问:
How to write a proof directly from the definition of derivative as a limit without using any of the differentiation rules?
I have a practice problem I have worked through but don't have an answer for it. How would you answer this?

Let a ∈ R. Let f be a function which is differentiable at a. Assume that f is always
positive. We define a new function g by g(x) = 1/
f(x)
. Prove that g is differentiable
at a and that
g
0
(a) = −f
0
(a)
(f(a))2
.
Write a proof directly from the definition of derivative as a limit without using any
of the differentiation rules (such as quotient rule or chain rule).

答:
Find a common denominator. Definition of $f'(x)$ is $\displaystyle\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

$$g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} = \lim_{h \to 0} \frac{\frac{1}{f(x+h)} - \frac{1}{f(x)}}{h} = \lim_{h \to 0} \frac{\frac{f(x)-f(x+h)}{f(x)f(x+h)}}{h} = \lim_{h \to 0} \frac{\left(\frac{f(x)-f(x+h)}{h}\right)}{f(x)f(x+h)}$$

Use the quotient limit law and factor out -1 from the numerator. Then the above is equivalent to

$$\frac{\displaystyle-\lim_{h \to 0}\left(\frac{f(x+h)-f(x)}{h}\right)}{\displaystyle\lim_{h \to 0}\left(f(x)f(x+h)\right)}$$

Replacing x with a we have

$$\frac{\displaystyle-\lim_{h \to 0}\left(\frac{f(a+h)-f(a)}{h}\right)}{\displaystyle\lim_{h \to 0}\left(f(a)f(a+h)\right)}$$

The numerator is equivalent to $-f'(a)$. The denominator is equivalent to $(f(a))^2$ as the second factor is a limit of a composite function and $\lim_{h \to 0} (a + h) = a$ so $\lim_{h \to 0}\left(f(a+h)\right) = f(a)$ (since $f$ is continuous at $a$ since it is differentiable at a). As the quotient of two limits that exist $g'(a)$ exists and is equivalent to $\frac{-f'(a)}{(f(a))^2}$


问:
When is a second derivative test considered inconclusive?
Just to check in, when a second derivative test is inconclusive does that mean $f^{'}(x)=0$ and $f^{"}(x)=0$ and there is a saddle point?

答:
The 2nd derivative test is inconclusive when you evaluate the 2nd derivative at your critical numbers and you get either $0$ or undefined.

NOTE: You'll only apply the 2nd derivative test when $f$ is continuous and differentiable and $c$ is a number such that $f^{'}(c)=0$ and $f^{''}$ exists near $c$

EXAMPLE: $f(x)=-3x^{5}+5x^3$ you find critical numbers: $x=-1,0,1$

We find that $f^{''}(0)=0$ so we know the second derivative test is inconclusive: in other words, the change in the slope is not known, so we don't know what the 'curve' will be.

We have received word that there can be a saddle point. But I don't know how to find that.

ContentLicense: CC BY-SA 4.0


问:
Integral of $\sec(3x)$ between certain variable limits.
$$\int_x^{x+\frac{\pi }{8}} \frac{1}{\cos (3 t)} \, dt$$

I get:
$$\frac{1}{3} \, \log\left(\sec\left(3x + \frac{3\pi}{8} \right)+\tan\left(3x+ \frac{3\pi}{8} \right) \right) - \frac{1}{3} \, \log(\sec(3x) + \tan(3x))$$

I can't make that look nice, though, and I'm supposed to be drawing a graph :/

答:
Consider first the antiderivative
$$I=\int \sec (3t)\,dt =\frac 13\int \sec (u)\,du  \qquad \text{with} \qquad u=3t$$  Now, the tangent half-angle substitution
$$\int \sec (u)\,du=2\int \frac {dv}{1-v^2}\qquad \text{with} \qquad u=2 \tan ^{-1}(v)$$ Partial fraction decomposition to get
$$2\int \frac {dv}{1-v^2}=\log (1+v)-\log (1-v)=-\log \left(\frac{1-v}{1+v}\right)$$
Back to $u$ $(v=\tan \left(\frac{u}{2}\right))$
$$\frac{1-v}{1+v}=\frac{1-\tan \left(\frac{u}{2}\right)}{1+\tan \left(\frac{u}{2}\right)}=\frac{\tan \left(\frac{\pi }{4}\right)-\tan \left(\frac{u}{2}\right) } {1+\tan \left(\frac{\pi }{4}\right)\tan \left(\frac{u}{2}\right) }=\tan \left(\frac{\pi }{4}-\frac u2 \right)$$ Back to $t$ $(u=3t)$
$$I=-\frac 13\log \left(\tan \left(\frac{\pi }{4}-\frac{3 t}{2}\right)\right)=\frac 13\log \left(\cot \left(\frac{\pi }{4}-\frac{3 t}{2}\right)\right)$$ Now, using the bounds (do not forget the absolute values - being lazy, I did not ype them in the preliminary calculations)
$$\int_x^{x+\frac{\pi }{8}} \frac{1}{\cos (3 t)} \, dt=\frac 13\log \left(\Big|\tan \left(\frac{\pi }{16}-\frac{3 x}{2}\right) \tan \left(\frac{\pi }{4}-\frac{3
   x}{2}\right)\Big|\right)$$


问:
Why do knowers of Bayes's Theorem still commit the Base Rate Fallacy?
Source: 6 July 2014, "Do doctors understand test results?" by William Kremer, BBC World Service.

A 50-year-old woman, with no prior symptoms of breast cancer, participates in routine mammography screening. She tests positive, is alarmed, and wants to know from the physician whether she has breast cancer  or what the chances are. Apart from the screening results, the physician knows nothing else about this woman. How many women who test positive actually have breast cancer?

a. nine in 10
     b. eight in 10
     c. one in 10
     d. one in 100

Additional (needed) information: The probability that a woman has breast cancer is 1% ("prevalence" or base rate probability). If a woman has breast cancer, the probability that she tests positive is 90% ("sensitivity" or reliability rating). If a woman does not have breast cancer, the probability that she nevertheless tests positive is 9% ("false alarm rate" or false-positive rate).

I have used a similar example when teaching introductory  probability and statistics for about 30 years now, and feel most  instructors use some variation of this problem (if for no reason other than) as an example of Bayes's Theorem.

What amazes me is that only 21% of gynaecologists responded (as reported by the author) with the correct answer (let us please leave the solution of this problem for a student).

My more cosmic questions: 1. Why do our students not understand this basic application of Bayes's Theorem? 2. What can we as instructors do to explain this prior and posterior probability concept more clearly? 3. Why do the trained experts (gynaecologists) get this question wrong?

答:
My guess is, expressed using a few symbols: People confuse $P(A|B)$ with $P(B|A)$.

This guess is based on paw88789’s answer, and the following quote from the article linked in the question:

In one session, almost half the group of 160 gynaecologists responded that the woman's chance of having cancer was nine in 10.

I think, in general, knowing which wrong answer people gave is very useful in trying to understanding why they gave that wrong answer.

ContentLicense: CC BY-SA 3.0


问:
Number of ways to get a flush from shuffled cards
How many different ways are there to draw a flush from a well shuffled full deck of cards?

If correctly counting, I suppose the answer is $\frac{13\times 12\times 11 \times 10 \times 9}{1\times 2 \times 3 \times 4 \times 5}$, but would like to know your thoughts on it.

答: You can select $5$ cards from a suit in $\binom{13}5$ ways, and since there are $4$ suits, multiply it by $4$, which you did not.


问:
Help: Need to prove that $|G| = 18$.
if $f : G → \mathbb{Z}_6$ is a surjective morphism and $|Ker(f)| = 3$ prove that $|G| = 18$.

答: By the first isomorphism theorem, $f(G)\cong G/\rm{ker}f$.  Therefore $|G|/3=6\implies |G|=18$.

Title: Help: Need to prove that $|G| = 18$.


问:
Line Integral of a Vector Field in $\Bbb{R}^{3}$
Let $\mathbf{F}(\mathbf{r})=(x^{2}z,xy^{2},z^{2})$ and define a closed curve $C$ in $\Bbb{R}^{3}$ comprised of three straight line segments $C_{1},C_{2}$ and $C_{3}$ formed by the intersection of the plane $x+y+z=1$ with the planes $y=0,z=0$ and $x=0$ (respectively).

Sketch the curve $C=C_{1}\cup C_{2}\cup C_{3}$ in the $(x,y,z)$-space, and calculate $\displaystyle\int_{C}\mathbf{F}\cdot d\mathbf{r}$ by considering each line segment separately. Be careful to make sure the orientation of each segment of $C$ is aligned with the other segments.

Below is an image that I have plotted for all four planes: orange represents $x+y+z=1$, yellow represents $y=0$, blue represents $x=0$, and red represents $z=0$. I am not too sure how to evaluate the line integral, and how to relate the curve $C$ back to $\mathbf{F}(\mathbf{r})$ - appreciate any pointers I can get, thanks.

Body:
Intersection points of $x + y + z = 1$ plane with $x = 0, y = 0, z = 0$ planes are,

$A (1,0,0), B (0, 1, 0), C (0, 0, 1)$.

So our lines (all oriented in the same direction) are

$AB = (1, 0, 0) + (-1, 1, 0)t \, $ ($t = 0$ to $t = 1$)

$BC = (0, 1, 0) + (0, -1, 1)t \, $  ($t = 0$ to $t = 1$)

$CA = (0, 0, 1) + (1, 0, -1)t \, $  ($t = 0$ to $t = 1$)

$i)$ Now for our line integral along path $AB$

Please note $r(t) = (1-t, t, 0) \,$ and $\, r'(t) = (-1,1,0)$

$C_1 = \displaystyle \int_{t=0}^{1} (x^{2}z,xy^{2},z^{2}) dr = \int_{t=0}^{1} ((1-t)^2.0,(1-t)t^2, 0) r'(t) dt$

$C_1 = \displaystyle \int_{t=0}^{1} (0,(t^2-t^3), 0) \cdot (-1,1,0) dt = \int_{t=0}^1 (t^2-t^3) dt = \frac{1}{12}$

You can similarly find $C_2$ and $C_3$.


问:
R.v. $X$, $E[X^2]<\infty$. How to prove that $\mathrm{Var}[X\land a] \le \mathrm{Var}[X]$ for any real number $a$?
As the title says, It looks simple, however not very intuitive. I finished it with the step function converging to it. However, it seems too complicated and I wonder would there be some elementary proof.</p>

答:
If $a\geq \mu=\mathrm E(X)$ we have
$$
\mathrm {Var} (X\wedge a) \leq \mathrm E[(X\wedge a-\mu)^2]
\\ = \mathrm E[(X-\mu)^2\mathbf 1_{X<a}] +\mathrm E[(a-\mu)^2\mathbf 1_{X\geq a}]
\\ \leq \mathrm E[(X-\mu)^2\mathbf 1_{X<a}]+ \mathrm E[(X-\mu)^2\mathbf 1_{X\geq a}]
=\mathrm{Var}(X).
$$

If $a\leq \mu$ we have
$$
\mathrm {Var} (X\wedge a) \leq \mathrm E[(X\wedge a-a)^2]
\\ = \mathrm E[(a-X)^2\mathbf 1_{X<a}] +\mathrm E[(a-a)^2\mathbf 1_{X\geq a}]
\\ \leq \mathrm E[(\mu-X)^2\mathbf 1_{X<a}]  \leq \mathrm {Var}(X).
$$


问:
infinite sum of inverse binomial coefficient encountered in Bayesian treatment of the German tank problem
in the Bayesian treatment of the German tank problem in Wikipedia here, they use:

$\displaystyle \sum_{n=m}^\infty \dfrac{1}{\binom{n}{k}}=\dfrac{k}{k-1}\dfrac{1}{\binom{m-1}{k-1}}$

how can I prove this in a clever combinatorics fashion?

I found this paper, see eqn. (9), which uses Gauss' hypergeometric function-- a bit beyond me.

there must be some way via a recursion relation, like I found in this old paper. theorem 1 in that reference has a similar infinite sum of an inverse binomial coefficient.

Body:
Let $$S=\sum_{n=m}^{\infty} {n \choose k}^{-1}~~~~(1)$$
Use $${n \choose k}^{-1}=(n+1)\int_{0}^{1} x^k (1-x)^{n-k} dx~~~~(2)$$
Then $$S=\int_{0}^{1} \sum_{n=m}^{\infty} x^k(1-x)^{-k} \sum_{n=m}^{\infty} [(n+1) (1-x)^n]~~~~(3)$$
Use sum of infinite GP: $$\sum_{j=m}^{\infty} (j+1)z^j= x^{-2} (1-x)^m(1+mx)~~~~(4)$$
Then $$S=\int_{0}^{1} x^{k-2} (1-x)^{m-k}(1+mx) dx~~~~(5)$$
Use $\beta$ integral:
$$\int_{0}^{1} z^u (1-z)^v dz=\frac{\Gamma(1+u) \Gamma(1+v)}{\Gamma(2+u+v)}~~~~(6).$$
Then $$S=\frac{\Gamma(k-1) \Gamma(1+m-k)}{\Gamma(m)}+\frac{m\Gamma(k) \Gamma(1+m-k)}{\Gamma(1+m)}~~~~(7)$$
$$\implies S=\frac{k (k-2)! (m-k)!}{(m-1)!}= \frac{k}{k-1} {m-1 \choose k-1}^{-1}~~~~(8)$$


问:
Showing that a matrix raised to a power is equal to itself
Let $F$ be a field of cardinality $q=2^k$ for some $k \in \mathbb{N}$ and let

$$E= \begin{bmatrix}
1 & 0 & & & & & &\\
1 & 0 & & & & & &\\
& & \ddots & & & & & &\\
& & & 1 & 0 & & &\\
& & & 1 & 0 & & \\
& & & & & & 1 & 0 & a_{n-1}a_{n-2}\\
& & & & & & 1& 0 & a_{n-2} \\
& & & & & & & & 1+a_{n-1}
\end{bmatrix} \in M_n(F).$$
I want to show that $E^q=E.$

I was able to show for $q=2,4,8$ that
$$E^q= \begin{bmatrix}
1 & 0 & & & & & &\\
1 & 0 & & & & & &\\
& & \ddots & & & & & &\\
& & & 1 & 0 & & &\\
& & & 1 & 0 & & \\
& & & & & & 1 & 0 & a_{n-1}^qa_{n-2}\\
& & & & & & 1& 0 & a_{n-2} \\
& & & & & & & & 1+a_{n-1}^q
\end{bmatrix}$$
through brute force computation. This gives the desired result as $x^q=x$ in $F$. Note that this is not the case for $E^3, E^4, E^5, E^6, E^7.$

I want to generalize this for any $q=2^k$ and I have no idea how. Any help would be appreciated.

答:
Write $x=a_{n-1}$ and $y=a_{n-2}$ for notational ease. Let $M$ represent the matrix formed by the first $n-1$ columns of $E$ with zeroes in column $n$, and let $N=E-M$ be the $n$th column of $E$ with zeroes in the first $n-1$ columns. It is not hard to check that
$$M^2=M,\ N^2=(x+1)N,\ NM=0.$$
In particular, when we expand the sum $(M+N)^q$, all terms with an $NM$ vanish, so
\begin{align*}
(M+N)^q
&=\sum_{i=0}^q M^{q-i}N^i\\
&=M^q+N^q+\sum_{i=1}^{q-1} M^{q-i}N^i\\
&=M+(x+1)^{q-1}N+MN\sum_{i=1}^{q-1}(x+1)^{i-1}\\
&=M+(x+1)^{q-1}N+MN\left(\frac{(x+1)^{q-1}-1}x\right),
\end{align*}
where the last fraction is as a polynomial in $x$. We need to show that
$$(x+1)^{q-1}N+MN\left(\frac{(x+1)^{q-1}-1}x\right)=N,$$
which is equivalent to
$$MN\left(\frac{(x+1)^{q-1}-1}x\right)=N\left((x+1)^{q-1}-1\right).$$
If $x\neq 0,1$, this is true, since both sides are $0$. When $x=1$, we require that $MN=N$, which can be checked directly (we can find that $MN$ is $0$ everywhere except for the bottom of the last column, where it is $\dots 0,xy,xy,0$). When $x=0$, the right side is still $0$, so it suffices for $MN$ to be $0$, which is true.


问:
Discrete and Combinatorial Mathematics Ralph P. Grimaldi (fifth edition) Problem 18 Section 11.3
I'm working on problem #18 of section 11.3 from Ralph P. Grimaldi's textbook Discrete and Combinatorial Mathematics an Applied Introduction, fifth edition.

Let $k$ be a fixed positive integer and let $G=(V,E)$ be a loop-free undirected graph, where $deg(v)
\geq k$ for all $v \in V$. Prove that $G$ contains a path of length $k.$

I'm really having trouble understanding how to tackle the problem because the number of vertices is not given. At the end of section 11.1 states that when a graph is a multigraph it will be stated, however it is not stated in this problem that G cannot be a multigraph.

If G can be a multigraph then clearly the statement is false:

Start with $K_n$ the complete graph on $n$ vertices $v_1,v_2,...,v_n$ then add the edges $\{v_1,v_2\}$,$\{v_2,v_3\},...,\{v_{n-1},v_n\},\{v_n,v_1\}$, then every vertex has degree $n-1+2=n+1$, however there are only $n$ vertices, thus there is no path of length $n+1$.

For example with $K_5$ (the added edges being $\color{red}{red}$):

How do I go about working this problem?

答:
Welcome to MSE!

Hint:

This is an example of a "greedy algorithm". The idea is to just keep adding vertices into your path while avoiding "obvious" problems. In this case, you never want to repeat a vertex. In slightly more detail:

Pick a vertex $v_1$. It has degree at least $k$, so it has at least $k$ neighbors. Pick a neighbor $v_2$.

Now this vertex has degree at least $k$, but there's one vertex we don't want to go back to ($v_1$). Thankfully $k>1$, so we can pick a new vertex $v_3$.

But this new vertex has degree $k$ too. Of course, now there are two vertices which we want to avoid ($v_1$ and $v_2$). Thankfully $k > 2$, and we can pick a new vertex $v_4$.

Keep doing this until you can no longer guarantee a new vertex exists. How long must your path be? Can you make this argument precise?

I hope this helps ^_^


答:
Let $G$ have no self-loops and parallel edges. Then $G$ has atleast $k+1$ vertices. Start with any vertex $v_1\in V$. It has $k$ distinct neighbours, chose any one neighbour, say $v_2$. Now $v_2$ has $k$ distinct neighbours but one of them is $v_1$, which can't be selected as the next vertex of the path since we can't repeat vertices. Select $v_3\ne v_1$ neighbour of $v_2$. You can proceed in a similar fashion till $v_{k+1}$ since at $v_{k+1}$ it is possible that the $k$ neighbours of $v_{k+1}$ are the previously selected $k$ vertices of the path. The length of the path constructed is $k$.

Note that this theorem does not hold when $G$ has parallel edges or self-loops. Consider a graph of two vertices with two parallel edges between them, then $\deg(v)\ge2$ but you can't have a path of length $2$. Alternatively, to a graph of a single vertex, you can keep adding self-loops to make the degree of the vertex larger but you don't have any path of length $\ge1$ in the graph. Self-loops and parallel edges create trouble because for a vertex of degree $m$ you cannot assume $m$ distinct neighbours.

Title: Discrete and Combinatorial Mathematics Ralph P. Grimaldi (fifth edition) Problem 18 Section 11.3

Body:
I'm working on problem #18 of section 11.3 from Ralph P. Grimaldi's textbook Discrete and Combinatorial Mathematics an Applied Introduction, fifth edition.

Let $k$ be a fixed positive integer and let $G=(V,E)$ be a loop-free undirected graph, where $deg(v)
\geq k$ for all $v \in V$. Prove that $G$ contains a path of length $k.$

I'm really having trouble understanding how to tackle the problem because the number of vertices is not given. At the end of section 11.1 states that when a graph is a multigraph it will be stated, however it is not stated in this problem that G cannot be a multigraph.

If G can be a multigraph then clearly the statement is false:

Start with $K_n$ the complete graph on $n$ vertices $v_1,v_2,...,v_n$ then add the edges $\{v_1,v_2\}$,$\{v_2,v_3\},...,\{v_{n-1},v_n\},\{v_n,v_1\}$, then every vertex has degree $n-1+2=n+1$, however there are only $n$ vertices, thus there is no path of length $n+1$.

For example with $K_5$ (the added edges being $\color{red}{red}$):

How do I go about working this problem?


问:
Removing the subtractive cancellation of $(a+x)^n - a^n$ as x approaches zero
$$(a+x)^n - a^n = \\
e^{n\ln(a+x)}-e^{n\ln(a)} = \\
?$$

This looks simple but I'm stumped.

I've tried to solve this another way too:

$$ f(a) = a^n \\ \frac{\delta f}{\delta a} = \lim_{x\rightarrow 0}\frac{(a+x)^n-a^n}{x}  \\ \\
(a+x)^n - a^n = \frac{(a+x)^n-a^n}{x}x = \frac{\delta f}{\delta n}x=a^n\ln(a)x:x\rightarrow0\\$$

How do I finish the first one, and is the second one correct?

答:
$$S_n=(a+x)^n-a^n=\sum_{k=1}^n  \binom{n}{k}  a^{n-k} x^k=a^n\sum_{k=1}^n  \binom{n}{k}\left(\frac{x}{a}\right)^k$$ Truncate where ever you want.

If $x \ll a$, the first approximation would be
$$S_n=n  a^{n-1} x$$ but you can have better with
$$S_n=\frac{2 n x a^n}{2 a-(n-1)x}$$


问:
A question about pullback of a smooth form.
In Lee book, "Introduction to Smooth Manifold, 2nd edition" there is a proof about the existence and Uniqueness of the Exterior differentiation (page 365). He define the exterior differentiation of a smooth $k$-form $\omega\in \Omega^k(M)$ locally on each smooth chart $(U, \phi)$ by using pullbacks:
$$d\omega = \phi^*d(\phi^{-1*}\omega).$$
My question is how do we pullback $\omega$ by $\phi^{-1}$? why I ask because by definition of pullback we could pull back by $\phi^{-1}$ only for forms on $U$, but here $\omega$ is a smooth form on $M$ not $U$. Something is not clear here for me please help.

答: Properly speaking, Lee is not pulling back $\omega$ but its restriction to $U$. We can define the restriction to $U$ of $\omega$ via $\omega|_U = i^*\omega$ where $i: U \longrightarrow M$ is the inclusion map. So really, this defines $(d\omega)|_U$. You can show that these glue together to properly define $d\omega$.


问:
Show that if $\text{int}(E) = \text{int}(F) = ∅$ and $E$ is closed, then $\text{int}(E\cup F) = \emptyset $.
I have the following doubt, they ask me the following:

Let $(X, d)$ a metric space and $E, F \subseteq X$. Show that if $\mathrm{int}(E) = \text{int}(F) = ∅$ and $E$ is closed, then $\text{int}(E\cup F) = \emptyset $.

I know that $x \in A$ is interior if there exists an $ r>0$ such that $B_r(x) \subset A$, but I cannot think how to start.

I appreciate your time reading this.

答: If $B$ is any open ball then $B$ is not contained in $E$. Take some $x \in B \setminus E$. There is a smaller open ball $B’ \subset B$ around $x$ that does not intersect $E$ (since $E$ is closed). But also $B’$ is not contained in $F$. So there is a point in $B’$ that is not in $E\cup F$.


问:
Prove that compact sets are closed under arbitrary intersections and finite unions.
I want to show that compact sets are closed under arbitrary intersections and finite unions. I tried the problem and then I noticed that I might be able to take it a step further. Please let me know if I have any mistakes, thanks!

Proof I: Compact sets are closed under finite unions.

Let $\displaystyle A = \{A_1, \dots, A_N\}$ be a set of finitely many compact sets. We know that the elements of $\displaystyle A$ are compact if and only if they are both closed and bounded. It follows that,
\begin{equation*}
\bigcup_{k=1}^N A_k
\end{equation*}
is closed because finite unions of closed sets are closed. Thus we have that the finite union of compacts sets is closed. More over, we have that $M_1, \dots, M_N$ are bounds of $A_1, \dots, A_N$,  respectively. Let $M = \max\{M_1, \dots, M_N\}$, then it follows that
\begin{equation*}
\forall A_k \in A, \forall x\in A_k, |x| \leq M
\end{equation*}
therefore $M$ is a bound of $\bigcup_{k=1}^N A_k$.

We have that $\bigcup_{k=1}^N A_k$ is both closed and bounded, therefore it is compact. Compact sets are compact under finite unions.

Proof II: Compact sets are closed under arbitrary intersections.

Let $\displaystyle A = \{A_1, A_2, A_3, \dots \}$ be an infinite set of compact sets. We know that the elements of $\displaystyle A$ are compact if and only if they are both closed and bounded. It follows that,
\begin{equation*}
\bigcap_{k=1}^\infty A_k
\end{equation*}
is closed because arbitrary intersections of closed sets are closed. Thus we have that the arbitrary intersection of compact sets is closed.

Is there any way to show that $\bigcap_{k=1}^\infty A_k$ is bounded so that I can say it is compact?

答: You can simply say that $\cap_{k=1}^\infty A_k \subseteq A_1$. $A_1$ is compact, so it is bounded, and any subset of a bounded subset is bounded. By the way, strictly speaking you have not shown that an arbitrary intersection of compact subsets is compact - only that a countable intersection is. This is an immediate fix: just index your intersection by some $I$ rather than by $\mathbb N$. Other than this, your proof is entirely correct!


问:
Laplace transform of $f(\frac{t}{c})$
Laplace transform of $f(\frac{t}{c})$. So in order for me to able to do the Laplace transform of the following function, what I did was the following:

$L(f(\frac{t}{c})) = \int_{0}^{\infty}e^{-st}f(\frac{t}{c})dt$

Now because I don't know of an easier way to solve this, what I did was integration by parts(feel free to correct me if there is an easier way to do this other than integration by parts)

Anyway, then I said:

Letting $dv = e^{-st}dt \implies v = \frac{-1}{s}e^{-st}$ and $u = f(\frac{t}{c}) \implies du = \frac{1}{c} f(\frac{t}{c})$

But what to do now? Am I even going in the right direction? The algebra gets considerably more mucky after this, and I do not know of a better way to solve it.

答:
$$L \left \{f\left(\frac{t}{c}\right)\right \}= \int_{0}^{\infty}e^{-st}f(\frac{t}{c})dt$$
Substitute $t=uc \implies dt=cdu$:
$$L \left\{f\left(\frac{t}{c}\right)\right\} =c \int_{0}^{\infty}e^{-(sc)u}f(u)du=cF(sc)$$

Title: Laplace transform of $f(\frac{t}{c})$


问:
Web 2.0 and mathematics for high school students
First of all, if this is not the right place to ask, I am sorry and just close this topic.

I'm trying to introduce Web 2.0 into my Math class for teenagers in the range of 12-16 years old.  I've thought of creating forums for problem solving and discussing, del.icio.us for researching/sharing math webs, and personal blogs to make each student prove his/her evolution about topics talked in class.

However, I would like to know if somebody has had experience or has any idea of how we could use Social Web into a high school math class.

答:
I think introducing web-based content in a high school math course (or any level math course, for that matter) demands a lot of work for little to no return. I understand that you want to engage your students, but I think that trying to do so in a web- or computer-based environment will lead to them spending more time battling the interface than solving or thinking about problems. Just think about the amount of time it took you to learn Latex when starting out.

If you want to get your students to think about math, and talk about math between themselves, then find good, instructive problems for them to solve. Preferably ones that can be expressed in some real world or simple geometric terms, and that can be visualized or drawn. Visual aids are your friend. Rubics cubes, bits of string, pieces of paper - whatever you can find to bring your problem to life is helpful.

Then put them into a long, coherent problem sheet with a clearly identifiable result at the end, and assign it as a group project. Give them a week, have them work in groups of three or four on the problems, and make them hand it in as a team. At the end, get one or two teams to present their work to the class. Try and get the class to discuss the solutions. Then change up the groups and repeat.

Of course this demands an incredible amount of work from you as a teacher. But it'll be work put towards teaching your students mathematics, instead of teaching them to claw their way through a halfway functional web interface.


问:
To prove $EA = FB$ or that $CQ'$ is radical axis
Given disjoint circles $c_1 = \odot(P,PA), c_2 = \odot(O,OB)$ such that $B$ and $A$ are in the same half-plane wrt $OP$ and that $PA \parallel OB \perp OP$.

Line $CDQ$ is the perpendicular bisector of $AB$, $D \in AB,Q \in OP$.

Point $Q'$ is the reflection of $Q$ wrt to $M$, the midpoint of $OP$.

Point $C$ on line perpendicular to $OP$ in $Q'$.

$E = CA \cap c_1, F = CB \cap c_2$.

I want to prove $EA = FB$.

We know that $DC = DQ$ because $DM$ is midsegment of trapezium $OBAP$.

答:
Since $OB \parallel AP$, the line $BA$ goes through the external homothetic center $X$ of the circles $(O),(P)$. The other intersections of $BA$ with the circle are $B_2,A_2$. The common external tangents also go through $X$.  The points of contact of the the upper tangent are $J,K$.

The angles marked in red are all equal to $\angle{BXO}$. Let $H=B'B_2\cap OP$.  $\triangle{OHB}\sim\triangle{OBX}\implies OH/OB=OB/OX\implies OH\cdot OX=OB^2\implies H$ is the inverse of $X$ wrt $(O)$.  Thus $JJ'$, the polar of $X$ wrt $(O)$, contains $H$. $H'$ is the reflection of $H$ wrt $O$. $M=(O+P)/2,$ the midpoint of $O,P$.

Since $X$ is the external homothetic center, $(O)\sim (P)$.  Distances of corresponding elements all have the same ratio, i.e. $XO/XP=XB/XA=XB_2/XA_2=XJ/XK$ etc.  Let $h$ be the homothety that takes $(O)\rightarrow (P)$, i.e. $h(O)=P, h(B)=A$, etc.

Consider the homothety that takes a point $T$ to $(T+h(T))/2$, the midpoint of $T,h(T)$. Then, referring to the points in the OP, $O\rightarrow M, B\rightarrow (B+A)/2, H'\rightarrow Q,H\rightarrow Q',$ and $J\rightarrow M'=(J+K)/2$.  But the midpoint $M'$ of $J,K$ is on the radical axis of $(O),(P)$ because the tangents from $M'$ to the two circles are of equal length.  So $Q'$, which by similarity is on the line through $M'$ perpendicular to $OP$, is also on the radical axis.  And so is $C$, so we are done.

Note that the bisector of $B_2,A_2$ passes through $Q'$, so this is a more direct construction.


问:
Does existence of homomorphism imply there is $\psi$ with $\psi(ab) = \psi(b) \circ \psi(a)$
Let $G$ be a group and $A(G)$ its automorphism group. Suppose there is a homomorphism $\phi: G \to A(G)$. That is, for all $a, b \in G$, we have $\phi(ab) = \phi(a) \circ \phi(b)$. Must there exist $\psi: G \to A(G)$ such that, for all $a, b \in G$, $\psi(ab) = \psi(b) \circ \psi(a)$?

答:
Hint:

$g \mapsto g^{-1}$ is an isomorphism $G \cong G^\text{op}$ for any group $G$.

Here $G^\text{op}$ has the same underlying set as $G$ but with multiplication done backwards: $g \cdot^\text{op} h = h \cdot g$.

In this language, $\psi$ should be a homomorphism $G \to A(G)^\text{op}$ (I'll leave it to you to check this). Can you use the isomorphism above to build such a $\psi$ if you are a given a homomorphism $\phi : G \to A(G)$?

I hope this helps ^_^


问:
Changing from Discrete Random Variable into Continuous Random Variable
A random glass breaks at an exponentially distributed time, with mean 24 months. Probability that out of 500 glass, at most 210 breaks within a year? (Numerical answer, not just a formula)

First of all, using exponential distribution, the probability that a window breaks within a year is $e^{\frac{-1}{2}}$. I understand that we can simply find the CDF using binomial,

$$ \sum_{i=0}^{210} {500 \choose i}\left( e^{\frac{-1}{2}} \right)^{i}  \left({1-e^{\frac{-1}{2}}} \right)^{500-i}
$$

However, finding the numerical value for this is very impractical, and I figured the point of the exercise is to find the PDF in terms of a continuous random variable, that I can simply integrate? However, I can't figure out how to model the question with a continuous random variable.

答:
What you are essentially looking for is the CDF of the binomial distribution.

A quick glance at the wiki page (and the references therein) tell us that the CDF is defined by the regularized beta function which can be expressed as an integral as follows for $p=e^{-\frac{1}{2}}$, $n=500$ and $k=210$.

$$Pr(X \leq k) = (n-k) {n\choose{k}}\int\limits_0^{1-p} t^{n-k-1} (1-t)^k dt$$

You can evaluate the integral analytically.


问:
Prove that the sequence $\{a_n \}_{n \geq 1}$ is convergent.

Suppose that a bounded sequence $\{a_n \}_{n \geq 1}$ is such that $$a_{n + 2} \leq \dfrac {1} {3} a_{n+1} + \dfrac {2} {3} a_n,\ \ \ \ \text {for}\ n \geq 1$$ Prove that the sequence $\{a_n \}_{n \geq 1}$ is convergent.

What I find is that for all $n \geq 1,$ we have $$a_{n + 2} - a_2 \leq \dfrac {2} {3} (a_1 - a_{n+1}).$$

Also it can't be eventually monotone increasing since for otherwise for all $n \geq 1,$ we have $$a_{n+2} - a_{n+1} \leq \dfrac {2} {3} (a_n - a_{n+1}) \leq 0 \implies a_{n+2} \leq a_{n+1},$$ a contradiction. So if the sequence is eventually monotone it has to be eventually monotone decreasing.

Is it of any importance? Thanks.

答:
With $c>0$ determined below, let $b_n=a_{n+1}+ca_n$. Then $\{b_n\}_{n\ge1}$ is also bounded from below and we obtain
$$b_{n+1}=a_{n+2}+ca_{n+1}\le \left(\frac13+c\right)a_{n+1}+\frac23 a_n=\frac{(1+3c)a_{n+1}+2a_n}{3}.$$
If we smartly pick $c=\frac23$, this amounts to
$$ b_{n+1}\le b_n.$$
We conclude that sequence $b_n$ converges to some limit $b$.

By applying $\liminf$ to both sides of the equation $a_{n+1}=b_n-\frac23 a_n$, we find
$$\tag1\liminf a_n=b-\frac23\limsup a_n $$
and similarly
$$\tag2\limsup a_n=b-\frac23\liminf a_n.$$
Solving the linear equations  $(1)$ and $(2)$, we arrive at
$$ \liminf a_n=\limsup a_n=\frac 53 b,$$
i.e.,
$$\lim a_n=\frac 53 b. $$


问:
Finding an upper limit on a continuous function
I'm tasked with finding a smallest constant $c$ so that for a function $f\in C^4([0,1])$ and $x,y\in[0,1]$:
$$\left\vert\frac{f(x)-f(y)}{x-y}-\frac{f'(x)+f'(y)}{2}\right\vert\le c\vert x-y\vert ^2\max_\limits{\xi\in[0,1]}\vert f'''(\xi)\vert$$
Since this is an excercise I was given on the Topic of Taylor series, my first thought was making use of Taylor's theorem and somehow set the upper bound using the Lagrange remainder but I'm not quite sure that's the way to go. Alternatively I thought about using the fact, that $f(x)$ is lipschitz continuous. Any nudge in the right direction would be greatly appreciated. Thanks in advance.

答:
In fact the smallest constant for which the inequality holds is $c = \frac{1}{12}$.

For the proof, consider the map

$$\varphi(t) =f(x) - f(t) +\frac{t-x}{2}\left(f^\prime(t) + f^\prime(x)\right)-\frac{(t-x)^3}{12}A$$ where $A$ is such that $\varphi(y)=0$.

As $\varphi(x)=0$ Rolle's theorem ensures the existence of $c_1 \in (x,y)$ with $\varphi^\prime(c_1)=0$, i.e.
$$0 =  - f^\prime(c_1) +\frac{f^\prime(c_1) + f^\prime(x)}{2}+\frac{c_1-x}{2}f^{\prime\prime}(c_1)+\frac{(c_1-x)^2}{4}A.$$

Applying Taylor's theorem to $f^\prime$ it exists $c \in (x,c_1)$ with

$$f^\prime(x) - f^\prime(c_1)=(x-c_1)f^{\prime\prime}(c_1) + \frac{(x-c_1)^2}{2} f^{\prime \prime \prime}(c).$$

The two equalities above imply $A=f^{\prime \prime \prime}(c)$. Using $\varphi(y)=0$ in the definition of $\varphi$ we get for $x \neq y$

$$\frac{f(x)-f(y)}{x-y} - \frac{f^\prime(x) + f^\prime(x)}{2}= -\frac{(x-y)^2}{12}f^{\prime \prime \prime}(c)$$ and therefore the desired inequality

$$\left\vert\frac{f(x)-f(y)}{x-y}-\frac{f^{\prime}(x)+f^{\prime}(y)}{2}\right\vert \le \frac{\vert x-y\vert ^2}{12}\max_\limits{\xi\in[0,1]}\vert f^{\prime \prime \prime}(\xi)\vert.$$

For $f(x) = x^3$, the inequality becomes an equality for all $x \neq y$, proving that $c=\frac{1}{12}$ is the smallest constant for which the inequality is valid.


问:
How to prove that $|\tan z| \lt 2$ on the contour $x=\pm~ \pi n, y=\pm~ \pi n$?
We are in the complex plane. And I just cannot prove that  $|\tan z| \lt 2$ on the contour $x=\pm ~\pi n, y=\pm ~\pi n$.</p>

答:
Hint: Write

$$\tan(z) = \frac{e^{iz}-e^{-iz}}{i(e^{iz}+e^{-iz})}$$

So that:

$$|\tan(x+iy)|=\frac{|e^{ix}e^{-y}-e^{-ix}e^{y}|}{|e^{ix}e^{-y}+e^{-ix}e^{y}|}$$

As to what to do explicitly:

On the vertical sides, (since $e^{\pm in\pi}=(-1)^n)$:
$$|\tan(\pm n\pi+iy)|=\frac{|e^{-y}-e^{y}|}{|e^{-y}+e^{y}|}=|\tanh(y)|< 1$$

and on the right hand side:

$$|\tan(x+in\pi)|=\frac{|e^{ix}e^{-n\pi}-e^{-ix}e^{n\pi}|}{|e^{ix}e^{-n\pi}+e^{-ix}e^{n\pi}|} \leq \frac{|e^{-n\pi}+e^{n\pi}|}{|e^{-n\pi}-e^{n\pi}|}=|\coth(n\pi)|<1$$

Where the bound on the left hand side is estimated similarly.


问:
Trouble with Hatcher 1.1.6 (conjugacy and free homotopy)
Let $\Phi: \pi_1(X, x_0) \rightarrow [S^1, X]$. I wish to show that $\Phi([f]) = \Phi([g])$ iff $[f]$ and $[g]$ are conjugate in $\pi_1(X, x_0)$.

To me this this is almost trivial, since $\Phi([f]) = \Phi([g])$ gives us that we have a homotopy between $[f]$ and $[g]$, and since both are in the same fundamental group they both are loops around $x_0$. However, I know this is wrong and I am misunderstanding something. Furthermore, I know that the right approach is something along the lines of constructing $f = \overline{h}gh$, but why? All are loops centered around the same point, so what is this solving really? How is it "moving" the base point if all have the same base point? I have been beating my head against the wall for hours with this problem. Any help would be much appreciated.

Thank you.

答: If $\Phi([f]) = \Phi([g])$, let $F:I \times I \to X$ be a free homotopy of loops between $f$ and $g$ where the bottom edge of $I \times I$ is mapped to $f$ and the top edge to $g$. Let $h$ be the right edge of $I \times I$, i.e., the loop $h(t) = F(1, t)$. Note that $h^{-1}$ is exactly the loop $F(0, 1-t)$ (the left edge traversed downwards) because $F$ is a homotopy of loops so $F(0, t) = F(1, t)$ for all $t \in I$. The boundary loop of $I \times I$ traversed anti-clockwise is nullhomotopic in $I \times I$ while preserving the basepoint $(0, 0)$. This induces a basepoint-preserving nullhomotopy of $fhg^{-1}h^{-1}$, showing that $[f]$ and $[g]$ are conjugate in $\pi_1(X, x_0)$. Conversely if $[f] = [hgh^{-1}]$ let $F: I \times I \to X$ be a basepoint-preserving homotopy. Since the left and right edges of $I \times I$ are mapped to constant loops at $x_0$, we might as well consider them to be $h$ and $h^{-1}$. Then we are left with a free homotopy of loops between $f$ and $g$.


问:
Best Algebraic Topology book/Alternative to Allen Hatcher free book?
Allen Hatcher seems impossible and this is set as the course text?

So was wondering is there a better book than this? It's pretty cheap book compared to other books on amazon and is free online.

Any good intro to Algebraic topology books?

I can find a big lists of Algebraic geometry books on here. On a very old thread on Maths overflow someone recommended that a person should read James Munkres Topology first, then you should read Allen Hatcher book.

It just seems like Rudin's book but crammed with ten times more material.

答:
If you want a more rigorous book with geometric motivation I reccomend  John M. Lee`s topological manifolds where he does a lot of stuff on covering spaces homologies and cohomologies. As a supplement you can next go to his book on Smooth Manifold to get to the differential case. I especially like his very through and rigorous introduction of quotient spaces/topologies and so on which are used very heavily and which hatcher explains mostly in a very pictorial and unsatisfying way.

However let me also note that Hatchers through examination of the covering space of the circle (which also lee does) has been a very helpful example for me to keep in mind whenever I am thinking of covering spaces in general. So I propose that you should read that part.


问:
About general strategies for game-problems: Adding 31 from numbers on a regular calculator
I am new to game related problems so, in particular, I'm starting with symmetry game problems. In this case, I was trying to solve the following problem:

Two players, A and B, use a special calculator with number buttons without the zero and
provided only with the "+" button as operation. Player A starts by pressing one
number button, then pressing "+" and passing the calculator to B. Then
B must choose a number that is either in the same column or the same row as the
number picked by A in their previous turn (B cannot choose the same number A
picked in their last turn). Then B presses "+", wich makes the
calculator show the current sum of all the numbers chosen so far. In the next
turn, A does the same thing. The first player that obtains a sum equal
or larger than 31 by pressing the "+" button loses the game. Which one
of the two players has a winning strategy?

I was thinking that maybe A would want to start with the greatest possible number, but in that case B would want to do the same. This however, doesn't seem to lead to anything interesting. As I stated before, I'm very new to this kind of problems and have no idea how to approach them. I have seen many examples but all of the solutions seem very specific to the problems. Do general strategies to approach this kind of problems exist? For example, could number theory be useful to solve this problem?

I would like to see some general hints not only for this problem but for general problems related to game strategies.

答:
Here, we can capture the relevant state information using $(S, M)$ where $S$ represents the current Sum displayed on the calculator and $M$ represents the number button pressed / Move to get that sum.

Note for e.g. that $(30, x)$ for any allowable $x \in \{1, 2, 3, \dots 9\}$ represents a winning state for the player who just achieved that state (i.e. the next player has no option but to lose).

Assuming both players are trying to win and are aware of their best strategy, let us define a function $W(S, M)$ which takes the following values - $0$ if there are no winning moves for the next player (i.e. it is a winning state for the previous player).  Similarly $W(S, M)=x >0$ would indicate the next player can press the number $x$ to get to a winning state.  Of course the choices of $x$ are limited by the moves permitted on the calculator keypad, and there may be more than one move which may take the next player to a winning state, so for the sake of uniqueness we pick the maximum allowable such move to be $x$.

Now given $W(30, 1)=0$ represents a winning state for the previous player, we can think of which last moves could have led to that state, and from which previous states could we have such moves. Clearly the last move was $1$, so the last sum was $29$, and the possible states which allow a move of $1$ are hence $(29, 2), (29, 3), (29, 4), (29, 7)$. We can hence set $W$ value of all these states to $1$, which is the best move at this state for any next player for reaching a winning position.

We can follow this process for all the winning states $(30, 2), (30, 3), \dots$ thereby identifying values of the $W$ function.  Further then we can progress to states of form $(29, x)$ and so on step by step.  Clearly given the constraints on the moves and possible sums, this is perhaps better done with a program.  Here is one written in python:

```<code>allowable_moves = {
    1: {2, 3, 4, 7},
    2: {1, 3, 5, 8},
    3: {1, 2, 6, 9},
    4: {1, 5, 6, 7},
    5: {2, 4, 6, 8},
    6: {3, 4, 5, 9},
    7: {1, 4, 8, 9},
    8: {2, 5, 7, 9},
    9: {3, 6, 7, 8},
    }
"""
allowable_moves is a dictionary of allowable next moves, if the last number
key pressed is given.  Note the same dictionary also captures the "reverse"
moves, i.e. given a move, what could have been allowable preceding moves.
"""

win = {(i, j+1) : 0 for i in range(31) for j in range(9)}
"""
Each state is represented by:
    c_sum   current sum on the calc display, from 0 to 30
    move    last number key pressed, from 1 to 9
The entries of the "win" dict represent:
    0           a winning position for the last player
    x           what number the next player should press to win, 1 to 9
"""

for c_sum in range(30, -1, -1):
    for move in range(1, 10, 1):
        if win[(c_sum,move)] == 0:
            last_set = allowable_moves[move]
            # last_set is the possible moves before the one just played

            for last_move in last_set:
                if c_sum >= move and win[(c_sum - move, last_move)] == 0:

                    win[(c_sum-move, last_move)] = move
                    # records that this is a possible winning move

    print(str(c_sum).zfill(2), " : ",
          [win[(c_sum,i)] for i in range(1, 10, 1)])
</code>```

The output is a table of $W$ values:

```<code>        Moves --->
Sum |   1, 2, 3, 4, 5, 6, 7, 8, 9
----------------------------------
30  |  [0, 0, 0, 0, 0, 0, 0, 0, 0]
29  |  [0, 1, 1, 1, 0, 0, 1, 0, 0]
28  |  [2, 1, 2, 1, 2, 0, 1, 2, 0]
27  |  [3, 3, 0, 0, 0, 3, 0, 0, 3]
26  |  [4, 0, 0, 0, 4, 4, 4, 0, 0]
25  |  [0, 5, 0, 5, 0, 5, 0, 5, 0]
24  |  [3, 5, 6, 6, 6, 5, 1, 5, 6]
23  |  [7, 3, 6, 7, 6, 4, 4, 7, 7]
22  |  [4, 8, 6, 6, 8, 5, 8, 5, 8]
21  |  [0, 8, 9, 0, 8, 9, 9, 9, 8]
20  |  [7, 5, 9, 7, 0, 9, 9, 9, 7]
19  |  [0, 8, 9, 0, 8, 9, 9, 9, 8]
18  |  [7, 8, 1, 7, 8, 0, 8, 7, 8]
17  |  [4, 0, 9, 0, 4, 9, 9, 9, 0]
16  |  [0, 0, 9, 0, 0, 9, 9, 9, 0]
15  |  [4, 5, 2, 5, 4, 5, 4, 5, 0]
14  |  [2, 0, 2, 0, 2, 0, 0, 2, 0]
13  |  [4, 0, 0, 0, 4, 4, 4, 0, 0]
12  |  [4, 0, 6, 6, 6, 4, 4, 2, 6]
11  |  [2, 5, 2, 5, 2, 5, 0, 5, 0]
10  |  [4, 3, 2, 0, 4, 4, 4, 2, 3]
09  |  [4, 0, 0, 0, 4, 4, 4, 0, 0]
08  |  [0, 0, 9, 6, 6, 9, 9, 9, 6]
07  |  [7, 1, 9, 7, 2, 9, 9, 9, 7]
06  |  [4, 3, 9, 0, 4, 9, 9, 9, 3]
05  |  [4, 8, 9, 0, 8, 9, 9, 9, 8]
04  |  [7, 0, 9, 7, 0, 9, 9, 9, 7]
03  |  [0, 0, 0, 0, 0, 0, 0, 0, 0]
02  |  [4, 1, 9, 1, 4, 9, 9, 9, 0]
01  |  [4, 8, 2, 0, 8, 4, 8, 2, 8]
00  |  [3, 3, 9, 0, 0, 9, 9, 9, 3]
</code>```

from which it is evident that the first player can get to $(3, 3)$ or $(9, 9)$ in the first move, both of which have a $W$ value of $0$, so the next player has no winning moves.  Whatever is played then, the first player can jump to another state with $W$ value $0$ because of the way we constructed the table, which gives the winning strategy.

P.S. The last row in the table does not mean much as there is no previous move when sum shown is $0$ at the start - except that it shows the only winning moves are $3$ or $9$.

Body:
Here, we can capture the relevant state information using $(S, M)$ where $S$ represents the current Sum displayed on the calculator and $M$ represents the number button pressed / Move to get that sum.

Note for e.g. that $(30, x)$ for any allowable $x \in \{1, 2, 3, \dots 9\}$ represents a winning state for the player who just achieved that state (i.e. the next player has no option but to lose).

Assuming both players are trying to win and are aware of their best strategy, let us define a function $W(S, M)$ which takes the following values - $0$ if there are no winning moves for the next player (i.e. it is a winning state for the previous player).  Similarly $W(S, M)=x >0$ would indicate the next player can press the number $x$ to get to a winning state.  Of course the choices of $x$ are limited by the moves permitted on the calculator keypad, and there may be more than one move which may take the next player to a winning state, so for the sake of uniqueness we pick the maximum allowable such move to be $x$.

Now given $W(30, 1)=0$ represents a winning state for the previous player, we can think of which last moves could have led to that state, and from which previous states could we have such moves. Clearly the last move was $1$, so the last sum was $29$, and the possible states which allow a move of $1$ are hence $(29, 2), (29, 3), (29, 4), (29, 7)$. We can hence set $W$ value of all these states to $1$, which is the best move at this state for any next player for reaching a winning position.

We can follow this process for all the winning states $(30, 2), (30, 3), \dots$ thereby identifying values of the $W$ function.  Further then we can progress to states of form $(29, x)$ and so on step by step.  Clearly given the constraints on the moves and possible sums, this is perhaps better done with a program.  Here is one written in python:

allowable_moves = {
    1: {2, 3, 4, 7},
    2: {1, 3, 5, 8},
    3: {1, 2, 6, 9},
    4: {1, 5, 6, 7},
    5: {2, 4, 6, 8},
    6: {3, 4, 5, 9},
    7: {1, 4, 8, 9},
    8: {2, 5, 7, 9},
    9: {3, 6, 7, 8},
    }
"""
allowable_moves is a dictionary of allowable next moves, if the last number
key pressed is given.  Note the same dictionary also captures the "reverse"
moves, i.e. given a move, what could have been allowable preceding moves.
"""

win = {(i, j+1) : 0 for i in range(31) for j in range(9)}
"""
Each state is represented by:
    c_sum   current sum on the calc display, from 0 to 30
    move    last number key pressed, from 1 to 9
The entries of the "win" dict represent:
    0           a winning position for the last player
    x           what number the next player should press to win, 1 to 9
"""

for c_sum in range(30, -1, -1):
    for move in range(1, 10, 1):
        if win[(c_sum,move)] == 0:
            last_set = allowable_moves[move]
            # last_set is the possible moves before the one just played

            for last_move in last_set:
                if c_sum >= move and win[(c_sum - move, last_move)] == 0:

                    win[(c_sum-move, last_move)] = move
                    # records that this is a possible winning move

    print(str(c_sum).zfill(2), " : ",
          [win[(c_sum,i)] for i in range(1, 10, 1)])

The output is a table of $W$ values:

        Moves --->
Sum |   1, 2, 3, 4, 5, 6, 7, 8, 9
----------------------------------
30  |  [0, 0, 0, 0, 0, 0, 0, 0, 0]
29  |  [0, 1, 1, 1, 0, 0, 1, 0, 0]
28  |  [2, 1, 2, 1, 2, 0, 1, 2, 0]
27  |  [3, 3, 0, 0, 0, 3, 0, 0, 3]
26  |  [4, 0, 0, 0, 4, 4, 4, 0, 0]
25  |  [0, 5, 0, 5, 0, 5, 0, 5, 0]
24  |  [3, 5, 6, 6, 6, 5, 1, 5, 6]
23  |  [7, 3, 6, 7, 6, 4, 4, 7, 7]
22  |  [4, 8, 6, 6, 8, 5, 8, 5, 8]
21  |  [0, 8, 9, 0, 8, 9, 9, 9, 8]
20  |  [7, 5, 9, 7, 0, 9, 9, 9, 7]
19  |  [0, 8, 9, 0, 8, 9, 9, 9, 8]
18  |  [7, 8, 1, 7, 8, 0, 8, 7, 8]
17  |  [4, 0, 9, 0, 4, 9, 9, 9, 0]
16  |  [0, 0, 9, 0, 0, 9, 9, 9, 0]
15  |  [4, 5, 2, 5, 4, 5, 4, 5, 0]
14  |  [2, 0, 2, 0, 2, 0, 0, 2, 0]
13  |  [4, 0, 0, 0, 4, 4, 4, 0, 0]
12  |  [4, 0, 6, 6, 6, 4, 4, 2, 6]
11  |  [2, 5, 2, 5, 2, 5, 0, 5, 0]
10  |  [4, 3, 2, 0, 4, 4, 4, 2, 3]
09  |  [4, 0, 0, 0, 4, 4, 4, 0, 0]
08  |  [0, 0, 9, 6, 6, 9, 9, 9, 6]
07  |  [7, 1, 9, 7, 2, 9, 9, 9, 7]
06  |  [4, 3, 9, 0, 4, 9, 9, 9, 3]
05  |  [4, 8, 9, 0, 8, 9, 9, 9, 8]
04  |  [7, 0, 9, 7, 0, 9, 9, 9, 7]
03  |  [0, 0, 0, 0, 0, 0, 0, 0, 0]
02  |  [4, 1, 9, 1, 4, 9, 9, 9, 0]
01  |  [4, 8, 2, 0, 8, 4, 8, 2, 8]
00  |  [3, 3, 9, 0, 0, 9, 9, 9, 3]

from which it is evident that the first player can get to $(3, 3)$ or $(9, 9)$ in the first move, both of which have a $W$ value of $0$, so the next player has no winning moves.  Whatever is played then, the first player can jump to another state with $W$ value $0$ because of the way we constructed the table, which gives the winning strategy.

P.S. The last row in the table does not mean much as there is no previous move when sum shown is $0$ at the start - except that it shows the only winning moves are $3$ or $9$.


问:
Harnack's inequality in Evans PDE book.
So in an excersise we derived a so called "explicit form" of the inequality.

Now later in the book the Harnack inequality is defined as

How are these two statements related to each other? I just dont see how you get from the upper statement to the bottom one and the book does not cover it either

答:
The first equation is a special case of the second one. Pick any x in $B⁰(0, r)$. Clearly $u(x) \leq \sup_V u(\cdot)$ holds and, due to Harnack’s inequality, $u(x) \leq \sup_V u(\cdot) \leq C \inf_V u(\cdot)$ is true. Since $\inf_V u(\cdot) \leq u(y)$ holds for any $y$, we can conclude $u(x) \leq C u(0)$. If we interchange the roles of $x$ and $0$, we can $u(0) \leq C u(x)$ or $\frac{1}{C} u(0) \leq u(x)$. Putting this together, we get $\frac{1}{C} u(0) \leq u(x) \leq C u(0)$. This is the first statement.

Addition: This is not exactly the first statement since the constants aren’t reciprocal to each other. The first equation can therefore be regarded as a more sophisticated variant. But I think the idea should be clear from the above.

Title: Harnack's inequality in Evans PDE book.


问:
Average slope of a curve
If $(x_0,y_0)$ and $(x_0 + \Delta x, y_0 + \Delta y)$ are two points on the curve, then the "average slope" of the curve between these two points is defined as the ration of change in $y$ to the change in $x$,

average slope= $\frac {\Delta y}{\Delta x}$

Why do we call it the average slope?

答:
It is literally an average slope.

Let $f$ be a continuous function on an interval $[a,b]$. Then, we define its average over $[a,b]$ to be the number
$$
f_{\operatorname{avg}} := {1 \over b-a} \int_a^b f(t) dt
$$
Let $y = F(x)$ be a curve, which I assume to be at least $C^1$. Fix the starting point $x_0$ and the ending point $x_1=x_0+\Delta x$. Then, the slope at a point $z$ is the derivative $F'(z)$. Thus, the "average slope'' should mean
\begin{equation*}
(F')_{\operatorname{avg}} = {1 \over x_1 - x_0}\int_{x_0}^{x_1} F'(t) dt = {F(x_1) - F(x_0) \over x_1-x_0}.
\end{equation*}
The final equality is the so-called Fundamental Theorem of Calculus.

Title: Average slope of a curve


问:
How do I show $y=\ \frac{1}{2}-\frac{1}{2^{x}+1\ }$ is odd?
Of course I know that $f(-x)=-f(x)$, but I get confused as to how $$\frac{1}{2}-\frac{2^{x}}{2^{x}+1\ }=-\frac{1}{2}+\frac{1}{1+2^{x}}$$ In the steps of algebra.

答:
So, I'd recommend turning the function into a proper fraction, so we get
$$ f(x) =  \dfrac{2^x-1}{2^{x+1}+2} $$

Therefore, $f(-x)$ is equal to
$$ f(-x) = \dfrac{2^{-x}-1}{2^{-x+1}+2}$$

Multiply by $\dfrac{2^x}{2^x}$ and we get
$$ f(-x) = \dfrac{1-2^x}{2^1+2\cdot 2^x} = \dfrac{1-2^x}{2^{x+1}+2} = -f(x)$$

Like we wanted.


问:
Use cylindrical coordinates to find the volume of the solid using triple integrals
The given rectangular equations are
$$x^2+y^2+z^2=64$$
$$(x-4)^2+y^2=16$$
Converting to cylindrical coordinates I get
$$r^2+z^2=64$$
$$r=4cos(\theta)$$
So my triple integral is
$$4\int_0^{\frac{\pi}{2}}\int_0^{4\cos(\theta)}\int_0^{\sqrt{16-r^2}}rdzdrd\theta$$
It's a long and tedious calculation but I worked through it and got $-\frac{1408\pi}{15}$ which doesn't make sense because you can't have a negative volume. The negative came from the second integral which is
$$4\int_0^{\frac{\pi}{2}}\int_0^{4\cos(\theta)}r\sqrt{16-r^2}drd\theta$$
where I used a u-sub, where $u=16-r^2$ and $du=-2r$ and moving that $-\frac{1}{2}$to the outside gives me
$$-2\int_0^{\frac{\pi}{2}}\int_0^{4\cos(\theta)}\sqrt{u}dud\theta$$
So could I just swap the limits of integration to get rid of it or have I messed up somewhere else?

答:
There are two mistakes I notice. You have written wrong bound of $z$ and the equation of cylinder that was already pointed out. The correct integral should be

$V = 4\displaystyle \int_0^{\frac{\pi}{2}}\int_0^{8\cos(\theta)}\int_{0}^{\sqrt{64-r^2}}rdzdrd\theta \approx 617.22$

$V = 4\displaystyle \int_0^{\frac{\pi}{2}}\int_0^{8\cos(\theta)} r \, \sqrt{64-r^2} \, dr \, d\theta$

On substition, $64-r^2 = u, rdr = -\frac{1}{2}du$.

The bound $r = 0$ becomes $u = 64$, $r = 8 \cos \theta$ becomes $u = 64 \sin ^2 \theta$.

$V = -2\displaystyle \int_0^{\frac{\pi}{2}}\int_{64}^{64\sin^2(\theta)} \sqrt u \, du \, d\theta = 2\int_0^{\frac{\pi}{2}}\int_{64\sin^2(\theta)}^{64} \sqrt u \, du \, d\theta$

$V = \displaystyle \frac{2048}{3} \int_0^{\frac{\pi}{2}} (1-sin^3 \theta) \, d\theta = \frac{2048}{3} (\frac{\pi}{2} - \frac{2}{3})$


问:
Learning math under time pressure
I'm taking my first and only university math course, discrete mathematics.

We have about a week and a half to learn a full chapter (around 50 pages) from the book and do the related assignment.

As I'm working full-time, I have about 2 hours a day to achieve the weekly mission.

I would like to know the best learning strategy for the current situation.

答:
The best recommendation I ever heard for such circumstances is:  read the problems FIRST... before your read the chapter.

That way:

you learn what the author thinks is important

you're alert to picking up and focusing on that material when you read the chapter

Title: Learning math under time pressure


问:
Formula to find the day of any date in Gregorian calendar derivation.
My teacher taught an expression to find the day of any date in the Gregorian calendar.

Expression:

$$ \bbox[5px,border:2px solid red]
{
D\equiv {d+c_m+c_y+c+\left\lfloor\dfrac c4\right\rfloor \pmod7}
}
$$

Where $D$ is the day code of the given date and day can be found from the following table,$d$ is the date,$c$ is the last two digits of the year(ones and tens place digit),$c_m,c_y$ are given below.

And

Day code($D$):
\begin{array}{|c|c|}
\hline \mathrm{\color{red}{Day}}& \color{red}{\text{Day Code}}\\
\hline \mathrm{Sunday}&0\\
\hline \mathrm{Monday}&1\\
\hline \mathrm{Tuesday} &2\\
\hline \mathrm{Wednesday} &3\\
\hline \mathrm{Thursday} &4\\
\hline \mathrm{Friday}&5\\
\hline \mathrm{Saturday} &6\\
\hline \end{array}

Month code($c_m$):

For ordinary year:
\begin{array}{|c|c|}
\hline \color{red}{\text{January}}& \color{red}{\text{February}}&\color{red}{\text{March}}\\
\hline 1&4&4\\
\hline \color{red}{\text{April}}& \color{red}{\text{May}}&\color{red}{\text{June}}\\
\hline 0 &2&5\\
\hline \color{red}{\text{July}}& \color{red}{\text{August}}&\color{red}{\text{September}}\\
\hline 0 &3&6\\
\hline \color{red}{\text{October}}& \color{red}{\text{November}}&\color{red}{\text{December}}\\
\hline 1&4&6\\
\hline \end{array}

For leap year:
\begin{array}{|c|c|}
\hline \color{red}{\text{January}}& \color{red}{\text{February}}&\color{red}{\text{March}}\\
\hline 0&3&4\\
\hline \color{red}{\text{April}}& \color{red}{\text{May}}&\color{red}{\text{June}}\\
\hline 0 &2&5\\
\hline \color{red}{\text{July}}& \color{red}{\text{August}}&\color{red}{\text{September}}\\
\hline 0 &3&6\\
\hline \color{red}{\text{October}}& \color{red}{\text{November}}&\color{red}{\text{December}}\\
\hline 1&4&6\\
\hline \end{array}

Century code($c_y$):
\begin{array}{|c|c|}
\hline \color{red}{\text{Century Leap Year}}& \color{red}{\text{Century Code}}\\
\hline 1300&3\\
\hline 1400&1\\
\hline  1500&6\\
\hline 1600 &5\\
\hline 1700 &3\\
\hline 1800&1\\
\hline 1900 &6\\
\hline 2000&5\\
\hline \end{array}

More generalised Century code:
\begin{array}{|c|c|}
\hline {\color{red}{\text{Leap Year Century}}}& \color{red}{\text{Century Code}}\\
\hline \text{Leap Year Century}&5\\
\hline \text{Leap Year Century+100}&3\\
\hline  \text{Leap Year Century+200}&1\\
\hline \text{Leap Year Century+300} &6\\
\hline \end{array}

He refused to give the derivation of the above expression(saying it won't be asked in the exam) but I can't use this expression without the derivation or without knowing how it works.

What I tried(think):

The denominator is $7$ because the numerator gives the total odd days(I think) and we have to find net odd days.

By observing, one can easily find that century code is two less than the number of odd days in a century.

For example:
Number of odd days in $100$ years are $5$ and century for (leap year century $+ 100$ years) is $3$(that is $5-2$).

Number of odd days in $200$ years are $3$ and the century code for (leap year century$+200$ years) is $1$(that is $3-2$).

Number of odd days in 300 years are $1$ and the century code for (leap year century$+300$ years) is $6$(that is $1-2=-1$ and $7-1=6$ ).

The number of odd days in 400 years are $0$ and the century code for (leap year century) is ($0-2=-2$ and $7-2=5$)

I am unable to find any relation between odd days in the month and month code.

What is my question?
How is this expression derived?

I am not able to derive the expression.

Thanks

Note:

By leap year century, I meant century year ($1200,1300,1400$) which is a leap year(example $400,800,1200,1600$).

答:
NEW ANSWER!

Let's choose our base starting point is Day 1 = Jan 1, 2000 (which was a leap year) and Day 0 = Dec 31, 1999.  Day 0 was a Fri $=5$.  So let $M_{2000} = 5$.

If we were asked what day of the week was the $k$th day of $2000$ that'd be easy.  As the days repeat ever seven days we'd have $D = k + M_{2000}\pmod 7$.

And if we were asked what day of the week was Month $d$, 2000, we could calculate that $k = d + $ sum of the number of days in the months before Month $\pmod 7$.

So Jan $d$, 2000: $D = d + M_{2000}\pmod 7$.

Feb $d$, 2000: $D= d+ 31 +M_{2000}\equiv d+3 +M_{2000}\pmod 7$

March $d$, 2000: $D = d + 31 + 29 + M_{2000} \equiv d+ 4 + M_{2000}\pmod 7$

... and so on.

In general. Month $d$, 2000: $D= d+ c_m + M_{2000}\pmod 7$ where $c_m = $ sum of then number of  in the months before Month $\pmod 7$.

Okay, so what about Month $d$, $20xx$?

In general a year has $365=52*7 + 1 \equiv 1 \pmod 7$ so if $D$ is the day of the week for Month $d$ $xxxx$ then the day of the week of Month $d$ $xxxx+1$ will be $365 + D \equiv 1+ D\pmod 7$.
So if we completely ignored leap years we'd have:

Month, d, 20$xx$ would be $D= d + c_m + xx + M_{2000}\pmod 7$.

But we can't ignore leap years.  We must add a day for every leap year encountered.

Okay.... bear with me.....

$2000$ was a leap year but we took that into account with the $c_m$s. So for $2000$ we don't make any adjustments. But to compare the dates of $2001$ to the dates of $2000$ the fact that $2000$ had $366 \equiv 2\pmod 7$ days means we must adjust and add $1$.  They next time we adjust will be when we go from $2004$ to $2005$ and $2004$ also has $366$, not $365$ days.

so the formula is Month $d$, 20$xx$ would be $D = d + c_m +xx +\lceil \frac {xx}4\rceil + M_{2000}\pmod 7$.

But note, the $c_m$ codes were calculated for the year 2000 that was a leap year in which February has $29$ days.  To adjust for years in which Feb has $28$ days we must increase $c_1$ and $c_2$ by $1$.  (This is a little counter intuitive... but the days of 2001 after February have been adjusted by $01 + \lceil \frac {01}4\rceil = 2$ already but it is the days before February 29 that need adjusting because there is no Feb 29 in 2001 so all the days need to be pushed forward to fill the gap).

So we are $90\%$ of the way there.

Month $d$, 20$xx$ would be $D = d + c_m +xx +\lceil \frac {xx}4\rceil + M_{2000}\pmod 7$.

But  what of other centuries?

Well, our formula $D = d + c_m +xx +\lceil \frac {xx}4\rceil + M_{mm00}\pmod 7$ would still work but for different values of $M_{mm00}$.

Now one ordinary century year  has 100 years out of which 76 years are ordinary year and 24 years are leap year so odd days are 124($76 \times 1 +24\times 2)$.
So $124\equiv 5\pmod 7$ days.

One leap century year has 75 ordinary years and 25 leap year so odd days are 125($75\times 1+25\times 2$) and $125\equiv 6\pmod 7$ days.

so as $M_{2000}=5$ we'd have $M_{2100} \equiv 5+5\equiv 3$, $M_{2200}\equiv 3+5\equiv 1\pmod 7$.  And $M_{2300}\equiv 1+5\equiv 6\pmod 7$ but, $M_{2400}\equiv 6+6\equiv 5$.  And we repeat.... (Very nice, the Gregorian calendar based make every four hundred years has a multiple of $7$ days so we always repeat.

So if $M_{xx00} = c_y = 5, 3,1,6$ for leap, leap +100, leap + 200, leap + 300$

We get the formula.

Month  $d$, $yyxx$ where $yyxx = 400*k + 100*y + xx$ then $D = d + c_m + c_y + xx+\lceil \frac {xx}4\rceil\pmod 7$

==========

Dec 31, 1299 was on a thursday$= 4$.

So Jan k, 1300 would be $k$ days later.  So that would be $k +4$.  But there are only seven days a week so they repeat every $7$ days so we will use $k+ 4\pmod 7$ to figure out the day of the week.

So for any date we will estimate the numbers of days since Dec 31, 1299 and take the remainder divided by $7$.  To get that Jan k, 1300 is on the date $k +4\pmod 7$ we have an offset of $4$.  If we let Jan account for offset of $1$ and $1300$ account for offset of $3$ the formula works so far for the Dates  Jan 1-31, 1300.  $D = d + c_m +c_y \pmod 7$ where $d=1-31$ and $c_m =1$ (completely arbitrary) and $c_y=3$ (ditto).

Now Feb m, 1300 would be $31  + m + $ days later. But $31\equiv 3 \pmod 7$ so instead off offsetting by $4$ we offset but $4+ 3 \equiv 0\pmod 7$. So if the offset code, $c_m$ of Jan was $1$, and Jan has $31 \equiv 3 \pmod 7$ days, then the offset code, $c_m$ for February must be $1+3\equiv 4\pmod 7$.

And so on for all the months.  Feb has $28\equiv 0 \pmod 7$ days so the $c_m$ for March is $4+0=0$.  And March has $31\equiv 3 \pmod 7$ days so the offset for april is $4 + 3 \equiv 0 \pmod 7$.  And so on.

So in this way we can figure out all the days from Jan 1, 1300 to Dec 31 1300.

Okay, now a year has $365=52*7 + 1\equiv 1 \pmod 7$.  So any day of a date in 1301 will be $1$ day latter than the same date in $1300$.  And any day of a date in 130$j$ will be $j$ days latter than the same date in $1300$.

But every fourth year is a leap year and that addes an extra day every four years.

So that tells us how to calculate any date in the $1300$.   Now a century has some many days $\pmod 7$.  So that is way we have the century code, $c_y$.  Some centuries have leap years on $xx$00 and others don't so the code takes that into account.

And that's that.

....

Although.... I don't think those values are right.  The ceiling rather than floor can only work if $xx00$ belongs to the century before (which properly it does)  And I did calculations and the century offsets didn't agree with my calculations.  (But it was late and I didn't do it too carefully, and maybe I was assuming all centuries began with 00 and maybe they do work for centuries begining on 01.)


答:
Starting from where you left off:

$$\frac12 - \frac{2^x}{2^x+1} = \frac12-\left(\frac{2^x+1}{2^x+1}-\frac1{2^x+1}\right) = \frac12-1+\frac1{2^x+1}$$

Title: How do I show $y=\ \frac{1}{2}-\frac{1}{2^{x}+1\ }$ is odd?

Body: Of course I know that $f(-x)=-f(x)$, but I get confused as to how $$\frac{1}{2}-\frac{2^{x}}{2^{x}+1\ }=-\frac{1}{2}+\frac{1}{1+2^{x}}$$ In the steps of algebra.


问:
Prove $\Vert T(x,y)\Vert\leq \Vert T\Vert\Vert x\Vert\Vert y\Vert$.
Let $T$ be an operator defined on $X\times Y$, the product of normed space $X$ and $Y$ such that

(1) $T(x+x',y+y')=T(x,y)+T(x,y')+T(x',y)+T(x',y')$, $\forall (x,y)\in X\times Y, (x',y')\in X\times Y$.

(2) $T(\alpha x,\beta y)=\alpha\beta T(x,y), \forall (x,y)\in X\times Y$ with $\alpha$ and $\beta$ being scalars, i.e. $T$ is bilinear.

Let

$$\Vert T\Vert = \sup\limits_{\Vert x\Vert\leq 1, \Vert y\Vert\leq 1}\vert T(x,y)\vert.$$

Show that $\|T(x,y)\| \leq \Vert T\Vert\Vert x\Vert\Vert y\Vert$.

I don't have an idea to prove it. I don't have an idea to modifying $\|T(x,y)\|$ such that get $\|T(x,y)\| \leq \Vert T\Vert\Vert x\Vert\Vert y\Vert$. Any hint to prove it?

答: If you exclude  the trivial cases ($x =0$ or $y=0$), then  for given $(x,y) \in X\times Y,$ with $x\neq 0$ and $y\neq 0,$ you can use the definition of the operator norm to estimate $$\left\| T\left(\frac{x}{\|x\|}, \frac{y}{\|y\|} \right)\right\|.$$ Then use (2).


问:
For an assertion $P(n)$ where $n\ge2$, the following information is given,
There is no formal Induction assertion given. But there is an assertion $P(n)$ for $n\ge2$.

(a)We know that $P(2)$ is true.

(b) we know for any $a\ge2$, truth of $P(a)$ implies truth of $P(2a)$.

(c) we know for any $b\ge3$, truth of $P(b)$ implies truth of $P(b-1)$.

We have to use induction to prove that $P(n)$ is true for all $n\ge2$.

My Solution: I use the Well-ordering property (WOP) of the natural numbers.

T = { k>= 2; p(k) is True}

F= {l>= 2, p(l) is False}

so T U F = N(all natural Numbers), and T∩F = ∅

Case 1: (F = ∅ Then T = N so P(n) is always True)

Case 2: (F ≠ ∅ Then ∅ ≠ F ⊆ N s.t there is an element m = min F (WOP))

now, m ≠ 2 because 2 ∈ T (by Hypothesis)

I cant figure out how to use p(a) and p(b) to prove the statement p(n).

答:
Using (a), (b) and induction you can derive that $P(2^n)$ is true for any $n\in\mathbb{N}.$

Now suppose you want to show the validity of $P(n)$ for some random positive integer $n\ge 2.$ Notice that $2^n\gt n,$ and $P(2^n)$ is true. Then use (c) to verify the result for the (decreasing) sequence $P(2^n-1), P(2^n-2),\cdots, P(n+1), P(n).$

This method is known as (Cauchy's) forward backward induction. See here for an example.


问:
How can we explicitly write the distribution of this function of a discrete random variable?
The context is the unusually complicated formula for determining the critical strike multiplier in Warhammer 40K: Inquisitor - Martyr: https://mome-borogove.github.io/40k/20200707/understanding-criticals.html

The relevant values are theoretically real numbers, but the game displays individual sources in terms of percentage with one decimal place, and the totals as the sum of the sources, rounded to the nearest integer. So we can safely assume the random variables are discrete.

To formulate this mathematically, let $s$ and $d$ be fixed nonnegative integers and $X$ uniformly distributed over $[100]$ (the notation $[n]$ denotes the set $\{i\in\mathbb Z : 1\leqslant i\leqslant n\}$). For convenience we will also write $[m:n]$ to denote $\{i\in\mathbb Z : m\leqslant i\leqslant n\}$ - note that $|[m:n]| = n-m+1$.  Write $X_s:=X+s$, then $X_s$ is uniformly distributed over $[s+1:s+100]$. Then define
\begin{align}
Y_s &= \frac75\cdot \mathsf1_{[45]}(X_s) + 2\cdot \mathsf1_{[46:85]}(X_s) + \frac52\cdot \mathsf1_{[86:115]}(X_s)\\
&+ 3\cdot \mathsf1_{[116:135]}(X_s)+ \frac72\cdot \mathsf1_{[136:150]}(X_s) + 4\cdot\mathsf 1_{[151,\infty)}(X_s).
\end{align}
The critical strike multiplier is then $Y+d=:Y_{s,d}$. I'd like to write the distribution for $Y_{s,d}$ explicitly, i.e. $\mathbb P(Y_{s,d} = k)$ for $k\in\left\{\frac75,2,\frac52,3,\frac72,4\right\}$. If $s\geqslant 150$ then $Y_{s,d}=4$ with probability one, so we need only consider $0\leqslant s<150$. I could write an R script (or even an Excel formula) to compute these probabilities for each value of $d$ to produce a lookup table from which we can find $\mathbb P(Y_{s,d}=k)$ for given $d$ and $k$, but this would be somewhat tedious.

I'm sure there's a more elegant way to write the distribution of $Y_{s,d}$ (as a function of $s$ and $d$, where of course $Y_{s,d} = Y_s + d$), but the number of cases involved, among other things, makes it difficult for me to see how. I'm sure I could figure it out with a good hint, so I'd prefer that over a detailed solution.

答:
You are over-complicating, there are two random variables involved in a chain: the first one, namely $X_1$, with Bernoulli distribution with say probability $p$ to get a crit, the second one, namely $X_2$, uniformly distributed in the set $\{1,\ldots ,100\}$, with the relations

$$\Pr [X_2=k|X_1=1]=\frac1{100},\quad\Pr [X_2=k|X_1=0]=0,\quad\text{ when }k\in\{1,\ldots,100\}$$

Therefore

$$
\Pr [X_2=k]=\Pr [X_2=k|X_1=1]\Pr [X_1=1]+\Pr [X_2=k|X_1=0]\Pr [X_1=0]\\[1em]
=\begin{cases}
\frac{p}{100},&\text{ when }k\in \{1,\ldots, 100\}\\
1-p,&\text{ when }k=0\\
0,&\text{ otherwise }
\end{cases}
$$

Above $k=0$ represent when you dont get a crit, you can change this part but the distribution will be the same as far as the function that map values to percentages is consistent with the game, by example we can set the map $f$ by

$$
\begin{align*}
&x\mapsto 100,\quad \text{ when }x=0\\
&x\mapsto 140,\quad\text{ for }x\in\{1,\ldots ,45\}\\
&x\mapsto 200,\quad \text{ for }x\in \{46,\ldots ,85\}\\
&\ldots\ldots \ldots \ldots  \\
&x\mapsto 400,\quad \text{ when }x\in \{151,\ldots \}
\end{align*}
$$

All other factors are not random, so to evaluate "what is the probability to get $350\%$ crit factor or more" we do

$$
\Pr [Y\geqslant 350]=\Pr [X_2 \in  \{x-\lfloor m_1 \rfloor:x\in f^{-1}([350-m_2,\infty))\}]
$$

where $m_2$ is the percentage modifier added from your player, $m_1$ is the modifier added to the base crit modifier, $f$ is the map described above, and $\lfloor \cdot  \rfloor$ is the floor function (you can change it by the nearest integer function if you consider its best suited).


问:
Flipping a Coin 10 Times and Getting a Sequence of Heads
I toss a fair coin $10$ times resulting in a sequence of heads and tails. Let $X$ be the number of times that the sequence $HH$ appears. $HH$ appears thrice here.

Solution:

The example goes, say we flip a coin $10$ times and you get all heads then we have $HH$ appears $10$ times. (I would write it out, but my LateX Skills are not up to par.)

And we see $P(HH) = \frac{1}{4}$ because we have $.5$ for heads and tails for the two coins.

$X = HH$ appears.

This is where I don't understand what happens: $\sum_{n=0}^9P(X)\cdot X = \frac{9}{4}$.

When I thought of it, do you factor out $P(X)$ because it is always $\frac{1}{4}$, but I am wrong. Any help would be awesome!

答: $X$ follows a bionomial distribution with success probability $p=1/4$ and $n=9$ the number of trials. Too see this let $X$ be the number of $HH$ appeared in a flip coin of 10 tosses. Clearly, as you said to get $HH$ twice in a row has probability equal to $p=1/4$. So, you look at your problem from the point of view that $HH$ is your success outcome and the rest $\{HK,KH,KK\}$ consists of the failure outcomes. Now you look for how many possible ways you have in order to get $HH$ in $10$ tosses. There are $9$ ways, in order to figure this out you look at the positions where the first $H$ can appear in a sequence of $HH$. It can appear in any of the first $9$ positions. So, we end up that $X\sim Bionomial(9,1/4)$. And this has $E[X]=np=9/4$.


问:
Filling at least $1$ box with a white ball from $1$ to $n$ numbered boxes

Given $n$ boxes numbered $1$ to $n$, each box is to be filled with either a white ball or a blue ball
such that at least one box contains a white ball and boxes containing white balls are consecutively
numbered.  What is the total number of ways this can be done?

Solution provided:

The task is similar to choosing two out of the $n + 1$ crosses to mark the start and
end of the consecutively numbered boxes that contain white balls.
This is ${n+1 \choose 2}$, which is equal to $\frac{n(n + 1)}{2}$.

My question:

I get that the picture shows $n$ boxes because we can have $1$ to $n$ white balls.

I have a feeling they are using the $r$-combination formula here: ${r + n -1 \choose r}$.

That's all about I get. I have no clue why they have $n+1$ crosses or why they are choosing only $2$ from those crosses or what those crosses represent even?

PS: I know there is sum of sequence approach to this but I'm required to provide a combinatorial working.

答:
To make a $7$-lettered string $$BBBBWWW$$ given the $7$ slots
$$ \square \square \square \square \square \square \square $$
first separate the squares where consecutive $W$'s are to be placed like this
$$ \square \square \square \square \times \square \square \square \times $$
and then place the W's into squares between crosses (and B's in rest).

The two crosses, which mark the beginning and end of substring of $W$'s, can be placed in any $n+1$ places around $n$ slots, hence $$ \binom{n+1}{2}$$


答:
Note that $\Delta x$ and $\Delta y$ can be thought of as the change in $x$ and the change in $y$, respectively, between the two points: $P = (x_0, y_0)$ and $Q = (x_0+Δx,y_0+Δy)$.

So dividing $\frac {\Delta y}{\Delta x}$ gives us the average slope when summing the slopes of each point $R$ on the curve, which lies between $P$ and $Q$. It is the value of the change in $y$ over that interval, with respect to the change in $x$ over that same interval.

Title: Average slope of a curve

Body:
If $(x_0,y_0)$ and $(x_0 + \Delta x, y_0 + \Delta y)$ are two points on the curve, then the "average slope" of the curve between these two points is defined as the ration of change in $y$ to the change in $x$,

average slope= $\frac {\Delta y}{\Delta x}$

Why do we call it the average slope?


Title: Use cylindrical coordinates to find the volume of the solid using triple integrals

Body:
The given rectangular equations are
$$x^2+y^2+z^2=64$$
$$(x-4)^2+y^2=16$$
Converting to cylindrical coordinates I get
$$r^2+z^2=64$$
$$r=4cos(\theta)$$
So my triple integral is
$$4\int_0^{\frac{\pi}{2}}\int_0^{4\cos(\theta)}\int_0^{\sqrt{16-r^2}}rdzdrd\theta$$
It's a long and tedious calculation but I worked through it and got $-\frac{1408\pi}{15}$ which doesn't make sense because you can't have a negative volume. The negative came from the second integral which is
$$4\int_0^{\frac{\pi}{2}}\int_0^{4\cos(\theta)}r\sqrt{16-r^2}drd\theta$$
where I used a u-sub, where $u=16-r^2$ and $du=-2r$ and moving that $-\frac{1}{2}$to the outside gives me
$$-2\int_0^{\frac{\pi}{2}}\int_0^{4\cos(\theta)}\sqrt{u}dud\theta$$
So could I just swap the limits of integration to get rid of it or have I messed up somewhere else?


问:
Prove $T$ is continuous if and only if there exist $k>0$ such that $\Vert T(x,y)\Vert\leq k\Vert x\Vert \Vert y\Vert.$
Let $T$ be an operator defined on $X\times Y$, the product of normed space $X$ and $Y$ such that

(1) $T(x+x',y+y')=T(x,y)+T(x,y')+T(x',y)+T(x',y')$, $\forall (x,y)\in X\times Y, (x',y')\in X\times Y$.

(2) $T(\alpha x,\beta y)=\alpha\beta T(x,y), \forall (x,y)\in X\times Y$ with $\alpha$ and $\beta$ being scalars, i.e. $T$ is bilinear.

Show that $T$ is continuous if and only if there exist $k>0$ such that
$$\Vert T(x,y)\Vert\leq k\Vert x\Vert \Vert y\Vert.$$

I have tried as below.

($\Rightarrow)$ Given $T$ is continuous. Then $T$ is bounded, i.e if there exists $M > 0$ such that $\Vert (x,y)\Vert\leq 1$ implies $\Vert T(x,y)\Vert\leq M$. Now, we will prove $M=k\Vert x\Vert \Vert y\Vert$, with $k>0$. Let $\alpha,\beta>0$.
\begin{align}
\Vert T(x,y)\Vert
&=
\left\Vert T\left(\dfrac{\alpha}{\Vert x\Vert} x  \dfrac{\Vert x\Vert}{\alpha},\dfrac{\beta}{\Vert y\Vert} y \dfrac{\Vert y\Vert}{\beta}\right)\right\Vert\\
&=
\dfrac{\Vert x\Vert\Vert y\Vert}{\alpha\beta}\left  \Vert T\left(\dfrac{\alpha}{\Vert x\Vert} x  ,\dfrac{\beta}{\Vert y\Vert} y \right)\right\Vert\\
&\leq \dfrac{1}{\alpha\beta}\Vert x\Vert\Vert y\Vert
\end{align}

So, $k=\dfrac{1}{\alpha\beta}>0$.

($\Leftarrow)$ Given there exist $k>0$ such that
$$\Vert T(x,y)\Vert\leq k\Vert x\Vert \Vert y\Vert.$$

We will prove $T$ is continuous, i.e. we will show $T$ is bounded, if there exists $M > 0$ such that $\Vert (x,y)\Vert\leq 1$ implies $\Vert T(x,y)\Vert\leq M$.

Obviously, $k\Vert x\Vert \Vert y\Vert>0$. So, if we take $M=k\Vert x\Vert \Vert y\Vert$ then we have $\Vert T(x,y)\Vert\leq M$. Thus, $T$ is bounded operator then imply $T$ is continuous operator.

Does my proof is correct? I doubt it.
Any hint to prove it? I don't know the definition of continuous operator (not linear).

答:
Suppose $T$ is continuous. Suppose there is no constant $k$ such that $\|(T(x,y)\| \leq k\|x\|\|y\|$. Then there exist $x_n$ and $y_n$ such that $\|(T(x_n,y_n)\| > n\|x_n\|\|y_n\|$. Let $u_n=\frac {x_n}{\sqrt n \|x\|_n}$ and $v_n=\frac {y_n}{\sqrt n \|y\|_n}$. Then $u_n \to 0$ and $v_n \to 0$. Hence $(u_n,v_n) \to (0,0)$ in $X \times Y$. By continuity of $T$ we get $T(u_n,v_n) \to (0,0)$ in $X \times Y$. This gives $\frac 1 { n \|x\|_n\|y\|_n}T(x_n,y_n) \to 0$ but $\|\frac 1 { n \|x\|_n\|y\|_n}T(x_n,y_n)\|>1$ for all $n$.

Converse: Suppose $\|(T(x,y)\| \leq k\|x\|\|y\|$. Let $(x_n,y_n) \to (x,y)$. Then $T(x_n,y_n)=T(x_n-x,y_n-y)+T(x,y)+T(x_n-x,y)+T(x_n,y_n-y)$. It is easy to check that this tends to $0$ from the inequality $\|(T(x,y)\| \leq k\|x\|\|y\|$ so $T$ is continuous.


问:
Where to find the paper of G.H. Hardy called "Sur les zéros de la fonction Zeta de Riemann" please?
Where to find the paper of G.H. Hardy called "Sur les zéros de la fonction Zeta de Riemann" please?

I've been looking for a while now, and I just don't find it, the basics of the hypothesis !

If anyone can help !

答: this paper is included in book by peter borwein "the riemann hypothesis"</p>


答:
The question requires a single string of consecutive boxes to contain white balls, while all the remaining boxes contain blue balls. As an example, if there are ten boxes, a possible situation satisfying the conditions would be BBWWWWWBBB.

The provided solution is saying that we can count this by just choosing the first and last box(es) that contain the white balls. In the above example, we would choose boxes 3 and 7. Thus the number of ways of filling the boxes would be equivalent to the number of ways of choosing the first and last box.

Normally, that would just be $\binom{n}{2}$. However in this case, the conditions specifically allow for just a single box to be white, and this would not be accounted for by $\binom{n}{2}$ as the first and last boxes would be the same box. That is why they threw in those little x's.

Now for the example I gave above, the configuration would be xBxBxWxWxWxWxWxBxBxBx; instead of choosing the first and last boxes, I choose the x before the first box with a white ball and the x after the last box with a white ball, i.e. the third and eighth x. By doing it this way, we can choose a single box to contain a white ball, by just choosing the x's right before and right after that single box. But since there will always be one more x than there are boxes, the number of ways will now be $\binom{n+1}{2}$.

Note that another simple way to get around the problem of single boxes would just be to add the number of ways of choosing a single box ($n$) to the earlier $\binom{n}{2}$ and get

$$ \binom{n}{2} + n = \frac{n^2-n}{2} + n = \frac{n^2+n}{2} = \binom{n+1}{2} $$

Title: Filling at least $1$ box with a white ball from $1$ to $n$ numbered boxes

Body:
Given $n$ boxes numbered $1$ to $n$, each box is to be filled with either a white ball or a blue ball
such that at least one box contains a white ball and boxes containing white balls are consecutively
numbered.  What is the total number of ways this can be done?

Solution provided:

The task is similar to choosing two out of the $n + 1$ crosses to mark the start and
end of the consecutively numbered boxes that contain white balls.
This is ${n+1 \choose 2}$, which is equal to $\frac{n(n + 1)}{2}$.

My question:

I get that the picture shows $n$ boxes because we can have $1$ to $n$ white balls.

I have a feeling they are using the $r$-combination formula here: ${r + n -1 \choose r}$.

That's all about I get. I have no clue why they have $n+1$ crosses or why they are choosing only $2$ from those crosses or what those crosses represent even?

PS: I know there is sum of sequence approach to this but I'm required to provide a combinatorial working.


问:
determine $f(r)$ so that $\nabla \bullet \mathbf{F} = 0$
I am self studying Divergence, Curl and All That: An Informal Text On Vector Calculus

I have some confusion regarding problem 22 from chapter 2, which uses information given in problem 21:

My problem is understanding how to even work with $f(r)$, and how to translate the vector function into spherical coordinates.

I do not think I want to just write $f(r) = f(\sqrt{x^2+y^2+z^2})$ and write $F(r)$ as $(ix+jy+kz)/r * f(\sqrt{x^2+y^2+z^2})$ because I do not know how to evaluate $r^2F_r$ from here. I thought perhaps $\partial/\partial{\phi} (sin{\phi}F_{\phi}) = 0$ and similarly for $\theta$, but then I thought I need to translate the vector function into spherical coordinates before I determine what the partial of phi and theta is. I do not feel confident about my ability to translate this into spherical coordinates.

I know I am a bit all over the place with this one, but some hints, a push forward or some insight would be greatly appreciated.

答: In problem $21$, $\mathbf{F}$ is represented as $F_r \mathbf{\hat{e}}_r + F_\theta \mathbf{\hat{e}}_\theta + F_\phi \mathbf{\hat{e}}_\phi$. So in problem $22$, $\mathbf{F}$ has only a radial component, $f(r)$. The divergence is then $\frac{1}{r^2}\frac{d}{dr}(r^2 f)$, which is zero iff $r^2 f(r)$ is constant. Thus $f(r) = A/r^2$ for some constant $A$.


问:
Computing the probability of a type-1 error
Let $X_1, X_2,\dots, X_{130}$ be a sample from $GAM(θ, k).$ This sample is used for a
hypothesis test. The null hypothesis is
$θ = 2$ and $k = 2$
and the alternative hypothesis is irrelevant for this question. The test statistic is
$\log(X_{0.3}),$
where log is the natural logarithm and $X_{0.3}$ is the 30th percentile of the sample. The
critical region for this hypothesis test and this test statistic is
(−∞, 0.477] .
Compute (approximately) α, the probability of a type-I error.

答: This question is clearly copyrighted by the University of Amsterdam teacher as this is one of the assignment questions which is only due on 6/11/2020. Don't cheat.</p>


问:
Intuition behind Cantor-Bernstein-Schröder
The book I am working from (Introduction to Set Theory, Hrbacek & Jech) gives a proof of this result, which I can follow as a chain of implications, but which does not make natural, intuitive sense to me. At the end of the proof, I found myself quite confused, and had to look carefully at the build-up to see how the conclusion followed. I get the steps, now - but not the intuition.

The authors took sets $X$ and $Y$ and assumed injections $f: X \rightarrow Y$, $g: Y \rightarrow X$. Since $X \sim g[f[X]]$ and $X \supseteq g[Y] \supseteq g[f[X]]$, and $Y \sim g[Y]$, the authors went to prove the lemma that $A \sim A_1, A \supseteq B \supseteq A_1 \implies A \sim B$.

For the lemma, they defined recursive sequences $\{A_n\}_{n \in \omega}, \{B_n\}_{n \in \omega}$ by $A_0 = A, A_{n+1} = f[A],$ $ B_0 = B, B_{n+1} = f[B].$ Since $A_0 \supseteq B_0 \supseteq A_1 \implies f[A_0] \supseteq f[B_0] \supseteq f[A_1]$, we get from induction that $A_{n+1} \subseteq A_n$. Putting $\{C_n\}_{n \in \omega} = \{A_n - B_n\}_{n \in \omega},$ they noted that $C_{n+1} = f[C_n]$ (since $A_n \supseteq B_n$ inductively, $f[A_n - B_n] = f[A_n] - f[B_n]$). Putting

$$ C = \bigcup_{n=0}^\infty C_n, \text{   } D = A - C,$$

they noted that $f[C] = \bigcup_{n=1}^\infty C_n$, and that $h(x): A \rightarrow f[C] \cup D$ can be defined by sending $x \in C \rightarrow f(x),$ and $x \in D \rightarrow x$. It is clear that $h$ is one-to-one and onto.

Inductively, for $n > 0$, $C_0 \cap C_n = \varnothing$; it follows that $C_0 \cap \bigcup_{n=1}^\infty C_n = C_0 \cap f[C] = \varnothing$. We know that $C_0 \cup f[C] \cup D$ = A, and since all three sets are disjoint, we may conclude that $f[C] \cup D = A - C_0 = A - (A - B) = B$. Thus, our bijection $h$ maps $A$ to $B$, as we wanted.

What is the intuition here? What were H&J, or Cantor, Bernstein, etc. thinking when they went to prove this - the "high-level" idea? Is there a clearer proof, in the sense of thought process (not necessarily shorter)?

答:
As you pointed out, to prove Cantor-Schroeder-Bernstein theorem, one needs to prove the following lemma:

If $A_1 \subset B \subset A$ and $|A_1|=|A|$, then $|B|=|A|$.

According to the hypothesis, there exists some one-to-one mapping $f$ from $A$ onto $A_1$. So, we need a one-to-one mapping $g$ from $A$ onto $B$. How can we find that?

One may think that we should embed the set $B$ in the set $A$ by inclusion so that we find the required mapping as the inverse of the inclusion mapping, that is, the mapping$$h:A \to B \\ h(x)=x.$$This mapping is one-to-one, but it cannot be defined on the whole set $A$ because for $x \in A-B$ $h(x)=x$ is not contained in the set $B$.

One may think that we should use the mapping $f$ as our required mapping, that is,$$k:A \to B \\ k(x)=f(x).$$This mapping is one-to-one, but it cannot cover the whole set $B$ because the range of the function $f$ is the set $A_1 \subset B$ and there may be some $x\in B$ which is not in the set $A_1$.

As we see there are two extreme approaches for finding the required mapping; the first one covers the whole set $B$, and the second one covers the whole set $A$. So, it is intuitively expectable that to find the required mapping we should use both the mappings (approaches) simultaneously so that both the sets $A$ and $B$ are covered in a one-to-one manner.

But, there exists a problem. If we use both the mappings simultaneously, for example $g: A \to B \quad g(x)=\begin{cases}f(x) & \text{if } x\in C; \\ x & \text{if } x \in A-C \end{cases}$ for some subset $C \subset A$, then we may miss either the onto property or the one-to-one property, because the ranges of the pieces of $g$ may overlap.

So, our original problem is reduced to finding some subset $C \subset A$ such that the ranges of the pieces of the mapping are disjoint and the union of them is equal to the set $B$. Now, how to find such a $C$?

Here is an idea. Since the function $h(x)=x$ cannot be defined on $C_0=A-B$, as explained above, let us map this subset of $A$ by the function $k(x)=f(x)$ into the set $B$. So, we obtain the mapping$$g_0(x)= \begin{cases}f(x) & \text{if } x \in C_0; \\ x & \text{if }x \in A-C_0 \end{cases}.$$But, we have missed the one-to-one property because the ranges of the pieces overlap (In fact, $f[C_0]$ is contained in the range of the second one, since $f[C_0] \subset f[A] \subset A-C_0$).

So, we need to remove the problematic points $C_1=f[C_0]$ from the domain of the second piece (since the domain and the range of the function $h(x)=x$ are the same) to retain the one-to-one property. However, since we need to define the mapping $g$ on the whole set $A$, we need to add such points to the domain of the first piece. So, we obtain the mapping$$g_1(x)= \begin{cases}f(x) & \text{if } x \in C_0 \cup C_1; \\ x & \text{if }x \in A-(C_0 \cup C_1) \end{cases}.$$But, we have missed the one-to-one property because the ranges of the pieces overlap (In fact, $f[C_1]$ is contained in the range of the second one, since $f[C_1]=f^2[C_0] \subset f^2[A] \subset A-(C_0 \cup C_1)$).

So, we need to remove the problematic points $C_2=f[C_1]$ from the domain of the second piece (since the domain and the range of the function $h(x)=x$ are the same) to retain the one-to-one property. However, since we need to define the mapping $g$ on the whole set $A$, we need to add such points to the domain of the first piece. So, we obtain the mapping$$g_2(x)= \begin{cases}f(x) & \text{if } x \in C_0 \cup C_1 \cup C_2; \\ x & \text{if }x \in A-(C_0 \cup C_1 \cup C_2) \end{cases}.$$
$$\vdots \qquad \vdots \qquad \vdots$$
But, we have missed the one-to-one property because the ranges of the pieces overlap (In fact, $f[C_{n-1}]$ is contained in the range of the second one, since $f[C_{n-1}]=f^n[C_0] \subset f^n[A] \subset A-(C_0 \cup C_1 \cup \cdots C_{n-1})$).

So, we need to remove the problematic points $C_n=f[C_{n-1}]$ from the domain of the second piece (since the domain and the range of the function $h(x)=x$ are the same) to retain the one-to-one property. However, since we need to define the mapping $g$ on the whole set $A$, we need to add such points to the domain of the first piece. So, we obtain the mapping$$g_n(x)= \begin{cases}f(x) & \text{if } x \in C_0 \cup C_1 \cup \cdots \cup C_n; \\ x & \text{if }x \in A-(C_0 \cup C_1 \cup \cdots  \cup C_n) \end{cases}.$$
$$\vdots \qquad \vdots \qquad \vdots$$
This pattern motivates us to define the mapping $g$ as follows.$$g(x)=\begin{cases}f(x) & \text{if } x \in C; \\ x & \text{if } x \in A-C \end{cases}, \qquad C= \bigcup_{n=0}^{\infty }C_n$$Noting that$f[C]=\bigcup_{n=1}^{\infty }C_n$, we can easily see that the mapping $g$ is one-to-one because each of its pieces is and the ranges of the pieces are disjoint and it is onto the set $B$.

Addendum

Looking at how the $C_n$'s are constructed, one may think that the existence of the set $C$ (and so the proof of the theorem) relies on the existence of some infinite set like $\mathbb{N}$ to be able to define the sets $C_n$'s recursively. However, in this section we show that such a view is not correct.

In fact, to obtain the bijective mapping $g$, we need some sets $C$ such that the values of the function $f$ at the points of $f[C]$ do not lie outside of $f[C]$. The existence of such a set can be guaranteed by applying some fixed-point theorem (Knaster-Tarski Theorem) to some monotone function of sets, as follows.

Let $F: \mathcal{P}(A) \to \mathcal{P}(B)$ be monotone, i.e., if $X \subset Y$, then $F(X) \subset F(Y)$ ($\mathcal{P}(A)$ is the power set of $A$). Consider the set $T= \{ X \subset A \mid F(X) \subset X \}$. It can be easily seen that $\overline{X}=\bigcap T$ is the least fixed point of $F$ (Proof: $A \in T$, so $T \neq \varnothing$ and so $\overline{X}=\bigcap T$ can be defined. Since $F$ is monotone and for any $X \in T$ we have $\bigcap T \subset X$, $F(\overline{X}) \subset F(X)$ for every $X \in T$, so $\overline{X} \in T$.

Since $F$ is monotone and $F(\overline{X}) \subset \overline{X}$, we have $F(F(\overline{X})) \subset F(\overline{X})$, so $F(\overline{X}) \in T$.

However, since $\overline{X} \subset X$ for every $X \in T$, we have $\overline{X} \subset F(\overline{X})$. Thus, $F(\overline{X})=\overline{X}$.

If $F$ has some other fixed points $X'$, i.e., $F(X')=X'$, then $X' \in T$. Since $\overline{X} \subset X$ for every $X\in T$, we conclude that $\overline{X}=\bigcap T$ is the least fixed point of $F$).

Consider the function $F(X)=(A-B)\cup f[X]$. Clearly, it is monotone, so the set $C=\overline{X}$ defined above is its least fixed point.

Now, we can easily see that the mapping $g:A \to B$ defined by$$g(x)=\begin{cases}f(x) & \text{if } x\in C; \\ x & \text{if } x \in A-C \end{cases}$$ is one-to-one and onto the set $B$ (We only need to note that $$\begin{align}f[C] \cup (A-C) & =f[C] \cup (A-((A-B) \cup f[C])) \\ & = f[C] \cup ((A-(A-B)) - f[C]) \\ & =f[C] \cup (B-f[C]) \\ & =B \end{align}$$(Please note that in the above calculation we have used the fact that $f[C] \subset A_1 \subset B \subset A$) and$$\begin{align}f[C] \cap (A-C) & =f[C] \cap (A-((A-B) \cap f[C])) \\ & = f[C] \cap ((A-(A-B)) - f[C]) \\ & =f[C] \cap (B-f[C]) \\ & = \varnothing \end{align}$$(Please note that in the above calculation we have used the fact that $f[C] \subset A_1 \subset B \subset A$)).

Now, the least fixed point of the function $F$ can be obtained recursively as follows.

Clearly the function $F$ is continuous, meaning that for any nondecreasing sequence of subsets of $A$, $\langle X_i \mid i \in \mathbb{N} \rangle$, $X_i \subset X_j$ whenever $i \le j$, we have$$F \left ( \bigcup_{i \in \mathbb{N}}X_i \right ) = \bigcup_{i \in \mathbb{N}} F \left ( X_i \right ).$$Let us define recursively $X_0=\varnothing$, $X_{i+1}=F(X_i)$ and then define $\overline{X}=\bigcup_{i \in \mathbb{N}}X_i$. Clearly, the $\langle X_i \mid n \in \mathbb{N} \rangle$ is a nondecreasing sequence of subsets of $A$. So we have$$\begin{align}F \left ( \bigcup_{i \in \mathbb{N}} X_i \right ) & = \bigcup_{i \in \mathbb{N}}F(X_i) \\ & = \varnothing \cup F(X_0) \cup F(X_1) \cup \cdots \\ & = X_0 \cup X_1 \cup X_2 \cup \cdots \\ & = \bigcup_{i \in \mathbb{N}} X_i. \end{align}$$Thus, $\overline{X}=\bigcup_{i \in \mathbb{N}}X_i$ is a fixed point of $F$.

Now, if $X'$ is another fixed point of $F$, since $F$ is monotone and $\langle X_i \mid i \in \mathbb{N} \rangle$ is a nondecreasing sequence of subsets of $A$, we have$$\varnothing \subset X' \quad \Rightarrow \quad X_1=F(\varnothing ) \subset F(X')=X' \\ X_1 \subset X' \quad \Rightarrow \quad X_2=F(X_1) \subset F(X')=X' \\ \vdots \qquad \vdots \qquad \vdots \\ X_{n-1} \subset X' \quad \Rightarrow \quad X_n =F(X_{n-1}) \subset F(X')=X' \\ \vdots \qquad \vdots \qquad \vdots$$So, $\overline{X}=\bigcup_{i \in \mathbb{N}} \subset X'$. Thus, $\overline{X}$ is the least fixed point of $F$.

Hence, we conclude that the fixed point of the function $F(X)=(A-B) \cup f[X]$, $C$, must be of the form$$\begin{align}C & =(A-B) \cup ((A-B) \cup f[A-B]) \cup ((A-B) \cup f[A-B] \cup f[f[A-B]]) \cup \cdots \\ & = C_0 \cup (C_0 \cup C_1) \cup (C_0 \cup C_1 \cup C_2) \cup \cdots \\ & = \bigcup_{i \in \mathbb{N}}C_n\end{align}$$(Please remember that $f$ is injective), which was already obtained from our original argument.

Therefore, the existence of the set $C$ can be confirmed without needing existence of some infinite set like $\mathbb{N}$. Now, if $A$ is a finite set, then $B$ must be equal to $A$ and so $C=A-B=\varnothing$. But, if $A$ is an infinite set (so the existence of some infinite set has been already assumed in our theory), then the set $C$ is constructed from an infinite chain of sets, as explained above.


问:
How to solve for $x$ in trigonometric (sine wave) equation with $x$ passing through the floor function?
I have the following equation:

$\displaystyle 1=\sin\left(2\pi\left(\frac{2^{\operatorname{floor}\left(x\right)-1}-1}{\operatorname{floor}\left(x\right)}+\frac{1}{4}\right)\right)$

Is it possible to solve for $x$, here? When removing the floor function, the sine wave changes entirely (so this is not ideal when solving for roots). I would grateful for any help.

答:
We have:
$$\sin\bigg(2\pi \bigg( \frac{2^{\lfloor x \rfloor-1}-1}{\lfloor x \rfloor}+\frac{1}{4} \bigg) \bigg)=\sin\bigg(\frac{\pi}{2}+2\pi \bigg( \frac{2^{\lfloor x \rfloor-1}-1}{\lfloor x \rfloor} \bigg) \bigg)=1$$
Taking $\sin \big( \frac{\pi}{2} + \theta) = \cos(\theta)$ yields:
$$\cos \bigg(2\pi \bigg( \frac{2^{\lfloor x \rfloor-1}-1}{\lfloor x \rfloor} \bigg) \bigg)=1$$

We know that $cos(2\pi k)=1$ if and only if $k \in \mathbb{Z}$. Hence, we have for $\lfloor x \rfloor = n$ i.e. $n \in \mathbb{Z}$ :
$$\bigg( \frac{2^{n-1}-1}{n} \bigg) \in \mathbb{Z} \implies n \mid (2^{n-1}-1)$$

We must $2^{n-1}-1$ to be an integer since the integer $n$ divides it. Thus, $n \in \mathbb{Z}^+$ i.e. $n$ is a positive integer. We can trivially see that $n=1$ works. For $n>1$, we have $2^{n-1}-1$ to be odd, and thus, $n$ is odd. It follows from Fermat's Little Theorem that all odd primes $n$ work as well. However, these are not the only solutions. Composite numbers $n$ which satisfy $n \mid (2^{n-1}-1)$ are known as base-$2$ Fermat Pseudoprimes (as they pass the Fermat Primality Test for base $2$). Some examples are $341, 561$ and $645$. There are infinitely many such composite solutions as well.

Now, we need $\lfloor x \rfloor = n$ which is equivalent to $x \in [n,n+1)$. Let us call the set of primes as $\mathbb{P}$ and let us call the set of base-$2$ Fermat Pseudoprimes as $S$. Hence, the set of solutions is:
$$x \in \bigcup_{n \in A} \space [n,n+1) \quad ; \quad A = \{1\}+(\mathbb{P}-\{2\})+S$$


问:
How to prove there's an upper bound on a series
There's a series as following.
$$
\sqrt{3}, \sqrt{3 + \sqrt{3}}, \sqrt{3 + \sqrt{3 + \sqrt{3}}} + \cdots
$$
Is there any way I can find the upper bound to prove the limit exist?
I've tried to write down the formula of this series as following:
$$
x_n = \sqrt{3 + x_{n - 1}}
$$
but still find no way to modify it to get to the upper bound.

答: Hint : Show inductively that all the terms are bounded by $$\frac{1+\sqrt{13}}{2}$$


问:
Number of sequences of the form $x_1<x_2>x_3<x_4>x_5$

Find the number of sequences containing $5$ distinct terms $x_1,x_2,x_3,x_4,x_5$ from the set $A=\left\{1,2,3,4....20\right\}$ such that $x_1<x_2>x_3<x_4>x_5$.

My try:

The total number of sequences with $5$ distinct terms is $5!\times \binom{20}{5}$, among which there are $\binom{20}{5}$ strictly increasing sequences and $\binom{20}{5}$ strictly decreasing sequences. But now I am stuck in finding the sequences with $x_1<x_2>x_3<x_4>x_5$ among $5!\times \binom{20}{5}-2\times \binom{20}{5}$

答:
Note that for any set of five distinct numbers, the number of ways of arranging them so that they satisfy the given condition must be the same as for any other five distinct numbers. Thus we really only need to count the number of ways of ordering $\{1,2,3,4,5\}$ to satisfy the given condition.

Note that $5$ has to be either $x_2$ or $x_4$; note also that whichever of $x_2$ or $x_4$ is not $5$, it will have to be either $3$ or $4$.

Suppose $(x_2, x_4) \in \{(4,5),(5,4)\}$; in each case, there are $3!$ ways of arranging the other three numbers. Now suppose $(x_2, x_4) \in \{(3,5),(5,3)\}$; then the $4$ has to go in the outside position next to the position containing the $5$, which leaves $2!$ ways of arranging the other numbers.

Thus the number of arrangements of $\{1,2,3,4,5\}$ satisfying the given condition is $2 \cdot 3! + 2 \cdot 2! = 16$. But then the number of ways of sequences from $\{1,2,3,4,...,20\}$ satisfying the condition will just be

$$16 \binom{20}{5} = \boxed{248,064}$$


答:
After choosing five numbers, relabel them ${1,\ldots,5}$ from smallest to largest.

Observe now

$5$ can only be placed at $x_2$ or $x_4$

When $5$ is placed at $x_4$, $4$ can go to $x_2$ or $x_5$
$$ \square < 4 > \square < 5 > \square$$
$$ \square < 3 > \square < 5 > 4$$

These are only two valid ways possible. This corresponds to number of ways $3! + 2! = 8$ (filling ${1,2,3}$ in remaining boxes).

Similarly when $5$ is placed at $x_2$, we have $8$ more ways.

Hence answer is $$16 \cdot \binom{20}{5}$$

Credits to @A.J. for considering all cases.

Title: Number of sequences of the form $x_1<x_2>x_3<x_4>x_5$

Body:
Find the number of sequences containing $5$ distinct terms $x_1,x_2,x_3,x_4,x_5$ from the set $A=\left\{1,2,3,4....20\right\}$ such that $x_1<x_2>x_3<x_4>x_5$.

My try:

The total number of sequences with $5$ distinct terms is $5!\times \binom{20}{5}$, among which there are $\binom{20}{5}$ strictly increasing sequences and $\binom{20}{5}$ strictly decreasing sequences. But now I am stuck in finding the sequences with $x_1<x_2>x_3<x_4>x_5$ among $5!\times \binom{20}{5}-2\times \binom{20}{5}$


问:
Riemann integrable function is bounded, without any type of choice
I know a proof that Riemann integrable positive function $f$ defined on $[a,b]$ is bounded.

Suppose $f\geq 0$ is unbounded. First, for each $n$, choose a point $a_n$ in $[a,b]$ such that $f(a_n) > n$. Let $a$ be an arbitrary positive real number. Choose $n$ with $a > \frac1n$, and we can have a partition $p$ with each interval length bigger than $\frac1n$ while mesh($p$) $< a$. Then $R(f, p) > f(a_{n^2}) \frac1n = n$  so the Riemann sums do not tend to a finite limit.

But I think while we choose $a_n$ in the first step, countable choice is used. Are there any other ways to prove it without using some kind of choice, since this is a very 'basic' theorem in 'elementary 'analysis so it seems weird that it requires some kind of choice axiom.

答:
You don't have to choose the whole sequence in advance. We assumed that $f$ is unbounded, so for any given $n$ we can find $x$ such that $f(x) > n^2$, that's just part of the definition.

Now for any fixed $a$ pick a suitable $n$, and $x$ such that $f(x)>n^2$, etc. We are really using the whole sequence here. It's a proof of convenience.