问:
Is this function defined over a rational rectangle integrable?
Let $Q_0 = \{ x : x\in \mathbb{Q} , 0\leq x \leq 1 \}$ and let $f:R_Q \to \mathbb{R}$ with $R_Q = Q_0 \times Q_0$. f defined as:

$$
f(x,y)= \left\{ \begin{array}{lc}
             \sin(x) &   \text{if}  \; 0\leq y \leq x^2 \\
             \\ \hspace{0.5cm} 0 &  \text{in other case} \\
             \end{array}
   \right.
$$

Is this function integrable over $R_Q$?

My attempt:

For $0\leq y \leq x^2 $:

$$ f(x,y) = g(x) \ h(y) \ \text{ with } \ g:\mathbb{Q} \to \mathbb{R} , \ \ g(x)=\sin(x), \ h:\mathbb{Q} \to \mathbb{R} \ \ h(y) = 1 $$

Restriction of continuous functions is continuous then $g$ and $h$ are continuous functions, consequently $f$ is continuous because is a product of continuous functions. Therefore $f$ is integrable over $R_Q$.

Is the above correct? If so, how is the integral computed?

答:
Your argument is not quite complete. You've shown that the function $\ f\ $ is continuous, and hence integrable, over the set $\ A=\left\{(x,y)\in\mathbb{Q}^2\right.\,\left|\,0\le x\le 1, 0\le y\le x^2\right\}\ $.  However, $\ f\ $ is not continuous over $\ R_Q\ $, so this is not sufficient to establish its integrability over that set. It is not continuous at the point $\ \left(\frac{1}{2}, \frac{1}{4}\right)\in R_Q\ $, for instance, because $\ f\left(\frac{1}{2}, \frac{1}{4}\right)\ne0\ $, but $\ \lim_\limits{\epsilon\rightarrow0^+}f\left(\frac{1}{2}, \frac{1}{4}+\epsilon\right)=0\ $.

The easiest way to establish the integrability of $\ f\ $ over $\ R_Q\ $ is to note that $\ R_Q\ $ is countable, and therefore has measure $\ 0\ $.  All real-valued functions are integrable over any set of measure $\ 0\ $, and their integrals over such a set are always $\ 0\ $.


问:
Can $n$ convex solids in $\mathbb{R}^3$ be mutually touching?
Say that a collection of $n$ convex solids in $\mathbb{R}^3$ are mutually touching if, for every pair of them, they share some positive-measure part of their boundaries. (That is, we're not permitting the solids to touch only via an edge or a corner.)

As an example, if we "thicken up" an $n=4$ configuration in 2 dimensions (the maximum, since $K_5$ is nonplanar), then placing a large cube on top of the thickened configuration yields $n=5$ mutually touching convex regions. (By thickening it into a sort of wedge shape, we can put convex solids on both the top and bottom, for $n=6$.)

Do such configurations exist for arbitrarily high $n$? For infinitely many solids at once? What if we require the solids be congruent to each other? (Note that if we relax the convexity requirement, the answer to all of these is pretty easily seen to be yes.)

I vaguely recall reading a paper which provided an affirmative answer to the arbitrarily high $n$ case with congruent solids; I think it used the Voronoi cells given by equally spaced points on a helix, though it's been several years since I read it. Any pointers to this paper, or an equivalent result, would be appreciated (if I'm not just imagining it)!

答:
Based on your description, I think I've found an argument that works. However many points there are, space them on a segment of helix defined by the following equation in cylindrical coordinates:

$$
(r, \theta, z) = (1, \theta, \frac{\theta}{2}) \hspace{1cm}\text{for } 0\leq\theta\leq\frac{\pi}{2}
$$

The relevant solids will be the Voronoi cells of these points, cut off at some large distance. We want to check the existence of a boundary between every chosen pair of points on our helix segment. It will suffice to show that for every pair of points on the helix (not just the ones we happened to choose), there will be some point $Q$ in space that is equidistant from the pair and is closer to that pair of points than to any other point on the helix. (The existence of such a point shows that there must be a boundary between the two corresponding cells, with $Q$ lying on that boundary.)

The general idea of the proof is that if you're standing a few steps behind the center of a circle and looking straight through the center and across to the opposite side, the points with a larger angle to your line of sight will be closer to you. On the other hand, if you're looking at a line, the points with a larger angle will be farther. In the helix, which is kind of a cross between a circle and a line, these effects can be balanced, and the nearest points to you will be at an intermediate angle to you line of sight.

Let $P_1 = (1, \theta_1, \theta_1/2)$ and $P_2 = (1, \theta_2, \theta_2/2)$. We'll want to find a point $Q$ in space that is closer to those two points than to any other point on the helix segment. We'll choose $Q$ to be the following, for some $a$, which we'll specify later:

$$(-a, \frac{\theta_1+\theta_2}{2}, \frac{\theta_1+\theta_2}{4})$$

Define $R(\phi)=(1, \phi+\frac{\theta_1+\theta_2}{2}, \frac{\phi}{2}+\frac{\theta_1+\theta_2}{4})$. (This is always a point on the helix.) Note that since the helix segment only spans an angle of $\pi/2$, we only need to worry about $\phi$ between $-\pi/2$ and $\pi/2$. Also, note that $R(\phi^*)=P_1$ and $R(-\phi^*)=P_2$ for $\phi^*=\frac{\theta_1-\theta_2}{2}$.

We can now compute the squared distance between $R(\phi)$ and $Q$:

$$
s^2 = \frac{\phi^2}{4} + (\sin\phi)^2 + (a+\cos\phi)^2 = \frac{\phi^2}{4} + 1 + a^2 + 2a\cos\phi
$$

Setting the derivative to 0, so we can find possible locations of minimal distance:

$$
\frac{ds^2}{d\phi} = \frac{\phi}{2} - 2a\sin\phi = 0
$$

This yields the solution $\phi=0$ for all $a$, and an additional pair of non-trivial solutions, $\phi = -\phi_a, \phi_a$ for $\frac{1}{4}<a<\frac{\pi}{4}$. (We're only counting solutions in $(-\frac{\pi}{2}, \frac{\pi}{2})$, which is why there's an upper limit on $a$ too.)

We have $\lim_{a\to \frac{1}{4}}\phi_a = 0$ and $\lim_{a\to \frac{\pi}{4}}\phi_a = \frac{\pi}{2}$, and $\phi_a$ varies continuously with $a$. So we can choose an $a$ to produce any desired value of $\phi_a$ between 0 and $\frac\pi2$.
Further, we can compute the second derivative of $s^2$, which is:

$$
\frac{d^2s^2}{d\phi^2} = \frac{1}{2} - 2a\cos\phi
$$

This is negative for $a>\frac{1}{4}$, indicating that the $\phi=0$ critical point is a local maxima. Therefore, the neighbouring $-\phi_a$ and $\phi_a$ solutions must be local minima, and since $s^2(\phi)$ is an even function, they must have the same value of $s^2$ for both. Also, since there are no other possible local minima for $\phi\in(-\frac{\pi}{2}, \frac{\pi}{2})$, $\phi_a$ and $-\phi_a$ must give the global minima on that interval. Therefore, for given values of $\theta_1$, $\theta_2$, we choose $a$ such that $\phi_a = \phi^* = \frac{\theta_1 - \theta_2}{2}$, ensuring that $P_1$ and $P_2$ will correspond to $\phi_a$ and $-\phi_a$ and will therefore be the closest points to $Q$.

ContentLicense: CC BY-SA 4.0


问:
Are the invariant sets of all iterated function systems necessarily fractal?
An iterated function system is defined as a finite set of contraction mappings, defined over a complete metric space $X$, and iteration is defined as sequential composition of these contraction mappings, where each mapping has some nonzero probability of being used at every iteration.  Hutchinson (1981) proved that such iterated function systems (IFS) have invariant sets that iteration converges to, call this set $S \subset X$, so that for $n$ different contraction mappings defining our IFS, we have
$$
S = \bigcup_{i=1}^n f_i(S)
$$
I am wondering if such invariant sets $S$ are necessarily fractal if they are generated by an IFS. It is well know that many fractals can be generated by an IFS, but it is not clear if IFS's necessarily generate fractal invariant sets.

答:
It slightly depends on your definition of fractal, but typically no it is not true that IFS always generate fractals. Let $X = [0,1]$ and consider $f_1$ and $f_2$ defined on $X$ by $f_1(x) = \frac{x}{2}$ and $f_2(x) = \frac{1+x}{2}$. These are both contractions with a contractive factor of $\frac{1}{2}$.
It's easy to check that
$$ [0,1] = f_1([0,1]) \cup f_2([0,1]),$$
so the invariant set is $X$ itself.

There's also the dumb example where the IFS consists of a single contraction, then the attractor is just a point.


问:
A projectile has been shot into the atmosphere. Find the polynomial that interpolates the speed and acceleration in function of time.
This is the table (tempo = time, velocidade = speed, aceleracao = acceleration)

I used Newton's interpolation to get the polynomial (5th degree).

I got
$$102 + 138(x-15) + (-124/15)(x-15)^2 + (289/225)(x-15)^2(x-30) + (-0.0689)(x-15)^2(x-30)^2 + (0.0048)(x-15)^2(x-30)^2(x-45)$$

I tested this in my calculator and the values for p(15) and p(30) are correct but p(45) is completely wrong.

Then I got the derivative to test the derivatives
$$14341.5 - 2226.13 x + 122.896 x^2 - 2.8676 x^3 + 0.024 x^4$$

p'(15) and p'(30) are correct, albeit with a slight error, but p'(45) is very off (I got 132 point something).

Is this normal? Is it possible to get a better estimate for this?

My calculations:

答:
For me, the problem comes for the conversion to decimals (notice that $45^5=184528125$ and any error on the last coefficient will have a dramatic impact.

Since I suppose that you used matrix calculations, you must have arrived at
$$v=-68178+\frac{69497 t}{5}-\frac{26902 t^2}{25}+\frac{88873 t^3}{2250}-\frac{11633
   t^4}{16875}+\frac{259 t^5}{56250}$$ which reproduces exactly the values of speed and acceleration.

Deriving, we have
$$\frac{69497}{5}-\frac{53804 t}{25}+\frac{88873 t^2}{750}-\frac{46532
   t^3}{16875}+\frac{259 t^4}{11250}$$ which ,concerted to decimals, gives
$$13899.4-2152.16 t+118.497 t^2-2.75745 t^3+0.0230222   t^4$$ which are quite different to the ones you posted.


问:
Is is true that $ \exists~ \lim(a_n+b_n)~\text{and}~\exists~\lim a_n \Rightarrow \exists ~\lim b_n. $
I"m interested in knowing if the following affirmation is true
$$
\exists~\lim(a_n+b_n)~\text{and}~\exists~\lim a_n \Rightarrow \exists~\lim b_n.
$$
It seems true, and I tried to prove it using the definition, but I couldn't.

答:
let $lim_{n \to \infty }b_n=b$ and $lim_{n \to \infty}(a_n+b_n)=u$

$lim_{n \to \infty }b_n=b \Rightarrow lim_{n \to \infty }-b_n=-b$

$ lim_{n \to \infty }-b_n=-b$ and $lim_{n \to \infty}(a_n+b_n)=u$ $\Rightarrow lim_{n \to \infty}(a_n+b_n+(-b_n))=lim_{n \to \infty }-b_n+lim_{n \to \infty}(a_n+b_n)=u-b$

$lim_{n \to \infty} a_n=u-b$


问:
If $f$ is a uniformly continuous function on a metric vector space, is there a continuous $g$ with $f=e^{{\rm i}g}$?
Let $E$ be a metric space and $$S^1:=\{z\in\mathbb C:|z|=1\}.$$ Say that $f\in C^0(E,S^1)$ is inessential if $$f=e^{{\rm i}g}\tag1$$ for some $g\in C^0(E,\mathbb R)$. We can show that

If $f\in C^0(E,S^1)$ with $f(E)\ne S^1$, then $f$ is inessential.

If $f_i\in C^0(E,S^1)$ with $$f_1(x)\ne-f_2(x)\;\;\;\text{for all }x\in E\tag2$$ and $f_1$ is inessential, then $f_2$ inessential.

In Appendix 2.5 of Treatise on Analysis the following corollary is shown:

If we take a look at the proof, we see that the compactness assumption is not needed as long as $f:E\times[0,1]\to S^1$ is uniformly continuous.

Now, the following is inferred from that corollary:

It seems like the only reason why the closed ball in $\mathbb R^d$ is considered
is that this is a compact metric vector space. Now I wonder, aren't we able to use precisely the same arguments to show that any uniformly continuous function $f:E\to S^1$ on a metric vector space is inessential?

EDIT

It was remarked that this can be solved by some facts about algebraic topology. However, I'm not familiar with algebraic topology. So, I'd really appreciate if someone could provide an elementary answer (maybe imposing suitable assumptions, which are satisfied in the special case of $\mathbb R^d$).

答:
Maybe I'm missing something, but I don't think that an algebraic topological argument is necessary to obtain the desired claim; at least under a suitable set of assumptions.

Lemma 1 Let $E$ be a metric space and $g_i\in C(E,\mathbb C)$ with $e^{g_1}=e^{g_2}$. Then $$\exists k\in\mathbb Z:g_1-g_2=2k\pi\tag1.$$

The proof easily follows from the continuity assumption and noting that $$e^{z_1}=e^{z_2}\Leftrightarrow\exists k\in\mathbb Z:2k\pi\tag2$$ for all $z_i\in\mathbb C$.

Theorem 2: Let $\mathbb K=\mathbb R$ or $\mathbb K=\mathbb C$, $E$ be a normed $\mathbb K$-vector space, $\tau$ be a vector topology on $E$ such that the closed unit ball $\overline B_1(0)$ around $0$ is $\tau$-compact and $f:E\to\mathbb C\setminus\{0\}$ be $\tau$-continuous and uniformly norm-continuous. Then there is an unique norm-continuous $g:E\to\mathbb C$ with $g(0)=0$ and $e^g=f$.

Fix $r>0$ for the moment. Since $\overline B_r(0)$ is $\tau$-compact and $f$ is $\tau$-continuous, $f(\overline B_r(0))$ is compact and hence (since $0\not\in f(E)$) $$\varepsilon:=\inf_{x\in\overline B_r(0)}|f(x)|>0\tag3.$$ Now, since $f$ is uniformly norm-continuous, there is a $\delta>0$ with $$|f(x)-f(y)|<\frac\varepsilon2\;\;\;\text{for all }x,y\in E\text{ with }\left\|x-y\right\|_E<\delta\tag4.$$ Let $k\in\mathbb N$ and $0=t_0<\cdots<t_k=1$ with $$\max_{1\le i\le k}|t_{i-1}-t_i|<\frac\delta r\tag5.$$ Then $$|f(t_{i-1}x)-f(t_ix)|<\frac\varepsilon2\;\;\;\text{for all }1\le i\le k\text{ and }x\in\overline B_r(0)\tag6.$$ Since $0\not\in f(E)$, this yields $$\left|1-\frac{f(t_ix)}{f(t_{i-1}x)}\right|<\frac\varepsilon{2|f(t_{i-1}x)|}\le\frac12\tag7$$ for all $1\le i\le k$ and $x\in\overline B_r(0)$. Let $\ln$ denote the principal branch of the complex algorithm and $$g(x):=\sum_{i=1}^k\ln\frac{f(t_ix)}{f(t_{i-1}x)}\;\;\;\text{or }x\in\overline B_r(0).$$ By $(7)$, $$\frac{f(t_ix)}{f(t_{i-1}x)}\in B_{\frac12}(1)\subseteq\mathbb C\setminus(-\infty,0]\;\;\;\text{for all }x\in\overline B_r(0)\tag8$$ and hence (since $\ln$ is continuous on $\mathbb C\setminus(-\infty,0]$) we conclude that $g$ is norm-continuous. Moreover, $g(0)=0$ and $$e^{g(x)}=f(x)\;\;\;\text{for all }x\in\overline B_r(0).\tag9$$ By

By Lemma 1, $g$ is unique. Now let's denote this $g$ by $g_r$. Then, by uniqueness, $$g_{r_1}=\left.g_{r_2}\right|_{\overline B_{r_1}(0)}\;\;\;\text{for all }0<r_1\le r_2\tag{10}$$ and hence $$g(x):=g_r(x)\;\;\;\text{for }x\in\overline B_r(0)\text{ and }r>0\tag{11}$$ is a well-defined function on $E$ satisfying the desired properties.


问:
prove the theorem without any reference to integration
show that  $A= \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} E_k$ in theorem $1.41$ and hence prove the theorem without any reference to integration

Here is the outline of theorem $1.41$

Theorem $1.41$: Let $\{E_k\}$ be a sequence of measurable sets in $X$,  such that
$$\sum_{k=1}^{\infty}\mu(E_k)<\infty. $$ Then almost all $x \in X$ lie in  at most finitely many of the sets  $E_k$

My attempt : Here  $A$ denote the set of all $x$ which lie  in infinitely many $E_k$

If $x \in A$  then $x \in \bigcup_{k=n}^{\infty} E_k$ for  all $k\le n \in \mathbb{N}$

If $x \notin A$ then  $x \in \bigcup_{k=r}^{n-1} E_k$ $ \implies   x$ is contained  in finitely many $E_k  $for example take $\{E_r,E_{r+1},...,E_{n-1}\}$

after that  i don't know how to proceed further

答:
If $x \in \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} E_k$ then for every $n$ there exist at least one $k \geq n$ such that $x \in E_k$. This implies that $x \in E_k$ for infinitely many values of $k$.

Conversely, if $x \notin \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} E_k$ then there exists $n$ such that $x \notin E_k$ for any $k \geq n$, so $x \in E_k$ for at most $n-1$ values of $k$.

Finally, $\mu (A) \leq \mu  \bigcup_{k=n}^{\infty} E_k)$ for every $n$ and this gives $\mu (A) \leq \sum\limits_{k=n}^{\infty} \mu (E_k)$. But convergence of the series $\sum\mu(E_k)$ implies that $\sum\limits_{k=n}^{\infty} \mu (E_k) \to 0$ as $ n\to \infty$. Hence, $\mu (A)=0$.


问:
Prove that the probability there exists at least 10 triangles in the random graph is greater in RG(p) than RG(q) if p > q
I came across this question and I don't know where to begin:

Assume you have a random 100-vertex graph and define the probability of a graph G to be $p^{E(G)}(1-p)^{100 choose 2 - E(G)}$ where $E(G)$ is the number of edges and p is the parameter (the probability a given edge exists). Prove that the probability there exists at least 10 triangles in the random graph is greater in RG(p) then RG(q) if p > q

Does anyone have any suggestions on how to start?

答:
This is a really good beginner question, and goes through one of the foremost notions in stochastic theory : that of stochastic domination. It avoids any large computation and provides a natural notion which is easy to prove via smart coupling.

$\underline{\mathit{Stochastic \ domination}}$

Definition : Suppose that $X,Y$ are real valued random variables. Then $X$ is said to (first order) stochastically dominate $Y$ if $P(X \geq t) \geq P(Y \geq t)$ for every real $t$.

This is not to be confused with $X \geq Y$ almost surely. It just means that for any $t$, it is likelier that $X$ is larger than $t$ than $Y$ is larger than $t$.

Now, if $G$ is picked as specified in the question, then one can talk about $T(p)$, the number of triangles present in $RG(p)$.

The claim is that $T(p)$ stochastically dominates $T(q)$. If this is true, then $P(T(p) \geq 10) \geq P(T(q) \geq 10)$, as desired.

$\underline{\mathit{A \ theorem\ for\ proving\ stochastic\ domination}}$

The most common method of proving stochastic domination is to create a coupling between the random variables in question. That is, provide a method of constructing random samples from both distributions, even if they may be dependent on each other.

To make this rigorous, we have the notion of coupling : given random variables $X$ and $Y$, we call a random variable $M = (M_1,M_2)$ a coupling of $X$ and $Y$ if the first marginal of $M$ i.e. $M_1$ has the same distribution as $X$, and $M_2$ has the same distribution as $Y$.

To give an example : take $X$ and $Y$ to be independent $Ber(\frac 12)$ random variables. We can couple them independently  : let $M$ be the tuple consisting of two independent coin tosses, then the first and second marginal are $X$ and $Y$. But there's another coupling too : flip the first coin , call it $M_1$, and then set $M_2 = M_1$. Then both $M_1$ and $M_2$ are $Ber(\frac 12)$ but clearly dependent on each other. So couplings can occur in many ways.

A popular theorem in this regard is the following :

Let $X,Y$ be random variables. Suppose that we can find a coupling $(M_1,M_2)$ of $X$ and $Y$ such that $M_1 \geq M_2$ almost surely. Then , $X$ stochastically dominates $Y$.

I will leave the proof as an exercise for you to look up or to try yourself, it is pretty much from definition.

$\underline{\mathit{Coupling \ RG(p)\  and \ RG(q)}}$

So how do we couple these random graphs? We go edge-by-edge, a very popular coupling for graphs which are constructed edge-by-edge (so called Erdos-Renyi graphs). Each edge is present or not with probability $p$ for $RG(p)$ and $q$ for $RG(q)$.

Here's our coupling. Do the following for every pair of vertices in the graph :

Sample a $Z_p = Ber(p)$ random variable.

For a certain $r$, sample a $Z_r = Ber(r)$ random variable, independent of $Z_p$, such that $Z_q = Z_pZ_r$ has the distribution $Ber(q)$. I leave you to calculate $r$ in terms of $p$ and $q$. See that $r > 0$ if $p>q$.

Now, for $RG(q)$, set an edge between these vertices if $Z_q = 1$, and for $RG(p)$ set an edge between these vertices if $Z_p = 1$.

That's your coupling! Indeed, for $RG(q)$, each edge is in or not with probability $q$, and similarly for $RG(p)$.

$\underline{\mathit{Finishing \ the \ problem}}$

The nice thing about this construction is that if $Z_q = 1$ then $Z_p = 1$! Therefore , if there's an edge in a random graph of $RG(q)$ from the above construction, then there's an edge in the coupled random graph produced by $RG(p)$ as well!

In particular, $RG(q)$ is a subgraph of $RG(p)$ with probability $1$ under this construction, which of course implies that a triangle in $RG(q)$ remains a triangle in $RG(p)$. Thus, $T(p) \geq T(q)$ almost surely, and we may conclude from the theorem.

$\mathit{Corollaries \ and\ Conclusion}$

Note that the above construction doesn't just restrict itself to triangles : the occurrence of any kind of subgraph , be it a cycle of a particular length or a clique, has increased chance if the parameter increases. Therefore, we get a vast generalization of the statement you make, with little or no effort. Especially without having to go into algebraic expressions and compare them, which is cumbersome and tiring.

There are many uses of stochastic domination in graph theory, especially the edge-to-edge coupling. Stochastic domination is also used in stochastic optimal control theory to regulate random variables so that they behave in a streamlined fashion, allowing the prediction of monotone optimal policies. Last but not the least, stochastic domination is very important in percolation theory and the theory of random walks, since domination by a random variable with certain tail bounds implies that the same tail bounds carry over to the dominated random variable.


问:
Find an interval of convergence and an explicit formula for $f(x)$
Let $f(x) = 1 + 2x + x^2 + 2x^3 +x^4+...$

If $c_{2n} = 1$ and $c_{2n+1} = 2$ $\forall n \ge 0$ find the interval of convergence and an explicit formula for $f(x)$.

The answers are $I = (-1,1)$ and $f(x) = \frac{1 + 2x}{1 - x^2}$

Can anyone please give me an idea how to get it...

Thanks in advance

答:
Hint: Use the geometric series

$$ \sum_{m=0}^{\infty}z^m=\frac{1}{1-z}. $$

Added: Following my comment, first, we need to prove that the series converges absolutely. Now, notice this

$$ 1 + 2|x| + |x^2| + 2|x^3| +|x^4|+...\leq 2 + 2|x| + 2|x^2| + 2|x^3| +2|x^4|+... $$

$$ = 2\sum_{k=0}^{\infty} |x^k| = 2\sum_{k=0}^{\infty} |x|^k = \frac{1}{1-|x|},\quad |x|<1. $$

Thus the series converges absolutely. Now, you ca rearrange the series as

$$ f(x)= \sum_{k=0}^{\infty}x^{2k}+ 2\sum_{k=0}^{\infty}x^{2k+1} .$$

Use the hint and you should be able to finish the problem.

ContentLicense: CC BY-SA 3.0


问:
Can composition of integer polynomial and rational polynomial with a non-integer coefficient result in integer polynomial?
Can we find two polynomials $p(x)$ and $q(x)$, where $p(x)$ is a non-constant monic polynomial over integers and $q(x)$ is a monic polynomial over rationals with at least one non-integer coefficient, such that their composition $p(q(x))$ is a polynomial over integers? If not, how to prove it?

For example let $q(x)=x^2+\frac{1}{2}x+1$ and $p(x)=x^3+a_2x^2+a_1x+a_0$, then $p(q(x))=x^6+\frac{3}{2}x^5+\dots$, so no matter what integers $a_i$ we choose, the resulting polynomial will have a non-integer coefficient. The monic condition is important, since otherwise we could multiply $p(x)$ with such integer that would guarantee all coefficients to be integers. I've tried to look at the coefficient in composition for general polynomials, which I believe should follow this formula:

\begin{align}
[x^r]p(q(x))=\sum_{k_1+2k_2+\dots+mk_m=r}\sum_{k_0=0}^{n-(k_1+\dots+k_m)}\binom{k_0+k_1+\dots+k_m}{k_0,k_1,\dots,k_m}a_{k_0+k_1+\dots+k_m}\left(\prod_{j=0}^{m}b_j^{k_j}\right)
\end{align}
(here $a_i$ and $b_i$ are the coefficients of $p(x)$ and $q(x)$ with degrees $n$ and $m$, respectively).
However it is not at all clear on which coefficient to focus to prove it will give the non-integer number.

This arose when trying to solve the Infinitely many solutions leads to existence of a polynomial, but it seems interesting enough by itself.

答:
For the non-monic case:

Consider the monic polynomial $p(y) - p(q(x)) \in \mathbb Z[x][y]$. It has a root $q(x) \in \operatorname{Frac}(\mathbb Z[x])$. By the rational root theorem applied to the UFD $\mathbb Z[x]$, it follows that $q(x) \in \mathbb Z[x]$.


问:
Is this the normal derivative of a measure? What does it mean?
I'm reading Filippo Santambrogio's Optimal transport for Applied Mathematicians and I've come across the following (on page 123). The context is that $\mathbf{v}_t$ is a vector-valued velocity function and $\varrho_t$ is a density measure.

We say that a family of pairs of measures/vector fields $(\varrho_t, \mathbf{v}_t)$ with $\mathbf{v}_t \in L^1(\varrho_t; \mathbb{R}^d)$  and $\int_0^T ||\mathbf{v}_t||_{L^1(\varrho_t)} dt < +\infty$  solves the continuity equation
$$ \partial_t \varrho_t \, + \, \nabla \cdot (\varrho_t\mathbf{v}_t) = 0 $$
on $(0,T)$ in the distributional sense if, for any bounded and Lipschitz test function $\phi \in C_c^1((0,T)\times \overline{\Omega})$, we have
$$ \int_0^T \int_\Omega (\partial_t\phi) d\varrho_tdt + \int_0^T \int_\Omega \nabla \phi \cdot \mathbf{v}_td\varrho_t dt = 0$$
This formulation includes no-flux boundary conditions on $\delta \Omega$ for $\mathbf{v}_t$, i.e. $\varrho_t \mathbf{v}_t \cdot \mathbf{n} = 0$

My question is this: what does the boundary condition at the end ($\varrho_t \mathbf{v}_t \cdot \mathbf{n} = 0$) mean? It seems like we are multiplying a measure by a two vector fields. Does this give us a measure? Do we find the dot product of the vector fields first?

答:
The zero scalar product $\mathbf{v}_t \cdot \mathbf{n} = 0$ would mean the velocity has no component normal to the boundary, so simply any flow at the boundary is precisely along the boundary, not across it.

With the density factor added, $\varrho_t \mathbf{v}_t \cdot \mathbf{n} = 0$ means the effective transport of mass accross the boundary is zero. That means you can have a non-zero cross-border velocity at some point or area, but the density there has to be zero. Generally, even iv $\mathbf{v}_t$ is not parallel to the boundary the mass flux across the border is zero.


问:
Let $z_n=(x_n,y_n)$ be a sequence in $\mathbb R^2$. Show that $\lim z_n=z=(x,y)$ if, and only if, $\lim x_n=x$ and $\lim y_n=y$
Pretty much what it says on the tin. I got a very initial idea for the reverse, but got stuck. Any help? No clue about the forward implication.

Reverse:

Since we know that $\lim x_n=x$ we know that there exists $N_x$ such that when $n>N_x$, then $d(x_n,x)<\frac{\varepsilon}{2}$.

Likewise, since we know that $\lim x_n=y$ we know that there exists $N_y$ such that when $n>N_y$, then $d(y_n,y)<\frac{\varepsilon}{2}$.

Set $N=max\{N_x,N_y\}$.

Then when $n>N$, by the Triangle Inequality:

$d(x_n,y_n)<d(x_n,x)+d(x,y)+d(y,y_n)$

Is this anything?

答:
$$d(z,z_{n}) = \sqrt{d(x,x_{n})^{2} + d(y,y_{n})^{2}} $$

So we get the inequalities :

$$d(x,x_{n})\leq d(z,z_{n})$$

$$d(y,y_{n})\leq d(z,z_{n})$$

So if $d(z,z_{n})\to 0$ then $d(y,y_{n}) \to 0$ and $d(x,x_{n}) \to 0$.

Also taking the max under the radical:

$$d(z,z_{n}) \leq \sqrt{2} max(d(x,x_{n}),d(y,y_{n})) $$

So if $d(y,y_{n}) \to 0$ and $d(x,x_{n}) \to 0$ then $max(d(x,x_{n}),d(y,y_{n})) \to 0$, so $d(z,z_{n})\to 0$.


问:
$E[g(X,Y)|Y=y] = E[g(X,y)|Y=y]$ - Exercise $3.77$ Ross Probability Models.
How would you show that $E[g(X,Y)|Y=y] = E[g(X,y)|Y=y]$ in the continuous case? This problem is in Probability Models by Ross. No solution is provided, and no information is given about the random variables, altough I suppose we could assume the functions involved are absolute convergent or something.

The penultimate step in the discrete case is that all $p(X=i, Y=j|Y=y)$ are $0$ unless $j = y$, which I can't translate into the continuous case. My real analysis skills are non-existent, but you aren't supposed to know that for this book either way.

答:
I tried to prove it in the simplified case g(X,Y) = XY. (Part a in Ross' book)

Let Z = XY. Then $$E[Z|Y=y] = \int z \cdot f_{Z|Y}(z|y)dz$$ $$= y \int x \cdot f_{Z|Y}(z|y)dx$$ through change of variable. And $$ f_{Z|Y}(z|y) = f_{XY|Y}(xy|y)$$ $$ = \frac{f(XY=xy, Y=y)}{f(Y=y)}$$ $$=\frac{f(X=x, Y=y)}{f(Y=y)}$$
which gives the desired conclusion in this case.

Is this any more rigorous than the comment made by Peter Morfe in the comment above? To convince ourselves about the last step, I suppose we could look at the cumulative distribution functions and take derivatives.


问:
the splitting field of $x^3 - x + 1$ over $\mathbb{F}_3$
I am trying to figure out the splitting field of $x^3 - x + 1$ over $\mathbb{F}_3$ .

I know $x^3 - x + 1$ over $\mathbb{F}_3$ is irreducible because there are no root in  $\mathbb{F}_3$.

The spitting field will be the field $\mathbb{F}_3[x]/\langle x^3 -x +1\rangle$（my intuition is that all roots are added to $\mathbb{F}_3$.）
So,the spitting field is isomorphic to $\mathbb{F}_3^3$.

My question is ;

How to formally prove that the spitting field is the field
$\mathbb{F}_3[x]/\langle x^3 -x +1\rangle$

Thank you for your help.

答:
Fix an algebraic closure an take $\omega$ a root of $f = X^3-X+1$ there. Since the Frobenius map is an automorphism, you can check that $\omega + i$ with $i \in \Bbb F_3$ are also roots of $f$.

Hence the splitting field of $f$ is $\Bbb F_3 (\omega)$, of degree $\deg f = 3$ over $\Bbb F_3$, and thus by uniqueness we get $\Bbb F_3(\omega) = \Bbb F_{3^3}$. Now, consider the ring map

$$
\Bbb F_3[X] \to \Bbb F_3(\omega)
$$

sending $X$ to $\omega$.

The kernel of the map is $(f)$. There are a couple of ways to see this but one is to note that $(f) \subset \ker \mathbf{ev}_\omega$ is a maximal ideal, and the kernel a proper submodule since since $X \mapsto \omega \neq 0$. Its
image is $\Bbb F_3[\omega]$, the smallest $\Bbb F_3$-algebra containing $\omega$, hence $\Bbb F_3[X]/(f) \simeq \Bbb F_3[\omega]$.

It is a general fact that if $\alpha$ is algebraic over $k$, then $k(\alpha) = k[\alpha]$. Namely, by the same argument as before one gets $k(\alpha) \supset k[\alpha] = k[X]/(m(\alpha,k))$, which is a field by the irreducibility of the minimal polynomial. Since this field contains $\alpha$, by definition of $k(\alpha)$ we have $k(\alpha) \subset k[\alpha]$, hence the equality.

Also, there's nothing special about $3$ here, it may very well be substituted for any prime $p$ to obtain that the splitting field of $X^p-X+1$ is $\Bbb F_{p^p}$.


问:
Show that differential equation has integrating factor $\eta (x, y) = \nu (x^2 + y^2)$ and find it
I have differential equation $$x + x^4 + 2x^2 y^2 + y^4 + yy' = 0$$

The problem is to find integrating factor $\eta (x, y) = \nu (x^2 + y^2)$ for such big equation :(

Does anybody have any ideas?

答:
$$\frac{d}{dx}(x^2 + y^2) = 2x + 2yy'$$

Let $v = x^2 + y^2$

The above equation becomes

$$v^2 + \frac{1}{2}\frac{dv}{dx} = 0$$


问:
Will $y=x$ and $y=\sin^n \left(\frac{\pi x}{2} \right)$ meet in the interval $(0,1)$ for all $n \ge 2$?
I am currently learning the basics of real analysis, and still haven't learned the concepts and properties of pointwise convergence and uniform convergence of functions.

Still, I was thinking of a function $f_n : (0,1) \to \mathbb{R}$,
$$ f_n(x) = \sin^n \left(\frac{\pi x}{2} \right) $$
and if we took the $n$ to infinity, it would give
$$ \lim_{n \to \infty} f_n(x) = 0 $$

From that, I thought that there might exist some $f_n$ that $y=f_n(x)$ does not meet $y=x$.

I thought it this way since for every $x \in (0,1)$, $\lim_{n \to \infty}f_n (x) = 0$, I could use $\epsilon - N$. To be specific, for given $x \in (0,1)$, I have taken $\epsilon = x$, then exists a natural number $N$ such that the following holds:
$$ n \ge N \Rightarrow |f_n(x)| < x $$

I thought that this means that for that point $x$, $(x,f_n(x))$ is below $(x,x)$, and hence will not meet $y=x$.

Since I can take such $N$ for every point in $(0,1)$, I thought that $y=f_n(x)$ and $y=x$ will not meet for some large $N$.

But my instructor told me it will always meet since the function is pointwise convergent, but not uniformly convergent. I do not have enough knowledge of these concepts, so I am asking help for them.

My own guess to the failure of my conjecture(?) is that taking the large $N$ is impossible.
To take $N$, it has to be bigger or equal to all the other $N$ taken from each $x \in (0,1)$. In other words,
$$ N := \sup\{N_x : x \in (0,1)\}$$
where $N_x$ is the $N$ for each $x$.
I thought that the set is not upper-bounded hence unable to take $N$.

Thank you in advance for your considerate help.

答:
When $n \geq 2$ increases, the root moves to a solution which is closer and closer to $x=1^-$ (ignoring the trivial solutions $x=0$ and $x=1$).

To see what happens, using Taylor and the binomial expansion, you face
$$\frac{x-\sin ^n\left(\frac{\pi  x}{2}\right)}{x-1}=1+\frac{\pi ^2 n}{8}  (x-1)+\left(\frac{\pi ^4 n}{192}-\frac{\pi ^4
   n^2}{128}\right) (x-1)^3+O\left((x-1)^5\right)$$

So, a first approximation is just
$$x_{(1)}=1-\frac{8}{\pi ^2 n}$$ For axample for $n=10$, this would give $x\sim 0.918943$ while the exact solution (given by Newton method) would be $0.915564$.

We could improve using Householder method, the first iterate of which being
$$x=1-\frac{24 n}{3 n \left(\pi ^2 n-4\right)+8}$$ which, for $n=10$ gives $0.915756$ which is much better.

The next iterative method (unnamed) would give
$$x=1-\frac{8 \left(3 n \left(\pi ^2 n-4\right)+8\right)}{\pi ^2 n \left(3 \pi ^2 n^2-24
   n+16\right)}$$ which, for $n=10$ gives $0.915626$ which is still better.

Another thing which coudl be done is to continue the Taylor expansion and use series reversion to get, as approximation
$$x=1+t+\frac 1 \pi \sum_{k=1}^p {a_k} (\pi t)^{2k+1}\qquad \text{with}\qquad t=-\frac{8}{\pi ^2 n}$$

the first $a_k$ terms being
$$\left\{\frac{3 n-2}{48} ,\frac{105 n^2-120 n+28}{11520},\frac{6615 n^3-10500
   n^2+4788 n-592}{3870720}\right\}$$


Title: Show that differential equation has integrating factor $\eta (x, y) = \nu (x^2 + y^2)$ and find it

Body:
I have differential equation $$x + x^4 + 2x^2 y^2 + y^4 + yy' = 0$$

The problem is to find integrating factor $\eta (x, y) = \nu (x^2 + y^2)$ for such big equation :(

Does anybody have any ideas?

Body:
$$x + \color {red}{x^4 + 2x^2 y^2 + y^4} + yy' = 0$$
$$ \color{red}{(x^2+ y^2 )^2 }+x+ yy' = 0$$
$$ (x^2+ y^2 )^2 +\dfrac 12(x^2)'+ \dfrac 12 (y^2)' = 0$$
$$ (x^2+ y^2 )^2 +\dfrac 12(x^2 +y^2)' = 0$$
$$1=- \dfrac 12 \dfrac {(x^2 +y^2)'}{(x^2+ y^2 )^2 }$$
$$ \left(\dfrac {1}{x^2+ y^2  }\right)'=2$$
Integrate.


问:
$\sum_{k=1}^n m(k)$, where $m(k)$ is defined by $2^{m(k)} || k$.
I'm looking at the sum:
$$f(n) = \sum_{k=1}^n m(k),$$
where $m(k)$ is defined by $2^{m(k)} || k$, i.e. $2^{m(k)}$ is the largest power of $2$ that divides $k$. For example, we have $f(8) = 0+1+0+2+0+1+0+3 = 7$. Here's a table of $n$, $m(n)$, and $f(n)$ for a few small $n$:

```<code>n     m(n)  f(n)
1     0     0
2     1     1
3     0     1
4     2     3
5     0     3
6     1     4
7     0     4
8     3     7
</code>```

The values $m(n)$ in the table are the last $m(k)$-value for the given $n$, i.e. the value of $m(k)$ when $k=n$. I haven't spent much time trying to figure this out, mostly because it's a funny little problem and I thought I'd share it. It seems to me that we have $f(2^j) = 2^j-1 \forall j$ and $f(n)\sim n$.

答:
There is a name for your "m" function, it is called the 2-adic valuation, and is usually denoted by $\nu_2$.

It is easy to show that

$$f(2^j)=2^{j}-1$$

Indeed, for $1 \leq k \leq j-1$ there are exactly $2^{j-k-1}$ numbers between $1$ and $2^j$ with $\nu_2(n)=k$, namely

$$\{ 2^k\cdot 1, 2^k \cdot 3, 2^k \cdot 5,..., 2^k\cdot (2^{j-k}-1) \}$$

When $k=n$ there is exactly $1$ number with $\nu_2(n)=k$.

Then, by rearranging the terms of your sum into groups of equal terms you get

$$f(2^j)=j+\sum_{k=1}^{j-1} k 2^{j-k-1}$$

It is easy to show that this sum is exactly $2^j-1$.

Edit With the new fix, you need to add a 1 to every single term, thus the sum increases by exactly $2^j$.


问:
Cartesian product of two collections of sets?
If we have
$\mathcal{G} = \{\phi, \{0\}, \{1\}, \{1,2\} \},$
would $\mathcal{G} \times \mathcal{G}$ have the form $\{\phi, (\{0\}, \{0\}), (\{0\}, \{1\}), (\{0\}, \{0, 1\}), \dots \}$?

In that case, what would an example of $(w_1, w_2) \in B \in \mathcal{G} \times \mathcal{G}$ be?

My working:
Take for example $B = (\{0\}, \{0, 1\})$. This is not a set; so, I don't understand how one could take an element $(w_1, w_2) \in B$.

P.S.: I am taking a Probability Theory class now and even though I did a Measure Theory class last term, I am finding things very difficult. But I really want to get good at this stuff. Would there be any reading material you'd recommend? (I find the lectures hard to follow.)

EDIT:
I encountered this statement in the book Theory of Probability and Random Processes by Koralov. In Definition 2.1, he states:

A finite-dimenstional cylinder is a set of the form $$A = \{\omega: (\omega_{t_1}, \dots, \omega_{t_k}) \in B \},$$ where $t_1, \dots, t_k
\geq 1$, and $B \in \mathcal{G} \times \dots \times \mathcal{G}$ (k
times).

答:
$\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\G}{\mathcal{G}}
$It might be easier to process if you name every set with a letter, say...

$$A = \varnothing \quad B = \set{0} \quad C = \set{1} \quad D = \set{1,2}$$

Then $\G = \set{A,B,C,D}$ and $\G^2$ is given by

\begin{align*}
\G \times \G = \Big\{
&(A,A),(A,B),(A,C),(A,D),\\
&(B,A),(B,B),(B,C),(B,D),\\
&(C,A),(C,B),(C,C),(C,D),\\
&(D,A),(D,B),(D,C),(D,D)\Big\}
\end{align*}

You can then replace $A,B,C,D$ with their explicit definitions:

\begin{align*}
\G \times \G = \Big\{
&\Big(\varnothing,\varnothing\Big) \; , \;\Big(\varnothing,\set{0}\Big) \; , \;\Big(\varnothing,\set{1}\Big) \; , \;\Big(\varnothing,\set{1,2}\Big),\\
&\Big(\set{0},\varnothing\Big) \; , \;\Big(\set{0},\set{0}\Big) \; , \;\Big(\set{0},\set{1}\Big) \; , \;\Big(\set{0},\set{1,2}\Big),\\
&\Big(\set{1},\varnothing\Big) \; , \;\Big(\set{1},\set{0}\Big) \; , \;\Big(\set{1},\set{1}\Big) \; , \;\Big(\set{1},\set{1,2}\Big),\\
&\Big(\set{1,2},\varnothing\Big) \; , \;\Big(\set{1,2},\set{0}\Big) \; , \;\Big(\set{1,2},\set{1}\Big) \; , \;\Big(\set{1,2},\set{1,2}\Big)\Big\}
\end{align*}

More loosely,

$$\G \times \G = \Big\{ \text{all possible pairs } (X,Y) \text{ where } X,Y \in \G \Big\}$$

What you end up with, then, is ordered pairs of sets. Be careful with the ordered pair $(\varnothing,\varnothing)$; you seem to think that might be equal to $\varnothing$, but recall that ordered pairs have a formal definition:

$$(x,y) = \Big\{ \set{x},\set{x,y} \Big\}$$

You should see why, then, $(\varnothing,\varnothing) \ne \varnothing$.

Of course, $\G^2$ is still a set, so you can take elements from it, and those elements are ordered pairs. If we say $(w_1,w_2) \in \G \times \G$, then we are just saying $(w_1,w_2)$ is some ordered pair in $\G \times \G$ (and, moreover, that means $w_1 \in \G$ and $w_2 \in \G$ from definitions).

But can we take elements of a $B \in \G \times \G$ instead? Like you saw: can we take $(w_1,w_2) \in B$ for $B \in \G \times \G$? (To avoid confusion, this need not be the same $B$ as earlier.)

Note that $B \in \G \times \G$ may be characterized as set in terms of ordered pairs. It might be best to work with an example, say $B = (\set{1,2},\set{1,2})$. Then

$$B = \Big( \set{1,2},\set{1,2} \Big)
= \Big\{ \set{1,2} \; ,  \; \big\{ \set{1,2},\set{1,2} \big\} \Big\}$$

but

$$(w_1,w_2) = \Big\{ w_1, \set{w_1,w_2} \Big\}$$

If $(w_1,w_2) \in B$, then it means that $(w_1,w_2)$ is represented by the set, larger set in $B$.


答:
$(a,b)$ is a way to denote an element of the Cartesian product $A \times B$ where $a \in A$ and $b \in B$.

There are several ways is set theory to define the ordered pair. A usual one is

$$(a,b)= \{\{a\}, \{a,b \}\}.$$

Based on that you can derive $\mathcal G \times \mathcal G$.

Title: Cartesian product of two collections of sets?

Body:
If we have
$\mathcal{G} = \{\phi, \{0\}, \{1\}, \{1,2\} \},$
would $\mathcal{G} \times \mathcal{G}$ have the form $\{\phi, (\{0\}, \{0\}), (\{0\}, \{1\}), (\{0\}, \{0, 1\}), \dots \}$?

In that case, what would an example of $(w_1, w_2) \in B \in \mathcal{G} \times \mathcal{G}$ be?

My working:
Take for example $B = (\{0\}, \{0, 1\})$. This is not a set; so, I don't understand how one could take an element $(w_1, w_2) \in B$.

P.S.: I am taking a Probability Theory class now and even though I did a Measure Theory class last term, I am finding things very difficult. But I really want to get good at this stuff. Would there be any reading material you'd recommend? (I find the lectures hard to follow.)

EDIT:
I encountered this statement in the book Theory of Probability and Random Processes by Koralov. In Definition 2.1, he states:

A finite-dimenstional cylinder is a set of the form $$A = \{\omega: (\omega_{t_1}, \dots, \omega_{t_k}) \in B \},$$ where $t_1, \dots, t_k
\geq 1$, and $B \in \mathcal{G} \times \dots \times \mathcal{G}$ (k
times).


答:
One simplification of the sum is $f(n)=\sum\limits_{k=1}^\infty \left\lfloor\frac{n}{2^k}\right\rfloor$

(obviously, only terms for $k\leq \log_2(n)$ are going to be non-zero). This can be evaluated very easily on the computer, since each consecutive term corresponds to dividing the previous one by 2 and ignoring any remainder.

Title: $\sum_{k=1}^n m(k)$, where $m(k)$ is defined by $2^{m(k)} || k$.

Body:
I'm looking at the sum:
$$f(n) = \sum_{k=1}^n m(k),$$
where $m(k)$ is defined by $2^{m(k)} || k$, i.e. $2^{m(k)}$ is the largest power of $2$ that divides $k$. For example, we have $f(8) = 0+1+0+2+0+1+0+3 = 7$. Here's a table of $n$, $m(n)$, and $f(n)$ for a few small $n$:

```<code>n     m(n)  f(n)
1     0     0
2     1     1
3     0     1
4     2     3
5     0     3
6     1     4
7     0     4
8     3     7
</code>```

The values $m(n)$ in the table are the last $m(k)$-value for the given $n$, i.e. the value of $m(k)$ when $k=n$. I haven't spent much time trying to figure this out, mostly because it's a funny little problem and I thought I'd share it. It seems to me that we have $f(2^j) = 2^j-1 \forall j$ and $f(n)\sim n$.


答:
Almost.

$\emptyset$ wouldn't be in the cartesian product but $(\emptyset,\emptyset)$ will be.

Don't overthink it.  $A\times B$ is nothing more than the set of ordered pairs and an ordered pair is just a pair of an element from the first set and an element for the second.... in order.

So if $\color{blue}{\mathcal{G} = \{\phi, \{0\}, \{1\}, \{1,2\} \}}$ is a set with four elements and $\color{red}{\mathcal{G} = \{\phi, \{0\}, \{1\}, \{1,2\} \}}$ is the same set of four elementts then  $\color{blue}{\mathcal{G}}\times \color{red}{\mathcal{G}}$ will be a set of the sixteen pairs that can be made:

$\color{blue}{\mathcal{G}}\times \color{red}{\mathcal{G}}= \{(\color{blue}{\emptyset},\color{red}{\emptyset}),(\color{blue}{\emptyset},\color{red}{\{0\}}),(\color{blue}{\emptyset},\color{red}{\{1\}}),(\color{blue}{\emptyset},\color{red}{\{1,2\}}),$

$\{(\color{blue}{\{0\}}$$,\color{red}{\emptyset}),(\color{blue}{\{0\}},\color{red}{\{0\}}),(\color{blue}{\{0\}},\color{red}{\{1\}}),(\color{blue}{\{0\}},\color{red}{\{1,2\}}),$

$\{(\color{blue}{\{1\}}$$,\color{red}{\emptyset}),(\color{blue}{\{1\}},\color{red}{\{0\}}),(\color{blue}{\{1\}},\color{red}{\{1\}}),(\color{blue}{\{1\}},\color{red}{\{1,2\}}),$

$\{(\color{blue}{\{1,2\}}$$,\color{red}{\emptyset}),(\color{blue}{\{1,2\}},\color{red}{\{0\}}),(\color{blue}{\{1,2\}},\color{red}{\{1\}}),(\color{blue}{\{1,2\}},\color{red}{\{1,2\}})\}$

Don't overthink it.

......

In that case, what would an example of (w1,w2)∈B∈G×G be?

Oh.....

Now I see why you were overthinking it.   That statement makes no sense.

$B\in \mathcal G \times \mathcal G$ would mean that $B$ is an ordered pair $(J,K)$ where $J$ is one of the four sets and $K$ is one of the four sets: $\emptyset, \{0\}, \{1\}, \{1,2\}$.

Then, if I were to say $w \in B = (J,K)$ then that means $w$ is one of either $J$ or $K$.

So that makes no sense. Are you sure the statement wasn't:

$(w_1, w_2) = B \in  \mathcal G \times \mathcal G$

or maybe

$(w_1, w_2) \in B \subset  \mathcal G \times \mathcal G$

?


问:
Evaluating a limit at infinity
I was doing a physics problem when I had to compute an integral of the form $$I=\int\frac{dx}{\sqrt{a+x^2}} $$ for $a>0$, which is easily eavluated by making the substitution $x=\sqrt{a}\tan^{-1}(\theta)$ to yield $$I=\frac{1}{a}\sin\big(\tan^{-1}\big(\frac{u}{\sqrt{a}}\big)\big)+C=\frac{1}{a}\cdot\frac{x}{\sqrt{a+x^2}}+C$$However, I needed to evaluate $$\lim_{x\to\infty} \frac{x}{\sqrt{a+x^2}}-\lim_{x\to-\infty}\frac{x}{\sqrt{a+x^2}}$$ in the process. I manipulated the fraction by noticing that $$\frac{x}{\sqrt{a+x^2}}=\frac{1}{\sqrt{\frac{a}{x^2}+1}}$$ and as $x\to\infty$, it is clear that the limit goes to $1$. But the limit as $x\to-\infty$ is less clear. I tried to make the substitution $x\mapsto-y$ for $y>0$, which seemed to make more sense, as we obtain $$\lim_{x\to-\infty}\frac{x}{\sqrt{a+x^2}}=-\lim_{y\to\infty}\frac{y}{\sqrt{a+y^2}}=-1$$ but I am not sure how valid that method is. Is there a better way to evaluate the limit as $x\to-\infty$?

答: What you did is correct. One simple way to see this is to observe that as $x$ becomes arbitrarily large, the contribution of $a$ to the limit is negligible... In other words, this is the idea behind the comparison test. Anyway, the point is that you can pretend the $a$ is not there... Now with the understanding that $\sqrt{}$ means positive square root, you can see why what you did is correct...

Title: Evaluating a limit at infinity


问:
This might be a very stupid question but is there some inner product like function associated with matrix multiplied by a vector?
Consider the standard inner product $\langle \cdot, \cdot \rangle$ on $\mathbb{R}^n$

$$\langle w, v \rangle : = \sum_{i = 1}^n w_i v_i$$
which coincides with $w^Tv$.

So my question is, can $A^Tv$ also be defined as some sort of extended inner product function?

i.e., extend $\langle \cdot, \cdot \rangle: \mathbb{R}^{m \times n} \times \mathbb{R}^n \to \mathbb{R}^n$

This question was inspired from someone who asked me if $A^\top v$ was the result of some sort of an inner product and of course I said "no...". But I do wonder if you can extend inner product to cover cases like this.

答:
Let $A$ have columns $a_1,\ldots,a_n$.

Then $A^T v$ is the column vector with entries $\langle a_i^Tv\rangle$, $i=1,\ldots,n$.


问:
Is Bayes' Theorem really that interesting?
I have trouble understanding the massive importance that is afforded to Bayes' theorem in undergraduate courses in probability and popular science.

From the purely mathematical point of view, I think it would be uncontroversial to say that Bayes' theorem does not amount to a particularly sophisticated result. Indeed, the relation
$$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B\cap A)P(A)}{P(B)P(A)}=\frac{P(B|A)P(A)}{P(B)}$$
is a one line proof that follows from expanding both sides directly from the definition of conditional probability. Thus, I expect that what people find interesting about Bayes' theorem has to do with its practical applications or implications. However, even in those cases I find the typical examples being used as a justification of this to be a bit artificial.

To illustrate this, the classical application of Bayes' theorem usually goes something like this: Suppose that

1% of women have breast cancer;

80% of mammograms are positive when breast cancer is present; and

10% of mammograms are positive when breast cancer is not present.

If a woman has a positive mammogram, then what is the probability that she has breast cancer?

I understand that Bayes' theorem allows to compute the desired probability with the given information, and that this probability is counterintuitively low. However, I can't help but feel that the premise of this question is wholly artificial. The only reason why we need to use Bayes' theorem here is that the full information with which the other probabilities (i.e., 1% have cancer, 80% true positive, etc.) have been computed is not provided to us. If we have access to the sample data with which these probabilities were computed, then we can directly find
$$P(\text{cancer}|\text{positive test})=\frac{\text{number of women with cancer and positive test}}{\text{number of women with positive test}}.$$
In mathematical terms, if you know how to compute $P(B|A)$, $P(A)$, and $P(B)$, then this means that you know how to compute $P(A\cap B)$ and $P(B)$, in which case you already have your answer.

From the above arguments, it seems to me that Bayes' theorem is essentially only useful for the following reasons:

In an adversarial context, i.e., someone who has access to the data only tells you about $P(B|A)$ when $P(A|B)$ is actually the quantity that is relevant to your interests, hoping that you will get confused and will not notice.

An opportunity to dispel the confusion between $P(A|B)$ and $P(B|A)$ with concrete examples, and to explain that these are very different when the ratio between $P(A)$ and $P(B)$ deviates significantly from one.

Am I missing something big about the usefulness of Bayes' theorem? In light of point 2., especially, I don't understand why Bayes' theorem stands out so much compared to, say, the Borel-Kolmogorov paradox, or the "paradox" that $P[X=x]=0$ when $X$ is a continuous random variable, etc.

答:
Let me start by a memory. From my undergraduate days, 30 years ago, I vividly remember the time when Bayes was introduced. We had spent a lot of time and effort on sampling theory and how to know if things could be proved. And to me, at the time, it always ended up that we needed to have a sample size of x (my remembrance was that a sample size of 7 often was the minimum).

To me Bayes represented a totally different approach which to me was more in alignment with my view of reality. In sampling we looked at groups, with Bayes we started with individual things. So for me this was a very eye-opening addition to the field of probability praxis (and theory of course, but that came later for me). The book we had, written by Raiffa I believe, was about decisition theory. 30 years later I still remember the discussion about whether to do one more test drilling in the oil field.

So, just maybe, in your curriculum the importance placed on Bayes is there to show that statistics does have several different branches, not only sampling theory or how present graphs as correct as possible.


答:
Let $m(k)$ be the exponent of $2$ occurring in $k$, i.e. $2^{m(k)}\|k$.
One quickly deduces $$\tag1f(2n+1)=f(2n)=n+f(n)$$ as only the even numbers contribute and $2k$ contributes one more to $2n$ and $2n+1$ than $k$ does to $n$, i.e. $m(2k)=m(k)+1$.
Since $f(1)=m(1)=0$, we find $f(3)=f(2)=1$, $f(5)=f(4)=3$, and so on.
The sequence goes
$$ 0, 1, 1, 3, 3, 4, 4, 7, 7, 8, 8, 10, 10, 11, 11, 15, 15, 16, 16, 18, 18, 19, 19, 22, 22, 23, 23, 25, 25, 26, 26,\ldots$$
And according to OEIS, this equals $m(n!)$ and also $n$ minus the number of $1$s in the binary expansion of $n$. One easily verifies that these two discriptions indeed follow the recursion formula $(1)$.</p>


问:
A series convergence
What is the closed-form answer for this series(if there is):
$$f(z)=\sum_{n=1}^{\infty}\dfrac{z^n}{n^2(n+1)}$$ assuming $z$ is a complex number and $|z|<1$.

I have looked for some approximating fitting functions, but I think there must be an exact answer too!

答:
Since
$$\frac 1{n^2(n+1)}=\frac{1}{n^2}+\frac{1}{n+1}-\frac{1}{n}$$
this is the infinite series of
$$f(z)=\frac{z-1 }{z}\log (1-z)+\text{Li}_2(z)-1$$

Title: A series convergence


问:
A bag of scrabble tile contains 5 Es, 4 As, 3 Ns and 2 Bs.
I draw 4 tiles one by one from the bag as an ordered sample without replacement. Assume equally likely outcomes. Let C be the event that I get 2 Es, 1 A and 1 N. Compute P(c).

So I started by taking P(14,4) which gave me 24,024. Then, to account for repeated elements in the sample space (5 Es, 4 As, 3 Ns, 2 Bs), I figured I'd have to divide by $5!\cdot4!\cdot3!\cdot2!$. The problem is that this gave me a value less than 1. Where am I going wrong?

答:
This is a question on probability so here is how you should go about it -

Sample space: $5 E, 4 A, 3 N, 2 B$

Event C of $4$ tiles:  $2 E, 1 A, 1 N$

So $P(C) = \displaystyle \frac{{5 \choose 2}{4 \choose 1}{3 \choose 1}}{14 \choose 4} = \frac{120}{1001}$

If you consider the order in your calculation, you will get to the same answer. In that case in total, you have $14$ choices for the first, $13$ for the next and so on ($ ^{14}P_4 $). For favorable cases, multiply your selection by $4!$ for possible arrangements of them.

$P(C) = \displaystyle \frac{{5 \choose 2}{4 \choose 1}{3 \choose 1} \ 4!}{14 \times 13 \times 12 \times 11} = \frac{120}{1001}$


问:
Show that set of words in NP
Let  ∈ NP. How can I show that the set of all cyclic permutations of words from  also lies in NP

The issue I have with this task is that I cannot understand the correlation of cyclic permutation as a component of A to the NP prove itself

答:
There are two common (and equivalent) definitions of the NP class: one involving non-deterministic Turing Machines, and another one using certificates. I will work with the second one here.

I suppose we work on an alphabet $\Sigma$.
A language $L$ is in the class NP if there exists a deterministic TM $M$, working in polynomial time, and a polynomial $p$ such that

$\forall x \in \Sigma^\ast,\; x \in L \Leftrightarrow \exists y \in \Sigma^{p(|x|)},\; M$ accepts the input $(x, y)$

Said differently, there exists a certificate $y$ of polynomial size that "helps" the machine decide whether your word is in the language or not. For example, in the 3-COLOR problem (deciding whether or not a graph is 3-colourable), such a $y$ could be the encoding of a valid colouring.

In your case, you know that all the aforementioned objects exist ($M, p \dots$). You simply have to slightly modify them: to the certificate $y$, you add some information to tell which permutation of your input is in $A$ (for example, if we note by $\sigma$ the simple cyclic permutation moving each letter to the left, you just add $n$ at the end of the certificate if $\sigma^n(x) \in A$. Now, you modify the machine so that on input $(x, y|n)$, it first computes $\sigma^n(x)$ and then runs as before on this modified input. Everything thing is still polynomial, and so $\{\sigma^n(x) | n \in \mathbb N, x \in A\} \in $NP

Title: Show that set of words in NP


问:
How to integrate the function $\lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}e^{tx}$
How to integrate the following family of functions? I really do not have too many ideas. The context of this is to find moment generating functions but the context is not too important here.

$$I_n:=\int_{0}^{\infty}n\lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}e^{tx}\,dx$$

I wanted to try this inductively/iteratively but I got nowhere (by the way I don't have to do this iteratively).  I should also stress that $\lambda>t$ so that I think $I_n$ converges for each $n$. Could someone please given me  a hint or two? I know how to do $n=1$ btw.

Many thanks!

答:
If you are familiar with the exponential distribution, its trivial to notice that $\lambda e^{-\lambda x}$ is the PDF and $1 - e^{-\lambda x}$ is the CDF. Therefore,
a substitution $u = 1 - e^{-\lambda x}$ would simplify it.

$$ u = 1 - e^{-\lambda x}$$
$$du = \lambda e^{-\lambda x} dx \tag{1}$$
$$ 1 - u = e^{-\lambda x}$$
$$ \frac{-\ln(1 - u)}{\lambda} = x$$
$$ \frac{-t \hspace{0.05cm} \ln(1 - u)}{\lambda} = tx$$
$$ e^{\frac{-t \hspace{0.05cm} \ln(1 - u)}{\lambda}} = e^{tx}$$
$$ (1 - u)^{\frac{-t}{\lambda}} = e^{tx} \tag{2}$$
Plugging $(1)$ and $(2)$ into this expression,
$$ I_n = \int_0^{\infty} n (1 - e^{-\lambda x})^{n-1} e^{tx} \lambda e^{-\lambda x} dx$$
and changing the limits, we get,
$$ I_n = \int_0^{1} n u^{n-1} (1 - u)^{\frac{-t}{\lambda}} dt$$
This can be re-written in terms of the Beta function as follows-
$$ I_n = n \operatorname{B}(n, 1 - \frac{t}{\lambda})$$


问:
A be a normal operator and has all its eigenvalues collinear in the complex plane iff it is of the form aH + bI
A be a normal operator on a finite dimensional Hilbert space H of dimension n . and has all its eigenvalues collinear in the complex plane iff it is of the form aH + bI , where H is Hermitian and I is the identity and a,b are complex numbers.
$\mathbf Idea$: 'only if' part : the eigen values of the matrix A lies in a line . Then we can shift the line passing through the origin and then rotate it . Then we can get the eigen values of the new matrix lies in the real axis . so the new matrix is Hermitian .
'if' part: if A = aH + bI then A is normal and eigen values of H are $a_1,a_2, ... ,a_n$ where $a_i$ is real then if we can find eigen values of A then we will see that those are lies in a line .
My idea is this but I can not argue this mathematically . Can anybody help me to prove it formally .

答:
If A = aH + bI
Then A is normal .
Now the eigen values of H are $\alpha_1, ... , \alpha_n$ where $\alpha_i$ is real.
the eigen values of A are $a\alpha_1 + b ,...,a\alpha_n+b $
there exists real number $\lambda_1,... ,\lambda_{n-1}$ such that
take$\alpha_1$ = $\alpha$
the eigen values of A are $a\alpha+b , ... ,a\lambda_{n-1}\alpha+b$
the eigen values are colinier .


问:
Variance for area and volume?
A machine manufactures cubes with a side length which varies Uniformly over the interval $[0.2, 0.3]$ in millimeters.

For the following problems, make sure you use the correct units. (Assume the
sides of the base and the height are all the same.)

a. What is the expected side length?

b. What is the standard deviation of the side length?

c. What is the expected area of one of the square bases?

d. What is the standard deviation of one of the square bases?

e. What is the expected volume of one of the cubes?

f. What is the standard deviation of the volume of one of the cubes?

g. What is the expected cost for making 1 cube?

h. What is the variance in the cost for making 1 cube?

i. What is the expected cost for making 10 cubes?

j. What is the variance in the cost for making 10 cubes?

a and b are simple

$E(X) = \cfrac{0.2+0.3}{2} = .25$ mm

$Var(X) = \cfrac{(0.3-0.2)^2}{12} = 1/1200$ so $sd(X) = \sqrt{1/1200} = .0289$ mm

I think I understand how to do c)

$E(area) = .25^2 = .0625 mm^2$ since we assume the sides are all the same so just square E(X)

I don't know how to do d)

I think I know how to do e)

$E(volume) = .25^3 = 1/64  mm^3$ since we assume the sides are all the same so just cube E(X)

I don't know how to do f)

g) $E(.12\cdot Volume + .06) = .12E(Volume) + .06 = .12 \cdot 1/64 + .06 = .061875$ cents

h) I don't know variance of volume so I can't do h)

i) $10 \cdot E(.12\cdot Volume + .06) = 10 \cdot (.12E(Volume) + .06) = 10 \cdot (.12 \cdot 1/64 + .06) = .61875$ cents

j) I don't know variance of volume so I can't do j)

So what I'm asking is how to find the variance for the area and volume of these cubes? Any hints would be appreciated thank you! Also I would appreciate it if someone could tell me if my methodology and answers for everything else is correct!

Thank You!

答:
a. ok

b. ok

c. Wrong.

. What is the expected area of one of the square bases?

$$\mathbb{E}[X^2]=\mathbb{V}[X]+\mathbb{E}^2[X]\approx 0.0633 \text{   mm}^2$$

d. First derive the distribution of the area

$$Y=X^2$$

$$f_Y(y)=\frac{5}{\sqrt{y}}\cdot \mathbb{1}_{[0.04;0.09]}(y)$$

The area is expressed in $\text{mm}^2$ so also its st dev is. (I think you can calculate any indicator you want given the density: mean, variance, st dev etc etc)

e. Wrong. Same solution as c. You have to calculate $\mathbb{E}[X^3]$

f. Same procedure as d. First derive the distribution of $Z=X^3$ then calculate whatever you want. Alternatively you can calculate $V[X^3]$ with the distribution of $X$ as in the previous e. The best choice is up to you.

g. and following: in the text you posted there are no information about costs...

Title: Variance for area and volume?


答:
The parametric equation of a line in the complex plane is $at + b$ where $t \in \mathbb{R}$ and $a,b \in \mathbb{C}$ with $a \neq 0$.

In one direction, if $A = aH + bI$ where $H$ is Hermitian then $H$ has real eigenvalues $\lambda_1, \dots, \lambda_n$ (possibly including multiplicities) and so (verify!) the eigenvalues of $A$ are $a \lambda_1 + b, \dots, a \lambda_n + b$. If $a = 0$ then $A = bI$ so the only eigenvalue of $A$ is $b$ (of multiplicity $n$) and in particular all the eigenvalues of $A$ are collinear (and lie on infinitely many lines). If $a \neq 0$ then they all lie on the line $at + b$.

For the other direction, let $A$ be a normal operator whose eigenvalues lie on the line $at + b$ where $a \neq 0$. Denote the eigenvalues of $A$ by $a t_1 + b, \dots, a t_n + b$ where $t \in \mathbb{R}$ and set $H = \frac{1}{a}(A - bI)$. Then $H$ is normal with real eigenvalues $t_1,\dots,t_n$, hence Hermitian and so $A = aH + bI$.

Title: A be a normal operator and has all its eigenvalues collinear in the complex plane iff it is of the form aH + bI

Body:
A be a normal operator on a finite dimensional Hilbert space H of dimension n . and has all its eigenvalues collinear in the complex plane iff it is of the form aH + bI , where H is Hermitian and I is the identity and a,b are complex numbers.
$\mathbf Idea$: 'only if' part : the eigen values of the matrix A lies in a line . Then we can shift the line passing through the origin and then rotate it . Then we can get the eigen values of the new matrix lies in the real axis . so the new matrix is Hermitian .
'if' part: if A = aH + bI then A is normal and eigen values of H are $a_1,a_2, ... ,a_n$ where $a_i$ is real then if we can find eigen values of A then we will see that those are lies in a line .
My idea is this but I can not argue this mathematically . Can anybody help me to prove it formally .


问:
Let $f: \Bbb R^n\to\Bbb R$, the Hessian of $f$ is postive definite everywhere. Show that $\mathrm{grad} \, f: \Bbb R^n\to\Bbb R^n$ is bijective.
Let $f: \Bbb R^n\to\Bbb R$, the Hessian of $f$ is postive definite everywhere. Show that $\mathrm{gradient} \, f: \Bbb R^n\to\Bbb R^n$ is bijective.

Argue by contradtion, if $\mathrm{grad} \, f(x_1)=\mathrm{grad} \, f(x_2)$ for some $x_1\neq x_2$, then by Rolle's theorem, it is easy to derive a contradition. But how to prove the surjectivity of $\mathrm{grad} \, f=(f_1',\cdots,f_n')$?

答: It is not true, consider $f(x)=e^{x_1}+\dots + e^{x_n}$, then the Hessian of $f$ at $x$ is $diag(e^{x_1}, \dots, e^{x_n})$ and thus positive definite. However, the gradient $\nabla f(x)=(e^{x_1}, \dots, e^{x_n})$ is not surjective (e.g. the origin is not in the image).


答:
An alternative approach is to differentiate $$zf'(z)=\sum_{n=1}^\infty\frac{z^n}{n(n+1)}$$
$$(zf')'=\sum_{n=1}^\infty\frac{z^{n-1}}{n+1}=-\frac{\ln(1-z)+z}{z^2}$$
Hence integrate twice to get an integral formula for $f(z)$; of course the polylog function will still appear in integral form.

Title: A series convergence

Body:
What is the closed-form answer for this series(if there is):
$$f(z)=\sum_{n=1}^{\infty}\dfrac{z^n}{n^2(n+1)}$$ assuming $z$ is a complex number and $|z|<1$.

I have looked for some approximating fitting functions, but I think there must be an exact answer too!


问:
which of these fair gambles gives your the greatest chance of winning?
I came to know about a gambling game with following conditions:

a>
Betting option for 3/4 winning chance.

1> You will have to bet 1 of 4 numbers.

2> If the out come of game doesn't match with your bet, You WON 100 coins.

3> If the out come of game matched with your bet, You LOSS 300 coins.

b> Betting option for 1/4 winning chance.

1> you will have to bet 1 of 4 numbers.

2> If your bet matched out come of game, you got 3 times. i.e., if you bet 100 coins, you get 300 coins if you won.

3> In case it you loss, you pay to dealer 100 coins.

c> Betting option for 2/4 winning chance.

1> For same game, You will have to bet any two 2 out of 4 numbers.

2> If you won you get double.

In case a , your probability of winning is 3/4 and loss 1/4. Ratio of of money you get when win to money you pay when loss is 1:3.

Here is my questions:

1> I think in big picture, the player have more chance to win because odd of player winning is higher. How do dealer make money at all?

2> If player triple his stake when player loss the bet, next time player win the bet, he should got his money back. The probability of loosing 3 game consecutively is very low (0.25*0.25*0.25 = 0.015). So there is more chance for dealer to went bankrupt. Again How do dealer make money at all?

3> Do you think there is another way to win this game or increase probability of winning the game?

P.S. : This question is truly for academic purpose. I am not trying promoting gambling here.

Thanks in advance

答:
Try this simple simulation and see who wins at the end.

```<code>play = function(asset){
outcome = sample(c(1,2,3,4),1)
yourpick = sample(c(1,2,3,4),1)

if(outcome != yourpick)
asset = asset + 100

else
asset = asset - 300

return (asset)
}

asset = 5000
times = 1

while (times < 100 & asset > 0){
asset = play(asset)
print(asset)
times = times + 1
}
</code>```

It is generally true that the dealer won't make any money if the winning chance is greater than 50% (the dealer, however, can charge a fee for playing the game and make money that way).

You can play around with my code and test your strategy in part 2).

For 3), read a bit more about Gambler's Ruin.

OwnerDisplayName: someone_ smiley

Body:
Try this simple simulation and see who wins at the end.

play = function(asset){
outcome = sample(c(1,2,3,4),1)
yourpick = sample(c(1,2,3,4),1)

if(outcome != yourpick)
asset = asset + 100

else
asset = asset - 300

return (asset)
}

asset = 5000
times = 1

while (times < 100 & asset > 0){
asset = play(asset)
print(asset)
times = times + 1
}

It is generally true that the dealer won't make any money if the winning chance is greater than 50% (the dealer, however, can charge a fee for playing the game and make money that way).

You can play around with my code and test your strategy in part 2).

For 3), read a bit more about Gambler's Ruin.


问:
Solving $\cos(2\theta)=\cos(\theta)$
The question :

Solve Equation $\cos(2\theta)=\cos(\theta)$ for $0 \le \theta \le 2\pi$ in terms of $\pi$.

My solution: -

Using trigonometry identity , I quickly get
$$2\cos^2(\theta)-1=\cos(\theta)$$
replcae $$\cos(\theta)=a.$$
I'll get quadratic equation of
$$2a^2-a-1=0,$$
where $a$ is $$a = -\frac12 , 1.$$
The result I got is$$\theta=0,{2\over3}\pi,2\pi.$$
The scheme said I'm missing is $\frac43\pi$.

Where am I missing? How to achieve it?

Title: Solving $\cos(2\theta)=\cos(\theta)$

Body: $\cos (2\pi-\frac {2 \pi} 3)=\cos \frac {2 \pi} 3=-\frac 1 2$ and $2\pi-\frac {2\pi} 3=\frac {4\pi} 3$.


问:
Riemann hypothesis and prime distribution
What exactly does the Riemann hypothesis imply for the prime numbers?
Since the explicit formula is independent of the Riemann hypothesis, what would it actually mean for the primes if all the nontrivial zeros of the zeta function had real part 0.5? Is there some sort of a "simple" explanation for that?</p>

答:
The key link here is the explicit formula
$$
\psi(x)=x-\sum_{\rho} \frac{x^{\rho}}{\rho}-\frac{\zeta'(0)}{\zeta(0)}-\frac{1}{2}\log\Big(1-\frac{1}{x^2}\Big)\ ,
$$
where $\sum_{\rho}$ denotes summing over all the zeroes $\rho$ of $\zeta$ with $0<\text{Re}(\rho)<1$ and $$\psi(x)=\sum_{p^k\leq x,\ p\text{ prime}}\log p$$ is the second Chebyshev function. This is proved using complex analysis, contour integration and Rouche's theorem, which links the zeroes of a function $f$ to an integral involving $f'/f$.

Now, it can be shown that if RH is true, then the sum $\sum_{\rho}$ in the explicit formula can be controlled to keep it small (because $\text{Re}(\rho)=1/2$ for all $\rho$) so that we get
$$\psi(x)=x+O\big(\sqrt{x}(\log x)^2\big),$$
which in turn can be used to show that
$$\pi(x)=\text{Li}(x)+O\big(\sqrt{x}\log x\big).$$

You can try reading Ram Murty's Problems in Analytic Number Theory for more details.


问:
Area of domain of a complex variable function

Suppose that f is a one-to-one analytic function mapping the disc $|z|<1$ onto a bounded domain D. Show that the area of D is given by $$A(D)=\int \int_{|z|<1} {|f'(z)|}^2 dxdy$$

This is a practice problem from Fisher's complex variables chapter 3.5, which is on Riemann's mapping thm and Shwarz tranformation.

I don't see how i can apply either concept to this problem however--the double integral looks more like Green's thm to calculate domain areas but as a complex variable z. I am stuck on deriving the actual result (which seems somewhat intuitive); any help or guidance is appreciated.

答:
If we identify $f=u+iv$ with the corresponding mapping $(x, y) \mapsto (u(x, y), v(x,y))$ then
$$
 A(D) = \int_{x^2+y^2<1} | \det J_f(x, y)| \, dx dy
$$
where $J_f$ is the Jacobian matrix of $f$. Now
$$
 \det J_f(x, y) = \begin{vmatrix} u_x & u_y \\ v_x & v_y \end{vmatrix}
= \begin{vmatrix} u_x & -v_x\\ v_x & u_x \end{vmatrix} = u_x^2 + v_x^2
$$
using the Cauchy-Riemann equations. On the other hand,
$$
 |f'(z)|^2 = |u_x + iv_x|^2 = u_x^2 + v_x^2 \, .
$$


答: HINT: When $cos(\theta) = \frac{-1}{2}$ which means  $\theta = \pm\frac{2}{3} \pi + 2k\pi \ \ k \in \mathbb{Z}$

Title: Solving $\cos(2\theta)=\cos(\theta)$

Body:
The question :

Solve Equation $\cos(2\theta)=\cos(\theta)$ for $0 \le \theta \le 2\pi$ in terms of $\pi$.

My solution: -

Using trigonometry identity , I quickly get
$$2\cos^2(\theta)-1=\cos(\theta)$$
replcae $$\cos(\theta)=a.$$
I'll get quadratic equation of
$$2a^2-a-1=0,$$
where $a$ is $$a = -\frac12 , 1.$$
The result I got is$$\theta=0,{2\over3}\pi,2\pi.$$
The scheme said I'm missing is $\frac43\pi$.

Where am I missing? How to achieve it?


问:
Remove degree $4$ term from quintic?
If I have a quintic polynomial $$p(x) = x^5 + bx^4 + cx^3 + dx^2 + ex + f$$what variable substitution could I make that would allow me to remove the $bx^4$ term, producing a different polynomial in the form $$q(t) = gt^5 + ht^3 + jt^2 + kt + m$$ where $g$, $h$, $j$, $k$, and $m$ can be expressed in terms of $b$, $c$, $d$, $e$, and $f$?

答: Do $x=y-\frac b5$. Since $\left(y-\frac b5\right)^5=y^5-by^4+\cdots$ and $b\left(y-\frac b5\right)^4=by^4+\cdots$, when you expand everything no term with degree $4$ will remain.

Title: Remove degree $4$ term from quintic?


问:
Why the distance of any point on a ellipse is equal to an equation of a line?
I don't see the geometric reason of the following equation $$\sqrt{(x-c)^2+y^2}=a-\frac{c}{a}x$$ Why the distance of any point $$P(x,y)$$ on a ellipse is equal to an equation of a line?
Thank you.

答:
A conics is  defined as collection or locus of points $P(x,y)$ which move such that there distance from a fixed point $S(ae,0)$ always equals $e$ times, its perpendicular distance from a fixed line $x=a/e.$ This gives a connection/equation of $x,y$ which represents ellipse if $e<1$, hyperbola is $e>1$ and parabola if $e=1$.
$$\sqrt{(x-ae)^2+y^2}=(x-a/e) \implies \frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}=1$$
An equation of a curve is a relationship between $x,y$ which cannot be dissected as line or some thing. For example $x^2+y^2=2x+2y-1$ is equation of circle even if RHS appears to be  an equation of a line. Do not dissect and equation like you have thought.


答:
The first thing missing in your solution is the '$^2$' on the LHS of your equation, as @Gary has spotted out.

$$2cos(\theta)-1=cos(\theta)$$

It should be $$2\cos^2(\theta)-1=\cos(\theta).$$
Then you've let $a = \cos\theta$ and corrected factorized it to $(2a+1)(a-1) = 0$, so $a = -\dfrac12$ or $a = 1$.

Here's a more visual interpretation of @Semiclassical and @Christoph's comments.

Note that $\cos \theta$ is the $x$-coordinate of the green point on the unit circle.  The vertical line $x = -\dfrac12$ intersects the unit circle at two points (symmetric about the $x$-axis), giving solutions $\theta = \pi \pm \dfrac\pi3 = \dfrac{2\pi}3$ or $\dfrac{4\pi}3$.

Even though the vertical line $x = 1$ touches the unit circle at only one point $(1,0)$, as $\theta\in[0,2\pi]$, we have $\theta = 0$ or $2\pi$.

Image source: Khan Academy

Title: Solving $\cos(2\theta)=\cos(\theta)$


问:
Undergraduate Analysis Book
I am a first-year Math undergraduate and have been searching online for books in Analysis and have found two which seem to be good for an introduction.

Ross' Elementary Analysis:The Theory Of Calculus.

Abbott's Understanding Analysis.

I did a quick skim at the content and it seems that Abbott's Understanding Analysis has more content（breadth) but Ross' seems to go deeper?

I am not sure which book to choose (self-study) and any other recommendations/suggestions are welcome and highly appreciated!

答:
Real Analysis By S Kumaresam.
It is very comprehensive, but it covers only some topics.

Mathematical Analysis by Rudin.
However, this is a little bit tough.

Mathematical Analysis by Tom Apostal.
This one is best.

Introduction to Real Analysis Book by Robert G. Bartle
Very comprehensive book, good for beginners.

Title: Undergraduate Analysis Book


问:
How can a set be infinite and also compact?
I'm struggling a bit to understand some aspects of compactness in infinite sets.

We say that $[0,1]$ is compact by Heine-Borel, but that means that for all open covers of $[0,1], \exists$ a finite subcover of $S$. Does that mean that no infinite collection of open sets between $0$ and $1$ inclusive exists? Furthermore, if that's true does that mean there's a minimum non-empty size/interval for an open set? How is that possible?

What about the case of $$\mathscr{F} = \{ (\frac{1}{(n+1)}, \frac{1}{n})  \mid n \in \mathbb{N} \} \cup \{ (\frac{1}{n} - \frac{1}{n^3 + 1}, \frac{1}{n} + \frac{1}{n^3 + 1})  \mid n \in \mathbb{N} \}$$?

It seems to me that here that the only subcover that would exist for $[0,1]$ would be infinite.

答:
No, it does not mean that "no infinite collection of open sets between 0 and 1 inclusive exists". It means that for every (also infinite) cover (of open sets) that contains $[0,1]$, we can find a finite number of them (open sets) that contains $[0,1]$.

Important remark: the condition for compactness is that every open cover has a finite subcover, not just that a single given open cover. A proof that [0,1] is compact with respect to the euclidean topology can be found here.

I suggest that you try to understand the main idea of the proof: it contains all the ingredients to understand what you have to do when dealing with compactness.


问:
How does a linear rescaling alter a polytope equation?
Suppose a have a polytope $P$  defined by $Ax \leq b$. Suppose $x \in \mathcal{X} = \{ x \in \mathbb{R}^d \mid l_i \leq x_i \leq r_i, i = 1,\ldots, d\}$ i.e a hyperrectange in $\mathbb{R}^d$. How would $A$, $b$ change if I were to apply a linear rescaling to $x$ so that $x_i^{*} = \theta_0 x_i + \theta_1$?

In particular, I have a sampling scheme in mind for the design and analysis of a computer experiment - I want to compute a maximin design with boundaries defined by a polytope (see section 5.5 of this book). My parameter space is defined by a polytope $Ax \leq b$ but it is much more conveniet to construct the design with all the $x_i$ scaled to the unit interval (solves the problem of inputs 'dominating' each other. Hence, I want to construct my design points on $x^{*} \in [0,1]^d$ then simply rescale: $x = \text{diag}(r-l)x^{*} + l$ to generate inputs to a computer model .

In essence, if I linearly rescale my hyperrectangle to be a unit hypercube, how does this change the inequality which defines my polytope?

答:
After a bit of thought I'd realised that the solution to this is quite straightforward.

If $x = diag(r-l)x^{*} + l$ then I can simply substitute this into the inequality which defines my polytope.

$$Ax \leq b$$
$$\implies A \left(\text{diag}(r-l)x^{*} + l\right) \leq b  $$
$$\implies A \,\text{diag(r-l)}x^{*} + Al \leq b $$
$$\implies A \,\text{diag(r-l)}x^{*} \leq b - Al $$

Hence we can say $A^{*} = A \,\text{diag(r-l)}$ and $b^{*} = b - Al$ define an equivalent polytype on $[0,1]^d$ by $A^{*} x^{*} \leq b^{*}$


问:
Find an interval of convergence and an explicit formula for $f(x)$
Let $f(x) = 1 + 2x + x^2 + 2x^3 +x^4+...$

If $c_{2n} = 1$ and $c_{2n+1} = 2$ $\forall n \ge 0$ find the interval of convergence and an explicit formula for $f(x)$.

The answers are $I = (-1,1)$ and $f(x) = \frac{1 + 2x}{1 - x^2}$

Can anyone please give me an idea how to get it...

Thanks in advance

答:
The radius of convergence is $\dfrac1{\limsup\limits_{n\to\infty}|a_n|^{1/n}}$ and since $1\le a_n\le2$, the Squeeze Theorem says the radius of convergence is $1$.

$$
\begin{align}
&\hphantom{(}1+2x\hphantom{)}+\hphantom{(}x^2+2x^3\hphantom{)}+\hphantom{(}x^4+2x^5\hphantom{)}+\hphantom{(}x^6+2x^7\hphantom{)}+\dots\\[8pt]
=&(1+2x)+(1+2x)x^2+(1+2x)x^4+(1+2x)x^6+\dots\\[8pt]
=&(1+2x)(1+x^2+x^4+x^6+\dots)\\[4pt]
=&\frac{1+2x}{1-x^2}
\end{align}
$$

$$
\begin{align}
&1+2x+x^2+2x^3+x^4+2x^5+x^6+2x^7+\dots\\[12pt]
=&1+\hphantom{2}x+x^2+\hphantom{2}x^3+x^4+\hphantom{2}x^5+x^6+\hphantom{2}x^7+\dots\\
&\hphantom{1}+\hphantom{2}x\hphantom{\,+\;x^2}+\phantom{2}x^3\hphantom{\,+\;x^4}+\phantom{2}x^5\hphantom{\,+\;x^4}+\phantom{2}x^7\dots\\[4pt]
=&\frac1{1-x}+\frac{x}{1-x^2}\\[9pt]
=&\frac{1+2x}{1-x^2}
\end{align}
$$


问:
Can a function have a relative extreme at a removable singularity?
I was presented this free response problem set in AP Calculus BC today regarding Taylor series polynomials. The given function $g(x)=\frac{\cos(2x)-1}{x^2}$ contains a removable singularity at $x=0$, so it can be (and is according to the paper) defined for all $x\ne0$. After finding a 4th degree Taylor polynomial for $g(x)$, the following prompt is provided:

Determine whether $g$ has a relative minimum, a relative maximum, or neither at $x=0$. Justify your answer.

I claimed that $g$ cannot have a relative minimum or maximum at $x=0$ because $x=0$ is not in the domain of $g(x)$.

My teacher claimed that this is wrong because the restricted domain of $g(x)$ is irrelevant. He answered the prompt by utilizing the 4th degree Taylor polynomial for $g(x)$:
$$g(x)=-2+\frac{2^4x^2}{4!}-\frac{2^6x^4}{6!}$$
$$g'(x)=\frac{2^5x}{4!}-\frac{2^8x^3}{6!}$$
$$g'(0)=\frac{2^5(0)}{4!}-\frac{2^8(0)^3}{6!}=0$$
$$\text{Therefore, }g(x)\text{ has a relative extreme at }x=0$$
$$g''(x)=\frac{2^5}{4!}-\frac{3*2^8x^2}{6!}$$
$$g''(0)=\frac{2^5}{4!}-\frac{3*2^8(0)^2}{6!}=\frac{2^5}{4!}$$
$$g''(0)>0$$
$$\text{Therefore, }g(x)\text{ has a relative minimum at }x=0$$
I assume it is okay to use that Taylor polynomial to approximate $g(x)$ since the derivative of the Taylor polynomial should be the same sign as $g(x)$ at any degree given that the series is centered at the point being analyzed. I never learned that in class though, and we've basically just started learning about Taylor series, so maybe that observation is false. I imagine that my assumption at least works for the problem above. Ignoring the efficacy of the method, $g'(x)<0$ immediately before $x=0$, and $g'(x)>0$ immediately afterwards, satisfying the conclusion.

Despite all of this, I fail to see how the singularity at $x=0$ is irrelevant. If it means anything, a removable singularity is actually considered a discontinuity in AP Calculus. That said, the AP Calculus courses also claim that a function is non-differentiable where it is discontinuous, so I assume that it is impossible to find $g'(0)$ to determine the point's status as an extreme anyway since there would have to at least be a "discontinuity" there. Furthermore, the free response prompt explicitly states that $g(x)$ is undefined at $x=0$.

While writing this post, I became even more confused when I noticed the first sentence in the free response problem set. It says:

The function $g$ is continuous for all real numbers $x$ and is defined by $g(x)=\frac{\cos(2x)-1}{x^2}\text{ for }x\ne0.$

Now I am wondering how $g$ could possibly be continuous for all real numbers $x$ if there is a value of $x$ at which $g(x)$ is not defined.

Am I just reading way too deep into all of this and/or making errors in logic, or is my calculus teacher wrong? This post contains multiple questions, so I want to make it clear that the title is my main issue.

答:
A function cannot be extremal at a singularity, since by definition they are undefined at singularities, while the definition of an extremum requires us to give a value taken by the function at the extremum. Though also note that singularities are not discontinuities, for the very same reason: wether a function is discontinuous at a point or not depends on the value it takes there, and if it's undefined at a point, the definition of continuity makes no sense for that point.

But: singularities pose the interesting question wether a function $f$ can be continuously extended to the singularity. Can we define a new function $\tilde f$ which agrees with the old function wherever it is defined, and which is also defined at the singularity, and which is continuous there? If the question can be answered with "yes", we call the singularity removable. So for removable singularities it makes sense to ask wether $\tilde f$ is extremal at the singularity of $f$. That's the mathematical content of the question "Is $f$ extremal at its removable singularity?". It really asks about $\tilde f$, not about $f$. But since there is only one such $\tilde f$, we can kind of imagine that $\tilde f$ is what $f$ should have been from the beginning, since it's the only natural way to extend $f$. This often leads to people dropping the distinction entirely, which is what your teacher did.


问:
Why can't a neighbourhood contain the entire plane?
Context:

Let $S$ be the spherical surface of radius 1 with center at the origin of $R^{3}$, let $p=(0,0,1)$ be the north pole of $S,$ and let $f: S-p \rightarrow R^{2}$ be the stereographic projection from $p$ of $S-p$ onto the equatorial plane. Construct a diagram which shows that $f$ is continuous. Over what part of $S-p$ is $f$ a contracting function? Show that $f$ is $1-1,$ and that $f^{-1}: R^{2} \rightarrow S-p$ is also continuous. What is the image under $f$ of the deleted neighborhood $N(p, r, S)-p$ for $r$ -values less than $1 ?$ Why is it impossible to define $f p$ so that the extended function is continuous?

Solution given:

Contracts distances for any pair of points in the lower hemisphere. $f^{-1}$ sends $y \in R^{2}$ into the intersection of the ray py with S. See Fig. S3.
The same picture shows that the inverse is continuous. The image of $N(p, r, S)-p$ is the exterior of a circle with center at the origin. If $f p$ were defined and $f$ were continuous at $p$, then any $N(fp,\epsilon)$ would have to contain the entire exterior of some circle; but this is impossible.

My question:

Why is it impossible for a neighbourhood to contain the entire exterior or the entire plane, excluding the circle itself?

Is it because :

A neighbourhood cannot have infinite radius?

A neighbourhood is an open set, thus in the plane where it exists, there will always be points on the plane that is not in the neighbourhood, in this case any point $q$ such that $d(O,q)=\epsilon=\infty$, assuming there exists the center of this neighbourhood, $O$? (Though the authors have not introduced the concept of open sets)

A neighbourhood is an open disk, while the plane has no definite shape, thus it cannot extend over the whole plane?

Thanks in advance.

Source

Section 1.3, Exercise 3, Chinn, Steenrod: First Concepts of Topology

答:
There's a simple argument: $S^2$ is compact and so every continuous image of $S^2$ is compact. And thus $\mathbb{R}^2$ cannot be an image of $S^2$ because it is not compact. In particular no continuous function mapping $S^2\backslash\{p\}$ onto $\mathbb{R}^2$ can be extended to full $S^2$.

If $f p$ were defined and $f$ were continuous at $p$, then any $N(fp,\epsilon)$ would have to contain the entire exterior of some circle; but this is impossible.

Introducing interior and exterior of neighbourhoods is already not an easy task. Also it seems to strongly depend on the fact that $f p$ takes circles to circles. I think that the argument here is that there is a neighbourhood $U$ of $f(p)$ that is bounded (every point in $\mathbb{R}^2$ has such a neighbourhood), but in $S^2$ every neighbourhood of $p$ is mapped onto an unbounded subset of $\mathbb{R}^2$ via the stereographic projection.

The compactness argument is straightforward and easy.

I'll try to address your confusing questions now.

A neighbourhood cannot have infinite radius?

Depends on how you define "neighbourhood" and "radius". Are you asking whether every neighbourhood is bounded? No, it does not have to be. In fact every open subset of $S^2$ or $S^2\backslash\{p\}$ is bounded, unlike in $\mathbb{R}^2$.

A neighbourhood is an open set, thus in the plane where it exists, there will always be points on the plane that is not in the neighbourhood, in this case any point $q$ such that $d(O,q)=\epsilon=\infty$, assuming there exists the center of this neighbourhood?

I really don't understand what you are trying to say here. What is "the center of this neighbourhood"? What do you mean by $d(0,q)=\epsilon=\infty$? What do you mean by "$d(0,q)$"?

A neighbourhood is an open disk, while the plane has no definite shape, thus it cannot extend over the whole plane?

I don't understand this question as well. What do you mean by "extending a neighbourhood over the whole plane"? Typically we extend functions, not subsets. Also what do you mean by "no definite shape" or "shape" alone? That doesn't make much sense.

Mathematics is a science of precision. Most of confusion can be avoided if we focus on mastering definitions.


问:
Is the Lebesgue measure of Borel subsets "regular" w.r.t Jordan-measurable subsets?
Let $\lambda^d$ denote the $d$-dimensional Lebesgue measure. Is it true that for any Borel subset $A$, there exists a sequence of Jordan-measurable subsets $B_1, \dots$

$$\lim_{n \to \infty} \lambda(A \Delta B_n) = 0$$

My guess is that the answer to the above is "yes", as it holds for the examples I can think of, such as the rational numbers and the fat cantor set. If the answer to the above is "no", does it hold for open sets?

答:
Because $\lambda$ is outer regular, for every $\epsilon > 0$ there is an open $U$ such that
$$\lambda(A \triangle U) < \epsilon/2.$$
There is a sequence of $V_1,V_2,...$ of open balls such that $U = \bigcup_j V_j$. The boundaries of open balls have measure $0$. Because $\partial(A \cup B) \subset \partial (A) \cup \partial (B)$, the boundary of $W_n=\bigcup_{j=1}^n V_j$ has measure $0$ for every $n$. Then,

$$\lambda(A \triangle W_n) \leq  \lambda(A \triangle U) + \lambda(U \triangle W_n) < \epsilon$$

for large enough $n$ because $\lambda(U \triangle W_n) \to 0$.


答:
To take one of your questions.

Suppose we take as a cover of the closed unit interval the set containing every every open sub-interval.

Note that the endpoints have to be included in a cover so there will be sets of the form $[0, a)$ and $(b, 1]$ included as well as the whole set itself $[0,1]$. These are open in the topology we are considering.

This has a finite sub-cover by Heine Borel.

Note that any such sub-cover can be refined to a cover with smaller intervals. eg Take $(a, b)$ with $a \lt b$ (where it may involve $[0, .$ or $.,1]$ to the two intervals $(a, \frac {2a+3b}5)$ and $(\frac {3a+2b}5, b)$. The cover remains finite, but the length of the largest interval is reduced. We can make the cover as fine as we choose while it remains finite.

The ability to refine covers in this way becomes significant in the theory of measure/integration.

Title: How can a set be infinite and also compact?

Body:
I'm struggling a bit to understand some aspects of compactness in infinite sets.

We say that $[0,1]$ is compact by Heine-Borel, but that means that for all open covers of $[0,1], \exists$ a finite subcover of $S$. Does that mean that no infinite collection of open sets between $0$ and $1$ inclusive exists? Furthermore, if that's true does that mean there's a minimum non-empty size/interval for an open set? How is that possible?

What about the case of $$\mathscr{F} = \{ (\frac{1}{(n+1)}, \frac{1}{n})  \mid n \in \mathbb{N} \} \cup \{ (\frac{1}{n} - \frac{1}{n^3 + 1}, \frac{1}{n} + \frac{1}{n^3 + 1})  \mid n \in \mathbb{N} \}$$?

It seems to me that here that the only subcover that would exist for $[0,1]$ would be infinite.


问:
Orders of moves in the Rubik's cube
Playing with my Rubik's cube, I was thinking of facts about it that are immediate to mathematicians but novel to others. Here's one:

Given a Rubik's cube in the solved state, any sequence of moves, if repeated long enough, will eventually return the cube back to the solved state

To a mathematician, that follows from "The moves of a Rubik's cube form a finite group. Therefore all the elements have finite order."

Each atomic move of Rubik's cube has order 4. The simplest compound move is R U (rotate the right face 90° clockwise, then rotate the top face 90° clockwise). What's the order of R U?

Secondly, what's the greatest order of an element in the Rubik's cube group?

I tried repeating R U for a long while. I lost count (~50 repetitions), but it did eventually return home.

答:
For some time ago I wrote my bachelor thesis about the element of greatest order:

http://www.math.kth.se/~boij/kandexjobbVT11/Material/rubikscube.pdf

Actually, I came up with a new theorem for the generalized symmetric group that confines, as a special case, the orders inside the Rubik's cube and the result is quite interesting. The greatest order is 1260.

Title: Orders of moves in the Rubik's cube


问:
Cone problem - function of height vs time
in one of my courses I need to write a problem and I thought about one I just don't know how can I apply it so I will thank to anyone who can help.
In this problem you have a cone with the capacity of 100 milliliter. and a height of 15 cm. A man pours liquid in a constant rate of 5 milliliter per second (in 20 seconds the class will be full) find the height of the liquid as a function of time.

now, I'm sure it's not perfect, and I think there's missing data to find the function but I'm not sure what. Right now all I need to solve it is to find a cone radios witch is a part of a big cone that I know it's data. and a has a volume of 5t (the smaller cone).

答:
The key is to write a formula for the volume inside your cone with respect to height. You can start with the formula for the volume of a cone:

$$
V=\frac{\pi r^2 h}{3}
$$
But note here that the radius of your base is actually a linear function of $h$, and so you can write $r=ch$ for some constant $c$, and then we have
$$
V=\frac{\pi c^2h^3}{3}
$$
Now, you can solve for $c$ using the parameters you know (total volume and height of the cone).

You also know that the volume of liquid in the cone is $5t$, so you replace $V$ with $5t$, and rearrange to solve for $h$ in the formula.


问:
Unique bounded function satisfying $f(s)=1+\int_{0}^{s} e^{-t^{2}} f(s t) d t$
Prove that there is a unique bounded continuous real valued function $f:[0, \infty) \rightarrow \mathbb{R}$ such that
$$
f(s)=1+\int_{0}^{s} e^{-t^{2}} f(s t) d t
$$
for all $s \in[0, \infty)$

First as far as I searched $C[0,\infty)\cap B[0,\infty) $ is complete under sup metric.

So I tried to use Banach Fixed Point Theorem

similar to here https://math.stackexchange.com/a/1350747/342943

But I was confused about using the bound, which is not similar in the above question. $e^{-t^2}$ is going to overcome $f(t)$ for sufficiently bigger $n$.

Can you give me hint?

Or is there another approach that I can show above $f$ is unique and existent.

答:
The Banach fixed-point theorem can be used. As pointed out in the comments to On the solution of a Volterra type of integral equation, $T: C[0,\infty) \to C[0,\infty)$ defined by
$$
Tf(s)= 1+ \int_{0}^s e^{-t^2} f(st)dt
$$
maps bounded functions to bounded functions:
$$
|Tf(s)| \le 1 + \int_{0}^s e^{-t^2} \, dt \cdot \Vert f \Vert_\infty \\
\le 1 + \int_{0}^\infty e^{-t^2} \, dt \cdot \Vert f \Vert_\infty = 1 + \frac{\sqrt \pi}{2 } \Vert f \Vert_\infty
$$
and is a contraction:
$$
|Tf(s) - Tg(s)| \le  \int_0^s e^{-t^2} \, dt \cdot \Vert f-g \Vert_\infty \\
\le  \int_0^\infty e^{-t^2} \, dt \cdot \Vert f-g \Vert_\infty
= \frac{\sqrt \pi}{2 }\cdot \Vert f-g \Vert_\infty
$$
because $\frac{\sqrt \pi}{2 } \approx 0.886 < 1$.

For the value of $\int_0^\infty e^{-t^2} \, dt $ see for example Proving $\int_{0}^{\infty} \mathrm{e}^{-x^2} dx = \frac{\sqrt \pi}{2}$.


问:
Is it possible to calculate the input variable of a specific encryption function without brute-forcing it?
I didn't know where to post this question, here or SO but I feel it's more of a math problem.

Lately I have been reverse engineering a game, and figured out how it takes user input, encodes it, and verifies it against a specific value to see if the user input is a correct cheat code or not. The game has a unique id for every player and a specific key that it uses for this function. The algorithm looks like this:

```<code>userInput = ReadUserInput();
userInputEncrypted = userInput MOD encryptionKey;
for i = 0; i<16; i++{
    userInputEncrypted = userInputEncrypted^2 MOD encryptionKey;
}
userInputEncrypted = userInputEncrypted * userInput;
userInputEncrypted = userInputEncrypted MOD encryptionKey;

if (userInputEncrypted == correctValue)
    //Bingo!
</code>```

Now my problem is this:
I know what the correct values are, and I know the encryptionKey, is there a way mathematically to do this in reverse and instantly calculate what the starting userInput should be?

Or am I doomed to bruteforce it as the modulo that occurs 16 times gives me no other choice?

As you can see in the last step, the algorithm re-uses the starting userInput, so my hopes lie there.

Thanks in advance!

答:
It's essentially solving
$$x ^ {2^{16} + 1} \equiv a \pmod n$$
($x$ is the input, $a$ is the correct value, $n$ is the encryption key). It is called the  discrete root problem, and it can be found quite efficiently if the factorization of $\phi (n)$ is known, if not - this problem is probably hard.


问:
Shuffle a poker deck between 4 players, with least required entropy
We are shuffling a standard poker deck of 52 cards between 4 players, each getting 13 cards. The order of cards for a particular player does not matter.

A naive algorithm is to first shuffle the whole deck, then split it into 4 parts. However, this requires $log_2(52!) \approx 226$ bits of entropy.

Combinatorics says that such a shuffle should only require $log_2(52! / 13! / 13! / 13! / 13!) \approx 96$ bits of entropy, more than half less.

What is the algorithm that would produce such an efficient shuffle, using a random generator that produces integers in a range?

I was thinking of this algorithm:

Start with a sorted deck, for each card:

generate a number between 1 and 4, to determine which player the card goes to.

once one of the players gets 13 cards, only generate numbers between 1 and 3 (and so on)

Stop when only one player remains, give them all undealt cards

Would such algorithm produce an unbiased deal, and if not, what is the correct one?

答:
You can use a kind of binary search process to decide which player receive each card:

The cards can be in any order, including totally sorted.

Keep a tally of how many cards each player can still receive. Initially, each player can receive 13 cards apiece; these numbers will change as you deal out cards, and won't always be equal.

You can write out those tallies in a diagram like this:

□□□□ | □□□ | □□□□□ | □□

This is a diagram of a situation where Player 1 needs four more cards, Player 2 needs three more cards, Player 3 needs five more cards, Player 4 needs two more cards.

To assign the next card to a player, you could pick a random position in this diagram list; each position is owned by one of the players. To pick a position in this list, you could use binary search: Flip a coin to decide whether the position is in the first or second half of the list. Flip another coin to pick whether the position is in the first or second half of that sublist. And so on, until the range narrows to a single position. Then, assign the card to whichever player owns that position in the list.

However, because order within each partition doesn't matter, you can terminate the binary search early. You can terminate it as soon as you know for sure which player's partition the binary search will land in. The left and right endpoints of the sublist will both lie within a single player's partition. So you assign the card as soon as you know which player it belongs to.

After assigning the card, update the diagram list and repeat with the next card. Repeat until all cards are assigned.

By giving each player a number of chances equal to the number of cards remaining, you ensure that the partitions are uniformly random. By using binary search and terminating as early as possible, you guarantee that you use up the minimal amount of entropy.


问:
Unbounded linear operator on Hilbert spaces that is everywhere defined
Let $H$ be a Hilbert space.

I know a lot of examples of linear operator $T$ on $H$ such that $D(T) \subsetneq H$ and $T$ is not bounded, and such kind of operators are very important in analysis of PDEs.

However, I don't know any example of linear operator $T$ on $H$ such that $D(T)=H$
and $T$ is not bounded. Does anyone knows?

答: Let $(e_n)$ be an orthonormal basis for a Hilbert space $H$ or any linearly independent sequence of unit vectors. Extend this sequence to  a (Hamel) basis $(e_n) \cup (x_i)_{i \in I}$. Define$T(e_n)=ne_n$ and $T(x_i)=0$ for all $i$. This extends to an unbounded linear map from $H$ into $H$.


问:
Excircle and parallelogram
Can you provide a proof for the following claim:

Claim. Given any parallelogram $ABCD$ and excircle of triangle $\triangle ABC$ oposite to vertex $A$. An arbitrary tangent $t$ is constructed to the excircle. Let $n_1$ , $n_2$ , $n_3$ , $n_4$ be a signed distances from vertices $A$,$B$,$C$,$D$ to tangent line respectively, such that distances to a tangent from points on opposite sides are opposite in sign, while those from points on the same side have the same sign . Then $n_1+n_3=n_2+n_4$ .

I don't have idea how to start the proof. My first guess is to use barycentric coordinates but I am not sure which triangle to choose as the reference triangle. Any hints are welcomed.

EDIT

As DR SK MOBINUL HAQUE suggested in his comment the more general form of this claim holds as well. You can find a GeoGebra applet that demonstrates his generalization here.

答:
I will use $'$ to denote the feet of the perpendiculars. So, for example, $A'$ is the foot of the perpendicular from $A$ to $t$. Let the center of the parallelogram be $O$. Recall that the diagonals of a parallelogram bisect each other. Notice that $OO'$ is the common midsegment of trapezoids $AA'C'C$ and $DD'B'B$, completing the proof for the case where the line doesn't cut the parallelogram.

For the case where the line cuts the parallelogram, translate the line high enough so that the theorem proved above can be used. Finally, add the new heights to complete the proof.

Title: Excircle and parallelogram