答:
You need, first, to show that $N\le G$.

I will use the two-step subgroup test.

Fix $H\le G$.

Since $e\in H$ because $H\le G$ and, for any $a\in G$, we have $a^{-1}ea=e$, we can conclude that

$$e\in\bigcap_{a\in G}a^{-1}Ha=N,$$

so that $N\neq\varnothing$.

Since for all $a\in G$,

$$a^{-1}Ha=\{a^{-1}ha\in G\mid h\in H\},$$

we have that $N\subseteq G$.

Let $x,y\in N$. Then $x,y\in a^{-1}Ha$ for all $a\in G$. So, fixing $a\in G$, we get $x=a^{-1}h_{a,x}a$ and $y=a^{-1}h_{a,y}a$ for some $h_{a,x},h_{a,y}\in H$.

Then

$$\begin{align}
x^{-1}&=(a^{-1}h_{a,x}a)^{-1}\\
&=a^{-1}h_{a,x}^{-1}(a^{-1})^{-1}\\
&= a^{-1}h_{a,x}^{-1}a\\
&\in a^{-1}Ha
\end{align}$$

as $h_{a,x}^{-1}\in H$, as $H\le G$. Hence $x^{-1}\in N$ since $a$ is arbitrary.

Consider

$$\begin{align}
xy&=(a^{-1}h_{a,x}a)(a^{-1}h_{a,y}a)\\
&=(a^{-1}h_{a,x})(aa^{-1})(h_{a,y}a)\\
&=a^{-1}(h_{a,x}h_{a,y})a,
\end{align}$$

which is in $a^{-1}Ha$ since $H\le G$ implies $h_{a,x}h_{a,y}\in H$. Hence $xy\in N$ as $a$ is arbitrary.

Hence $N\le G$.

For normality, it suffices to show that $$z^{-1}Nz\subseteq N$$ for all $z\in G$. To this end, fix $z\in G$. Then

$$\begin{align}
s\in z^{-1}Nz &\implies s\in z^{-1}\left(\bigcap_{a\in G}a^{-1}Ha\right)z\\
&\implies s\in\bigcap_{a\in G}z^{-1}(a^{-1}Ha)z\\
&\implies s\in\bigcap_{a\in G}(az)^{-1}H(az)\\
&\implies s\in\bigcap_{b\in G}b^{-1}Hb\tag{1}\\
&\implies s\in N,
\end{align}$$

where $(1)$ holds since right multiplication by a group element on a group is bijective. Hence $z^{-1}Nz\subseteq N$.

Hence $N\unlhd G$.

Body:
If $H$ is any subgroup of $G$ and $N = \bigcap_{a\in G}a^{-1}Ha$, prove that $N \triangleleft G$.

The notation here $N = \bigcap_{a\in G}a^{-1}Ha$ is tripping me up and I am not sure how to handle it. I know that if I can show that left cosets of $N$ equal right cosets of $N$ or that all $a \in G$ are abelian, then that is sufficient enough to conclude that $N \triangleleft G$, but I suspect there another way that I should be approaching this problem that uses the fact that uses the intersection.

Any tips or hints in the right direction would be helpful, thanks


问:
Is the projective linear group $PGL(3,\mathbb{R})$ a Lie group?
I have never come across a statement that it is a Lie group, although
the group has 8 one-parametric subgroups  and we can asign to $PGL(3,\mathbb{R})$ a Lie algebra.

So, is the $PGL(3,\mathbb{R})$  a Lie group  and what is a connection between $PGL(3,\mathbb{R})$ and the above mentioned Lie algebra?

Thanks.

答:
So you already know that $GL(3, \mathbb{R})$ is a Lie group. $PGL(3, \mathbb{R})$ is a quotient of $GL(3, \mathbb{R})$ by the subgroup $\mathbb{R}^*$ sitting inside $GL(3, \mathbb{R})$ as the scalar multiples of the identity.

The interesting general question here is then: suppose we know that $G$ is a s Lie group and $H$ is a normal subgroup of $G$, how 'well-behaved' must $H$ be in order for the quotient $G/H$ to be a Lie group as well?

I hope some other commenter with more time can answer this general question. However for $GL(3, \mathbb{R})$ or more general $GL(n, \mathbb{R})$ with $n$ odd we are in a very special case where $H$ is not just well-behaved but very well behaved: here the scalar multiples of the identity are not just a normal subgroup but a factor in a direct product:

$$GL(3, \mathbb{R}) \cong \mathbb{R}^* \times SL(3, \mathbb{R})$$

It is easy to see that this group isomorphism is continuous, so it is not just an isomorphism of groups but also of Lie groups.

As a result we have that $PGL(3, \mathbb{R}) \cong SL(3, \mathbb{R})$ both as a group and as a manifold.

So in this case it is much easier to check that $PGL(3, \mathbb{R})$ is a Lie group: we can just check it for $SL(3, \mathbb{R})$, which is easy since it is 'concrete' with 'actual' elements that are matrices rather than 'abstact' in the sense of 'the quotient of something else'.

Subsequently the Lie algebra of $PGL(3, \mathbb{R})$ is the same as the Lie algebra of $SL(3, \mathbb{R})$. Usually this Lie algebra is denoted $\mathfrak{sl}(3, \mathbb{R})$ and represented as the set of all 3-by-3 trace zero matrices.

In a sense the question whether $PGL(n, \mathbb{R})$ is a Lie group is more interesting for even $n$: the answer is still yes, but the reason is more subtle.

ContentLicense: CC BY-SA 4.0


问:
Does this property of limits hold when we compose functions?
Suppose we have two functions $f,g:\mathbb{R}\to{\mathbb{R}}$ such that:

$$\lim_{x\to{}\alpha}{f(x)}=L \text{ and }\lim_{y\to{\beta}}{g(y)}=\alpha.$$

Is then true that:

$$\lim_{y\to{}\beta}{f(g(y))=L}?$$

My intuition tells me yes. Using an $\epsilon-\delta$ approach seems very long and tedious. Is that the case? Or is there a nice way of showing this to be true. Or, is there a nice counterexample?

Body:
Here is a counterexample :

$$g(x)=1\;if\; x\ne 0\; and\;g(0)=2$$

$$f(x)=3\;if\;x\ne 1\; and\;f(1)=4$$

thus

$$f(g(x))=4\;if\;x\ne 0\;and\;f(g(0))=3$$
take $$\alpha=1\;and \;\beta=0$$

we have
$$\lim_{x\to 1}f(x)=3=L$$
$$\lim_{x\to 0}g(x)=1$$
but
$$\lim_{x\to 0}f(g(x))=4\ne 3$$


问:
What are examples of DISCRETE probability distributions (a) with a mean but no finite variance, (b) without a mean?
What are examples of  DISCRETE probability distributions (a) with a mean but no finite variance, (b) without a mean?

For continuous distributions, good answers are (a) the pareto distribution with alpha parameter between 1 and 2, and (b) the Cauchy distribution. (Actually, does anybody know a better common  example than the pareto, since it has have that particular parameter range?)

It should be pretty easy to construct a discrete example with fat tails too, where the infinite series converges for probability but not for higher moments, but my googling didn't find one.

答:
I'm not sure if this is what you're looking for, but for $n \in \mathbb{N} $
$$ p_n = A n^{-(1+m)}$$
With $A$ a normalizing constant, and $m=1,2,...$ being the first moment you want to diverge. This works because the sum of $1/n$ diverges but the sum of $1/n^2$ converges. There are of course many other such distributions.


问:
An Invertible n by n matrix can be reduced to the identity matrix using elementary row operations - A proof
I'm quite new to Linear Algebra and just wondered if someone could take me through the proof above? I realise some people have asked this before but I'm afraid some parts weren't understandable for me. I have learnt row reduced echelon forms and etc but just can't piece a proof together. Thank you very much!

As Ben suggested, this is the one I've seen -

Why can the row-reduced echelon matrix $R$ only be identity matrix?

Upon closer inspection, it's actually answered by you Ben!

So essentially, the bit I couldn't understand was why can't the RREF have a single row full of zeros? And what makes having a RREF matrix with a row full of zeros not invertible?

Apologies if I'm missing something very obvious :)

答:
Why can't the RREF have a single row full of zeros? And what makes having a RREF matrix with a row full of zeros not invertible?

Answering the second question addresses the first, so let's start there. Suppose that $R$ is an $n \times n$ matrix whose last row is zero. It follows that for any vector $x \in \Bbb R^n$, the vector $Rx$ will have a zero as its last entry; I recommend that you try to convince yourself that this is true using the definition of matrix multiplication.

With the above observation, we see that the equation $Rx = b$ where
$$
b = \pmatrix{0\\ \vdots \\ 0 \\ 1}
$$
cannot have a solution. However, if $R$ were invertible, then the equation would have a solution because
$$
Rx = b \iff
R^{-1}Rx = R^{-1}b \iff x = R^{-1}b.
$$
So because $R$ has a final row of zeros, the equation $Rx = b$ does not have a solution, which implies that $R$ is not invertible.


问:
Consider the sequence $a_n$ where $a_0=0$ and $a_{n+1} = \sqrt{a_n+6}$. Show that $a_n$ is contracting.
Consider the sequence $a_n$ where $a_0=0$ and $a_{n+1} = \sqrt{a_n+6}$. Show that $a_n$ is contracting. Conclude that the sequence $a_n$ converges and compute its limit.

I have been able to prove its bounded $0 < a_n <3$ and increasing, and have found its limit, but am not able to show its contracting.

答: Hint: Prove that $|f'(x)|<1$ for $f(x)=\sqrt{x+6}$ when $x >0$.


问:
How does the length of the tangent in this picture relate to $\sin(x)/\cos(x)$?

I understand the justificaton behind the length of $\sin(\theta)$ and $\cos(\theta)$. But I don't understand why AE is the length of $\tan(\theta)$, I understand that it satisfies the pythagorean identities but how exactly is that length related to $\frac{\sin(x)}{\cos(x)}$? I also don't understand the lengths of the other trigonometric functions $\sec(\theta)$, $\csc(\theta)$ and $\cot(\theta)$.

答: Right-angled triangles with one angle equal to $\theta$ include $\triangle OAE,\,FAO$, each with $\theta$ (the right-angle) at the vertex listed first (second). Now use similarity to $\triangle OCA$.


问:
Weierstrass $M$-test problem, $f_n(x)=(nx^2)/(n^3+x^3)$

Use the Weierstrass M-test to show $$f(x)=\sum_{n=1}^\infty \frac{nx^2}{n^3+x^3}$$ converges uniformly on any finite interval $[-R,R]$.

This was an exam question I had. My attempt was to find an upper bound for $\frac{nx^2}{n^3+x^3}$ by taking the derivative and finding critical points. One issue I had was that the function isn't defined if $x = -n$ for any $n \in \mathbb{N}$. I suppose we could just take the sum to start at $\mathrm{Ceil}(R)$.

My attempts to find critical points for the derivative led me to concluding $f_n'(x)=0$ if $x=0$ or $x$ has
$$-nx^3-n^4x+2n^4=0.$$

The class is over, is there any kind of hints to do this or was I on the right track? Thanks very much

答:
For $n>R\geq x>0$, $$\frac{nx^2}{n^3+x^3}\leq \frac{x^2}{n^2}\leq \frac{R^2}{n^2}$$

and a similar bound can be found when $x<0$, by the fact that $-R\leq x$.

which gives you uniform convergence on any interval $[-R,R]\backslash (-\mathbb{N})$. As you rigthfully point out, $x$ cannot be a negative integer. Notice that uniform convergence essentially only cares about how the tail of the series behaves and not on a fixed (i.e. not depending on $x$) number of finite terms , in this case $R$ of them.

ContentLicense: CC BY-SA 3.0


问:
Complement of closed convex set in Hilbert space
Let $H$ be a real Hilbert space and $C$ be a closed convex subset of $H$.

Is it true that if $x$ is a boundary point of $C$ then there exists $y\in H\setminus C$ such that the set $\{x+t(y-x);\;t>0\}$ lies in $H\setminus C$ ?

Thank you for any hint.

答:
We prove this fact by contradiction. The negation of your assertion is: For every $y \in H \setminus C$, there is $t > 0$, such that $x + t \, (y - x) \in C$. It is also clear that this property holds for all $y \in C$ (choose $t = 1$).

Hence, for all $d \in H$, there is $t > 0$ with $x + t d \in C$.
This means that the point $x$ belongs to the core (or: algebraic interior) of $C$. For closed, convex sets in Hilbert spaces, we have that the core coincides with the interior. Hence, $x$ belongs to the interior. This is a contradiction since $x$ is assumed to be a boundary point.


问:
Determine a generator of a subextension in a non-Kummer-setting
Let $K = \mathbb{Q}_3$ and $L=K(\alpha)$ where $\min_K(\alpha) = x^4 - 3x^2 + 18$. Furthermore, let $F/K$ be the unique unramified extension of degree $4$. It can be shown that $F/K$ is generated by a primitive $5$-th root of unity $\zeta_5$.

Now consider the automorphism $\varphi \in Gal(FL/K)$ given by $\zeta_5 \mapsto \zeta_5^3$ and $\alpha \mapsto \frac{(2 \alpha^2 - 3)u}{\alpha}$ where $u = \sqrt{-2/7}$ (this is another root of $x^4-3x^2+18$, cf. this post). The order of $\varphi$ is $4$.

Let $L' = (LF)^{\langle \varphi \rangle}$. By construction, the extension $LF/L'$ is cyclic of degree $4$.

Question: What is a generator $\gamma$ of $L'$?

This $\gamma$ must be an eigenvector of $\varphi$ with eigenvalue $1$ but it must be $\gamma \not\in K$.

I suppose that a basis of $LF$ as a $K$-vector space is given by $(1,\zeta_5, \zeta_5^2, \zeta_5^3, \alpha, \alpha \zeta_5, \alpha \zeta_5^2, \alpha \zeta_5^3)$. If I consider the representation matrix of $\varphi$ wrt. this basis,  we will receive a $8 \times 8$-matrix with two quadratic block matrices of size $4$ on the top left and the bottom right. I think that the bottom right matrix will have eigenvalue $1$ which will also give us the desired $\gamma$. But I am stuck with my computations here (the numbers seem to be really messy).

It would have been a bit easier if we would be in a Kummer-theoretic setting (i.e. the base field has a root of unity of degree $4$) with which I am a bit more familiar. But $\mathbb{Q}_3$ does not have a $4$-th root of unity.

Could you please guide me through my computation (or at least say if my approach is correct)?

答:
$\varphi$ is a lift of the Frobenius so $L'/K$ is totally ramified, of degree $2$, there are two such extensions $K(\sqrt3)$ and $K(\sqrt{-3})$, a check shows that it is

$$L'=K(\sqrt{ 3})=K(r\alpha), r=\sqrt{3/\alpha^2}\in F$$


答:
Regarding (a), the comment here suggests a random variable $X$ with probability mass function
$$p_{X}(x) = \dfrac{1}{\zeta(3)x^3}$$
for $x = 1, 2, \dots$.

We have $\sum_{x=1}^{\infty}p_{X}(x) = 1$ by definition.

It is also known that
$$\mathbb{E}[X] = \sum_{x=1}^{\infty}xp_{X}(x) = \sum_{x=1}^{\infty}\dfrac{1}{\zeta(3)x^2} = \dfrac{1}{\zeta(3)} \cdot \dfrac{\pi^2}{6}$$
and
$$\mathbb{E}[X^2] = \sum_{x=1}^{\infty}x^2p_{X}(x) = \sum_{x=1}^{\infty}\dfrac{1}{\zeta(3)x}\text{.}$$
We have
$$\dfrac{1/[\zeta(3)x]}{1/x} = \dfrac{x}{\zeta(3)x} = \dfrac{1}{\zeta(3)} < \infty$$
so since $\sum_{x=1}^{\infty}\dfrac{1}{x}$ diverges, it follows that $\mathbb{E}[X^2]$ is not finite. Thus, $\text{Var}(X)$ does not exist.

Regarding (b), the answer here gives the example of a random variable $X$ with probability mass function
$$p_{X}(x) = \dfrac{1}{x} - \dfrac{1}{x + 1}$$
for $x = 1, 2, \dots$.

We observe that
$$\sum_{x=1}^{\infty}p_{X}(x) = \sum_{x=1}^{\infty}\left(\dfrac{1}{x}-\dfrac{1}{x+1} \right) = \lim_{n \to \infty}\sum_{x=1}^{n}\left(\dfrac{1}{x}-\dfrac{1}{x+1} \right) = \lim_{n \to \infty}\left(1 - \dfrac{1}{n + 1} \right) = 1$$
but
$$\begin{align}
\mathbb{E}[X] &= \sum_{x=1}^{\infty}xp_{X}(x) \\
&= \sum_{x=1}^{\infty}x\left(\dfrac{1}{x} - \dfrac{1}{x+1} \right) \\
&= \lim_{n \to \infty}\sum_{x=1}^{n}x\left(\dfrac{1}{x} - \dfrac{1}{x+1} \right) \\
&= \lim_{n \to \infty}\left[1\left(1 - \dfrac{1}{2} \right) + 2\left(\dfrac{1}{2} - \dfrac{1}{3} \right) + 3\left(\dfrac{1}{3} - \dfrac{1}{4} \right) + \cdots + n\left(\dfrac{1}{n} - \dfrac{1}{n+1} \right)\right] \\
&= \lim_{n \to \infty}\left(1 - \dfrac{1}{2} + 1 - \dfrac{2}{3} + 1 - \dfrac{3}{4} + \cdots + 1 - \dfrac{n}{n+1} \right) \\
&= \lim_{n \to \infty}\left(\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \cdots + \dfrac{1}{n+1} \right) \\
&= \lim_{n \to \infty}\sum_{x=1}^{n}\dfrac{1}{x+1}
\end{align}$$
which diverges since $$\int_{1}^{\infty}\dfrac{1}{x+1}\text{ d}x $$
is divergent, and then applying the integral test.

Body:
What are examples of  DISCRETE probability distributions (a) with a mean but no finite variance, (b) without a mean?

For continuous distributions, good answers are (a) the pareto distribution with alpha parameter between 1 and 2, and (b) the Cauchy distribution. (Actually, does anybody know a better common  example than the pareto, since it has have that particular parameter range?)

It should be pretty easy to construct a discrete example with fat tails too, where the infinite series converges for probability but not for higher moments, but my googling didn't find one.


问:
Non-uniformly continous function
I have found a topic here (https://www.sciencedirect.com/topics/mathematics/uniformly-continuous-function) in Theorem 3.3.10 that: There is a continuous function $f$ on $[0, 1]$ which is unbounded, and
therefore not uniformly continuous.

Anyone, please help me to understand it.

答:
The proof you link defines

Let $(I_n)_n$ be a covering of $[0, 1]$ by open intervals $(r_n, s_n)$ (...)

and then it says

But by the property of this cover, it follows that no finite part of it covers [0, 1]

Which is false: every open cover of $[0,1]$ has finite subcover, because $[0, 1]$ is compact. The proof is just wrong.

Title: Non-uniformly continous function


答:
$\angle OAE = \frac{\pi}{2} \ (= 90^\circ)$,

$\angle CAE = \angle OAE - \angle OAC = \frac{\pi}{2} - \left(\frac{\pi}{2} - \angle COA\right) = \theta$,

So now $\Delta CAE$ and $\Delta COA$ share two of the same angles and so must be similar,

Hence the ratio of their sides must be equal, and the result follows.

Title: How does the length of the tangent in this picture relate to $\sin(x)/\cos(x)$?

Body: I understand the justificaton behind the length of $\sin(\theta)$ and $\cos(\theta)$. But I don't understand why AE is the length of $\tan(\theta)$, I understand that it satisfies the pythagorean identities but how exactly is that length related to $\frac{\sin(x)}{\cos(x)}$? I also don't understand the lengths of the other trigonometric functions $\sec(\theta)$, $\csc(\theta)$ and $\cot(\theta)$.


问:
$k$-dimensional manifold in $k^2$-dimensional space that passes through 3 points
I'm very unsure about the mathematical terminology, so any help with that would be appreciated too.

$d$ is a $k$-dimensional space and $D$ is a $k^2$-dimensional space. $x$, $y$, and $z$ are points in $D$. $P: D \to d$ is an orthogonal projection.

I am looking for an $f: d \to D$ function that passes through $x$, $y$, and $z$ and $P(f(p)) = p$. So $f$ is kind of an inverse of $P$ that picks a manifold in $D$ that goes through $x$, $y$, and $z$.

I think many such manifolds are possible. I'm looking for something simple and smooth.

The application is that a user can see and manipulate a simplified view ($d$) in an application and I want to map it back to the full thing ($D$) while making sure some special points in $D$ are accessible.

答:
I spoke to a mathematician! (My wife got home.)

So I think $f(p_1 ... p_k) = (p_1 ... p_k, g_{k+1}(p) ... g_{k^2}(p))$ where $g_{k+1} ... g_{k^2}$ are quadratic (?) polynomials that make sure $x,y,z$ are covered.


问:
If $f: \mathbb{C}\to\mathbb{C}$ is bounded, then is it a constant?
If a function $f: \mathbb{C}\to\mathbb{C}$ is bounded, then it is a constant. Is it true or false?</p>

答:
If you don't require that $f$ is analytic then this is false, as Martin Argerami says, but you can find a continuous function which is bounded and not constant:

$$f(z)=\begin{cases} z & |z|\leq1 \\\\ \frac z{|z|} & \text{otherwise}\end{cases}$$


问:
Do Regular Functions Separate Closed Sets in Affine Variety?
I'm focusing on the language of classic algebraic geometry.

Let $X$ be an affine variety and $A(X)$ be its coordinate ring. Let $E,F\subseteq X$ be $2$ disjoint closed subsets of $X$. Does there exist an $f\in A(X)$ such that $f(E) = 0$ and $f(F) = 1$?

If this is true, is there any generalization of this or in the language of scheme?

答: Yes, this is true. Let $R,S$ be the coordinate rings of $E,F$ respectively. Then $R\times S$ is the coordinate ring of $E\cup F$ as $E$ and $F$ are disjoint, and $A(X)\to R\times S$ is a surjection because $E\cup F$ is a closed subvariety/subscheme of $X$. So there is an element of $A(X)$ which has image $(0,1)\in R\times S$ and does exactly what you want. This proof works just fine for both affine schemes and affine varieties.


问:
Minimal Maximum $m$-sum
Let $n$ be a positive integer and $m\leq n$. Let $\sigma$ be any rearrangement of $1,2,\dots,n$, then define a $m$-sum of $\sigma$ to be a sum of $m$ consecutive numbers in the list. Define $g_m(\sigma)$ to be the maximum $m$-sum of $\sigma$, then define $f(n,m)$ be the minimal $g_m(\sigma)$ across all $\sigma$. What can we say about $f(n,m)$?

For example, let $n\geq3$ and $\sigma$ be $1,2,3,4,5$, then the $3$-sums of $\sigma$ are $1+2+3,2+3+4,3+4+5$, or $6,9,12$, and $g_3(\sigma)=12$. My program shows that $f(5,3)=8$ with the permutation $5,2,1,4,3$.

Now the question is to find a "simple" expression for $f(n,m)$.

Here is some fairly easy observations.

Trivially $f(n,n)=\frac{1}{2}n(n+1)$ and $f(n,1)=n$.

We can show that $f(n,2)=n+1$. For any permutation $\sigma$, $n$ must be next to a positive integer, and this gives a $2$-sum of $\sigma$ that is at least $n+1$ so $g_2(\sigma)\geq n+1$ for all $\sigma$ and thus $f(n,2)\geq n+1$. Now by looking at the permutation $n,1,n-1,2,n-2,3,\dots$, we see that the minimum $2$-sum is exactly $n+1$.

$f(n,n-1)=\frac12n^2-\frac12n+1$. $n$ has to be in a $(n-1)$-sum, and then the smallest we can make the remaining numbers to be is $1,2,\dots,n-2$, leaving out $n-1$, and this shows that $g_{n-1}(\sigma)\geq1+2+\cdots+n-2+n=\frac12n^2-\frac12n+1$ for all $\sigma$ and thus $f(n,n-1)\geq\frac12n^2-\frac12n+1$. Then the two $(n-1)$-sums of the permutation $n-1,1,2,\dots,n-2,n$ are $\frac12n(n-1)$ and $\frac12n(n-1)+1=\frac12n^2-\frac12n+1$, so we see that $f(n,n-1)$ is exactly $\frac12n^2-\frac12n+1$.

$f(n,n-2)=\frac12n^2-\frac32n+3$. Again, $n$ has to be in a $(n-2)$-sum, and the smallest we can make the remaining numbers to be is $1,2,\dots,n-3$, leaning out $n-1,n-2$, and this shows that $g_{n-2}(\sigma)\geq\frac12(n-2)(n-3)+n$ for all $\sigma$ so $f(n,n-2)\geq\frac12(n-2)(n-1)+n$. The three $(n-2)$-sums of the permutation $n,n-3,n-4,\dots,2,1,n-1,n-2$ are $\frac12(n-2)(n-3)+n$, $\frac12(n-2)(n-3)+n-1$ and $\frac12(n-2)(n-3)+n-1$. Therefore $f(n,n-2)=\frac12(n-2)(n-3)+n=\frac12n^2-\frac32n+3$.

However, for $f(n,n-3)$ we can no longer just leave out $n-1,n-2,n-3$. Using a python program, I found out that
\begin{array}{|c|c|c|c|}
\hline
n & f(n,n-3) & \sigma \\ \hline
 6 & 11 & 6, 4, 1, 5, 3, 2 \\ \hline
 7 & 15 & 7, 5, 2, 1, 6, 4, 3 \\ \hline
 8 & 20 & 8, 6, 3, 2, 1, 7, 5, 4 \\ \hline
 9 & 26 & 9, 7, 4, 3, 2, 1, 8, 6, 5 \\ \hline
 10 & 33 & 10, 8, 5, 4, 3, 2, 1, 9, 7, 6 \\ \hline
\end{array}

Of course, the $\sigma$ that gives the minimal $g_m(\sigma)$ is not unique, since you can just simply reverse the permutation. Here I choose the largest permutation with dictionary order.

So the optimal solution seems to be $n,n-2,n-5,n-6,\dots,2,1,n-1,n-3,n-4$ and gives $f(n,n-3)=\frac12n^2-\frac52n+8$. Moreover this only works for $n\geq6$ and not $n=5$. For $f(n,n-4)$ the optimal solution seems to be $n,n-1,n-6,n-7,\dots,1,n-2,n-3,n-4,n-5$ which gives $f(n,n-4)=\frac12n^2-\frac72n+14$, and this only works for $n\geq7$ and not $n=6$. It seems like by doing something similar for $f(n,n-1)$ and $f(n,n-2)$ there is always a simple formula for $f(n,n-a)$ when $n$ is large.

If we try to find a formula for $f(n,3)$, the situation is much more strange. Using the same python code, I found out that
\begin{array}{|c|c|c|c|}
\hline
n & f(n,3) & \sigma \\ \hline
3 & 6      & 3,2,1  \\ \hline
 4 & 7      & 4, 2, 1, 3 \\ \hline
 5 & 8      & 5, 2, 1, 4, 3 \\ \hline
 6 & 11     & 6, 4, 1, 5, 3, 2 \\ \hline
 7 & 12     & 7, 4, 1, 6, 3, 2, 5 \\ \hline
 8 & 13     & 8, 4, 1, 7, 3, 2, 6, 5 \\ \hline
 9 & 15     & 9, 5, 1, 8, 4, 3, 7, 2, 6 \\ \hline
 10 & 16    & 10, 5, 1, 9, 4, 3, 8, 2, 6, 7 \\ \hline
\end{array}

A few observations: $n$ is always at position $1$ and $1$ is always at position $3$.
Moreover, the numbers at position $1,4,7,\dots$ always appear to be $n,n-1,n-2,\dots$, and if we assume this we can make the program run much faster:
\begin{array}{|c|c|c|c|}
\hline
n & f(n,3) & \sigma \\ \hline
11 & 18 & 11,6,1,10,5,3,9,4,2,8,7 \\ \hline
 12 & 20 & 12,7,1,11,6,3,10,5,4,9,2,8 \\ \hline
 13 & 21 & 13,7,1,12,6,3,11,5,4,10,2,8,9 \\ \hline
 14 & 22 & 14,7,1,13,5,4,12,6,3,11,8,2,10,9 \\ \hline
 15 & 25 & 15,9,1,14,8,3,13,7,5,12,6,4,11,10,2 \\ \hline
 16 & 26 & 16,9,1,15,8,3,14,7,5,13,6,4,12,10,2,11 \\ \hline
\end{array}

There is one more pattern: the numbers at positions congruent to $1$ modulo $3$ are always greater than the numbers at positions congruent to $2$ modulo $3$, which are in turns greater than the numbers at positions congruent to $0$ modulo $3$. So these three sets of positions roughly contain the largest $1/3$, the middle $1/3$, and the smallest $1/3$ of the numbers. However, these are all just speculations and I don't really have the proof.

答:
This is not a full answer but should give you a good basis for exploring your interesting idea.

I've looked at $f(n,3)$ when $n$ is a multiple of 3. The total of all the numbers is $$\frac{1}{2}n(n+1).$$
Splitting these up into their successive blocks of three numbers shows that a lower bound for $f(n,3)$ is given by the ceiling function for $$\frac{3}{2}(n+1).$$
This bound appears to be attained. This, in fact, agrees with your results except for $n=15$. You give $25$, but the bound given by the ceiling function is $24$ and this is actually attained for
$$15,8, 1,13,9,2,11,3,10,5,7,12,4,6,14.$$

To prove that a given lower bound is attained simply requires an algebraic demonstration.

For example - Proof for $n=6k$

The numbers can be arranged as shown, where the centre of the list is marked:-

$$6k,3k,1,...,5k+1,2k+1,2k-1||2k,3k+1,4k+1, ..., 2,4k,5k.$$

The pattern here is that as you move 3 places out from the centre you change the number by $-2,+1,+1,-2, +1,+1,-2,...$.

A quick check should show that the min-max is $9k+2$.

A general lower bound

In general, a lower bound for $f(n,m)$ can be found by splitting numbers up into  successive blocks of $m$ numbers with an additional smaller block if $m$ is not a factor of $n$.

Title: Minimal Maximum $m$-sum


答:
For (b)

For $n\geq1,P(X=2^n)=(\frac{1}{2})^n$ and $P(Xa)=0$ otherwise

Title: What are examples of DISCRETE probability distributions (a) with a mean but no finite variance, (b) without a mean?


答: Note that $$f(x) - f(y) = \sqrt{x+6} - \sqrt{y+6} = \frac{x-y}{\sqrt{x+6} + \sqrt{y+6}}$$ so that $$|f(x) - f(y)| = \frac{|x-y|}{\sqrt{x+6} + \sqrt{y+6}} \le \frac{|x-y|}{2\sqrt 6}$$ provided that $x,y \ge 0$.

Body:
Consider the sequence $a_n$ where $a_0=0$ and $a_{n+1} = \sqrt{a_n+6}$. Show that $a_n$ is contracting. Conclude that the sequence $a_n$ converges and compute its limit.

I have been able to prove its bounded $0 < a_n <3$ and increasing, and have found its limit, but am not able to show its contracting.


问:
Is there a reasonably accurate and easy way to approximate lbs-stones or kg-stones (and vice versa)?
For example, when I need to talk with my American friends about body weight:

$\mathrm{KG}\times2+10\%$

($100\times2=200$, $200+10\%=220$... Pretty accurate and easy to do mentally, both ways)

However, when I need to talk with my British friends about body weight, how do I approximate from KG to Stones?

Additionally, if I have a similar formula for Stones-lbs, that will be useful too!

I mean, I know a Stone is 14 lbs, but how to approximate 247lbs in Stones without pulling out a calculator?

For your reference, a Stone is about 6.35 KG

I am thinking that a "good, practical" approximation follows these rules:

5% accuracy is kinda bad... Imagine weighing 100kg and wanting to lose 5kg - the error alone is making discussion about it difficult! I am thinking up to 2% error

Maximum of 3 steps

Only addition, subtraction, multiplication and division allowed (% also allowed)

Multiplication only by "easy" factors: 2, 3, 4, 5, 10

Division only by "easy divisors": 2, 10

Only "easy" percentages like: 1%, 10%...

答:
To go from kilograms to stone, double the number of kilograms 4 times, then move the decimal point two places to the left. This is effectively the same as dividing by 6.25, which is only about 1.6% different from 6.35.

For example, how much is 98 kg in stone? Double 98 four times (-> 196 -> 392 -> 784 -> 1568), then move the decimal point 2 places to the left: about 15.68 stone.

The actual answer is about 15.43 stone, so 15.68 stone is a pretty good estimate.


问:
If $H$ is any subgroup of $G$ and $N = \bigcap_{a\in G}a^{-1}Ha$, prove that $N \triangleleft G$.

If $H$ is any subgroup of $G$ and $N = \bigcap_{a\in G}a^{-1}Ha$, prove that $N \triangleleft G$.

The notation here $N = \bigcap_{a\in G}a^{-1}Ha$ is tripping me up and I am not sure how to handle it. I know that if I can show that left cosets of $N$ equal right cosets of $N$ or that all $a \in G$ are abelian, then that is sufficient enough to conclude that $N \triangleleft G$, but I suspect there another way that I should be approaching this problem that uses the fact that uses the intersection.

Any tips or hints in the right direction would be helpful, thanks

答:
A bit surprised to see all the lengthy answers here. I would certainly be discouraged if I needed to read one of those!

Note that $N$ is an intersection of subgroups, so $N$ is a subgroup itself.

If $g \in G$, then
$$g^{-1}N g = g^{-1}\left(\bigcap_{a \in G}a^{-1} H a\right) g = \bigcap_{a \in G} (ag)^{-1} H (ag) = N$$
so $N$ is a normal subgroup of $G$.

There are some details to be filled in in the above calculation:

Second equality: Define $\phi_g: G \to G: h \mapsto g^{-1}hg$. This is a bijection, so it has the property $\phi_g(\bigcap \mathcal{C}) = \bigcap  \phi_g(\mathcal{C})$.

Third equality: Use that $G \ni a \mapsto ag$ is a bijection, so $\{ag: a \in G\}=G$.


问:
ex falso quodlibet in natural deduction (ND)?
My understanding of inference rules is that they should be intuitively acceptable (like axioms). But my only intuition for ex falso quodlibet in natural deduction is not immediate but comes from its proof, which rests on disjunctive syllogism (which apparently holds in classical propositional logic and intuitionistic logic).
I am probably missing something obvious. Is the principle of explosion really basic/intuitive and why?

The closest justification I see, based on the current answers, is that Ex Falso encodes a (metalogical) desire that the proof system be consistent. But also see the references to intuitionism in the answers.

Edit: as pointed out below, it may have been a distraction for me to raise "intuitionism" in the original question, although Ex Falso arises in classical and intuitionistic logic. This does not exclude that I would like axioms and rules to be intuitively acceptable (to me).

" There  is  surprisingly  little  agreement  about  the  exact  status  of EFQ in  the literature.  Gentzen groups the ex  falso rule alongside the other operational rules(1969,  p.  77).   Prawitz  follows  this  practice.   He  treats ⊥ as  a  ‘0-place  sentential operation’ (Prawitz 1978, p. 38), one ‘for which there is no canonical proof’ (Prawitz1977, p. 26). EFQ is then understood as the elimination rule for ⊥. Others, including Dummett, present ⊥ as an elimination rule for ¬ (e.g. Dummett 1991, p. 291).  I contend that both of these approaches are mistaken...." ref.

Ex Falso has an ‘anomalous position [. . . ]  inside the scheme of introduction/elimination rules’, tarnishing the neat symmetry of the intuitionistic system" (Weir 1986, p. 461).

According to here: "Kolmogorov’s criterion whether to keep an axiom was whether a proposition has an “intuitive foundation” or “possesses intuitive obviousness” (van Heijenoort 1967: 421, 422) ... Kolmogorov said that, just like PEM, Ex Falso “has no intuitive foundation” (van Heijenoort 1967: 419). In particular, he says that Ex Falso is unacceptable for the reason that it asserts something about the consequences of something impossible (van Heijenoort 1967: 421)."

... "I submit that these difficulties can be overcome simply by treating⊥as a punc-tuation marker, as Tennant suggests.  Of course this means thatEFQis left withouta  logical  constant  which  it  serves  to  eliminate.   But  on  the  view  I  am  proposingno such constant is needed becauseEFQis not an operational rule at all.  Its roleis  best  accounted  for  by  according  it  the  status  of  astructural  rule.   It  simply  isthe structural rule that tells us that any sentence whatsoever follows from a con-tradiction.  Such a license isnotspecific to any logical constants, but amounts to ablanket policy.  Our reclassification ofEFQthus fits neatly with our characterizationof structural rules as global rules that assign properties to our deducibility relation." ref.

Edit: I came across this reformulation of my initial uneasiness.

"... adopting disjunctive syllogism as an axiom or as a primitive rule of inference is not an option.  Hence, it is ex  falso that is naturally treated as primitive, not the principle of disjunctive syllogism. Therefore, rather than thinking of ex  falso as the product of the mean-ings of disjunction and negation, we must treat disjunctive syllogism as a product of the ex falso rule.  This suggests that the law of disjunctive syllogism is not a con-sequence only of the meanings of the constants involved, but also hinges on one’s stance concerning the general principle expressed by EFQ." ref.

答:
In this answer, I'm going to try to convince you on very general grounds that the Ex Falso Quodlibet rule should be uncontroversial. The short version is that since $\bot\models \varphi$ for every sentence $\varphi$ (under any reasonable interpretation of these symbols), we also want $\bot\vdash \varphi$ for every sentence $\varphi$.

For me, a logic should always come with a notion of semantic entailment between sentences: $\varphi\models\psi$.

Depending on the logic, this notion could be defined in a variety of different ways, but the intuitive meaning should be the following.

(1) $\varphi\models \psi$ means: whenever $\varphi$ is true, $\psi$ is also true.

Let's suppose that the logic has a sentence $\bot$, called "false". Again, the intuitive semantics of $\bot$ should be the following.

(2) $\bot$ is never true.

Of course, we probably want to make the words "whenever" and "never" more precise. For this, we need a definition of "model" and what it means for a sentence to be "true in a model". But remember, I'm trying to be as flexible as possible here - I don't mean to suggest that a "model" needs to be a set-theoretic model like in first-order logic. If the word "model" makes you uncomfortable for some reason, feel free to substitute "world" or "situation". Anyway, we get the following more precise versions.

(1') $\varphi\models\psi$ means: For every model $M$, if $\varphi$ is true in $M$, then $\psi$ is true in $M$.

(2') For every model $M$, $\bot$ is not true in $M$.

Now it follows immediately from (1') and (2') that for any sentence $\psi$, $\bot\models \psi$. Indeed, for any model $M$, $\bot$ is not true in $M$ by (2'). So for any model $M$, if $\bot$ is true in $M$, then $\psi$ is true in $M$. Thus $\bot\models \psi$ by (1').

We can also introduce a proof system for our logic, which allows us to define the derivability relation: $\varphi\vdash\psi$.

Now the primary criterion for a proof system is that it should be sound and complete for our intended semantics. This means that for any sentences $\varphi$ and $\psi$, $$\varphi\vdash \psi \text{ if and only if } \varphi\models \psi.$$

By the discussion above, this means that we must have $\bot\vdash \psi$ for every sentence $\psi$. This is exactly what the rule Ex Falso Quodlibet gives us.

Now the desiderata of soundness and completeness completely determine the derivability relation $\vdash$: it must be equal to the entailment relation $\models$. But our logic may admit many different proofs systems which are equivalent in the sense that they produce the same derivability relation. Different (but equivalent) proof systems may be preferred in different situations on aesthetic and practical grounds. This is a subjective matter, and no logic has a "correct" proof system - different people may disagree about whether natural deduction or sequent calculus is more aesthetically pleasing, a Hilbert system may be convenient for certain proof theoretic arguments, but less convenient for others, etc. The crucial thing at the end of the day is not the particular rules of the system, but the soundness and completeness of the derivability relation.

So I have no opinion about whether Ex Falso Quodlibet should be included in a proof system as a basic rule. Personally, I find it to be both aesthetically pleasing and intuitive, but you may disagree. But if it is not included as a basic rule, it must be an admissible rule! That is, under the basic assumptions (1) and (2) on our logic, for any satisfactory proof system, we must have $\bot\vdash \psi$ for any sentence $\psi$.


答:
Consider the intersection point $T$ of line $OA$ with the tangent to the circle at $D$. By definition the length $DT$ is equal to $\tan \theta$ since we have a unit circle.

Now the right triangles $OTD$ and $OAE$ are isometric, by the second congruence case, and therefore, the corresponding edges $DT$ and $AE$  have the same length.


问:
Micropolar equations model
In most of the articles about the 3D micropolar equations, these equations are written as
\begin{equation}
\begin{cases}
\partial_t \textbf{u}-(\nu+\kappa)\Delta \textbf{u} -2\kappa\nabla\times \textbf{w}+\textbf{u}\cdot\nabla \textbf{u}+\nabla\Pi=0,\quad x \in \mathbb{R}^3,\; t\geq 0,\\
\partial_t \textbf{w}-\gamma\Delta \textbf{w}+4\kappa \textbf{w}-\mu\nabla\nabla\cdot \textbf{w} -2\kappa\nabla\times \textbf{u}+\textbf{u}\cdot\nabla \textbf{w}=0, \\
\nabla \cdot \textbf{u}=0,
\end{cases}
\end{equation}
where

$\textbf{w} =\textbf{w}(x,t)=(w_1(x,t),w_2(x,t),w_3(x,t))$ the field of microrotation representing the angular velocity of the rotation of the fluid particles, $\textbf{u} =\textbf{u}(x,t)=(u_1(x,t),u_2(x,t),u_3(x,t))$ denotes the fluid velocity,  $\Pi(x,t)$ the scalar pressure, $\kappa$ denotes the micro-rotation viscosity, $\nu$ the Newtonian kinematic viscosity, and $\gamma$ and $\mu$ the angular viscosities.

I want to know if the above written equations are considered in the general form.

That is, for example, if we consider the following ones (we change the signe in front of $2\kappa\nabla\times \textbf{u}$)
\begin{equation}
\begin{cases}
\partial_t \textbf{u}-(\nu+\kappa)\Delta \textbf{u} -2\kappa\nabla\times \textbf{w}+\textbf{u}\cdot\nabla \textbf{u}+\nabla\Pi=0,\quad x \in \mathbb{R}^3,\; t\geq 0,\\
\partial_t \textbf{w}-\gamma\Delta \textbf{w}+4\kappa \textbf{w}-\mu\nabla\nabla\cdot \textbf{w} +2\kappa\nabla\times \textbf{u}+\textbf{u}\cdot\nabla \textbf{w}=0, \\
\nabla \cdot \textbf{u}=0,
\end{cases}
\end{equation}
do the last equations are also considered as micropolar equations?

Thanks in advance.

答:
No, you cannot simply flip the sign of $\kappa$ in the second equation.  The proper physical formulation of the (isotropic, incompressible) micropolar equations is
\begin{equation}
\begin{cases}
\rho(\partial_t u + u \cdot \nabla u) = \nabla \cdot \mathbb{T} + f \\
\operatorname*{div} u =0 \\
j(\partial_t \omega + u \cdot \nabla \omega) = \nabla \cdot \mathbb{M} +2 \mathbb{T}^\ast + g
\end{cases}
\end{equation}
where $\rho >0$ is the constant fluid density, $j >0$ is the constant microinertia, $f$ is the external force, $g$ is the external microtorque,  $\mathbb{T}$ is the stress tensor, $\mathbb{M}$ is the couple stress tensor, and the vector $\mathbb{T}^\ast$ is given by
\begin{equation}
\mathbb{T}^\ast_i = \sum_{a,b} \epsilon_{aib} \mathbb{T}_{ab}
\end{equation}
where $\epsilon$ is the Kronecker symbol.   The stress tensor is given by
\begin{equation}
\mathbb{T} = \mu(\nabla u + (\nabla u)^T) - p I + \kappa \left(\frac{1}{2} \nabla \times u - \omega \right)_\ast
\end{equation}
where for a vector $v$ the tensor $v_\ast$ is defined via
\begin{equation}
(v_\ast)^\ast = v,
\end{equation}
i.e. the lower star inverts the upper star mapping defined above.  The couple stress tensor is given by
\begin{equation}
\mathbb{M} = \alpha (\nabla \cdot \omega) I + \beta (\nabla \omega + (\nabla \omega)^T) + \gamma (\nabla \times \omega)_\ast.
\end{equation}
There are then various constraints placed on the coefficients $\mu,\kappa,\alpha,\beta,\gamma$ from the thermodynamic requirement that the Clausius-Duhem inequality holds:
$$
\mathbb{T} : (\nabla u - \omega_\ast) + \mathbb{M} : \nabla \omega \ge 0.
$$

All of this can be found in the old papers deriving the micropolar model, say those of Eringen.  The essential point for your question is that the parameter $\kappa$ couples to both $u$ and $\omega$ via $\mathbb{T}$ and so you can't flip the sign in one equation without breaking the structure in terms of the tensors.  Also, for the  physical reason mentioned above, as written we must have $\kappa >0$ in the micropolar model.

Title: Micropolar equations model


问:
Finding the marginal PDF with dependent variable bounds
Consider the following joint PDF:
$$
f_{X,Y}(x,y)=
\begin{cases}
\frac{1}{8}(x-y) & -x<y<x, \; 0<x<2 \\
0 & \text{else}
\end{cases}
$$
We know that $f_{Y}(y)=\int_{-\infty}^{+\infty} f_{X,Y}(x,y)\;dx$. but since the variables are dependent ($-x<y<x$), merely integrating over the interval $(0,2)$ would not suffice. What is the approach to find the marginal PDF in such cases?

答:
The inequality $-x < y < x$ is the same as $x > |y|$. Also $-x < y < x$ along with bounds $0 < x < 2$ suggests $-2 < y < 2$ as the bounds for the support of $Y$. So pdf of $Y$ is
\begin{align*}
f_Y(y) = \int_{-\infty}^{+\infty} f_{X,Y}(x,y)\;dx &= \int_{-\infty}^{|y|} f_{X,Y}(x,y)\;dx + \int_{|y|}^{2} f_{X,Y}(x,y)\;dx + \int_{2}^{+\infty} f_{X,Y}(x,y)\;dx \\
&= 0 + \int_{|y|}^{2} \frac{1}{8}(x - y)\;dx + 0, \quad -2 < y < 2
\end{align*} and $0$ otherwise. You can easily work out the integral.


答:
If you sketch the region bounded by the inequalities $-x<y<x, 0<x<2$, it's the interior of the triangle with vertices $(0,0), (2,-2), (2,2)$.

Algebraically, by changing the inequality $-x<y$ to $-y < x$, the lower bounds of $x$ given $y$ has to satisfy all $3$ of

$$-y<x,\quad y<x,\quad0<x$$

i.e. $|y|<x$. Combining with the upper bound of $x$,

$$|y|<x<2$$

The range of $y$ that has some corresponding $x$ is

$$|y|<2\\
-2<y<2$$

Title: Finding the marginal PDF with dependent variable bounds

Body:
Consider the following joint PDF:
$$
f_{X,Y}(x,y)=
\begin{cases}
\frac{1}{8}(x-y) & -x<y<x, \; 0<x<2 \\
0 & \text{else}
\end{cases}
$$
We know that $f_{Y}(y)=\int_{-\infty}^{+\infty} f_{X,Y}(x,y)\;dx$. but since the variables are dependent ($-x<y<x$), merely integrating over the interval $(0,2)$ would not suffice. What is the approach to find the marginal PDF in such cases?


问:
Subgroup of $\mathbb{Z}\ast \mathbb{Z}$
I want to see $\mathbb{Z}\ast\mathbb{Z}$ has a subgroup of index 4 which is not normal and I want to calculate the rank of this subgroup.

One way is to classify all the subgroups of $\mathbb{Z}\ast\mathbb{Z}$. However, I want to know if there is easier way to see this.

答:
Somewhat interestingly, many results about free products are proved geometrically, and if I'm being honest, this technique is so useful I'm not familiar with purely algebraic ways to do what you're asking.
As such, this answer is going to rely on some Algebraic Topology, which you may have to take on faith depending on your background.

The idea is based on the following observations/theorems:

$\mathbb{Z} \ast \mathbb{Z}$ is the
fundamental group
of a wedge of $2$ circles.

Subgroups of a fundamental group are in bijective correspondence with
covering spaces
of the underlying space.

The subgroups of index $n$ are in bijection with the $n$-sheeted covering spaces.

The normal subgroups are in bijection with the "maximally symmetric" covering spaces.

So, to find a non-normal subgroup of index $4$, we want to find a covering of the wedge of $2$ circles with $4$ sheets that isn't symmetric.

The wedge of $2$ circles looks like this:

(the loops $a$ and $b$ correspond to the generators of $\mathbb{Z} \ast \mathbb{Z}$)

We want a covering space (for our purposes, another graph) which has $4$ loops corresponding to $a$ and $4$ loops corresponding to $b$ (that gives $4$-sheetedness). We also want to know that some (labelled) edges cannot be taken to some other edges (of the same label) by a graph automorphism. This is what is meant by "not maximally symmetric" (formally, we want to know the group of (labelled) graph automorphisms isn't transitive).

I would draw one, but Hatcher actually provides an example on page 58 of his Algebraic Topology book (which is freely available online):

Figuring out exactly how we get this subgroup from this graph, or why this graph works, are the fundamentals of an algebraic topology class, so I won't try to describe what's going on in any detail. The quick version is

There are $4$ edges labelled $a$ and $4$ edges labelled $b$. So this is a $4$-sheeted covering, and its fundamental group has index $4$ in $\mathbb{Z} \ast \mathbb{Z}$.

If we start at the thick point and we follow $ab$, we get back where we started. However, if we look at the bottom right point instead, then $ab$ takes us to the top left point (so we don't get back where we started). This witnesses the "not maximally symmetric"ness that we need to show that the subgroup we get isn't normal.

So the subgroup that we get (which, loosely, corresponds to looking at all the ways to get from the thick point to itself):

$$\langle a^2, b^4, ab, ba^2b^{-1}, bab^{-2} \rangle$$

is non-normal of index $4$. As desired.

Edit: I totally forgot you also want the rank.

You might suspect this subgroup is of rank $5$, seeing as that's how many generators are listed. That suspicion is correct, as shown by another geometric fact:

The rank of a group that you get in this way is $1 - \chi$ where $\chi$ is the euler characteristic of the graph. Here $\chi = -4$ is the number of vertices minus the number of edges, so the rank is

$$1 - \chi = 1 - (-4) = 5.$$

I hope this helps ^_^

Title: Subgroup of $\mathbb{Z}\ast \mathbb{Z}$


答:
A real $(a_n)$ sequence is called contractive if there exists a constant $0 < C < 1$ such that $$|a_{n+2} − a_{n+1}| ≤ C |a_{n+1} − a_n |$$ for all $n \geq 1$.

In case of your problem, we can drop the absolute values and derive the following.

$$a_{n+2} - a_{n+1}=\sqrt{a_{n+1}+6} -\sqrt{a_{n}+6} = \frac{a_{n+1}-a_n}{\sqrt{a_{n+1}+6} + \sqrt{a_{n}+6}} \leq \frac{a_{n+1}-a_n}{\sqrt{6}+\sqrt{6}}.$$

Please justify the chain of the inequalities.

Title: Consider the sequence $a_n$ where $a_0=0$ and $a_{n+1} = \sqrt{a_n+6}$. Show that $a_n$ is contracting.


问:
Generating function: Find a closed form of $\sum_{k=0}^n (-3)^k(k+1)$
Find the closed form of $\sum_{k=0}^n (-3)^k(k+1)$.

So the generating function would be: $$A(x)=1-6x+18x^2-108x^3...$$ So what I did notice is that its closed form is perhaps some variation of $1\over {1+x}$ but I didn't manage to find a general formula.

Thanks for any help in advance!

Body:
We have that $$A(x)=\sum_{n=0}^\infty a_nx^n,\qquad a_n=\sum_{k=0}^n(-3)^k(k+1).$$
Note that
$$a_{n+1}-a_n=(-3)^{n+1}(n+2),$$
which implies that
$$\begin{align*}
(1-x)A(x)=A(x)-xA(x)&=\sum_{n=0}^\infty (a_nx^{n}-a_nx^{n+1})\\\\
&=a_0+\sum_{n=1}^\infty (a_n-a_{n-1})x^n\\\\\\
&=1+\sum_{n=1}^\infty(-3)^{n+1}(n+2)x^n\\\\\\
&=1-3\sum_{n=1}^\infty n(-3x)^n-6\sum_{n=1}^\infty (-3x)^n
\end{align*}$$
Can you take it the rest of the way?


问:
If A is a positive matrix . Then how do we show that $e^{At}\geq 0$.
If A is an $n \times n$ matrix consisting of $a_{ij}$ with $a_{ij}\geq 0 , \hspace{2mm} \forall \hspace{1mm} 1\leq i,j\leq n$, $i\neq j$. Then how do we show that  for small enough $t\geq0$ we would have $e^{At}\geq 0$?

I tried to argue that since
$$ e^{At}=I + At+ \cdots +\frac{A^nt^n}{n!}+\cdots$$
and we are given that $a_{ij}\geq 0 , \hspace{2mm} \forall \hspace{1mm} 1\leq i,j\leq n$, $i\neq j$ hence $e^{At}$ would be non negative.

Is this argument enough AND can there be an alternative argument to prove the above. Any help would be greatly greatly appreciated.

答: This argument is enough since you can show recursively that the coefficients of $A^n$ are positive.


问:
Can the existence of some positive integers satisfying some equations ever be independent of ZFC?
In a subject like number theory there are often open questions along the lines of 'Are there positive integers $a_1 , \ldots , a_n $ satisfying the following (finite number of) equations'. For example Fermat's last theorem for a fixed exponent is such a statement. I was wondering if such a question could ever be independent from ZFC, meaning there is no proof for or against the statement.

I know that the incompleteness theorem states that in our language of arithmetic there has to be some statement that is independent of the axioms. But to me - while often surprisingly hard to prove - these questions seem simple enough, in the sense that they only ask for the existence of some integers satisfying algebraic equations, meaning you could write a computer program checking through all tuples of integers $(a_1 , \ldots , a_n)$. Of course if there is no such tuple your program won't terminate, but in that case I would still find it ridiculous if you could not prove that there are none.

I don't see an argument for what I am suspecting which is why I am asking this question. I would appreciate any information on the matter because this has been confusing me for a while.

答:
meaning you could write a computer program checking through all tuples of integers $(a_1 , \ldots , a_n)$.

You certainly could not! There are infinitely many of them! (That is to say, you could attempt to do this, and your program could run for hundreds of years without returning a result, and you still wouldn't know whether or not there were solutions.)

Anyway, the answer to your question is yes, suitably interpreted. This follows from Matiyasevich's theorem, which says that the sets of tuples of integers that can be represented as solutions to Diophantine equations (these equations include extra parameters and ask for the existence of those parameters, e.g. "there exists $n$ such that $x = n^2$" counts) are exactly the sets of tuples of integers which are recognizable by a Turing machine.

In particular, it follows that there is a specific (very large) Diophantine equation one can write down which has solutions if and only if a Turing machine searching for a proof of a contradiction in ZFC terminates; in other words, if and only if ZFC is inconsistent. So the existence of solutions to this Diophantine equation is independent of ZFC.


问:
find the volume of the solid of intersection of the two spheres of radii a and b (with b $<$ a)
Also the center of the smaller one lies on the surface of the larger one.

This is what I have thus far.

Utilize the Cartesian plane we have for the smaller and larger spheres respectively:
$(x-a)^2 + y^2 = b^2 $ and $x^2 + y^2 = a^2$

I solve for the $x$ value where these 2 intersect. I get $x_r = \frac{2a^2 - b^2}{2a}$

To get the volume of the larger cap in the intersection I used the method of cross sections.
$\int^{a}_{x_r} \pi(a^2 - x^2)dx$

Similarly for the smaller sphere: $\int^{b}_{x_r} \pi(b^2 - (x-a)^2)dx$

Afterwards, I sum the result of the two to get the answer? Did I make a mistake somewhere

答:
Your approach is fine. You are basically taking $x$ as your $z$ axis and hence in cylindrical coordinates $y^2 + z^2 = r^2$ and so you may want to write them as,

$r^2+x^2 = a^2$ and $r^2 + (x-a)^2 = b^2$.

Your first integral is fine but your second integral should be

$\displaystyle \int_{a-b}^{\frac{2a^2 - b^2}{2a}} \pi \, \big(b^2 - (a-x)^2\big) \,dx = \frac{\pi b^6}{24a^3} - \frac{\pi b^4}{2a} + \frac{2 \pi b^3}{3}$

The lower bound is the minimum value of $x$ for the smaller sphere, which is $(a-b)$ and it is part of the intersection volume.


问:
Simulating probability based on percentage occurrence
I have a question similar to this one.

If for a group of 85 riders, 40 riders fall over 24 races, then the probability of a rider falling in any single race is:

$$ 1 - (1 - p)^{24} = \frac{40}{85} $$

Where $ p \approx 0.0262 $.

However, using this probability, if I run 24 simulations (races) of 85 binomial draws (riders) with $p$ probability (replicated 10k times), a fall occurs, on average, $ \approx 53 $ times, not the expected $ \approx 40 $ times.

Linking back to this question, however, using their example parameters, in which there are 98 riders, with 4 falling over 52 races. If I run 52 simulations (races) of 98 binomial draws (riders) based on the probability of a rider falling in their example ($p \approx 0.0008$) (replicated 10k times), a fall does occur, on average, $ \approx 4 $ times, as expected.

I'm really struggling to understand what about my interpretation of $p$ is creating this issue with the simulation.

Thanks

R code for simulations:

```<code>> mean(replicate(10000, sum(replicate(24, rbinom(85,1,0.0262)))))
[1] 53.4684

> mean(replicate(10000, sum(replicate(52, rbinom(98,1,0.0008)))))
[1] 4.0213
</code>```

答:
Let $X_{i,j}$ denote the indicator of $i$'s failure in $j$'s race. I assume that $X_{i,j}$ are i.i.d. across $i$ and $j$. You want to estimate $p=\mathsf{E}X_{i,j}$ using $S:=\sum_{i}1\{\sum_jX_{i,j}>0\}$. First,
$$
\mathsf{E}[S]=\sum_i\mathsf{P}\!\left(\sum_{j}X_{i,j}>0\right)=n(1-(1-p)^m),
$$
where $n$ is the total number of riders and $m$ is the total number of races. A natural estimator of $p$ in this case is
$$
\hat{p}=1-(1-\hat{\mathsf{E}}[S]/n)^{1/m}.
$$
Therefore, the precision of $\hat{p}$ is directly related to the precision of $\hat{\mathsf{E}}[S]$. If you observe only one realization of $S$ as in your case (40), the resulted estimate of $p$ can be very far from its actual value. For example, you can get $\hat{p}=0$ or $1$ (when $S=0$ or $n$) with positive probability.

If you repeat the same experiment many times, say $N$, and use
$$
\hat{\mathsf{E}}[S]=\frac{1}{N}\sum_{k=1}^N S_k,
$$
where $S_k$ is the result of the $k$-th experiment, you get more and more precise estimates as $N$ increases. Specifically, by the law of large numbers $\hat{p}\to p$ a.s. as $N\to\infty$.


问:
Is it possible to calculate Hamming distance with only parity check matrix?
let's say I have this parity matirx, is there a way to calculate minimal distance between two valid words without writing all possible code words?
$$
H= \begin{bmatrix}
0 & 1 & 0 & 0 & 1 & 0 & 1\\
1 & 0 & 0 & 1 & 0 & 1 & 0\\
0 & 0 & 1 & 1 & 1 & 1 & 1
\end{bmatrix}
$$

答:
As hinted in the comments, a codeword $(c_1,\ldots,c_n)$ satisfies
$$
c_1 h_1+c_2 h_2+\cdots +c_1 h_n=0,
$$
where the right hand side is the zero vector and $h_i$ is the ith column of $H.$ So find a nontrivial linear combination (coefficients must be $0$ or $1$ since we are in the binary field $GF(2)$ and the only nonzero scalar is $1$) with the fewest number of columns that add up to the zero vector (mod 2) to obtain the minimum weight.

Now the matrix is binary so the only way two columns have a nontrivial linear combination adding up to zero is if they are equal. Since this is not the case here, you can establish the minimum distance is $3$ if $3$ columns add up to zero. For this $H$, columns $1,3,4$ add up to zero so the minimum weight (hence the minimum distance) is $3.$
So we have
$$
h_1+h_3+h_4=
\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}+
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}+
\begin{bmatrix}
0 \\
1 \\
1
\end{bmatrix}=
\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}.
$$


问:
Prove that $\lim\limits_{n \to \infty}\sum\limits_{k=1}^{n}\frac{1}{k \cdot\ln(n)}$ converges
I need to prove that $\lim\limits_{n \to \infty}\sum\limits_{k=1}^{n}\frac{1}{k \cdot \ln(n)}$ converges, and I don't know how can I do it.${}{}{}$

答:
Let $ n\in\mathbb{N}^{*} $, we have for all $ k\in\mathbb{N}^{*} $ : $$ \int_{k}^{k+1}{\frac{\mathrm{d}x}{x}}\leq\frac{1}{k}\leq\int_{k-1}^{k}{\frac{\mathrm{d}x}{x}} $$

Thus : \begin{aligned} \int_{1}^{n+1}{\frac{\mathrm{d}x}{x}}\leq\sum_{k=1}^{n}{\frac{1}{k}}\leq 1+\int_{1}^{n}{\frac{\mathrm{d}x}{x}}\ \ \ \  \\ \iff \ \ \ \ \ \ln{\left(1+n\right)}\leq\sum_{k=1}^{n}{\frac{1}{k}}\leq1+\ln{\left(1+n\right)}\\ \iff \frac{\ln{\left(1+n\right)}}{\ln{n}}\leq\frac{1}{\ln{n}}\sum_{k=1}^{n}{\frac{1}{k}}\leq 1+\frac{1}{\ln{n}}\ \ \ \ \ \end{aligned}

Hence : $$ \frac{1}{\ln{n}}\sum_{k=1}^{n}{\frac{1}{k}}\underset{n\to +\infty}{\longrightarrow}1 $$


问:
How to find the generating function of a recurrent series: $b_{0} = 4$, $b_{n}=3b_{n-1}-2^{n-1}$
The series is defined as such:
$$
b_{0}=4
$$$$
b_{n}=3b_{n-1}-2^{n-1}\text{ for }n\geq1
$$

How do I go about finding its generating function?
I started like this:
$$
f(x)=\sum_{n=0}^{\infty} b_{n} \cdot x^n =4+\sum_{n=1}^{\infty}(3b_{n-1}-2^{n-1})\cdot x^n
$$
And now I am lost. How should I proceed?

Body:
hint

$$b_n=3b_{n-1}-2^{n-1}\implies$$

$$b_n=3b_{n-1}-3.2^{n-1}+2.2^{n-1}\implies$$
$$b_n-2^n=3(b_{n-1}-2^{n-1})$$

So, with
$$c_n=b_n-2^n$$
you have
$$c_n=3.c_{n-1}$$
and
$$c_n=c_0.3^n\;\; where\;\;c_0=4-1=3$$

You can finish.


问:
Evaluating $ \int^{\infty}_{-\infty}\sin\left({\pi}^{4}x^{2}+\frac{1}{x^2}\right) dx$
$$\int^{\infty}_{-\infty}\sin\left({\pi}^{4}x^{2}+\frac{1}{x^2}\right) dx$$

This is a problem from the Pi Mu Epsilon Journal, and I'm having great trouble answering it. I've tried some substitutions and any trick I could think of to find some multiple of the integral, but everything has led to a dead end. Perhaps it's not even solvable with methods of real analysis?

Hints and other comments will be much appreciated!

答:
Not a different answer but rather generalizing the technique Peter Tamaroff used in his answer. Notice the fact that $(u - u^{-1})^2 + 2 = u^2 + u^{-2}$, we can generalize this for any convergent definite integral.

Say if we want to compute:
$$
\int^{\infty}_{-\infty} f(x)\,dx,
$$
knowing this is convergent. Using the trick we can see the integral is:
$$
\int^{\infty}_{-\infty} f(x)\,dx =  \int^{\infty}_{-\infty} f\left(x - \frac{a}{x}\right)\,dx \tag{1}
$$
Proof: We can make the substitution:
$$ u = x - \frac{a}{x}, \quad \text{for some } a>0.$$
Notice $u$ is again from $-\infty$ to $\infty$ for both $x>0$ and $x<0$. Then we have:
$$
x = \frac{u + \sqrt{u^2 + 4a}}{2} > 0  \;\text{ or }\; \frac{u - \sqrt{u^2 + 4a}}{2} < 0
$$
hence
$$
\frac{dx}{du} = \left\{\begin{aligned}
\frac{1}{2}+ \frac{u}{2\sqrt{u^2 + 4a}} \quad \text{when }x>0
\\
\frac{1}{2} - \frac{u}{2\sqrt{u^2 + 4a}} \quad \text{when }x<0
\end{aligned}\right.
$$
Hence:
$$
\int^{\infty}_{-\infty} f(u)\left(\frac{1}{2}+ \frac{u}{2\sqrt{u^2 + 4a}}\right)  du = \int^{\infty}_{0} f\left(x - \frac{a}{x}\right) \,dx \tag{2}
$$
and
$$
\int^{\infty}_{-\infty} f(u)\left(\frac{1}{2} -  \frac{u}{2\sqrt{u^2 + 4a}}\right)  du = \int^{0}_{-\infty} f\left(x - \frac{a}{x}\right)\,dx \tag{3}
$$
(2)+(3) yields (1).

The integral in OP is
$$
\int^{\infty}_{-\infty}\sin\left({\pi}^{4}x^{2}+\frac{1}{x^2}\right) dx
= \int^{\infty}_{-\infty}\sin\left({\pi}^4\Big(x - \frac{1}{\pi^2 x}\Big)^2 + 2\pi^2  \right) dx.
$$
Now using (1):
$$
\int^{\infty}_{-\infty}\sin\left({\pi}^4\Big(x - \frac{1}{\pi^2 x}\Big)^2 + 2\pi^2  \right) dx = \int^{\infty}_{-\infty}\sin ({\pi}^4 x^2 + 2\pi^2) \,dx
$$
We can easily see this is the same integral as Peter Tamaroff got in his answer:
$$
\frac{2}{\pi }\int_0^\infty  {\sin } \left( {{{\left( {\pi x} \right)}^2} + 2{\pi ^2}} \right)dx,
$$
and then Fresnel integral kicks in.


Title: How to find the generating function of a recurrent series: $b_{0} = 4$, $b_{n}=3b_{n-1}-2^{n-1}$

Body:
The series is defined as such:
$$
b_{0}=4
$$$$
b_{n}=3b_{n-1}-2^{n-1}\text{ for }n\geq1
$$

How do I go about finding its generating function?
I started like this:
$$
f(x)=\sum_{n=0}^{\infty} b_{n} \cdot x^n =4+\sum_{n=1}^{\infty}(3b_{n-1}-2^{n-1})\cdot x^n
$$
And now I am lost. How should I proceed?

Body:
$$f(x)=4+3x\sum_{n=1}^\infty b_{n-1} x^{n-1} - x \sum_{n=1}^\infty (2x)^{n-1}\\
= 4+3xf(x)-\frac{x}{1-2x}$$

Can you start from here?


问:
Compute triple integral $ \iiint\limits_D \sin(xy) - \sin(xz) + \sin(yz) \, dx \, dy \, dz $
Let $D = \{ (x,y,z): 1 \leq x^2 + y^2 + z^2 \leq 4 \}$ then compute

$$
\iiint\limits_D \sin(xy) - \sin(xz) + \sin(yz) \, dx \, dy \, dz
$$

I tried to use spherical coordinates but then the integrand becomes super ugly if I try to use cartesian coordinates then the domain gets ugly, it looks like there must be something better to do here.

Body: $$\int\limits_{x=-2}^2 \int\limits_{y = -\sqrt{4 - x^2}}^{\sqrt{4-x^2}} \int\limits_{z= -\sqrt{4 - x^2 - y^2}}^{\sqrt{4 - x^2 - y^2}} \left( \sin (x y) - \sin (x z) + \sin (y z) \right) dx\ dy\ dz - \\ \int\limits_{x=-1}^1 \int\limits_{y = -\sqrt{1 - x^2}}^{\sqrt{1-x^2}} \int\limits_{z=-\sqrt{1 - x^2 - y^2}}^{\sqrt{1 - x^2 - y^2}} \left( \sin (x y) - \sin (x z) + \sin (y z)\right) dx\ dy\ dz \\ = 0 - 0 = 0$$


问:
How to represent $ \frac{1}{a_n}$ in terms of the generating function of $a_n$?
Suppose I have the ordinary generating function of $a_n$,

$$ A(x) = \sum_{i} a_i x^i$$

Then from the above expression, is there a way to write the generating function of $\frac{1}{a_n}$?

If there is a way to do it using exponential generating function or dirchlet generating function, then that too is fine.

答:
In general the two have very little to do with each other. When $A(x) = \sum a_n x^n$ and $B(x) = \sum b_n x^n$ are two generating functions we call $\sum a_n b_n x^n$ their Hadamard product. So $\sum a_n^{-1} x^n$ could be called the Hadamard inverse. Hadamard products are hard to calculate in general (the only general result I know about them is that Hadamard products of rational functions are rational and the proof is not very direct) and Hadamard inverses even more so.

For example when $a_n = \frac{1}{n!}$ we have $A(x) = e^x$ but the Hadamard inverse $\sum n! x^n$ has zero radius of convergence. Interesting things can still be said about it but it doesn't have much of anything to do with $e^x$, as far as I know.

As another example which is even a polynomial, when $a_n = {m \choose n}$ we have $A(x) = (1 + x)^m$ but the Hadamard inverse $\sum {m \choose n}^{-1} x^n$ is much stranger. It can be represented using the Beta function integral but this doesn't have much to do with $A(x)$, again as far as I know.


问:
Proving sum of negative powers of 2 equals one before becoming greater than one
Let $x$: $x_1 \geq x_2 \geq x_3\geq...\geq x_n$ be negative powers of 2 with sum greater than one. Then $\exists$ $l$ s.t. $x_1 + x_2 + ... + x_l = 1$. It seems obvious from examples, but I'm finding it a bit difficult to prove it. Any help would be highly appreciated.

答:
Your comment is close.

Let $S_i = x_1+x_2+\cdots + x_i$.

For the first case $l=1$, the first sum $S_1 = x_1<1$. For the last case $l=n$, $S_n > 1$ (as given). For the increasing sum $S_l$, there must be a largest $l$ that satisfy $S_l < 1$.

Assume that $k$ is the greatest integer that satisfy $$S_k = \sum_{i=1}^kx_i < 1$$

Then the goal is to prove that the next sum

$$S_{k+1}=\sum_{i=1}^{k+1}x_i = 1.$$

As you pointed out, $S_k = \sum_{i=1}^k x_i$ is an integer multiple of $x_{k+1}$, so let $S_k = Ax_{k+1}$. Then from the hypothesis,

$$\begin{align*}
Ax_{k+1} &< 1\\
A &< x_{k+1}^{-1}
\end{align*}$$

$x_{k+1}^{-1}$ is an integer for being a positive power of $2$, and since both sides are integers,

$$\begin{align*}
A + 1 &\le x_{k+1}^{-1}\\
Ax_{k+1} + x_{k+1} &\le 1\\
S_{k+1} &\le 1
\end{align*}$$

Also by the definition of $k$, $S_{k+1} \not< 1$.

Then the only possibility is that $$S_{k+1} = \sum_{i=1}^{k+1}x_i = 1$$


问:
On the uniform convergence of the series of the following functions
prove that $$\sum_{n=1}^{\infty}\int_{0}^{x}t^n\sin\pi tdt$$Uniform convergence on$[0,1]$.
I try to use Weierstrass discriminant
$$\left|\int_{0}^{x}t^n\sin\pi tdt\right|\leqslant\int_{0}^{x}t^ndt=\frac{x^{n+1}}{n+1}$$
Uniform convergence when $x < 1$.I don't know if this is wrong,I want to know how to judge the convergence and divergence at 1.Thanks.

答:
Step 1.
We first notice that
\begin{align}
f_n(x) & = \sum_{i=1}^{n}\int_{0}^{x}t^i\sin\pi tdt\\
       & = \int_{0}^{x} \sin\pi t \left( \sum_{i=1}^{n}t^i \right)dt.
\end{align}
By Dominated convergence theorem, we get
\begin{align}
\lim_{n\rightarrow \infty} f_n(x)
& = \int_{0}^{x} \sin(\pi t) \left( \sum_{i=1}^{\infty}t^i \right)dt\\
& =  \int_{0}^{x} \sin(\pi t) \left( \frac{1}{1-t}-1\right)dt=: f(x).\\
\end{align}

Step 2. We now prove that $f_n$ converges uniformly to $f$, i.e.
$$\sup_{x\in [0,1]} |f(x)-f_n(x)|\xrightarrow{n\rightarrow \infty} 0.$$
Indeed,
\begin{align}
|f(x)-f_n(x)|
& = |\int_{0}^{x} \sin\pi t \left( \sum_{i=n+1}^{\infty}t^i \right)dt|\\
& \leq \int_{0}^{x} t^{n+1} \left( \sum_{i=0}^{\infty}t^i \right)dt.\\
& =  \int_{0}^{1} \frac{t^{n+1}}{1-t} dt.
\end{align}
Thus,
$$\sup_{x\in [0,1]} |f(x)-f_n(x)|\leq \int_{0}^{1} \frac{t^{n+1}}{1-t} dt
\xrightarrow{n\rightarrow \infty} \int_{0}^{1} \lim_{n\rightarrow \infty} \frac{t^{n+1}}{1-t}  dt=0.$$


问:
Proving results on derivatives of absolute value of a function
I need to prove these two results:

$f(a) = 0$ and $f'(a)\neq 0$.  You can use the definition of the derivative to show the left and right derivatives of $|f|$ at $a$ are
unequal.

$f(a) = 0$ and $f'(a) = 0$.  You can use the definition of the derivative to show that $|f|$ is differentiable at $a$ with $|f|'(a) =0$.

EDIT- My attempt:
For the second one, I tried to use what was given.
By the definition of derivative of $f$ at the point $a$, we have:
$$ f'(a) =  \lim\limits_{x \to a} \frac{f(x)-f(a)}{x-a}$$
It gives me from the given data: $$  0 = \lim\limits_{x \to a} \frac{f(x)}{x-a}$$
Let $g(x)= \frac{f(x)}{x-a}$. It is easy to observe that: $g(x)\to 0 \Leftrightarrow |g(x)|\to 0$. So, $$  \lim\limits_{x \to a} |g(x)|= 0\lim\limits_{x \to a} \frac{|f(x)|}{|x-a|} = 0 $$ By definition of absolute value:
$$ |g(x)| =
\begin{cases}
\frac{|f(x)|}{x-a},  & \text{if $x \gt a$} \\[2ex]
\frac{|f(x)|}{-(x-a)}, & \text{if  $x \lt a$}
\end{cases}$$
So the right-hand derivative of $|f(x)|$ at $x=a$ is:
$$ \lim\limits_{x \to a^+} \frac{|f(x)|}{x-a} = 0 $$
and the left-hand derivative of $|f(x)|$ at $x=a$ is:
$$ \lim\limits_{x \to a^-} \frac{|f(x)|}{x-a} = 0 $$
$$ \therefore |f'(a)| = \lim\limits_{x \to a} \frac{|f(x)|-|f(a)|}{(x-a)} = 0 $$

Please suggest if my solution works.

答:
Assume that $$f(a)=0\;\;and\;\;f'(a)>0$$

then

$$\lim_{x\to a^+}\frac{f(x)}{x-a}>0 \implies$$

$$\exists \eta>0 :\forall x\in(a,a+\eta) \frac{f(x)}{x-a}>0$$
$$\implies \forall x\in(a,a+\eta)\;\; f(x)>0$$

$$\implies \lim_{x\to a^+}\frac{|f(x)|}{x-a}=f'(a)$$

By the same, we will have

$$\lim_{x\to a^-}\frac{|f(x)|}{x-a}=-f'(a)$$


问:
How to use Leibniz notation properly
I think that I didn't understant properly how to use Leibniz notation for derivatives and partial derivatives. I know that:
$$\frac{df}{dx}=f'$$
And here I don't have any problem. But then if $g,f$ are two functions, how should I intend:
$$\frac{df}{dg}$$
Is it $f'\circ g$?

Things get worse when we have to work in more variables.Let:

$\mathbf{f}:\mathbb{R}^n\to\mathbb{R}^m$

$\mathbf{g}:\mathbb{R}^m\to\mathbb{R}^p$

$\mathbf{\Phi}:=\mathbf{g}   \circ  \mathbf{f}$

$\mathbf{\Phi}(\mathbf{x})=(\Phi_1(\mathbf{x}),...,\Phi_p(\mathbf{x}))$

Here's how I would write the $j$-th  partial derivative of the $i$-th component function of $\mathbf{\Phi}$:
$$\frac{\partial \Phi_i}{\partial x_j}(\mathbf{x})=\sum_{k=1}^{m} \left[ \frac{\partial g_i}{\partial x_k}(\mathbf{f}(\mathbf{x}))\right ]\left[\frac{\partial f_k}{\partial x_i}(\mathbf{x})\right ] $$
Or if I want to omit the argument
$$\frac{\partial \Phi_i}{\partial x_j}=\sum_{k=1}^{m} \left[ \frac{\partial g_i}{\partial x_k}\circ \mathbf{f}\right ]\left[\frac{\partial f_k}{\partial x_i}\right ] $$
But my book writes it in this way:
$$\frac{\partial \Phi_i}{\partial x_j}=\sum_{k=1}^{m}  \frac{\partial g_i}{\partial f_k} \frac{\partial f_k}{\partial x_i} $$

So am I supposed to understand by magic that:
$$\frac{\partial g_i}{\partial f_k}:=\frac{\partial g_i}{\partial x_k}\circ \mathbf{f}$$
I know that the formula given by the book is more elegant and synthetic, but when I read it the first time I didn't understand anything.
My question is: Is there a standard convention for this kind of notation?
Because I'm seriously hating this notation, not only because of how unreadable is(to me), but also because all of the "differential cancellation" that we make in ODE(but this maybe will be part of a future question).

Thank you :)

答:
Here is how I would state the chain rule in Leibnizian notation:

Let $y=f(u)$, and $u=g(x)$. Then, if $f$ is differentiable at $u$, and $g$ is differentiable at $x$, then $f \circ g$ is differentiable at $x$, and
$$
\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx} \, .
$$

This can be easily translated back into Lagrange notation in the following way. The first term, $dy/dx$, is simply a shorthand for
$$
\frac{df(g(x))}{dx}=(f \circ g)'(x) \, .
$$
We know that $y=f(u)$, and so
$$
\frac{dy}{du}=f'(u)=f'(g(x)) \, .
$$
Finally, we know that $u=g(x)$, meaning that
$$
\frac{du}{dx}=g'(x) \, .
$$

The only perplexing feature of this notation is how $u$ is both treated as a variable (in $dy/du$), and as a function (in $du/dx$). However, while many people might argue against this notation for this very reason, it does have its advantages. Since $u=g(x)$, we may as well write $dy/du$ as
$$
\frac{df(g(x))}{dg(x)} \, ,
$$
but since $g(x)$ is just a dummy variable that shows where we are evaluating the function at, this can be rewritten as
$$
\frac{df(u)}{du} \Biggr|_{u=g(x)} \, .
$$
I go into more detail about this here.

For $y=f(u)$, where $u=g(t)$, and $t=h(x)$, the chain rule becomes
$$
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dt} \cdot \frac{dt}{dx} \, .
$$
Again, since $dy/du = f'(u)$, this is equal to $f'(g(h(x))$, and the same translation can be made for the other terms.

Title: How to use Leibniz notation properly


问:
Set which can't be transformed into a cyclic group

A set $M$ on which a composition law $*$ cannot be defined so that $(M,*)$ it is a cyclic group is $\mathbb{R,N,Z,Q}$?

I received this question from a colleague and I think it is $\mathbb{Q}$ but I am not 100% sure I do not want to tell him something wrong.

答:
If a group $G$ is cyclic, i.e. $G=\langle g\rangle$ for some $g\in G$, then $G$ is either finite or countable. That's because there's a group epimorphism $\mathbb{Z}\to G$, $n\mapsto g^n$. This automatically tells us that $\mathbb{R}$ cannot be a cyclic group.

On the other hand if $X$ is a countable set, meaning infinite countable or finite, then indeed it can be turned into a cyclic group. First of all, if $G$ is any group, $X$ a set and $f:G\to X$ a bijection, then $X$ can be turned into a group such that $f$ becomes an isomorphism. This group structure on $X$ is given by $x\cdot y:=f(f^{-1}(x)f^{-1}(y))$. With that you only need to realize that for a countable set $X$ there exists a bijection $\mathbb{Z}/n\mathbb{Z}\to X$ for some $n\in\mathbb{Z}$,  turning $X$ into a cyclic group. In particular $\mathbb{N}$, $\mathbb{Z}$ and $\mathbb{Q}$ can be turned into cyclic groups.


问:
Reliability of a circuit board with components
I have a list of components with their reliability, for example

Resistor [Fixed, Metal Film], Number used 168, Rating 50%, Failure rate 0.005, weighting factor 1.5.

I have calculated the total reliability for all of these components, for the example I think it is 1.26. The next step is M.T.B.F, reliability over 5 days and over 28 days of 8 hour use.

I am currently a little confused as to what I should be doing with these figures I have worked out, do I add them? Multiply them? and then what?

Please help.

答:
Hint for MTBF: The minimum of exponential random variables with rates $\lambda_1,\ldots,\lambda_n$ is an exponential random variable with rate $\lambda_1 + \cdots + \lambda_n$ (see Wikipedia).

As for reliability, I can't help since I've never heard of this parameter.


问:
Task with combination of spectrums of matrices
Here (Represent the common term of the sequence as minus one in polynomial degree), I start to post unsolved problems from my old notebook.

So, here is the next one:
Let $A,B$ be heremitian matrices. Combination of spectrums of matrix $A$ and $B$, besides eigenvalues that equal zero , is equal to spectrum of matrix $A+B$, besides eigenvalues that equal zero.
Here what is this mean: Let $\sigma_A=({0,0,1,2})$ - spectrum of matrix $A$ and $\sigma_B=({0,2,3,0})$ - spectrum of matrix $B$. Then spectrum of matrix $A+B$ due to the problem must be $\sigma_{A+B}=(1,2,2,3)$.
So, the question is to prove that if this property is performed, then $AB=0$.

My atetemts was made in the direction, that such matrices $A$ and $B$ will always be zero matrices, so it always will be that $AB=0$. Trying to show this analytically, I've used the fact, that hermitian ,atrices has real eigenvalues and spectrum of matrices is invariant, so we can work only with diagonal matrices $A,B$ and $A+B$. But I can't get a strictly reasoned solution.
This problem I've found at NCUMC competition in 2015(http://mathdep.ifmo.ru/ncumc/wp-content/uploads/2017/01/NCUMC_15_problems.pdf)
Any help will be appreciated! Thanks in advance!

答:
Attempt:

Suppose that $A,B$ are Hermitian matrices of size $n$ such that the spectrum of $A + B$ is the concatenation of the non-zero elements of the spectra of $A$ and $B$. Because all eigenvalues of $A+B$ are non-zero, it must hold that $A + B$ is invertible. However, because $A$ and $B$ have $n$ non-zero eigenvalues all together, we have
$$
\operatorname{rank}(A) + \operatorname{rank}(B) = n.
$$
Let $U = \operatorname{im}(A)$ and $V = \operatorname{im}(B)$. We note that $\Bbb C^n = \operatorname{im}(A+B) \subseteq U + V$, which means that $U + V = \Bbb C^n$. With that, we note that the dimension formula yields
$$
\dim(U) + \dim(V) = \dim(U + V) + \dim(U \cap V) \implies\\
\operatorname{rank}(A) + \operatorname{rank}(B) = \dim(\Bbb C^n) + \dim(U \cap V) \implies\\
n = n + \dim (U \cap V) \implies \dim(U \cap V) = 0.
$$
That is, we have
$$
U+V = \Bbb C^n, \quad U \cap V = \{0\}.
$$
Now, because $A$ and $B$ are Hermitian, we have $U^\perp = \ker(A)$ and $V^\perp = \ker(B)$. From the above, conclude that
$$
U^\perp + V^\perp = (U \cap V)^\perp = \Bbb C^n, \quad
U^\perp \cap V^\perp = (U + V)^\perp = \{0\}.
$$


问:
Logarithmic function that starts at origin and reaches for 1 as x goes to infinity
I'm trying to obtain a logarithmic function with characteristics such as when $\lim_{x \to\infty} f(x)$, the function diverges at 1.

Any ideas?

Update:
I was playing around and I think I accidentidly created such function, but I'm not sure if it actually fulfills my functional demands.

答:
"Diverges at $1$" as $x$ grows makes no sense. As $x \to \infty$ the logarithm grows without bound.

If you want a convex function with a horizontal asymptote of $1$ consider
$$
1 - e^{-ax}.
$$

You can adjust the value of the positive constant $a$ to change the rate at which you creep up to $1$.

Edit in response to comment.

You can use
$$
L - e^{-ax}.
$$
for any $L < 1$ you like.


问:
Finding a complete theory which contains the theory of sentences that hold for every finite field.
I am trying to make some exercises for a introductory course to mathematical logic.

Let $T$ be the theory consisting of all sentences that are true in every finite field. It is not that hard to see that this is a consistent, but not a complete theory. I am being asked to give a concrete example of theory which contains $T$, but which is complete.

I know that I will need to add a sentence for every <s>cardinality</s> characteristic ,but I am unsure if this is enough to make the theory complete. If it is not I do not know how to give a concrete theory which is even larger.

I also thought about taking the set of all sentences that are true in a specific model of $T$, but again I do not know how to show completeness.

答:
You wrote:

I also thought about taking the set of all sentences that are true in a specific model of $T$, but again I do not know how to show completeness.

This is the easiest way to pick a completion of a consistent theory $T$. Let $M\models T$ and consider $$T' = \text{Th}(M) = \{\varphi\mid M\models \varphi\}.$$ The completeness of $T'$ is obvious - there's almost nothing to show. $T'$ is consistent, since $M\models T'$. And for any sentence $\psi$, either $M\models \psi$ or $M\models \lnot \psi$, so either $\psi$ or $\lnot\psi$ is in $T'$, so $T'$ is complete.

In fact, every completion of $T$ has this form. If $T'$ is a complete consistent theory extending $T$, then letting $M\models T'$, we have $T' = \text{Th}(M)$.

Ok, but your exercise asks you to give a concrete example of a completion of $T$. This is the problem with the strategy above. For a given model $M$, it may be hard (if not impossible) to give a concrete axiomatization of $\text{Th}(M)$.

Let's specialize now to the situation in your exercise, where $T$ is the theory consisting of all sentences which are true in every finite field. As pointed out in the comments, there's an easy way to axiomatize a complete consistent theory extending $T$: For every prime $p$ and natural number $n>0$, there is a unique finite field of cardinality $p^n$ up to isomorphism. So letting $T'$ be the field axioms together with a sentence asserting that a model has exactly $p^n$ elements, $T'$ has a unique model up to isomorphism. And it is clear that a theory with exactly one model up to isomorphism is consistent and complete.

Ok, can we solve your exercise in a less trivial way, by writing down a complete theory extending $T$ that has infinite models? It turns out the answer is yes, but it's pretty complicated and relies on some deep work by James Ax: The Elementary Theory of Finite fields (1968).

To begin with, Ax axiomatized the theory $T$: the models of $T$ are called pseudo-finite fields. These are characterized as fields which are perfect (i.e., have characteristic $0$, or have characteristic $p$ and every element has a $p^{\text{th}}$ root), have absolute Galois group $\hat{\mathbb{Z}}$ (equivalently, have a unique algebraic extension of degree $n$ for each $n$), and are pseudo algebraically closed. It is not immediately obvious that these conditions can be written down as first-order sentences in the language of fields, but with a little work they can be.

Now Ax also proved that two pseudo-finite fields $K$ and $L$ are elementarily equivalent if and only if $\text{Abs}(K) \cong \text{Abs}(L)$, where $\text{Abs}(F)$ is the relative algebraic closure of the prime field inside $F$, i.e., those elements of $F$ which are algebraic over $\mathbb{Q}$ in characteristic $0$ and over $\mathbb{F}_p$ in characteristic $p$. Now the isomorphism type of an algebraic extension $E$ of a perfect field $F$ is determined by which polynomials in $F[x]$ have roots in $E$ (see here). So we can axiomatize the complete theory of a pseudo-finite field $K$ by writing down:

The axioms of pseudo-finite fields.

Axioms fixing the characteristic ($0$ or a prime $p$).

Letting $F$ be the prime subfield of $K$, axioms specifying which polynomials in $F[x]$ have roots in $K$.


答:
Consider the partial series, up until the value passes $1$. Considering the last value used, we can calculate the minimum remaining number of steps ($MRS_i$) until the series reaches $1$, and this will always be an integer.

We can start with a partial series value of $0$ and the assumption that previous value $x_0 = 1$. This gives us $MRS_0 = 1$.

Now examining the next value, we can recalculate $MRS$ in the light of this. If $x_i<x_{i-1}$, we known that $x_{i-1}$ exactly divides $x_{i}$ and we multiply the previous $MRS$ by $\dfrac{x_i}{x_{i-1}}$, an integer, and either way we then subtract one, to again get an integer $MRS_i$. So as we apporach a partial series value of $1$ it is always an integer number of terms away, until $MRS_\ell = 0$ as required for the full series to pass $1$.

Note that this results holds whenever $x_i \mid x_{i-1}$ is a property of the sequence - we don't actually need the successive $x_i$ to be (negative) powers of two. Also, if the series passes other integers, it will also hit them exactly.

Title: Proving sum of negative powers of 2 equals one before becoming greater than one

Body: Let $x$: $x_1 \geq x_2 \geq x_3\geq...\geq x_n$ be negative powers of 2 with sum greater than one. Then $\exists$ $l$ s.t. $x_1 + x_2 + ... + x_l = 1$. It seems obvious from examples, but I'm finding it a bit difficult to prove it. Any help would be highly appreciated.


问:
Kummer extension correspondence without roots of unity (Serge Lang)
I'm trying to solve the following problem.

Let $k$ be a field of characteristic $0$. Assume that for each finite extension $E$ of $k$, the index $(E^* : E^{*n})$ is finite for every positive integer n. Show that for each positive integer $n$, there exists only a finite number of abelian extensions of $k$ of degree $n$.

If $k$ contains a primitive n-th root of unity, one could use the one-to-one correspondence of abelian extension of $k$ of exponent n and subgroups of $k^*$ containing the n-th powers of the nonzero elements of $k$. For this case one of the ways to solve is as in the answer of this post: Find the bijection between Kummer's field and Galois subgroup.

But for $k$ not containing n-th roots of unity, do we have any kind of correspondence between, say, abelian extension of $k$ of exponent m and abelian extension of $k(\zeta)$ of exponent n, whence $\zeta$ is a primitive n-th root of unity?

I observed that an abelian extension of $k$ of exponent n has extension degree no more than the extension degree over $k(\zeta)$ of the abelian extension of $k(\zeta)$ of exponent n generated by the same set, multiplied by $\varphi(n)$, whence $\varphi(n)$ denotes the Euler function.

Another observation: Assume $k$ does not contain n-th roots of unity. Let H be a subgroup of $k^*$ containing the n-th powers of the nonzero elements of $k$, then $H$ and $\zeta^j$ together generates a subgroup of $k(\zeta)^*$ containing the n-th powers of the nonzero elements of $k(\zeta)$.

答: Let $L/k$ be the compositum of all the abelian extensions of degree at most $n$ over $k(\zeta_n)$. Since $k$ has characteristic zero, $L/k$ is separable. Then, since $k(\zeta_n)$ has all $n$-th roots of unity, you already know that $L/k$ is finite. If $E/k$ is an abelian extension of degree $\leq n$, then $E(\zeta_n)$ is an abelian extension of $k(\zeta_n)$ of degree $\leq n$, hence $E\subset E(\zeta_n) \subset L$. Since $L/k$ is separable, it contains at most finitely many subextensions. Hence the set of possible $E$ is finite.


问:
plug-in estimator of variance
The answer to this question should be easy and a comment will be enough, but I am just struggling with it...

Let $F$ be a cumulative distribution function and $\mu$ the mean corresponding to a random variable which follows $F$. Let $\sigma^2=\int\limits_{-\infty}^\infty (x-\mu)^2dF(x)$ be the variance and $\hat{\sigma}^2=\frac{1}{n}\sum\limits_{i=1}^n (X_i-\overline{X})^2$ be the plug-in estimator of $\sigma^2$, where $X_i\sim F$ are iid.

Why is it true that $\mathbb E[(\bar X -\mu)^2]=\frac{1}{n}\sigma^2$?

I was thinking I need the distribution of $\bar X$ to figure out what the answer is but this seems too complicated

答:
It is true because

$$\mathbb{E}[\overline{X}_n-\mu]^2=\mathbb{V}[\overline{X}_n]=\mathbb{V}\Bigg[\frac{1}{n}\Sigma_i X_i\Bigg]=\frac{1}{n^2}\mathbb{V}[\Sigma_i X_i]=\frac{1}{n^2} n\sigma^2=\frac{\sigma^2}{n}$$

Title: plug-in estimator of variance


问:
Only solution to Diophantine equation with primes?
Find all solutions to:

$$p_1 ^2 - p_2^2 = p_3^2 + 1$$

where $p_i$ is a prime number (positive of-course)
The only solution that seems to exist is $(3,2,2)$ for $(p_1, p_2, p_3)$ accordingly. But I have no clue how to prove that no other solutions exist..
The only thing that jumps to my mind is assuming that another solution exists and somehow get to a contradiction...

I would highly appreciate your help, thank you!

答: The square of a prime can only be $0$ or $1$ modulo $4$; the difference of two such squares cannot be $2$ modulo $4$. However, the RHS is $2$ modulo $4$ if $p_3$ is an odd prime. Thus $p_3=2$; this implies (again by the modular argument) that $p_2=2$, hence $p_1=3$ and the only solution is $3^2-2^2=2^2+1$.


问:
How would you position the UK Open University's math BSc programme in the realm of math BSc's?
I hope this question fits the format here; at least there is an "education" tag defined. The Open University has a "BSc Honours" degree in mathematics, having its programme of study detailed through here. The courses, divided into modules, are grouped there and listed by three stages roughly corresponding to three years of study.

To me it looks like an applied mathematics curriculum, but I could be wrong about that.

What would be your take, if you are intimately familiar with other mathematics BSc programmes or perhaps even with this particular one? Are there key pedagogic aspects that you find neglected in this programme?

答:
I am currently studying level 2 modules towards BSc in Mathematics at OU. My background is in economics / finance / real estate from UCL and Cambridge and I am doing this as a 2nd UG along work. I am now half-way through M208 (Pure Mathematics) and MST224 (Mathematical methods). From Level 1 I needed to take only MST125 (Essential Mathematics 2) as rest was covered by credit transfer from my previous studies.

The below is based on my observations and research so far and I would love to if someone more knowledgeable could correct me.

Level 1 courses are quite easy and aimed at bringing students without A-levels up to the required level. Saying that MST125 is at the A-Level Maths / Further Maths is probably accurate. The questions are more straight-forward and concentrate on the method rather than problem solving. You are also allowed to take an extensive handbook into the exam.

Let me use the Oxford BA in Mathematics as a comparison. I do this because the structure and syllabus appears to be quite similar. I am not able to comment on the difference in difficulty or depth. This is not my intention as both degrees are aimed at different demographics.

https://courses.maths.ox.ac.uk/year/2019-2020#44186

By and large Level 2 at OU: M208 (Pure Maths), MST224 (Mathematical methods) and M248 (Data Analysis) cover more or less the Prelims (1st year) at Oxford: analysis, group theory, linear algebra, multi-variable calculus, some vector calculus, proof etc. You will be short on probability and dynamics.

On Level 3 you can take only 4 courses, so if you stick to pure math: M303 (Further Pure Math - double unit covering number theory, metric spaces, groups, rings and fields), Complex Analysis and add Probability, then you kind of cover the 1st term of the second year (excl. ODE and some Linear Algebra). If you would take additional level 3 courses available (above the limit) then you cover also some subjects equivalent to Hilary (2nd term) courses. But the choice is quite narrow and you don't have i.e. Topology or Integration.

https://courses.maths.ox.ac.uk/year/2019-2020#43995

So by and large with the OU BSc you might cover the first 2 years of a Math degree at a standard uni. You still cover more ground then some unis (i.e. Uni of Essex). I am still trying to figure out how to "bridge" the Year 3 (Functional Analysis, Galois, Measure Theory, Manifolds, Topology etc) in case that I would like to do an MSc at a brick uni afterwards. Advice would be welcome.

On studying at OU:

Key point is that you don't have lectures. You are supposed to work through the Open University textbooks and then attend tutorials and submit course-work that is graded by the tutor (for M208 which is 50% of the Y2 credits you have 7 such courseworks). During lockdown some tutors seem to have gravitated more towards lecture-style slides with some explainers on theory. Which is great for me.

During lockdown the uni performed quite well with almost no disruption. Exams took place in an open-book format, but with time limit.

The new textbooks for M208 (starting from 2019) are very well written. They present the ideas in a very clear way with good examples, proofs and exercises. The idea is that you read and work through them. I personally love them.

-In exams you can use the handbook with formulas and Theorems (but no proofs) which makes it much easier than i.e. Oxford where you need to remember everything. I am also not able to comment yet on the difficulty of Level 2 and 3 exams.

Out of curiosity I bought 2nd hand textbooks for a lot of Level 3 courses and they seem also to be quite good. My biggest headache that I will need to choose 4 modules out of 8 that I would love to do.

So overall, I am very satisfied with OU modules so far and plan to continue with OU as I can do it along full-time work. The aim of OU is distance learning and they do it very well. Their biggest strength are the well-written textbooks. The biggest drawback is that you can pick only a few modules on level 3.

I am currently wondering how to "bridge the gap" in terms of Year 3 courses (Functional Analysis, Galois, Measure Theory, Manifolds, Topology etc) in case that I want to pursue an MSc afterwards. I am certain that the BSc should be accepted where "maths, physics or engineering UG degree is required", but a Pure Maths degree might be more difficult due to the missing prerequisites.

Best regards,
Marek


问:
If $x-y$ were written as $-y+x$, then would its conjugate be $-y-x$?
The conjugate of $x-y$ is $x+y$. But, if I wrote it as $-y +x$, would the conjugate be $-y-x$?

Furthermore, is the conjugate of $x + y +z$

$(x+y)+z \to (x+y) - z $

$(x+z)+y \to (x+z) - y $

$(z+y)+z \to (z+y) - x $

Is there a way to formalize the idea of a conjugate?

答:
You can indeed formalize the notion of "conjugate", in several ways. But it's not "change the sign of the second term".

If you have a complex number $z = x+iy$ then its conjugate is $x-iy$. If you write $z = iy + x$ then the conjugate is still
$$
\bar z = x -iy = -iy +x .
$$

When you are dealing with radical expressions like
$a + b\sqrt{2}$ you find the conjugate by replacing the square root by its negative. That's a different conjugate, also often useful.


问:
Taylor formula, notation for remainder
Hi I am wondering how this Taylor formula came about, because I don't really see a remainder term here. Also not sure where the $\theta$ is coming here from. Is this some different way of notation for a remainder?

[...] among the fundamental functions $\varphi(x)$ (see Section 8, below). If $\beta\leqslant1$, then all the functions $\varphi(x)\in S^\beta$ are already analytic and there are, thereby, no functions of bounded support among them. Indeed, the remainder term of the Taylor formula
$$\frac{h^q}{q!}\varphi^{(q)}(x+\theta h)=\varphi(x+h)-\varphi(x)-h\varphi'(x)-\cdots-\frac{h^{q-1}}{(q-1)!}\varphi^{(q-1)}(x)$$

答:
The remainder for the expansion of the Taylor's formula at order $q-1$ of $f(x+h)$ is the left-hand side. Lagrange's form of the remainder is that
$$R_{q-1}(x+h)=\frac{h^q}{q!}\varphi^{(q)}(\xi),\quad x<\xi<x+h$$
This supposes $h>0$; if $h<0$, the condition is $x+h<\xi<x$. Both cases are covered by a single formula: any number between $x$ and $x+h$, can be written as $\xi=x+\theta h,\enspace 0<\theta<1,\:$  independently of the sign of $h$.

Title: Taylor formula, notation for remainder


问:
Problem calculating the integral $\int_{0}^{\infty}x^ne^{-xt}dx$
I want to calculate the Laplace transform of the function $ x ^ n $ but I have a problem with the limit that I get from applying integration by parts:

\begin{align*}
\int_{0}^{\infty}x^ne^{-xt}dx&=\int_{0}^{\infty}x^n\left(-\cfrac{1}{t}e^{-xt}\right)'dx\\
&=\left[-\cfrac{x^n}{t}e^{-xt}\right]_{0}^{\infty}-\int_{0}^{\infty}-\cfrac{1}{t}e^{-xt}(nx^{n-1})dx\\
&=\left[-\cfrac{x^n}{t}e^{-xt}\right]_{0}^{\infty}+\cfrac{n}{t}\int_{0}^{\infty}x^{n-1}e^{-xt}dx\\
&=\left[-\cfrac{x^n}{t}e^{-xt}\right]_{0}^{\infty}+\cfrac{n}{t}y_{n-1}(t)\\
&=\left[-\cfrac{x^n}{t}e^{-xt}\right]_{0}^{\infty}+\cfrac{n}{t}y_{n-1}(t)\\
&=\lim_{x\to\infty}-\cfrac{x^n}{t}e^{-xt}-\left[-\cfrac{0^n}{t}e^{-(0)t}\right]+\cfrac{n}{t}y_{n-1}(t)\\
&=\lim_{x\to\infty}-\cfrac{x^n}{t}e^{-xt}+\cfrac{n}{t}y_{n-1}(t)\\
\end{align*}

But the problem is that I don't know how to take this limit, my intuition tells me that it is $ 0 $ because lowering the term $ e ^ {- xt} $ presents an indeterminacy of the type $ \cfrac{\infty}{\infty} $ so if we apply l'Hôpital's rule of thumb enough times to eliminate the term "$ x $" from above the fraction and since the term below is never affected by the derivatives other than by the term $ t ^ m $ that multiplies it, then the limit would be 0 but I don't know how to prove it properly.

Although I also tried that it should be possible to demonstrate this limit by induction, but in that case I do not know how to make such a demonstration, especially I do not know how to use the fact that it is true for $ k $ and with it demonstrate that it is true for $ k + 1 $. I appreciate any help.

答:
You may avoid dealing with the limits by integrating as follows

$$\int_{0}^{\infty}x^ne^{-xt}dx
=(-1)^n \frac{d^n}{dt^n} \int_{0}^{\infty}e^{-xt}dx
= (-1)^n \frac{d^n}{dt^n} \left(\frac1t\right)=\frac{n!}{t^{n+1}}
$$


问:
Simulating probability based on percentage occurrence
I have a question similar to this one.

If for a group of 85 riders, 40 riders fall over 24 races, then the probability of a rider falling in any single race is:

$$ 1 - (1 - p)^{24} = \frac{40}{85} $$

Where $ p \approx 0.0262 $.

However, using this probability, if I run 24 simulations (races) of 85 binomial draws (riders) with $p$ probability (replicated 10k times), a fall occurs, on average, $ \approx 53 $ times, not the expected $ \approx 40 $ times.

Linking back to this question, however, using their example parameters, in which there are 98 riders, with 4 falling over 52 races. If I run 52 simulations (races) of 98 binomial draws (riders) based on the probability of a rider falling in their example ($p \approx 0.0008$) (replicated 10k times), a fall does occur, on average, $ \approx 4 $ times, as expected.

I'm really struggling to understand what about my interpretation of $p$ is creating this issue with the simulation.

Thanks

R code for simulations:

```<code>> mean(replicate(10000, sum(replicate(24, rbinom(85,1,0.0262)))))
[1] 53.4684

> mean(replicate(10000, sum(replicate(52, rbinom(98,1,0.0008)))))
[1] 4.0213
</code>```

答:
The reason why you are not getting consistent results is because your simulation does not model what the parameter estimate $p$ actually represents.  The original question is this:

Given a cohort of $n$ riders, each of whom ride some fixed number of
races $r$, and each rider has an individual, independent, and
identical probability $p$ of falling in any given single race, what is
the number of distinct riders that fall over $r$ races?
Conversely, given that $X \in \{0, 1, \ldots, n\}$ riders are observed
to have fallen at least once over $r$ races, what is the estimate
of the probability $p$ of a given rider falling in any single race?

As such, your simulation does not model this situation because yours counts falls, not riders.  If a single rider falls multiple times over $r$ races, you count each instance of a fall by the same rider.

So the way you would properly simulate this scenario is to generate $n$ iid geometric random variables $G_1, \ldots, G_n$ with parameter $p$.  $G_i$ represents the random number of races that rider $i$ will participate in until they experience their first fall.  Under this model, the appropriate parametrization of $G_i$ is $$\Pr[G_i = g] = p (1-p)^{g-1}, \quad g \in \{1, 2, 3, \ldots \}.$$  Then count the number of $G_i$ that are less than or equal to $r$, meaning they fell at least once over $r$ races.  This is your $X$, and you would simulate many observations of $X$.  The quantity $X/n$ gives an estimated proportion of riders falling over $r$ races.

In your case, you have $n = 85$, $r = 24$, and you observed $X = 40$.  Then your estimate of $p$ is $$\hat p = 1 - (1 - X/n)^{1/r} = 1 - (9/17)^{1/24} \approx 0.0261515.$$  When I simulate this in Mathematica using the code

```<code>Histogram[ParallelTable[Length[Select[
   1 + RandomVariate[GeometricDistribution[0.0261515], 85], # <= 24 &]], {10^5}]]
</code>```

I get the following:

Body:
The reason why you are not getting consistent results is because your simulation does not model what the parameter estimate $p$ actually represents.  The original question is this:

Given a cohort of $n$ riders, each of whom ride some fixed number of
races $r$, and each rider has an individual, independent, and
identical probability $p$ of falling in any given single race, what is
the number of distinct riders that fall over $r$ races?
Conversely, given that $X \in \{0, 1, \ldots, n\}$ riders are observed
to have fallen at least once over $r$ races, what is the estimate
of the probability $p$ of a given rider falling in any single race?

As such, your simulation does not model this situation because yours counts falls, not riders.  If a single rider falls multiple times over $r$ races, you count each instance of a fall by the same rider.

So the way you would properly simulate this scenario is to generate $n$ iid geometric random variables $G_1, \ldots, G_n$ with parameter $p$.  $G_i$ represents the random number of races that rider $i$ will participate in until they experience their first fall.  Under this model, the appropriate parametrization of $G_i$ is $$\Pr[G_i = g] = p (1-p)^{g-1}, \quad g \in \{1, 2, 3, \ldots \}.$$  Then count the number of $G_i$ that are less than or equal to $r$, meaning they fell at least once over $r$ races.  This is your $X$, and you would simulate many observations of $X$.  The quantity $X/n$ gives an estimated proportion of riders falling over $r$ races.

In your case, you have $n = 85$, $r = 24$, and you observed $X = 40$.  Then your estimate of $p$ is $$\hat p = 1 - (1 - X/n)^{1/r} = 1 - (9/17)^{1/24} \approx 0.0261515.$$  When I simulate this in Mathematica using the code

Histogram[ParallelTable[Length[Select[
   1 + RandomVariate[GeometricDistribution[0.0261515], 85], # <= 24 &]], {10^5}]]

I get the following: