrac14$, the restriction $X>\tfrac12$ has PDF $\tfrac83x$. Hence$$P(X>\tfrac12|Y=4)=\frac{P(Y=4|X>\tfrac12)P(X>\tfrac12)}{P(Y=4)}=\frac{\int_{1/2}^1\tfrac34x^4\tfrac83xdx}{\int_0^12x^5dx}=\frac{[x^6]_{1/2}^1}{[x^6]_0^1}=\frac{63}{64}.$$This slow code agrees to within experimental error, finding $632$ of $640$ random cases with $Y=4$ satisfy $X>\tfrac12$.


问:
Is the Ricci tensor of a symmetric space proportional to the killing form?
In Besse's book on Einstein manifolds, one can read the following theorem

7.73 Theorem: The Ricci curvature of a Riemannian symmetric space satisfies:
$$r=-\frac{1}{2}B_{\vert\mathfrak p}$$

Here $r$ is the Ricci tensor, $B$ is the killing form of $\mathfrak g=T_eG$ (where $G=\text{Isom}_0(M)$ is the connected component of the isometry group of $M$) and $\mathfrak g=\mathfrak k+\mathfrak p$ is the usual decomposition in $\pm1$ eigenspaces. Their proof is just one sentence:

Proof: Since, if $X\in \mathfrak p$, $\text{ad } X$ interchanges $\mathfrak k$ and $\mathfrak p$, we have $$\text{tr}((\text{ad } X\circ \text{ad } Y)_{|\mathfrak p})=\text{tr}((\text{ad } X\circ \text{ad } Y)_{|\mathfrak k})=\frac{1}{2}B(X,Y).$$

I understand that if the first equality holds than the second equality holds as well simply because
$$B(X,Y)=\text{tr}((\text{ad } X\circ \text{ad } Y)_{|\mathfrak p})+\text{tr}((\text{ad } X\circ \text{ad } Y)_{|\mathfrak k}).$$
Also it was shown previously in the book that $r(X,Y)=-\text{tr}((\text{ad } X\circ \text{ad } Y)_{|\mathfrak p})$ so the proof would be complete. However I don't understand why the first equality holds. I've looked in other textbook such as Kobayashi and Nomizu or Helgason but I haven't seen this result mentioned anywhere. Can somebody explain to me why this equality holds ?

答:
I'm not sure exactly what Besse had in mind, but I managed to prove your equality using a canonical inner product on $\mathfrak{g}$. It's possible I missed a way to shorten this to Besse's one-line proof.

Let $\theta_p \colon \mathfrak{g} \to \mathfrak{g}$ be the Cartan involution associated to the decomposition $\mathfrak{g}=\mathfrak{k} \oplus \mathfrak{p}$. There is also an associated inner product on $\mathfrak{g}$ that I'll denote
$$ B_p(X,Y) = -B(\theta_p X,Y) .$$
For $X \in \mathfrak{p}$, $\mathrm{ad} X$ is symmetric relative to the inner product $B_p$:
$$ B_p( \mathrm{ad}X (Y),Z) =  B_p( Y, \mathrm{ad}X(Z)) .$$
Denoting the transpose corresponding to $B_p$ by $(\cdot)^T$, we have that $\mathrm{ad} X |_{\mathfrak{k}}^T= \mathrm{ad} X |_{\mathfrak{p}}$.

Since $\mathrm{ad} X$ and $\mathrm{ad}Y$ interchange $\mathfrak{k}$ and $\mathfrak{p}$, we have that $(\mathrm{ad} X \circ \mathrm{ad}Y) |_\mathfrak{p} = \mathrm{ad} X |_{\mathfrak{k}} \circ \mathrm{ad}Y |_\mathfrak{p}$ which has the same trace as $ (\mathrm{ad}Y |_\mathfrak{p} \circ \mathrm{ad} X |_{\mathfrak{k}})^T = \mathrm{ad} X |_{\mathfrak{k}}^T \circ\mathrm{ad}Y |_\mathfrak{p}^T = \mathrm{ad} X |_{\mathfrak{p}} \circ\mathrm{ad}Y |_\mathfrak{k} = (\mathrm{ad} X \circ\mathrm{ad}Y) |_\mathfrak{k}$.


问:
Finding the intersection of a cubic parametric spline with a rotated ellipsoid
Given a parametric cublic spline in 3D $\vec{S}(t)$ and the position of an object $\vec{P}$ both in earth-centered, earth-fixed cartesian coordinates (ECEF), I would like to find the intersections of a rotated ellipsoid around the object (around its center of gravity). The ellipsoid is supposed to be rotated s.t. its axis match the axis of the body-fixed coordinate system called k.

My approach is:
To find the intersections, the general ellipsoid equation must be solved

$$ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} -1 = 0 $$

where a, b and c denote the semiaxis lenghts. Given the 3D spline and 3D position, to find the intersections

$$\vec{S}(t) - \vec{P}$$

is applied s.t. x, y and z become:

$$ x_{ECEF}(t) = a_x + b_x \, t + c_x \, t^2 + d_x \, t^3 - P_x $$
$$ y_{ECEF}(t) = a_y + b_y \, t + c_y \, t^2 + d_y \, t^3 - P_y $$
$$ z_{ECEF}(t) = a_z + b_z \, t + c_z \, t^2 + d_z \, t^3 - P_z $$

All the coefficients a,b,c,d and position vector entries are known. In order to rotate the ellipsoid, the two required rorational matrices are: $\underline{R}_{g, ECEF}$ to rotate from ECEF coordiantes to a local (body fixed) north east down (NED) coordinate system and $\underline{R}_{k, g}$  to rotate from this local coordinate system g to the required coordinate system k. All entries of these matrices are known.

The rotations are then performed by multiplying the matrices with the vector containing x,y and z:

$$  \begin{pmatrix}
    x(t) \\
    y(t) \\
    z(t) \\
    \end{pmatrix}_k = \underline{R}_{k, g} \, \underline{R}_{g, ECEF} \, \begin{pmatrix}
    x(t) \\
    y(t) \\
    z(t) \\
    \end{pmatrix}_{ECEF}   $$

All I did from now on was to simply plug $x_k(t)$, $y_k(t)$, $z_k(t)$ into the ellipsoid equation

$$ \dfrac{x_k(t)^2}{a^2} + \dfrac{y_k(t)^2}{b^2} + \dfrac{z_k(t)^2}{c^2} -1 = 0 $$

and to solve this numerically for $t$, which is not an issue.

I am unsure, if my approach is correct in general and I'm especially unsure about the coordinate system transformations.

Is it true that the ellipsoid will be aligned with the k-coordinate system when I apply the transformation as shown above?

Is the order of rotations apllied correct? Applying all the rotations the other way around (from k to g to ECEF) also works in the bigger context of code I'm using this in which confuses me quite a bit.

Can the resulting $t$ be simply plugged in to $\vec{S}(t)$ in order to find the corresponding intersection in ECEF coordinates ($\vec{S}(t)$ is defined in ECEF coordinates)?

答:
I don't understand all of your formulation. Anyway, I suggest that you counter-rotate the control points of your cubic instead of rotating the ellipsoid. If the ellipsoid is not centered at the origin, you can also translate to make it so. You can also rescale the coordinates to get a sphere.

Then plug the parametric equation of the cubic into the implicit equation of the unit sphere. Now you need to solve

$$(\mathbf a+\mathbf bt+\mathbf ct^2+\mathbf dt^3)^2=1$$ which is a sextic equation in $t$. It "suffices" to solve for $t$.


答:
The resulting shape is a sector of the circle.  It would be helpful to find the central angle of this sector.  This angle is equal to the following sum:

$$\theta=\frac{\pi}{2}+\frac{\pi}{8}+\frac{\pi}{18}+\ldots=\frac{\pi}{2}\left(\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\ldots\right)$$

Now here is a rather astonishing fact: the term inside the parentheses is equal to $\pi^2/6$.  This means we'd like to find the area of the sector of a unit circle ($r=1$) with central angle $\theta=\pi^3/12$.  The area is then given by:

$$A=\frac{1}{2}\cdot r\cdot\theta=\frac{1}{2}\cdot 1\cdot\frac{\pi^3}{12}=\frac{\pi^3}{24}$$


问:
Ridge Regression with two estimators
Given a loss $E$:
$$E = (Y - X\beta)^T(Y - X\beta) + \lambda\left \| \beta \right \|^2$$
The value of $\beta$ that minimizes the loss can be obtained by setting $\frac{\partial E}{\partial \beta} =0$

However I am interested in something like this:

$$E = (Y - X_1\beta - X_2\alpha)^T(Y - X_1\beta - X_2\alpha) + \lambda\left \|\alpha+ \beta \right \|^2$$

Can just solve for $\alpha$ and $\beta$ by setting $\frac{\partial E}{\partial \alpha} =0$ and $\frac{\partial E}{\partial \beta} =0$ so that I get two equations for two variables ? Or there is some other way?

答:
Let
$$\gamma = \begin{pmatrix}
\alpha \\
\beta
\end{pmatrix}$$
and define the matrices
$$\Gamma = \begin{pmatrix}
X_2 \; | \; X_1
\end{pmatrix}, \qquad \Lambda=\begin{pmatrix}
\mathbb{I} \; | \; \mathbb{I} \
\end{pmatrix},$$
where $\mathbb{I}$ is the identity matrix with order equal to the length of $\alpha$ (or $\beta$). Then, you can write your expression as
$$E = (Y - \Gamma\gamma)^\top (Y - \Gamma\gamma) + \lambda ||\Lambda\gamma||^2$$
and find the optimal values of $\gamma$ solving the equation
$$\frac{dE}{d\gamma} = 0.$$

Title: Ridge Regression with two estimators


问:
On the common compact support for a convergent sequence
Let $f\in L^1$. I guess it is true that there are $f_n$ s.t. $f_n\rightarrow f$ in $L^1$ and $\mathcal{F}(f_n)$ has compact support for each $n$.   Can I conclude somehow that there is a compact set $K$ s.t. $supp {\mathcal{F}(f_n)}\subset K$ for all $n>N$ for some N? Here $\mathcal{F}$ denotes Fourier Transform.

答:
Let $g$ be defined on $\mathbb R$ by
$$
  g(x) = e^{-x^2}, \quad \forall x\in  \mathbb R,
  $$
and let $f$ be the inverse Fourier transform of $g$.  Since $g$ is in the
Schwartz space $\mathscr S(\mathbb R)$, we have that $f$ is also in $\mathscr S(\mathbb R)$ and, in particular $f$ lies
in $L^1(\mathbb R)$.

We will show that there is no sequence $\{f_n\}_n$ converging to $f$ in $L^1(\mathbb R)$, and such that the support of all
the Fourier transforms
$\hat f_n$ lie in a single compact set $K$.

Assuming otherwise, we have that $\hat f_n\to \hat f$, at least pointwise,  so one concludes that $\hat f$
vanishes off $K$. However $\hat f=g$, which vanishes nowhere, giving a contradiction.


问:
A hint to the differential equation $xy'' + y' + \dfrac{1}{4\sqrt x}=0$
I am a bit unsure about how one would tackle this differential equation; I ran the problem in Maple and it gives a nice solution.

I am quite new to differential equations of this "sort" so I am asking for some kind of hint how one would approach a solution? Very appreciated.

答:
There are two ways we could approach this problem. Specific to this problem, we can rearrange the equation and integrate directly

$$xy''+y'=(xy')' = -\frac{1}{4\sqrt{x}} \implies xy' = -\frac{\sqrt{x}}{2}+C_1$$

being careful about adding a constant each time we integrate. Assuming $x>0$ we can then divide by $x$ and integrate again

$$y' = -\frac{1}{2\sqrt{x}}+\frac{C_1}{x} \implies y = -\sqrt{x}+C_1\log x +C_2$$

On the other hand, the systematic approach to this problem would be to solve for the homogeneous solution then the particular. The equation

$$xy''+y' = 0 \implies x^2y''+xy'+0\cdot y=0$$

is a Cauchy-Euler equation with the known ansatz $y=x^r$

$$r(r-1)x^r+rx^r+0\cdot x^r= (r^2)x^r = 0$$

The characteristic polynomial has a repeated root at $r=0$, meaning the homogeneous solutions are $1$ and $\log x \cdot 1$. The $\log x$ is tacked on to the solution for the same reason $xe^x$ is the repeated root solution for a constant coefficient differential equation, it is the "eigenvalue" equation for a generalized eigenvector i.e. if I have a generalized eigenvector such that

$$Av^{(2)} = \lambda v^{(2)} + v^{(1)}$$

where $v^{(1)}$ is an eigenvector of $A$ with eigenvalue of $\lambda$, then we have that

$$f(A)v^{(2)} = f(\lambda)v^{(2)}+f'(\lambda)v^{(1)}$$

In other words the repeated root solution in our case would be (with eigenvalue $r=0$)

$$C_1x^r + \frac{d}{dr}(x^r)C_2 = C_1x^r + C_2 \log x x^r = C_1 + C_2\log x$$

From here you can guess that the particular solution is of the form $A\sqrt{x}$ and use undetermined coefficients.


问:
Does $\int_{0}^1 i\times I\{x_i<\nu\} di$ exist?
$$
\int_{0}^1 i\times I\{x_i<\nu\} di
$$
where $x_i$ is uniformly distributed over $[0,1]$ and iid across $i$. Parameter $\nu\in(0,1)$.

By law of large numbers, we know $\int_{0}^1 I\{x_i<\nu\} di=\nu$ because $x_i$ is iid. But $i\times I\{x_i<\nu\}$ is no longer iid.

答:
What you wrote here doesn't make sense: a continuum of independent continuously distributed random variables on a fixed probability space doesn't really exist in any useful sense; in particular, it can't have Riemann integrable or even Lebesgue measurable sample paths.

What does make sense however is

$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{k}{n} I(X_k<\nu) = \lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^n k I(X_k<\nu)$$

where $X_k$ is a sequence of independent identically distributed random variables uniform on $[0,1]$. This is basically the "naive conversion" of the expression you wrote into the notation of Riemann sums with uniform partitions.

One way to look at this is through the lens of convergence in probability: the summands have mean $k\nu$ and variance $k^2 \nu(1-\nu)$ so the whole thing has mean $\frac{\sum_{k=1}^n k}{n^2} \nu \to \nu/2$ and variance $\frac{\sum_{k=1}^n k^2 \nu(1-\nu)}{n^4} \to 0$, so it converges in probability to the constant $\nu/2$ by Chebyshev's inequality.

Note that $\nu/2=\nu \int_0^1 x dx$, and this is the typical situation for "reasonable" functions: you'll get $\nu$ (the probability that any one summand "counts") times the corresponding deterministic integral.

I am less confident offhand how to deduce a.s. convergence; if I had to venture a guess I would try the Borel-Cantelli lemma.


问:
Show the left shift $T$ and $T^{-1} $ are measurable.
Let $X=\{0,1 \}^{\mathbb{Z}}$. So an element of $X$ is given by a sequence $(x_i)_{i\in \mathbb{Z}}$, where $x_i \in \{0,1 \}$.

Let $\mu_i$ be a probability measure on $(\{0,1 \}, \mathcal{P}(\{0,1 \}))$, where $\mu_i(\{0\})= \mu_i (\{1 \})= \frac{1}{2}$.

Define $\mu = \Pi_{i\in \mathbb{Z}} \mu_i$, and $B = \bigotimes_{i\in \mathbb{Z}} \mathcal{P}(\{0,1 \})$.

Let $T: X \rightarrow X$ be the Bernoulli shift defined by $T(x_i)=x_{i-1}$ for $i\in \mathbb{Z}$.

Show $T$ and $T^{-1}$ are measurable and $T$ is measure-preserving.

Attempt:

Let $A \in B$. To show that $T$ is measurable, I want to show that $T^{-1}(A)= \{x_{i+1} :  (x_i)_{i\in \mathbb{Z}} \in A \} \in B$.

Define $H= \{A \subset X : T^{-1}(A) \in B \}$. I showed that $H$ is a $\sigma$-algebra. So $B \subset H$. Thus, $T$ is measurable.

To show that $T^{-1} $ is measurable, can I do the same thing using $\sigma$-algebras? I'm not sure about this part.

To show that $T$ is measure-preserving, I need to show that $\mu(A)= \mu(T^{-1}(A))$ for all $A\in B$. I don't have any clues for this part either.

Any help will be appreciated.

Thank you!

答:
A few notes:

I don't see a problem with showing measurability of $T^{-1}$ similarly to $T$, it should work analogously since $T$ appears to be bijective on $X$

in order to show that $\mu(A)= \mu(T^{-1}(A))$ for all $A\in B$, it suffices to show equality for all cylinder sets because they form an $\cap$-stable generator of $B$


问:
Transformation of random variable and domain change
So i was wondering about this for a while. Given a random variable $X$ with density $f_X(x)=e^{-x}$ where $x\geq 0$. consider another random variable, $Y=\log X$. Is the domain of $Y$ given by $[0,\infty)$? If yes, why? thanks :)!

Y

答:
Do a drawing of the function $Y=\log X$ and see that Y domain is the X imagine, thus $Y \in \mathbb{R}$

Using the fundamental transformation theorem you get

$$f_Y(y)=f_X(g^{-1}(y))\cdot |\frac{d}{dy}g^{-1}(y)|=e^{y-e^y}$$


问:
Finding an expression for $M$ in terms of $t$. differential equations

The mass $M$ at time $t$ of the leaves of a certain plant varies according to the differential equation $$\frac{\mathrm{d}m}{\mathrm{d}t}= M-M^2$$
Given that at time $t=0$, $M=0.5$, find an expression for $M$ in terms of $t$

I wasn't sure how to start so I began by solving the differential equation to get $$\frac{1}{M-M^2}\mathrm{d}M=\mathrm{d}t$$

Then simplifying, I get $$ln\frac{M}{1-M}=t$$ using log rules. To cancel $ln$ I get $$\frac{M}{1-M}=e^t$$ and then I get $M=e^t-me^t$

However the answer is $$\frac{e^t}{1+e^t}$$
I don't understand how they got that?

答: I believe you have done the integration step wrong, you should end up with -ln((1/M) - 1) = t on that line and the rest should work from there. Let me know if you have any problems after resolving that!</p>


问:
Prove that $A$ is on the radical axis of two circles
Let $ABC$ be a triangle and the circle $(I,r)$, be the incircle of the triangle,which is tangent to $BC$ at $D$.Let $w$ be a circle which passes through $B$ and $C$ and that is tangent to $(I,r)$ at $L$.Prove that $A$ is located on the radical axis of the circles $w$ and $(IDL)$

I was thinking about using Protasov's Lemma because it is pretty similar to this problem and the tangency of the circles.

I am pretty sure that $LD$ is the angle bisector of $LD$ but I haven't proved it yet.This is also very similar to Protasov's Lemma.

答:
Comment

May be this idea helps: LE is radical axis and tangent points F, J and I are on a circle, this is possible only when A locates on LE.


问:
How to show for a linear ODE system, the solution can be given as the linear relation $p(t)=K(t)x(t)+s(t)$??
Suppose we have the following ode system:
$$\begin{bmatrix} \dot x(t) \\ \dot p(t) \\ \end{bmatrix} = \begin{bmatrix} a_{11} && a_{12} \\ a_{21} && a_{22} \\ \end{bmatrix}\begin{bmatrix} x(t) \\ p(t) \\ \end{bmatrix}
+\begin{bmatrix} r_1(t) \\ r_2(t) \\ \end{bmatrix};$$
is it always correct to suppose $p(t)=K(t)x(t)+s(t)$? How can we show this relation holds?

答:
Assuming $p(t) = Kx(t)+s(t)$. If your coefficients are constant, then:
$$\frac{dp}{dt} = K\frac{dx}{dt} + s'(t) = a_{21}x+a_{22}p + r(t)$$
implies that:
$$Ka_{11} = a_{21}, Ka_{12} = a_{22}.$$
If your coefficient matrix is not constant, you can work the same.


问:
How do I permute a list of integers so that as many possible integers will move to a place where a smaller integer used to stand?
I'm trying to understand the solution to an algorithm/logic problem. The problem statement is as follows:

You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers. For instance, if we are given an array [10,20,30,40], we can permute it so that it becomes [20,40,10,30]. Then on the first and the second positions the integers became larger (20>10, 40>20) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals 2. Help Vasya to permute integers in such a way that the number of positions in a new array, where integers are greater than in the original one, is maximal.

Input The first line contains a single integer  (1≤≤105) — the length of the array. The second line contains  integers 1,2,…, (1≤≤109) — the elements of the array.

Output Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.

Sample I/O: example input/output

I've looked at the solution but couldn't really understand the intuition behind it. I understand how it works, but I don't understand WHY it works. Here's the solution:

The answer is  minus maximal number of equal elements. Let the maximal number of equals be . Let's prove that − is reachable. It's clear that for every permutation of the array the answer will be the same, so let's sort the array in non-decreasing order. Now we should just make a left shift on . After it, the − right elements will move to a position of a smaller element. Now let's prove that the answer is no more than −. Let's consider some permutation. It's known that every permutation breaks into cycles. Let's look at two occurrences of the same number in the same cycle. Then there is at least one number between them which will move on a position of a non-smaller element. Even if it the same occurrence and even if the length of the cycle is 1, we can say that for every occurrence of this number there is at least one number that moves on a position of a non-smaller one. So if some number occurs  times, there are at least  bad positions and therefore no more than − good positions.

Please break this explanation down to a level I can understand. I've tried tons of dry-runs but couldn't figure out the logic behind it. To me, this is black magic.

答:
Let's start by simplifying a bit. Notice that the initial order of the numbers do not matter, so we can reorder them as we please.

Then, you need to both prove that there is a permutation such that the elements in n-x positions become greater after the action of the permutation, and that there is no permutation doing that for more than n-x positions.

To prove the first one, let's write the numbers you get in a column in order, but with all duplicates of the same number in the same line. For example, if the numbers are [1,1,1,2,3,4,4], you'd write:

1,1,1

2,

3,

4,4,

Then, the permutation obtained by sending each number to the number below's position (looping around as necessary) does the job. In the example's case, It'd give [2,4,1,3,4,1,1]

Now you need to show that there is no better solution. To do that, let's pick any permutation. Then for the x identical numbers, look at their successive images through the permutation until the antecedent is larger or equal than its image. Since the initial numbers are equal, it is impossible that one of them is part of another's sequence of images. Therefore, for each initial number, there is exactly one number whose image is smaller or equal to itself. In other words, it's impossible to do better than our solution.

In conclusion, we have shown that the maximum number of positions whose value can be increased with a permutation is n-x, with n being the total number of elements, and x being the maximum number of identical numbers.

Feel free to ask any questions you may have.


问:
Modular arithmetic question...
What is the value of $187^2 \mod 100$?

Title: Modular arithmetic question...

Body: What is the value of $187^2 \mod 100$?

Body: $187\equiv-13\pmod{100}$, $(-13)^2\equiv69\pmod{100}$


问:
Why are we able to find the closed form of each number of the form $ζ(2n)$, but struggling with numbers of the form $β(2n)$?
It is known that we are able to find the closed form of each number of the form $ζ(2n)$, but why are we struggling with numbers of the form $β(2n)$ (Dirichlet Beta function), even though they are very similar series?

答:
It has a closed form for positive odd integers $2k+1$,
namely
$$
\beta (2k+1)={{{({-1})^{k}}{E_{{2k}}}{\pi ^{{2k+1}}} \over {4^{{k+1}}}(2k)!}},$$
where

$E_{{n}}$ represent the Euler numbers.

For even integers, a representation is more difficult, as follows.

For every positive integer $k$ we have
$$
{\displaystyle \beta (2k)={\frac {1}{2(2k-1)!}}\sum _{m=0}^{\infty }\left(\left(\sum _{l=0}^{k-1}{\binom {2k-1}{2l}}{\frac {(-1)^{l}A_{2k-2l-1}}{2l+2m+1}}\right)-{\frac {(-1)^{k-1}}{2m+2k}}\right){\frac {A_{2m}}{(2m)!}}{\left({\frac {\pi }{2}}\right)}^{2m+2k},}
$$

where
${\displaystyle A_{k}}$ is the Euler zigzag number.

So the situation is exactly the same as for $\zeta(s)$, only with "even" and "odd" interchanged, where we have a closed formula for $\zeta(2n)$.


问:
Transitivity of $nth$ tuple vectors of Lexicography ordering
I have read about
Definition Of Lexicographic Ordering,
Lexicographic Order,
Generalized lexicographic order,
Lexicographical order,
Lexicographic ordering,
and many other related questions with their corresponding provided answers. It seems there's none about my question here:

Let $\left({S_1, \preceq_1}\right)$ and $\left({S_2, \preceq_2}\right)$ be ordered sets such that $\preccurlyeq$ be the lexicographic order on $S_1 \times S_2.$ By definition we have that $$\left({x_1, x_2}\right) \preccurlyeq \left({y_1, y_2}\right) \iff \left({x_1 \prec_1 y_1}\right) \lor \left({x_1 = y_1 \land x_2 \preceq_2 y_2}\right).$$
Then $\preccurlyeq$ is an ordering on $S_1 \times S_2.$ For vectors with only two entries: $(x_1, x_2), (y_1, y_2),$ the transitivity property of lexicography ordering is such that if $\left({x_1, x_2}\right) \preccurlyeq \left({y_1, y_2}\right)$ and $\left({y_1, y_2}\right) \preccurlyeq \left({z_1, z_2}\right)$ then $\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right)$ which is proved as follows.
By definition of lexicographic order on $S_1 \times S_2$, we need to show four cases:

Case (1): Let $x_1 = y_1 = z_1.$ Then, $x_2 \preceq_2 y_2 \ \text{  and } \ y_2 \preceq_2 z_2$ and it follows that: $\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right)$

Case (2): Let $x_1 = y_1$ and $y_1 \ne z_1.$ Then, $y_1 \prec z_1 \ \text{ but as } x_1 = y_1 \implies x_1 \prec z_1$ and it follows that $\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right).$

Case (3): Let $x_1 \ne y_1$ and $y_1 = z_1.$ Then, $x_1 \prec y_1 \text{ but as } \ y_1 = z_1, \implies  x_1 \prec z_1 $ and it follows that $\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right).$

Case (4): Let $x_1 \ne y_1$ and $y_1 \ne z_1.$  Then,  $x_1 \prec y_1$ and $y_1 \prec z_1 \implies x_1 \prec z_1$ and $\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right)$

Thus, in all cases it can be seen that: $\left({x_1, x_2}\right) \preccurlyeq \left({z_1, z_2}\right) \quad\quad\Box$

QUESTION: for n-tuple vectors, $A = (a_1, a_2, \ldots, a_n), \ B = (b_1, b_2, \ldots, b_n), \ C = (c_1, c_2, \ldots, c_n)$, how would I prove such a transitivity property using all these four cases? Could there be a better way than using the four cases to prove transitivity?

答:
I'm assuming "ordered set" means "partially ordered set". If $A=B$ or $B=C$, this is clear, so assume that $A\prec B\prec C$ (that is, none of them are equal). Then by definition of lexicographic order, there is some $i$ with $1\leq i\leq n$ such that $a_i\prec_i b_i$ and $a_j=b_j$ for all $j<i$. Similarly, there is some $k$ with $1\leq k\leq n$ such that $b_k\prec_k c_k$ and $b_j=c_j$ for all $j<k$. We need only consider two cases.

Case (1): Assume $k\leq i$. Then for all $j<k$, we have $a_j=b_j=c_j$ and $a_k\preceq_k b_k\prec_k c_k$, hence $A\prec C$.

Case (2): Assume $i<k$. Then for all $j<i$, we have $a_j=b_j=c_j$ and $a_i\prec_i b_i\preceq_i c_i$, hence $A\prec C$.


问:
Is the compactness theorem (from mathematical logic) equivalent to the Axiom of Choice?
Or more importantly, is it independent of the axiom of choice. The compactness theorem states the given a set of sentences $T$ in a first order Language $L, T$ has a model iff every finite subset of $T$ has a model. So for any natural number $n, T(n)$ is a finite subset of $n$ sentences. Now if every finite subset has a model than adding a sentence $r$ to $T(n)$ gives us $T(n+1)$ which also has a model so there is some transfinite induction going on here when $T$ is countable. This seems to cry out for the application of Zorn's Lemma which is equivalent to the axiom of choice. So to summarize, is the compactness theorem consistent with ~AC (the negation of the axiom of choice which is independent of ZF set theory)? </p>

答:
What Qiaochu writes is true.

The compactness theorem for first-order logic is equivalent (in $\sf ZF$) to the completeness theorem, as well to the ultrafilter lemma, and several other interesting principles. The proof is due to Henkin [1], but I could not find the paper online, the proof appears in [2, Theorem 2.2].

It should be remarked that if we assume that both the compactness theorem holds, and Łoś theorem's hold, then we can prove the axiom of choice holds [4]. Therefore we have to be extra careful when we prove the compactness theorems using ultrafilters.

The ultrafilter lemma was proved to be independent from the axiom of choice (i.e. it cannot prove full choice) in 1965. In fact it is consistent that the axiom of countable choice fails, but the ultrafilter lemma holds (see [3], [2, Theorem 5.21]. Interestingly, this was about a decade after it was known that the ultrafilter lemma and the compactness theorem are equivalent.

To slightly generalize on Qiaochu's last point, if $\cal L$ is a well-orderable language (i.e. the cardinality of the language is an ordinal) then we can prove the compactness theorem for $\cal L$ without using the axiom of choice at all. The problem begins when the languages are not well-ordered. While it may seem strange, remember that if you use a language which includes a constant for every real number, then you can no longer prove that the language is well-orderabe.

Bibliography.

Henkin, Leon. "Metamathematical theorems equivalent to the prime ideal theorems for Boolean algebras." Bull. Amer. Math. Soc 60 (1954): 387-388.

Jech, Thomas J. The Axiom of Choice. Courier Dover Publications, 2008.

Halpern, James D., and Azriel Lévy. "The Boolean prime ideal theorem does not imply the axiom of choice." Proc. of Symposium Pure Math. of the AMS. Vol. 13. (1971): 83-134.

Howard, Paul E. "Łoś’ theorem and the Boolean prime ideal theorem imply the axiom of choice." Proceedings of American Math. Society 49 (1975): 426-428.


问:
Show that $T(X)=(R,V)=\left( X_{(n)}-X_{(1)},\frac{X_{(n)}+X_{(1)}}{2} \right)$ is a minimal sufficient statistic for $\theta$.
Let $X_{1}, X_{2}, ..., X_{n}$ be a random sample from $\text{Uniform}(\theta,\theta+1)$ population with $-\infty<\theta<\theta+1< \infty$ show that $T(X)=(X_{(1)},X_{(n)})$ is a minimal sufficient statistic for $\theta$. Also, show that $T(X)=(R,V)=\left( X_{(n)}-X_{(1)},\frac{X_{(n)}+X_{(1)}}{2} \right)$ is a minimal sufficient statistic.

For the first part I did the following
$$f(x|\theta,\theta+1)=I_{(\theta,\theta+1)}(x_{(1)},x_{(n)})$$
then
$$\frac{f(x|\theta,\theta+1)}{f(y|\theta,\theta+1)}=\frac{I_{(\theta,\theta+1)}(x_{(1)},x_{(n)})}{I_{(\theta,\theta+1)}(y_{(1)},y_{(n)})}$$

This is a constant function in $\theta$ iff $x_{(1)}=y_{(1)}$ and $x_{(n)}=y_{(n)}$ s.t. $T(X)=(X_{(1)},X_{(n)})$ is a minimal sufficient statistic for $\theta$.

However, I am not sure how to proceed to show that $T(X)=(R,V)=\left( X_{(n)}-X_{(1)},\frac{X_{(n)}+X_{(1)}}{2} \right)$ is a minimal sufficient statistic. Can some help me with this?

答:
If you know$X_{(1)}$ and $X_{(n)}$, then you can certainly compute $\frac{X_{(1)} + X_{(n)}}{2}$ and $X_{(n)}- X_{(1)}$ from them. Likewise given the latter two, you can calculate the first two.  So if either pair is sufficient, the other is as well, because you are losing no data switching from one to the other.  Having a sufficient statistic for a parameter means that going back to the sample for more details won't improve your parameter estimate.

Intuitively, $\theta < X_{(1)} <$(the other data)$ < X_{(n)} < \theta + 1$. If you are trying to find an upper bound for $\theta$, and you have $X_{(1)}$, then going back to the sample for more data will not improve matters.  Likewise, the lower bound for $\theta + 1$ is most restricted by $X_{(n)}$.


问:
Intuition behind the construction of a fixed point from Kleene's fixed point theorem
Below is an explicit construction of a fixed point the existence of which is guaranteed by Kleene's fixed point theorem. I was wondering if there's any intuitive explanation of why the fixed point should be what it is (namely, $[s](r)$)? This construction looks like a magic to me (I don't think I would ever be able to come up with this construction on my own), so I wanted to see what intuition lies behind this construction, if any. If there's no intuition behind the initial stages of the proof, then at least suppose I've arrived at $(2)$. Is there any reason why I should think that exactly $[s](r)$ should be the fixed point? Or can this knowledge only be obtained by looking at all the formulas that I have for a long time and trying different options?

Kleene's fixed point theorem: If $H:N\to N$ is a total computable
function and $U$ is a Godel universal computable function (defined
here), then there is $n$ such that $U(n,x)=U(H(n),x)$.

Here's an explicit construction of a fixed point.

Convention: capital letters denote computable functions, and the
corresponding lowercase letters denote the numbers of programs that
compute those functions. Also, $[p](x)$ (alternatively, $U(p,x)$) denotes the result of
application of program number $p$ on input $x$.

Consider $F(x)=U(x,x)$. This is a computable partial function. Define
$V(n,x)=U(F(n),x)$, which is also a partial computable function. We
can write $$V(n,x)=[[n](n)](x)$$

Now consider the total computable function $S:N\to N$ such that
$U((S(n),x)=V(n,x)$ (which exists because $U$ is Godel universal). We
can write this equality as $$V(n,x)=[[s](n)](x)$$

Thus $$\tag{1} [[s](n)](x)=[[n](n)](x)$$

Now consider the total computable $H\circ S$, say the composition is
computable by program number $r$. Then $$\tag{2}[r](x)=[h]([s](x))$$

Now $[s](r)$ should be the fixed point. To prove this, we need to
prove that $$U([s](r),x)=U([h]([s](r)),x)$$ or equivalently $$
 \tag{3}[[s](r)](x)=[[h]([s](r))](x)$$

First, we have $[[h]([s](r))](x)=[[r](r)](x)$ from $(2)$, and then
$[[r](r)](x)=[[s](r)](x)$ from $(1)$. This proves $(3)$.

答:
One intuition that would be helpful for coming up with this sort of thing yourself is that these fixed point constructions are related to diagonalization. A very general case of this is Lawvere's fixed point theorem, but that is even essentially just using a very general setting to interpret the sort of fixed point function one would define in (untyped) lambda calculus:

$$λf. (λx. f(x\ x))(λx. f(x\ x))$$

Applied to a given $f$, this satisfies the equivalence:

$$(λx. f(x\ x)) (λx. f(x\ x)) \\
= \\
f ((λx. f(x\ x))(λx. f(x\ x)))
$$

So is a fixed point of $f$. This involves two cases of self-application: $x$ is applied to itself and the subterm $λx. f(x\ x)$ is applied to itself. This double use of self-application is the essence of diagonal arguments.

The fixed point theorem you cite is doing something very similar, but it needs to talk about mediating between functions and their encoding as natural numbers. If we ignore some of that, we get:

$$
f\ x = x\ x \\
v\ n\ x = f\ n\ x = n\ n\ x \\
s\ n\ x = v\ n\ x = n\ n\ x \\
r\ n = h\ (s\ n) = h (λx. n\ n\ x)
$$

So, at this point, $r$ should look very similar to $λn. h(n\ n)$, which is one of the self-applications above. To get the fixed point the usual way, $r$ needs to be applied to itself, which is the reason you should suspect $[s](r)$, as $[s]$ is the self-application function. It is essentially:

$$
s\ r = r\ r = (λn. h(λy. n\ n\ y))\ (λn. h(λy. n\ n\ y))
$$

which is almost the term I wrote above, modulo $f = λx. f\ x$ (the difference in this case has to do with the encoding, I think).

Often these kind of diagonal arguments are used to deduce contradictions. For instance, Russell's paradox and the proof of undecidability of the halting problem carry out this sort of diagonalization to produce a fixed point of logical negation. Since logical negation doesn't have a fixed point, you deduce that the assumptions used to construct the fixed point are false. However, in the case of total computable functions, the premises for carrying out the diagonalization are true, and enable you to construct fixed points that do exist.


答:
$$\frac{dM}{M-M^2}=\mathrm{d}t$$
$$\int \frac{dM}{M(1-M)}=t+c$$
$$\int \frac{dM}{M}+\int \dfrac {dM}{1-M}=t+c$$
$$\int \frac{dM}{M}-\int \dfrac {dM}{M-1}=t+c$$
$$\ln |M|-\ln {|M-1|}=t+c$$
$$\dfrac M{M-1}=Ce^t$$
$$M={(M-1)}Ce^t$$
$$M(1-Ce^t)=-Ce^t$$
This gives us:
$$M(t)=\dfrac {-Ce^t}{1-Ce^t}$$
Then apply initial condition:
$$M(0)=\dfrac 12 \implies C=-1$$
$$M(t)=\dfrac {e^t}{1+e^t}$$

Title: Finding an expression for $M$ in terms of $t$. differential equations

Body:
The mass $M$ at time $t$ of the leaves of a certain plant varies according to the differential equation $$\frac{\mathrm{d}m}{\mathrm{d}t}= M-M^2$$
Given that at time $t=0$, $M=0.5$, find an expression for $M$ in terms of $t$

I wasn't sure how to start so I began by solving the differential equation to get $$\frac{1}{M-M^2}\mathrm{d}M=\mathrm{d}t$$

Then simplifying, I get $$ln\frac{M}{1-M}=t$$ using log rules. To cancel $ln$ I get $$\frac{M}{1-M}=e^t$$ and then I get $M=e^t-me^t$

However the answer is $$\frac{e^t}{1+e^t}$$
I don't understand how they got that?


问:
Prove that addition under "real numbers" is well defined.
Problem:

If we denote real numbers as Cauchy sequences and:

$$[\{a_i\}]+[\{b_i\}] = [\{a_i+b_i\}] ; i∈ N$$

Show that "$+$" is well defined under real numbers.

My try:

Assume that:

$$
\begin{cases}
[\{a_i\}]=[\{a_i'\}]\\ \\
 [\{b_i\}]=[\{b_i'\}]
\end{cases}   (I)$$

We must show that: $$[\{a_i+b_i\}] = [\{a_i'+b_i'\}] $$

From LHS:

$$[\{a_i+b_i\}]=[\{a_i\}+\{b_i\}]$$   $$=[\{a_i\}]+[\{b_i\}]$$  From (I)$$[\{a_i\}]+[\{b_i\}]=[\{a_i'\}]+[\{b_i'\}]$$  $$=[\{a_i'\}+\{b_i'\}]$$ $$=[\{a_i'+b_i'\}]$$

Is that correct?

I also checked this link but I wanted to try another approach.

答:
As written in the comments, your proof is not fine as it's using the statement to prove.

Let us denote the sequences by $a=\{a_i\}$ and similarly for $a',\ b,\ b'$.

We have $[x]=[y]$ iff the sequence of the differences of the elements of $x$ and $y$ converges to $0$, i.e. $y_n-x_n\to 0$.

Now, if $[a]=[a']$ and $[b]=[b']$, that means $a_n'-a_n\to 0$ and $b_n'-b_n\to 0$, so using the additive property of limits, we get $c_n:=(a_n'-a_n)\,+\,(b_n'-b_n)\ \to 0$.

But observe that $c_n=(a_n'+b_n')\,-\,(a_n+b_n)=(a'+b')_n\,-\,(a+b)_n$, and this shows $[a+b]=[a'+b']$.


问:
If $N$ is neighborhood of $x_0 \in M$ in $(M,d)$, then it is also neighborhood of $x_0 \in M$ in $(M,d')$?
I am trying to prove that the limit of a sequence is indifferent to the distance function used. (Meaning the values are equal to another). This is definitely true if "neighbourhood" is indifferent from the definition of distance.

Geometrically speaking, the proof seems quite intuitive. For every norm the property $d(ax, ay) = a \times d(x,y)$ shows why this must be true. But I am lacking the proof giving only the property of a distance function.

The problem is that I cannot find any definition of continuity that is not using a distance function. If I could show that every metric is continuous, then I could use that $d(x,y) > 0$ and $d(x,x) = 0$ and then I could convert between both.

Any Ideas?
Thank you.

Edit: I am also wondering if you could maybe use some algebra, defining a function $f: d,d' \mapsto g $, where $g(d) = d'$ and then show some properties about $f$...

答:
It is true that if $x_n \to x$ and $x_n \to y$ in a metric space, then $x=y$, which follows from the fact that all metric spaces are Hausdorff. The proof is that if $x \neq y$ then $d(x,y) > 0$, so there are disjoint balls around $x, y$. For $n$ large, the sequence cannot be in both balls since they are disjoint, so we have a contradiction.

It is also easy to construct different metrics $d, d'$ for which $x_n \to x$ wrt to $d$, but $x_n$ does not converge wrt to $d'$.
For example, in the discrete metric $d'$, $x_n \to x$ if and only if $x_n = x$ for all $n$ large, i.e. if and only if the sequence is eventually constant equal to $x$. So any sequence converging to $x$ in the usual way that is not eventually constant won't converge wrt $d'$.
For another example, take the usual metric $d$, but take $d'(x,y)$ to be $\min(|x-y|,1)$ if $x,y$ are either both rational or both irrational, and $d'(x,y) = 1$ otherwise. You can check that $d'$ is a metric. Now take rationals $x_n \to \sqrt{2}$ in the usual metric. By defintion $d(x_n, \sqrt{2}) = 1$ for all $n$ since $x_n$ is rational and $\sqrt{2}$ is not, so $x_n$ does not converge to $\sqrt{2}$ wrt $d'$. In this case $x_n$ does not converge at all with respect to $d'$.

What is more difficult is to find an example for which $x_n \to x$ wrt $d$ but $x_n \to y$ wrt $d'$ with $x \neq y$. A discussion of this topic can be found here: https://mathoverflow.net/questions/53971/example-of-sequences-with-different-limits-for-two-norms. You can find my own example there as well.

What is true is that any two metrics on $\mathbb{R}^n$ induced by a norm (i.e. $d(x,y) = ||x-y||$ for a norm $||\cdot||$) are equivalent, i.e. they generate the same topology and thus have the same convergent sequences with the same limits.


问:
First Course in Linear Algebra book suggestions?
I'm starting maths degree this September and hoping to do some reading on Linear Algebra before I start. What are some good introductory books?

Is there something that starts from the very beginning but also covers stuff in depth? (Is there a book that will be useful for both 1st year and 2nd year linear algebra modules?)

It would also be great if the book has lots of exercises (proof questions!) and solutions given as well.

Thanks in advance.

答:
I would recommend, arranged somewhat in order of difficulty/sophistication, that you take a look at:

1) Strang Linear Algebra and its Applications (don't forget the MIT OCW lectures)

2) Paul Halmos Finite Dimensional Vector Spaces (the recommendation for his problem book is also good)

3) Hoffman & Kunze Linear Algebra

4) Shafarevich Linear Algebra and Geometry

5) Roman Advanced Linear Algebra

A nice, cheap alternative to Halmos and Hoffman would be Shilov's text published by Dover. There are also nice books that have linear algebra integrated with other material such as Apostol's Calculus (which I highly recommend), Hubbard's vector calculus text or Artin's abstract algebra text.


问:
Strong deformation retraction.
Suppose $X$ is a path connected contractible space. So we know there is a homotopy $F:X\times I \rightarrow X$ with $F(x,0)=x$ and $F(x,1)=c$, for all $x\in X$ and some fixed $c\in X$. Since $X$ is path connected we could change the constant $c\in X$ as we wish, so $X$ deformation retracts to any $x\in X$.

My question is, if we include the hyposthesis of $X$ being a Hausdorff Normal space, and $x_o\in X$ is a non degenarate point (basically meaning that the inclusion $x_o\hookrightarrow X$ is a cofibration), then can we claim that $X$ strongly deformation retracts to $x_o$.

By "strongly" I mean that the homotopy $F:X\times I \rightarrow X$ would have the additional property that $F(x_o,t)=x_o$ for all $t\in I$, in other words it's an homotopy relative to $x_o$.

I know that, since $x_o$ is a non degenerate point, there is an open neighborhood $N$  of $x_o$ which strongly deformation retracts to $x_o$, but I see no reason for that also being the case for $X$ itself.

Since I don't know about the validity of this, ideas for proofs or counterexamples are both welcome.

答:
Suppose that $i:A\subseteq X$ is a subspace inclusion which is both a cofibration and a homotopy equivalence. We place no other assumptions on $X$ or $A$. The assumptions grant a map $r':X\rightarrow A$ such that $r'i\simeq id_A$ and $ir'\simeq id_X$. Using the homotopy extension property we obtain a map $r:X\rightarrow A$ such that $r\simeq r'$ and $ri=id_A$. Necessarily $ir\simeq id_X$.

$A$ is a deformation retraction of $X$.

Now fix a homotopy $F:id_X\simeq ir$. Define
$$G:X\times\{0,1\}\times I\cup A\times I\times I\rightarrow X$$
by putting
$$G(x,0,t)=x,\qquad G(a,s,t)=F(a,(1-t)s),\qquad G(x,1,t)=F(r(x),1-t).$$
This is well-defined and continuous. We need the following.

Lemma: Let $X$ be a space and $A\subseteq X$ a cofibration. Then $X\times\{0,1\}\cup A\times I\subseteq X\times I$ is a cofibration. $\square$

The proof makes use of the fact that there is a self-homeomorphism of $I^2$ which maps $I\times\{0,1\}\cup\{0\}\times I$ onto $\{0\}\times I$. I'll leave details to your textbook.

Above we have defined $G$. Notice that its restriction to $(X\times\{0,1\}\cup A\times I)\times\{0\}$ is just the map $F$. Because of the lemma there exists a homotopy $\widetilde G:X\times I\times I\rightarrow X$ with $\widetilde G(x,s,0)=F(x,s)$ and $H|_{X\times\{0,1\}\times I\cup A\times I\times I}=G$. Put
$$H_t(x)=\widetilde G(x,t,1).$$
This yields $H_0=id_X$, $H_1=ir$ and $H_t|_A=i$. Thus we have our conclusion.

A subspace inclusion $A\subset X$ which is both a cofibration and a homotopy equivalence is a strong deformation retract.

Of course if $X$ is contractible, then the inclusion $x_0\hookrightarrow X$ of any point is a homotopy equivalence. If this inclusion is a cofibration, then the outcome above applies. The Hausdorff and normality assumptions are not necessary.

Title: Strong deformation retraction.


问:
Proof that a certain expression is the Dirac delta in spherical coordinates
My professor during the lesson said that the following expression is the 3D delta function written in spherical coordinates:
$$\sum_{l=0}^{\infty}\sum_{m=-l}^l \frac{\delta(p'-p)}{p^2}Y_{lm}(\alpha',\beta')Y_{lm}^{\star}(\alpha,\beta) = \delta{}^{(3)}(\vec{p'}-\vec{p})$$

where $(\alpha,\beta),(\alpha',\beta')$ are the polar and azimuthal angle of $\vec{p}$ and $\vec{p}'$ and $Y_{lm}$ are spherical harmonics.
Now, he didn't give any explanation (and I'm unable to prove it by myself), but if it is easy to prove, then it would be interesting to see the proof.

Thank you

答:
Let $d_L(p,\alpha,\beta|p',\alpha',\beta')=\sum_{\ell=0}^L \sum_{m=-\ell}^\ell Y_{\ell m}^*(\alpha',\beta')Y_{\ell m}(\alpha,\beta)\frac{\delta(p-p')}{p'^2}$ and let $\phi(p,\alpha,\beta)$ be a suitable test function.  Then, we have

$$\begin{align}
\lim_{L\to\infty}\langle d_L,\phi)&=\lim_{L\to\infty}\int_0^\infty \int_0^{2\pi}\int_0^\pi d_L(p,\alpha,\beta|p',\alpha',\beta')\phi(p',\alpha',\beta')\,p'^2\sin(\beta')\,dp'\,d\beta'\,d\alpha'\\\\
&=\sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell Y_{\ell m}(\alpha,\beta)\underbrace{\int_0^{2\pi}\int_0^\pi Y_{\ell m}^*(\alpha',\beta') \phi(p,\alpha',\beta')\,\sin(\beta')\,d\beta'\,d\alpha'}_{\text{Spherical Harmonic Expansion Coefficients of }\phi}\tag1\\\\
&=\phi(p,\alpha,\beta)
\end{align}$$

Hence as a distribution, $\lim_{L\to\infty}d_L(p,\alpha,\beta|p',\alpha',\beta')=\frac{\delta(p-p')}{p'^2}\frac{\delta(\beta-\beta')}{\sin(\beta')}\delta(\alpha-\alpha')$ as was to be shown!

NOTE:  HERE
is a reference to recognizing $(1)$ as the expansion of $\phi$ in terms of sperical harmonic basis functions.


问:
Squeeze theorem doesn't fit here?
Let $(a_k)_{k \geq 1}$ and $(b_k)_{k \geq 1}$ be two strictly positive sequences such that $$\frac{a_{k+1}}{a_k} \leq \frac{b_{k+1}}{b_k}$$ I have to prove that if $\sum_{k=1}^\infty b_k$ converges, then $\sum_{k=1}^\infty a_k$ must converges as well.

I found that $a_k \leq C b_k$ for $C > 0$. So if $\sum_{k=1}^\infty b_k$ converge, then $\sum_{k=1}^\infty C b_k$ converges as well. It is not clear for me that $\sum_{k=1}^\infty a_k$ must converges using the squeeze theorem. The best I have so far is $$0 \leq \sum_{k=1}^n a_k \leq \sum_{k=1}^n C b_k$$ So $$0 \leq \sum_{k=1}^\infty a_k \leq \sum_{k=1}^\infty C b_k = L \in \mathbb{R}$$

But nothing specify $L=0$. Am I missing something?

答:
Let $A_n := \sum_{k=1}^na_k$ and $B_n := \sum_{k=1}^nb_k$. Since $(B_n)$ is Cauchy, for $\varepsilon>0$ there exists $N$ such that $|B_n - B_m|<\frac{\varepsilon}{C}$ for $n\ge m\ge N$. Hence, for $n\ge m$ we obtain
$$
|A_n-A_m| = \left|\sum_{k=1}^na_k - \sum_{k=1}^ma_k\right| = \sum_{k=m+1}^na_k\le C\sum_{k=m+1}^nb_k = C|B_n-B_m| < \varepsilon.
$$
Therefore, $(A_n)$ is a Cauchy sequence and thus converges.

Title: Squeeze theorem doesn't fit here?


问:
Problem finding this equality
Let $X$ be a metric space, $Y \subset X$ a subset, and $x \in X$ a point.
We define the distance of $x$ from $Y$ as
$$d(x, Y) = \inf d(x,y)$$
Proof that
$$\bar{Y} = \{x \in X : d(x, Y) = 0\}$$

I know that
$$\bar{Y}=Y \cup D(Y) $$
Where $D(Y)$ are $Y$ Accumulation Points. Obviously if $y \in Y$ you have $d(y,Y)=0$. My doubt is how to try that $d(y,Y)=0$ if $y \in D(Y)$. Having tried this I can say that the demonstration is over?

答:
Claim 1: $\mbox{LHS} \subseteq \mbox{RHS}$. Let $x\in\bar{Y}$,
then there exists a sequence $(x_{n})$ in $Y$ such that $x_{n}\rightarrow x$.
Note that $0\leq d(x,Y)\leq d(x,x_{n})\rightarrow0$ as $n\rightarrow\infty$.
This shows that $d(x,Y)=0$ and hence $x\in\mbox{RHS}$.

Claim 2: $\mbox{RHS}\subseteq\mbox{LHS}$. Let $x\in\mbox{RHS}$,
then $d(x,Y)=0$. For each $n\in\mathbb{N}$, $\frac{1}{n}$ is not
a lower bound of the set $\{d(x,y)\mid y\in Y\}$ (because the greatest
lower bound of that set is $0$), so there exists $x_{n}\in Y$ such
that $d(x,x_{n})<\frac{1}{n}.$ We obtain a sequence $(x_{n})$ in $Y$ (formally, we invoke the Axiom of Choice to assert the existence of such sequence). Clearly $d(x,x_{n})\rightarrow0\Rightarrow x_{n}\rightarrow x$.
Hence $x\in\bar{Y}$.

Title: Problem finding this equality


答: Well, if we have $$\frac{a_{k+1}}{a_k} \leq \frac{b_{k+1}}{b_k},$$ we can prove by induction $$a_k\le\frac{a_1}{b_1}\,b_k$$ for all $k\ge1$, and the rest is an easy consequence.

Title: Squeeze theorem doesn't fit here?

Body:
Let $(a_k)_{k \geq 1}$ and $(b_k)_{k \geq 1}$ be two strictly positive sequences such that $$\frac{a_{k+1}}{a_k} \leq \frac{b_{k+1}}{b_k}$$ I have to prove that if $\sum_{k=1}^\infty b_k$ converges, then $\sum_{k=1}^\infty a_k$ must converges as well.

I found that $a_k \leq C b_k$ for $C > 0$. So if $\sum_{k=1}^\infty b_k$ converge, then $\sum_{k=1}^\infty C b_k$ converges as well. It is not clear for me that $\sum_{k=1}^\infty a_k$ must converges using the squeeze theorem. The best I have so far is $$0 \leq \sum_{k=1}^n a_k \leq \sum_{k=1}^n C b_k$$ So $$0 \leq \sum_{k=1}^\infty a_k \leq \sum_{k=1}^\infty C b_k = L \in \mathbb{R}$$

But nothing specify $L=0$. Am I missing something?


问:
Propositional Logic - List of N integers
Today in our propositional-calculus class, we were asked this question:

Suppose you have a list of N integers: you write a computer program to
analyze the list. If every integer in the list is greater than 10,
then the returned output is the value TRUE. Else, the output is the
value FALSE. If N = 0, and therefore, the list is empty, what is the
output?

While explaining how to interpret the problem, my teacher said that the returned value is FALSE, because, since the list is empty, there aren't any counterexamples against that statement.

I didn't seem to grasp that. First of all, isn't the else statement a condition in itself? Namely, the condition:

If there's at least one number less than (or equal to) 10 [...]

There aren't any counterexamples to that either, so why favouring a condition over the other?
Honestly, my solution (as wrong as it may be) would be to claim that the expressed conditions simply do not cover the "empty list" case, so the output value cannot be determined with the information we possess.

Otherwise, is the "else" to be interpreted as "in every other case"? In this eventuality, the output would be FALSE, but to be fair, I'm not sure of this answer.

So, what is the solution?

答:
I suspect your teacher just got muddled with which output represents what. If there are is a counterexample to the statement "every integer in the list is greater than 10", it should output FALSE. Since there can't be a counterexample, it should output TRUE. So the reasoning is correct, but your teacher has linked it to the other output, presumably by mistake.

FWIW if I was to program this in python, it would look like

```<code>def all_greater(mylist):
    for x in mylist:
        if x<10:
            return False
    return True
</code>```

This will return True if the list is empty, because the loop will execute 0 times.

Body:
I suspect your teacher just got muddled with which output represents what. If there are is a counterexample to the statement "every integer in the list is greater than 10", it should output FALSE. Since there can't be a counterexample, it should output TRUE. So the reasoning is correct, but your teacher has linked it to the other output, presumably by mistake.

FWIW if I was to program this in python, it would look like

def all_greater(mylist):
    for x in mylist:
        if x<10:
            return False
    return True

This will return True if the list is empty, because the loop will execute 0 times.


问:
Surprisingly elementary and direct proofs
What are some examples of theorems, whose first proof was quite hard and sophisticated, perhaps using some other deep theorems of some theory, before years later surprisingly a quite elementary, direct, perhaps even short proof has been found?

A related question is MO/24913, which deals with hard theorems whose proofs were simplified by the development of more sophisticated theories. But I would like to see examples where this wasn't necessary, but rather the theory turned out to be superfluous as for the proof of the theorem. I expect that this didn't happen so often. [Ok after reading all the answers, it obviously happened all the time!]

答:
Chebyshev had first proven the Bertrand's postulate which states that for any integer $n > 3$, there always exists at least one prime number $p$ with $n < p < 2n − 2$. A weaker but more elegant formulation is: for every $n > 1$ there is always at least one prime $p$ such that $n < p < 2n$.

Later Paul Erdős had given an elementary yet elegant proof of this. Surprisingly he was only 20 years old when he had given this proof.


问:
Question regarding Limits and the use of L'hopital
I am having trouble with this limit, and wondered if this step on line $5$ was allowed, where I used L'Hopital agian but in the numerator of the limit. Thanks in advance.

I have attatched a picture of my calculations.

答:
By a single application of L'Hospital,

$$\frac{\sqrt x\sin\sqrt x}{1-e^{-x}}\to\frac{\dfrac{\sin\sqrt x}{2\sqrt x}+\dfrac{\cos\sqrt x}2}{e^{-x}}\to\frac{\dfrac12+\dfrac12}1.$$

Note that a second application of L'Hospital is not allowed (nor necessary), as the conditions for application are not fulfilled.

Without l'Hospital:

Let $x=t^2$ and $u=t^2$. Then

$$\lim_{x\to 0^+}\frac{\sqrt x\sin\sqrt x}{1-e^{-x}}=\lim_{t\to 0}\frac{t\sin t}{1-e^{-t^2}}=\lim_{t\to 0}\frac{t^2}{1-e^{-t^2}}\frac{\sin t}t=\frac{\lim_{t\to 0}\dfrac{\sin t}t}{\lim_{u\to 0}\dfrac{1-e^{-u}}u}=1$$

using well-known limits.


问:
Derivation of Product Rule for Finite Differences without Shift Operator
Let $h(x)=f(x)g(x)$, and let the $\Delta$ be the forward difference operator on functions be
$$\Delta f := \frac{f_{j+1}-f_j}{\Delta x}$$
where $f_j = f(j\Delta x)$

Now I can derive the product rule for finite differences in terms of $\Delta f$ and $\Delta g$ easily

$$\Delta h = \frac{h_{j+1}-h_j}{\Delta x} = \frac{f_{j+1}g_{j+1}-f_jg_j}{\Delta x} = \frac{f_{j+1}g_{j+1}\color{red}{-f_jg_{j+1} }{\color{blue}{+f_jg_{j+1}}}-f_jg_j}{\Delta x}=g_{j+1}\Delta f+f_j\Delta g$$

However, you can see in the first term that $g$ is shifted forward by $1$ step.

But on Wikipedia for the finite differences the rule is express with a second order term:

$$\Delta(fg) = f\Delta g + g \Delta f + \Delta f \Delta g$$

I have tried converting my expression to this with all sorts of algebra, and I even tried using the wiki's formula to go backwards to my equation, but nothing has worked. It's clear there must be some clever trick to convert the shift operator in terms of $f,\Delta f,\Delta g$ but I can't see how.

Any ideas?

答: You have $g_{j+1}\Delta f=(g_{j+1}-g_j)\Delta f+g_j\Delta f$, so this is the same as $g_j\Delta f+\Delta x\Delta g\Delta f$.


问:
What classification does this formula have?
Say I have formulas $P(a,b,c)$, $Q(a,b,c)$, each in $\Sigma^0_1$. The formula that I am interested in is

$\varphi(a,b,c) = $ $\forall a,b,c [\neg P(a,b,c) \vee Q(a,b,c)]$

(You could probably guess that it originally looked like $P \Rightarrow Q$). Because of how quantifiers work I can't really pull out a negative by de Morgan and say we are talking about a $d- \Sigma^0_2$ formula. So how do I deal with this formula? Where is $\varphi$ in the arithmetical hierarchy?

答:
Basically, multi-ary Boolean combinations inside of quantifiers should be thought of as "picking the worse option." Here we have a $\Sigma_1^0$ (the $Q$) and a $\Pi^0_1$ (the $\neg P$) inside a universal quantifier; the former suggests $\Pi^0_2$ while the latter suggests $\Pi^0_1$. Since the former is worse, this says that the complexity of the whole is $\Pi^0_2$.

To see why this works, note that $$\forall x(\forall y A(x,y)\vee\exists zB(x,z))$$ is equivalent to $$\forall x,y\exists z(A(x,y)\vee B(x,z)).$$


问:
Calculate the spectral radius of a matrix with two parameters
Let $A(x,y) = \begin{bmatrix}
1 & -y\\
-x & 1
\end{bmatrix}$, where $x,y\gt0$. Find the spectral radius $\rho (x,y)$ of $A$ as a function of $x,y$.

What I did is I went ahead and found the characteristic polynomial of $A$, being:
$$(1-\lambda)^2-xy$$ So in order for $\lambda$ to be an eigenvalue, it has to be a solution of
$$(1-\lambda)^2-xy=0 \iff 1-\lambda = \sqrt{xy} \iff \lambda = 1-\sqrt{xy}$$

At that point, I'm unsure how to proceed? In order to find the spectral radius, I have to find the highest $\lambda$ (in absolute value) that satisfies the above equation. Any input is appreciated.

答:
You missed out the negative root.

$(1-\lambda)^2=xy\implies1-\lambda=\pm\sqrt{xy}\implies\lambda=1\pm\sqrt{xy}$ and the spectral radius is $1+\sqrt{xy}$ since $|1-\sqrt{xy}|\le1+\sqrt{xy}$.


问:
Cambridge Admissions Exam Statistics 1999

My work:

Since $f$ is the pdf we must have $\int_{0}^{1} Ax\,\mathrm dx=1 \implies A=2$.  Let $Y$ be the number of currants in my portion. We have $Y\sim B(4,x)$. For the expectation
$$E(Y)=4x,$$
however I don't know how to continue. I had a thought of using the expectation for $x$ in this, however, I can't statistically justify it.

For the second part, we require
$$P\left(X\geq\frac{1}{2}\mid Y=4\right)=\frac{P\left(X\geq\frac{1}{2},Y=4\right)}{P(Y=4)}.$$

I am completely stuck on how to approach this. I checked the student room for solutions but they seem to disagree with themselves and with other solutions on other websites. If anybody could help me in understanding this problem, and the technique required for it I would be really thankful.

答:
Simulated version for ten million cakes, agreeing with Answers of @tommik (+1) and @J.G. (+1). This simulation should be adequate for three
decimal places of accuracy.

The original proportion $p \sim \mathsf{Beta}(2,1).$
Notice that its mean is $2/3.$

```<code>set.seed(2021)
p = rbeta(10^7, 2, 1)
summary(p)
     Min.   1st Qu.    Median      Mean   3rd Qu.      Max.
0.0000918 0.4999540 0.7072202 0.6667052 0.8659839 0.9999999
</code>```

Then the number of currents in my portion is $Y \sim \mathsf{Binom}(4, p),$ averaging $8/3.$

```<code> y = rbinom(10^7, 4, p)
 mean(y)
[1] 2.666702  # aprx 8/3
8/3
[1] 2.666667
</code>```

Finally, conditional on getting all four currents, the
probability that my portion was more than half is
consistent with $63/64:$

```<code>p.4 = p[y == 4]
mean(p.4 > .5)
[1] 0.9843074  # aprx 63/64
63/64
[1] 0.984375
</code>```

Body:
Simulated version for ten million cakes, agreeing with Answers of @tommik (+1) and @J.G. (+1). This simulation should be adequate for three
decimal places of accuracy.

The original proportion $p \sim \mathsf{Beta}(2,1).$
Notice that its mean is $2/3.$

set.seed(2021)
p = rbeta(10^7, 2, 1)
summary(p)
     Min.   1st Qu.    Median      Mean   3rd Qu.      Max.
0.0000918 0.4999540 0.7072202 0.6667052 0.8659839 0.9999999

Then the number of currents in my portion is $Y \sim \mathsf{Binom}(4, p),$ averaging $8/3.$

 y = rbinom(10^7, 4, p)
 mean(y)
[1] 2.666702  # aprx 8/3
8/3
[1] 2.666667

Finally, conditional on getting all four currents, the
probability that my portion was more than half is
consistent with $63/64:$

p.4 = p[y == 4]
mean(p.4 > .5)
[1] 0.9843074  # aprx 63/64
63/64
[1] 0.984375


问:
Trouble understanding the general polynomial of degree $n$
Herstein defines the general polynomial of degree $n$ over a field $F$ as the polynomial $p(x) = x^n + a_1x^{n-1} + \cdots + a_n \in F(a_1, \cdots, a_n)[x]$. From here, since it was previously shown that the group of automorphisms of this field is $S_n$, we show that polynomial is not solvable by radicals over $F$.

But this whole result doesn't make much sense to me. Before, we treated of $F(x_1, \cdots, x_n)$ where these "$x_i$" were just symbols, like the "$x$" in a polynomial $p(x)$. And now, we are taking the coefficients of $p(x)$, which can very well be inside the field $F$.

For example, the polynomial $x^5 - 7x + 2$ is in $\mathbb{Q}[x]$, and so $\mathbb{Q}(a_1, \cdots, a_5) = \mathbb{Q}$, and this seems to contradict the earlier result that the group was $S_5$.

I know I am making a huge mess, but I just can't seem to really grasp the concept, and would love some sort of enlightenment. Since I am self studying, I don't really have a professor to go to.

Thanks a lot in advance!

答:
My mistake above was to think I had calculated the splitting field of $p(x)$, when, in reality I had calculated the field over which we are working on. Let me elaborate.

We want to get to the splitting field $F(x_1, \cdots, x_n)$, where these are the roots of a polynomial. What we have to start with are, by the Newton-Girard identities, the elementary symmetric functions on these roots, namely, the coefficients of the polynomial $p(x)$, and their generated field, $F(a_1, \cdots, a_n)$.

Dietrich Burde was right in his comment - nothing was wrong, except my interpretation of the problem. When I said $\mathbb{Q}(a_1, \cdots, a_5) = \mathbb{Q}$ in the example, I was absolutely right. The thing is, this "collapsed" field (by which I mean, the result after plugging in the actual particular values) is the particular field of elementary symmetric functions over the roots of this polynomial, which remain unknown. So I didn't obtain $F(x_1, \cdots, x_n)$, as I thought I had, but rather $F(a_1, \cdots, a_n)$.

And the theorem I stated in the question says that $\operatorname{o}(\operatorname{Aut}_{\mathbb{Q}}\mathbb{Q}(x_1, \cdots, x_5)) = 5!$, whereas, by my original reasoning, I thought I had $\operatorname{o}(\operatorname{Aut}_{\mathbb{Q}}\mathbb{Q}) = 1 \neq 5!$ as it should be.

EDIT: I believe the same reasoning explains the apparent contradiction that there are indeed quintics solvable by radicals: when actually plugging in the values of the roots into the field $F(x_1, \cdots, x_n)$, some of them may have already been in the previous field, so the underlying result is much more complex than the theorem on rational functions predicted and the bound may end up being much smaller than $n!$


问:
Linear Algebra Transformations
I need some help on this problem.

Assume that $V$ is a complex vector space. Let $T,S: V \rightarrow V$ be two linear operators such that $TS - ST = S$. Show that:

(a) Let $W$ be the subspace spanned by the eigenvectors of $T$. Show that $W$ is invariant under both $T$ and $S$.

(b) Show that $S_{|W}$ is a nilpotent operator.

(c) Show that there exists a nonzero common eigenvector of $T$ and $S$.

答:
I suppose I should answer this.

(a) You should prove that it's enough to show that $Sv \in W$ for any eigenvector $v$ of $T$. Now just start going: write $Tv = \lambda v$, and try to say something interesting about $Sv$ using the formula.

(b) This relies on something you'll discover about eigenvalues in the course of proving (a). It's going to seem sort of surprising. Again, a simplification: it's enough to prove that there is some $n$ such that for each eigenvector $v$ for $T$, $S^nv = 0$. I think you need to use the finite dimensionality of $V$.

(c) Such a vector is lurking in your work for the first two parts.

It's hard to resist mentioning the Lie algebra of $\mathfrak{sl}_2$: take, as an important example, $V = \mathbb C^2$, $T = \frac12 \begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}$ and $S = \begin{bmatrix}0 & 1\\ 0 & 0 \end{bmatrix}$.

Title: Linear Algebra Transformations