问:
What is the difference between "expectation", "variance" for statistics versus probability textbooks?
It seems that there are two ideas of expectation, variance, etc. going on in our world.

In any probability textbook:

I have a random variable $X$, which is a function from the sample space to the real line. Ok, now I define the expectation operator, which is a function that maps this random variable to a real number, and this function looks like,
$$\mathbb{E}[X] = \sum\limits_{i = 1}^n x_i p(x_i)$$ where $p$ is the probability mass function, $p: x_i \mapsto [0,1], \sum_{i = 1}^n p(x_i) = 1$ and $x_i \in \text{range}(X)$. The variance is,
$$\mathbb{E}[(X - \mathbb{E}[X])^2]$$

The definition is similar for a continuous RV.

However, in statistics, data science, finance, bioinformatics (and I guess everyday language when talking to your mother)

I have a multi-set of data $D = \{x_i\}_{i = 1}^n$ (weight of onions, height of school children). The mean of this dataset is

$$\dfrac{1}{n}\sum\limits_{i= 1}^n x_i$$

The variance of this dataset (according to "science buddy" and "mathisfun dot com" and government of Canada) is,

$$\dfrac{1}{n}\sum\limits_{i= 1}^n(x_i - \sum\limits_{j= 1}^n \dfrac{1}{n} x_j)^2$$

I mean, I can already see what's going on here (one is assuming uniform distribution), however, I want an authoritative explanation on the following:

Is the distinction real? Meaning, is there a universe where expectation/mean/variance... are defined for functions/random variables and another universe where expectation/mean/variance... are defined for raw data? Or are they essentially the same thing (with hidden/implicit assumption)

Why is it no probabilistic assumption is made when talking about mean or variance when it comes to dealing with data in statistics or data science (or other areas of real life)?

Is there some consistent language for distinguishing these two seemingly different mean and variance terminologies? For example, if my cashier asks me about the "mean weight" of two items, do I ask him/her for the probabilistic distribution of the random variable whose realization are the weights of these two items (def 1), or do I just add up the value and divide  (def 2)? How do I know which mean the person is talking about?/

答:
The confusion actually comes from notation, where symbols mean different things in two formulas.

First, let's take a look at the "probabilistic" definition:
$$
\mathbb{E}[X] = \frac{1}{n}\sum_{i=1}^n x_i p(x_i)
$$
Here the random variable $X$ takes $n$ distinct values $x_1,\ldots, x_n$, each with a probability  mass function $p(x_i)$.

In the "statistical" definition we have an estimate of the expected value, based on the observed values of the random variable $z_1, \ldots, z_N$:
$$
\hat{\mathbb{E}}[X] = \frac{1}{N}\sum_{j=1}^N z_j
$$
Notice that I renamed the variables compared to your original formula, so as to avoid the confusion. Here $N$ is the number of observations, and $z_j$s are the actual observations. For example, if $X$ is a random variable representing rolls of a loaded die, then $n = 6$ and ${x_i} = \{1, 2, 3, 4, 5, 6\}$; whereas $N$ can be arbitrarily large, and $z_j$s will just be a long sequence of random draws from the set of ${x_i}$s.

Now, you can rewrite the "statistical" formula by collecting different values of $z_j$s into the groups according to which $x_i$ they correspond to (for example, first collect all 1s, then all 2s, etc):
$$
\hat{\mathbb{E}}[X] = \frac{1}{N}\big(x_1\cdot|\{z_j=x_1\}| + \cdots+x_n\cdot|\{z_j=x_n\}|\big) \\
=\frac{1}{N}\sum_{i=1}^n x_i \sum_{j=1}^N\mathbb{1}[z_j=x_i] \\
=\sum_{i=1}^n x_i \hat{p}(x_i)
$$
where
$$
\hat{p}(x_i) = \frac{1}{N}\sum_{j=1}^N \mathbb{1}[z_j=x_i]
$$
is the estimate of the probability mass function: the number of times a value $x_i$ was encountered in the sample divided by the sample size, i.e. the observed frequency of the value $x_i$.

Now you can see that the "probabilistic" and "statistical" definitions are actually the same, with the only difference that we replace the theoretical mass distribution function $p(x)$ (which may not be known) with the empirical (observed) mass distribution function $\hat{p}(x)$.


问:
Decision variable must lie in union of multiple disjoint intervals
In my Linear Program, I'm trying to express that a decision variable $x \in R$ can only lie in certain intervals, e.g. $x \in [0,2] \cup [5,8] \cup [9,15]$.

I'm aware of that you can model either constraint 1 OR constraint 2 with a Big M trick (e.g. explained here in section 7.3 and asked here, but do not see directly how this could solve my question. Any ideas?

答:
If $[a_i,b_i]$ is the $i$-th interval then for a binary variable $z_i$ the inequality

$$a_iz_i-(1-z_i)M\leq x\leq b_iz_i+(1-z_i)M$$

gives $x\in[a_i,b_i]$  when $z_i=1$ and is "free" ($x\in [-M,M]$) when $z_i=0$. So a family of such constraints together with

$$\sum z_i= 1$$

models the membership in a union of intervals $x\in\bigcup_i[a_i,b_i]$.

ContentLicense: CC BY-SA 4.0


问:
Show that $\mathbb{P}(\exp(\sum_{i=1}^n X_i) >1)\to\frac{1}{2}$ for $X_i\in L^2$ i.i.d. with zero mean
Let $X_1,X_2,\ldots\in L^2$ be i.i.d. random variables with $E(X_i)=0$. Show that $$\mathbb{P}\left(\exp\left(\sum_{i=1}^n X_i\right) >1\right)\to\frac{1}{2}$$ as $n$ tends to infinity.

My approach:

$$\mathbb{P}\left(\exp\left(\sum_{i=1}^n X_i\right) >1\right)=\mathbb{P}\left(\sum_{i=1}^n X_i > 0\right)=\mathbb{P}\left(\frac{1}{n}\sum_{i=1}^n X_i > 0\right)$$

From the law of large numbers, we know that $\frac{1}{n}\sum\limits_{i=1}^n X_i\to 0$ in probability, i.e.

$$\forall\epsilon>0: \mathbb{P}\left(\left|\frac{1}{n}\sum_{i=1}^n X_i\right|>\epsilon\right)\to 0$$

I now want to make some kind of statement about $P\left(\frac{1}{n}\sum\limits_{i=1}^n X_i<0\right)$ in order to proceed with the proof. How do I continue?

答: Apply CLT instead of SLLN. $P(\frac {S_n} {\sqrt n \sigma } >0) \to 1-\Phi (0)=\frac  1 2$ where $\Phi$ is the standard normal distribution. Hence $P(S_n >0) \to \frac  1 2$.


问:
Are there real numbers a and b such that $f(x,y,t) = x^a t^b$ satisfies the heat equation?
The question is in the title. The heat equation is as follows:
$$
\frac{\partial f}{\partial t} = k \left( \frac{\partial^2 f}{\partial x^2} +  \frac{\partial^2 f}{\partial y^2} \right),\quad k\in\mathbb{R}
$$

Attempt at solution

Plugging the requested form into the above equation yields:

$$
\frac{\partial f}{\partial t} = b\ t^{b-1} x^a\\
\frac{\partial^2 f}{\partial x^2} = a (a-1)\ t^b x^{a-2}\\
\frac{\partial^2 f}{\partial y^2} = 0
$$

Which leads to showing that:
$$
b\ x^2 = k\ a(a-1)\ t
$$

I'm not sure how to proceed from this point. Is this the correct procedure to solve this problem? Thanks!

答:
Note that for your condition to be satisfied for all $x,t$, since $x$ and $t$ are independent, the only way for the condition to hold would be for both sides to be equal zero, leading to:
$$b = 0, a = 0,1$$ </p>

ContentLicense: CC BY-SA 3.0


问:
Counting $5$-character long Passwords with a Fixed Starting and Ending Points
Students are required to create $5$-character long passwords to access the library. The characters must be from lowercase letters or digits. Each password must start with a lowercase letter and end with a digit and contain at most $2$ digits. How many valid passwords are there?

Is this solution correct?

$(26×(26^3) ×10) + (26×(26^2×10)×10)$

答:
26 possibilities for the first character

10 possibilities for the last character (a digit)

2 cases:
there is a 2nd digit or there isn't.

case 1 : there is a 2nd digit

digit can go in any of 3 spots.

$T_1 = 3 \times 10 \times 26^2.$

case 2 : there is no 2nd digit

$T_2 = 26^3.$

Final answer:

$$26 \times 10 \times [T_1 + T_2].$$


答: Your solution is not correct.  As noted in the comments, you have not accounted for the fact that the extra digit can go in one of three different slots.  Given the constraints, there are $26^4 \times 10$ passwords with exactly one digit, and there are $3 \times 26^3 \times 100$ passwords with exactly two digits, so the answer is the sum of those two numbers, $26^3 \times 560$.

Title: Counting $5$-character long Passwords with a Fixed Starting and Ending Points

Body:
Students are required to create $5$-character long passwords to access the library. The characters must be from lowercase letters or digits. Each password must start with a lowercase letter and end with a digit and contain at most $2$ digits. How many valid passwords are there?

Is this solution correct?

$(26×(26^3) ×10) + (26×(26^2×10)×10)$


问:
Solving $\lim_{x \to 0}\frac{x+ x\cos(x)}{\sin(x)\cos(x)} $ without L'Hopitals
$$\lim_{x \to  0}\frac{x+ x\cos(x)}{\sin(x)\cos(x)} $$

This question is very easy by just using L'Hopital's, but I'm trying to do the question without it but I'm stuck on what to do. Any help would be greatly appreciated.

答:
Another way:

$$\frac{x+x\cos x}{\sin x\cos x}=\frac{x(1+\cos x)}{\frac12\sin 2x}=\frac{2x}{\sin 2x}(1+\cos x)\xrightarrow[x\to0]{}1\cdot(1+1)=2$$


问:
Surface integral on $S=\{(x,y,z)|x^2+y^2+z^2=1,x+y+z\leq 1\}$
Let $S=\{(x,y,z)|x^2+y^2+z^2=1,x+y+z\leq 1\}$, $F(x,y,z)=(x,0,-x)$ and $n(x,y,z)$ be the unit normal vector of $S$ such that $n(0,0,-1)=(0,0,-1)$.
I want to evaluate $\displaystyle \iint_{S}F(x,y,z)\cdot n(x,y,z)dS$.

My Attempt

Let $f(x,y,z)=x^2+y^2+z^2-1$. Then $n$ can be calculated by $n=\frac{\nabla f}{|\nabla f|}=(x,y,z)$. This satisfies the condition stated in the problem.

Therefore we have $\displaystyle \iint_{S}F(x,y,z)\cdot n(x,y,z)dS=\iint_{S}(x^2-zx)dS$. Now we need to calculate this surface integral, but I'm encountering issues.

According to this website, I have two options. One option is to find an orthogonal projection of $S$. The other option is to find a parameterization of $S$. However, I couldn't do either of them. Is there a simple expression for them? Any help is appreciated.

答:
The question should have been more explicit but here is how I read the question -

Sphere $x^2 + y^2 + z^2 = 1$ is cut by the plane $x + y + z = 1$ and you are taking the part which is bigger (below the plane $x + y + z = 1$). Please note this surface has a spherical part and a planar part (circle) and applying divergence theorem will give the flux through the entire surface including the planar part.

Vector Field $\vec{F} = (x, 0, -x)$

$div \vec{F} = \frac{\partial}{\partial x} (x) + 0 + \frac{\partial }{\partial z} (-x) = 1$

$\iint_S \vec{F} \cdot \hat{n} \, dS = \iiint_S div{\vec{F}} \,dV = \iiint_S \,dV = V$

So all we are left with is to find the volume $V$ of our surface.

The unit normal vector to the plane $x + y + z = 1$ is $\frac{1}{\sqrt3}(1, 1, 1)$ which is also a point on the sphere. The line from the origin that is perp to the plane is $\frac{t}{\sqrt3}(1, 1, 1)$ and the foot of the perp on the plane is $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}) \,$ for $t = \frac{1}{\sqrt3}$. The length of the perp is $\frac{1}{\sqrt3} \,$ which comes from $\, (\sqrt{(\frac{1}{3})^2 + (\frac{1}{3})^2 + (\frac{1}{3})^2} \,).$

So the smaller part of the sphere cut by the plane is a spherical cap of height $(1 - \frac{1}{\sqrt3})$.

You can either use the formula for spherical cap volume, if you remember, or considering symmetry, the volume of the spherical cap can be easily calculated as -

$V_1 = \displaystyle \int_0^{2\pi} \int_0^{cos^{-1}(\frac{1}{\sqrt3})} \int_{\frac{1} {\sqrt3 \cos \phi}}^1 \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \approx 0.482129$

Volume you are interested in $V = \frac{4}{3} \pi - V_1 \approx 3.707$

EDIT: if the planar part is not part of the surface then we can subtract flux through the planar surface but please note that -

$\vec{F} \cdot n = (x, 0, -x) \cdot \frac{1}{\sqrt3}(1, 1, 1) = 0$ so the flux through the planar part is zero.


问:
Hasse-Minkowski for cubic forms
We know that an analogue of the Hasse-Minkowski theorem does not hold for all cubic forms, e.g. because Selmer's cubic:
$$
3x^3 + 4y^3 + 5z^3 = 0
$$
has solutions over $\mathbb{R}$ and $\mathbb{Q}_p$ for all $p$, but no solutions over $\mathbb{Q}$.

My questions are:

Can we find a (non-trivial) class of cubic forms where an analogue of the Hasse-Minkowski theorem does hold?

Is there any intuition for why the local-global principle holds for quadratic but fails for cubic forms?

Are there higher degree forms where the local-global principle holds again?

Are questions like these addressed anywhere in the literature?

Many thanks.

答:
Let $L/\mathbb{Q}$ a Galois cyclic extension of degree $3$. Then the norm form $N:x\in L\mapsto N_{L/\mathbb{Q}}(x))$ is a cubic form in the coordinates of $x$ in a fixed basis of $L$. Now fix $a\in\mathbb{Q}$

By a theorem of Hilbert,  Hasse principle holds for the affine cubic form $x\mapsto N(x)-a$ ( a rational number is globally a norm of a cyclic extension if and only if it is locally)

Note this is true if you replace $3$ by any integer, provided you restrict yourself to cyclic extensions. You can also replace $\mathbb{Q}$ ny a global field.

I don't know if the following will give you "intuition", but certainly it will explain a lot (but not all) counter examples to Hasse principle. The keywords are: Brauer-Manin obstruction, and you can find a lot of things about it in the literature.
This is an obstruction coming from Brauer-Hasse-Noether theorem for central simple algebras.

You can find a nice introduction to the definition here: http://homepages.warwick.ac.uk/staff/S.Siksek/arith/notes/brauermanin.pdf

I don't think so. Using Brauer-Manin obstruction, you can construct counterexamples of any prescribed degree.

For example, Bayer, Lee and Parimala computed the Brauer Manin obstruction for multinorm equations, and I'm pretty sure that we can use their work to construct counterexamples of arbitrary large degree. The full paper is available on Parimala's web page.

Title: Hasse-Minkowski for cubic forms


问:
Is $i = \sqrt{e^{\pi\sqrt{e^{\pi\sqrt\ldots}}}}$?
I was messing with the identity $e^{i\pi}=-1$ and I got that $i = \sqrt{e^{\pi\sqrt{e^{\pi\sqrt\ldots}}}}$ and on. I plugged it in to a calculator and it was infinite. It grew very fast. Does that make $i$ solvable? </p>

答:
$$e^{i\pi}=-1=i^2$$

Replace $i$ with $x$.

$$e^{x\pi}=x^2$$

where $x=\pm i$

$$x=\pm\sqrt{e^{x\pi}}$$

Here $i$ satisfies $x=+\sqrt{e^{x\pi}}$.

$$i=\sqrt{e^{i\pi}}=\sqrt{e^{\pi\sqrt{e^{i\pi}}}}=\sqrt{e^{\pi\sqrt{e^{\pi\sqrt{...}}}}}$$

Hence proved.


问:
Explanation why an abelian tower admits a cyclic refinement
Now that school is wrapping up, I'm trying to crack down and get better at algebra. This proposition from Lang's Algebra loses me at the end.

Proposition 3.1. Let $G$ be a finite group. An abelian tower of $G$ admits a cyclic refinement. Let $G$ be a finite solvable group. Then $G$ admits a cyclic tower, whose last element is $\{e\}$.

Proof. The second assertion is an immediate consequence of the first, and it clearly suffices to prove that if $G$ is finite, abelian, then $G$ admits a cyclic tower. We use induction on the order of $G$. Let $x$ be an element of $G$. We may assume that $x \neq e$. Let $X$ be the cyclic group generated by $x$. Let $G' = G/X$. By induction, we can find a cyclic tower in $G'$, and its inverse image is a cyclic tower in $G$ whose last element is $X$. If we refine this tower by inserting $\{e\}$ at the end, we obtain the desired cyclic tower.

Here is my understanding so far: (Please excuse me if a lot of the things I say are very obvious/wrong, I'm trying to be detailed for my own understanding.) If $|G|=1$, then $\{e\}$ is the desired cyclic tower. So suppose the result holds for $|G|\leq n-1$. Suppose $|G|=n$. Letting $G'$ be as above, $|G'|=|G|/|X|\lt|G|$, so by the induction hypothesis, there exists a cyclic tower in $G'$, say
$$
G'=G/X\supset G_1'\supset G_2'\supset\cdots\supset G_m'.
$$
I'm not quite sure what Lang means by "its inverse image is a cyclic tower in $G$ whose last element is $X$." Is there some assumed homomorphism $f\colon G\to G'$, and then the inverse image of the tower would be
$$
f^{-1}(G')\supset f^{-1}(G_1')\supset\cdots\supset f^{-1}(G_m')?
$$
Why is the last element of the tower $X$, as Lang claims? Also, Lang says a normal tower is cyclic if each factor group $G_i/G_{i+1}$ in the tower is cyclic. Does this mean all the $G_i$ themselves are cyclic, or is it possible for the factor group to be cyclic, but the normal subgroup being modded out is not? Thanks for any explanation.

答:
The assumed homomorphism $G\rightarrow G'=G/X$ is the quotient map: $g\mapsto gX$. One of the isomorphism theorems says that this map establishes a bijection between subgroups of $G/X$ and subgroups of $G$ that contain $X$. Moreover, this bijection preserves inclusions, normality, and quotients. In other words if $\bar{U}$ is a subgroup of $G'$ with subgroup $\bar{U}'$, and if the corresponding subgroups of $G$ are $U$ and $U'$, then $U'\leq U$, it is normal in $U$ if and only if $\bar{U}'$ is normal in $\bar{U}$, and if they are, then $\bar{U}/\bar{U}'\cong U/U'$. All this is easy to prove, and Lang is using all of this in his proof.

The subgroup of $G$ corresponding to the trivial subgroup of $G'$ is indeed $X$ - the smallest subgroup of $G$ containing $X$.

As for the last question, it means exactly what it says: all factor groups are cyclic. It would be pointless to introduce this notion if it was simply a reformulation of "$G$ is cyclic". E.g. $1\leq \mathbb{Z}/2\mathbb{Z} \leq \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ is a cyclic tower where the top term is not cyclic.


问:
Quadratic residues mod $n$ of $n-1$
While investigating something about square numbers, I started to wonder which numbers $n$ have quadratic residues of $n-1$. Obviously all $n$ of the form $m^2+1$ will work. But there are other numbers such as $13$, is there a pattern to them? Any insights on this would be great.</p>

答:
In very short:

The equation $\,x^2=-1\pmod p\,,\;p\,$ a prime, has solution iff $\,p=1\pmod 4\,$ or $\,p=2\,$ (this is already a very nice exercise), and from here, with the prime factorization of $\,n\in\Bbb N\,$ , you get the corresponding condition for $\,x^2=-1\pmod m\,$.

Title: Quadratic residues mod $n$ of $n-1$


问:
prove $\frac{x-x^{c+1}}{1-x^{c+1}}<1$
Can you help prove that $\frac{x-x^{c+1}}{1-x^{c+1}}<1$, for all positive $x\neq 1$ and $c$ a positive integer?

Title: prove $\frac{x-x^{c+1}}{1-x^{c+1}}<1$

Body: Hint: $$\frac{x-x^{c+1}}{1-x^{c+1}}<1 \\ \frac{x- x^{c+1} - 1 + x^{c+1}}{1-x^{c+1}} < 0 \\ \frac{x-1}{1-x^{c+1}} < 0$$


问:
Statistical Inference: Casella Berger - PDF of a random variable
I have been going through Casella Berger recently, and I am not sure I understand this concept. I get that one needs to split the intervals so that the function X^2 is monotone, but why cannot the intervals be (-1,0), {0}, (0,2)? I have seen this question asked here before, but I still do not understand the explanation.

The question

Merry christmas to those who celebrate, and happy holidays regardless.

答:
The problem is that $X-\text{support}$ is not symmetric. If you do a drawing of the transformation function $Y=X^2$ you see that in the interval $[1;2]$ the transformation function is monotonic thus you can apply directly the theorem.  In the interval $[-1;1]$ you must use the definition of CDF

Anyway you can use the intervals $[-1;0]$ and $[0;2]$. The result will be the same.  But you have to pay attention because in $[-1;0]$ and $[0;1]$ $Y-\text{support}$ is the same and thus you should divide the support in 3 intervals to apply the theorem. The proposed solution is the most suitable

I am not at home and I have not my Casella Berger with me but I think the theorem they refer to is

$$f_Y(y)=f_X[g^{-1}(y)]\Bigg|\frac{d}{dy}g^{-1}(y)\Bigg|$$


Title: prove $\frac{x-x^{c+1}}{1-x^{c+1}}<1$

Body: Can you help prove that $\frac{x-x^{c+1}}{1-x^{c+1}}<1$, for all positive $x\neq 1$ and $c$ a positive integer?

Body:
1, If $\;0<x<1\;$, then

$$x^c<1\implies 1-x^c>0\implies \frac{x-x^{c+1}}{1-x^{c+1}}<1\iff x-x^{c+1}<1-x^{c+1}\iff x<1\;\;\checkmark$$

2.- if $\;x>1\;$,  then

$$x^{c+1}>1\implies 1-x^{c+1}<0\implies\frac{x-x^{c+1}}{1-x^{c+1}}<1\iff x-x^{c+1}>1-x^{c+1}\iff x>1\;\checkmark$$


Title: prove $\frac{x-x^{c+1}}{1-x^{c+1}}<1$

Body: $$\frac{x-x^{c+1}}{1-x^{c+1}}=1-\frac{1-x}{1-x^{c+1}}=1-\frac{1}{1+x+x^2..+x^c}< 1$$


问:
Help with understanding Duffing's oscillator
I'm working through some notes regarding analysis of some weakly perturbed oscillators. Among these is Duffing's oscillator which is governed by $$ \frac{d^2y}{dt^2}+y+\epsilon y^3=0, \; \; 0<\epsilon << 1 $$

Now to leading we have
\begin{align*}
   \frac{d^2y_0}{dt^2}+y_0=0
 \end{align*}
with general solution

$$ y_0 = a\cos t + b \sin t$$

which can be written w.l.o.g as $ y_0 = c \cos(t+\phi), \; \; c>0$.

So far so good. But now to understand the nature of the perturbation the notes make use of the Poincare-Lindstedt method, which relies on the observation that the period of the perturbed oscillator is $$2\pi(1+o(1))$$

as $\epsilon \rightarrow 0$. Here is my first question: how exactly was this observation made?

Next we define $Y(\tau; \epsilon) = y(t; \epsilon)$ where $\tau = t/\chi(\epsilon)$ and $\chi(\epsilon)$ is such that $Y$ is $2\pi$ periodic in $\tau$. Then, expanding $Y$ and $\chi$ in powers of $\epsilon$ we have

$$Y(\tau;\epsilon) \sim Y_0(\tau;\epsilon)+\epsilon Y_1(\tau;\epsilon)+...$$

and

$$\chi(\epsilon) \sim 1+\epsilon\chi_1 + \epsilon^2\chi_2+...$$

Here the $Y_i$ are $2\pi$ periodic in $\tau$. To leading order the solution has the same form, $Y_0 = a\cos(\tau + \phi)$. Going up to first order in $\epsilon$, we have
\begin{align*}
\frac{d^2Y_1}{d\tau^2} + Y_1 &= -2\chi_1Y_0-Y_0^3 \\
&= -2\chi_1a\cos(\tau+\phi) -\frac{a^3}{4}\left \{3\cos(\tau+\phi)+\cos\left(3(\tau+\phi)\right)\right\},
\end{align*}

where we have substituted for $Y_0$ and used a Fourier expansion. Then here is where I really have trouble. At this point then notes say that in order for $Y_1$ to be $2\pi$ perioduc in $\tau$ we must eliminate $\cos(\tau+\phi)$ terms, which gives us an algebraic expression for $\chi_1$. But why must we eliminate such terms in order for $Y_1$ to be $2\pi$ periodic? I don't understand this.

答:
1.) The equation remains conservative for $ϵ\ne 0$ so that all solutions always lie on level surfaces of
$$
E=\frac12y'^2+\frac12y^2+\fracϵ4y^4.
$$
All solutions that do not lie on the level of singular points can not help but be periodic. For $ϵ>0$ there are no singular points except the origin. Any solution for $ϵ\ll 1$ will be close to a circle and at time $2\pi$ will be close to the initial point.

2.) The frequency-one-terms on the right side are in resonance to the homogeneous solution of the left side, and thus will give solution terms that are increasing in amplitude. This gives obviously only a valid approximation for a very local solution. With the perturbation approach one has introduced flexibility to shift these terms of the equation into the base frequency, or said otherwise, to compensate them with changes in the frequency. Thus by selecting $χ_0=1$ and
$$
2χ_1a+\frac{3a^3}4=0
$$
there is no longer resonance in this level of the perturbation equations, the resulting periodic solution is correct to a higher order in $ϵ$.

Essentially, if you solve the equation using a simple perturbation approach $y=y_0+ϵy_1+...$, you get the first perturbation  equation as
$$
y_1''+y_1=-y_0^3=-c^3\cos^3(t+ϕ)=-\frac{c^3}{4}[\cos(3t+3ϕ)+3\cos(t+ϕ)]
$$
with particular solution of the form
$$
y_1=A\cos(3t+3ϕ)+Bt\sin(t+ϕ)
\\
\implies
y_1''+y_1=-8A\cos(3t+3ϕ)+2B\cos(t+ϕ)
$$
so that the coefficients are $A=\frac{c^3}{32}$ and $B=-\frac{3c^3}{8}$. In the expression for $y$ the combination of terms
$$
c\cos(t+ϕ)-ϵ\frac{3c^3}{8}t\sin(t+ϕ)
$$
can be interpreted as linear Taylor polynomial for
$$
c\cos\left(\left(1+ϵ\frac{3c^2}{8}\right)t+ϕ\right)
$$
These are equivalent approximations for $t\approx 0$, but the second one gives a result that is also globally valid for a periodic solution.


问:
Find $r$ from three moduli
Determine $r$ in $$x\equiv  15 \pmod{77}\\
x\equiv 16 \pmod{78}\\
x\equiv r \pmod{14}$$

With lists in Excel, I have discovered that $x = 5944$.

$$5944\equiv  15 \pmod{77}\\
5944\equiv 16 \pmod{78}\\
5944\equiv r \pmod{14}\implies r=8$$

This is not an elegant solution. How can I take on the task in a more mathematical way?

答:
By Chinese Remainder Theorem, since $77$ and $78$ are coprime, there is only one solution for $x$ mod $77 \times 78$. (Note that $5944$ is not a unique solution)

Observe that $$x \equiv 15\pmod {77} \equiv -62 \pmod {77}$$
$$x \equiv 16\pmod {78} \equiv -62 \pmod {78}$$

hence $x \equiv -62 \pmod {77 \times 78}$.

Since $14 \mid 77 \times 78$, $x \equiv -62 \pmod {14} \equiv 8 \pmod {14}$.

Title: Find $r$ from three moduli


答:
I will present three ways of tackling this problem.

$\textbf{Option 1}$: Directly

Parametrizing spherical coordinates as usual we can find the bounds by examining the plane equation

$$x+y+z = 1 \implies \sin\phi\cos\theta+\sin\phi\sin\theta+\cos\phi = 1$$

which after a little manipulation becomes

$$\cos\theta+\sin\theta = \frac{1-\cos\phi}{\sin\phi} = \tan\left(\frac{\phi}{2}\right)$$

It's obvious from looking at the sphere with the planar cap cutoff that doing the $\theta$ integral first would require two integrals (one each for the upper and lower parts), but $\phi$ first would only require one. Writing the integral gets us

$$\int_0^{2\pi}\int_{2\cot^{-1}\left(\frac{1}{\sin\theta+\cos\theta}\right)}^\pi \sin^3\phi\cos^2\theta-\sin^2\phi\cos\phi\cos\theta\:d\phi d\theta$$

$\textbf{Option 2}$: Rotate then directly

Once we have the scalar surface integral $\iint_S x^2-xz\:dS$ we can consider a change of variables to rotate the plane so it is completely horizontal from the top. The closest distance between the plane and the origin is $\frac{1}{\sqrt{3}}$ so hopefully that motivates the following rotational coordinate change:

$$\begin{cases}u = \frac{x-y}{\sqrt{2}} \\ v = \frac{x+y-2z}{\sqrt{6}} \\ w = \frac{x+y+z}{\sqrt{3}}\\ \end{cases}$$

This change of variables is a pure rotation so it has Jacobian $1$. This was obtained by choosing a direction for the new '$z$' and '$x$' ($w$ and $u$, respectively) then taking their cross product to find the third orthogonal vector (then including the factor to make them unit vectors), which means

$$u^2+v^2+w^2 = x^2+y^2+z^2$$

Since this is a rotation matrix, inverting the system of equations is as easy as taking the transpose:

$$\begin{cases}x = \frac{u}{\sqrt{2}} + \frac{v}{\sqrt{6}} + \frac{w}{\sqrt{3}} \\ y = -\frac{u}{\sqrt{2}} + \frac{v}{\sqrt{6}} + \frac{w}{\sqrt{3}} \\ z =  -\frac{2v}{\sqrt{6}} + \frac{w}{\sqrt{3}} \\ \end{cases}$$

which gives us a new surface integral on the same sphere, only rotated

$$\iint_S \frac{u^2+v^2}{2}+\frac{2uv}{\sqrt{3}}+\frac{uw}{\sqrt{6}}+\frac{vw}{\sqrt{2}}\:dS$$

The nice thing about this is that with this rotated sphere cutoff at $w=\frac{1}{\sqrt{3}}$, we can now exploit symmetry. $u$ and $v$ are both odd functions, so any term with just an odd power of either will vanish, leaving us with

$$\iint_S \frac{u^2+v^2}{2}\:dS = \int_0^{2\pi}\int_{\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)}^\pi \frac{1}{2}\sin^3\phi\:d\phi d\theta$$

which easy enough to evaluate since the integrand will end up being only cosines.

$\textbf{Option 3}:$ Divergence theorem

We close off the surface with a plane and consider the equation

$$\iint_{\text{sphere}}F\cdot dS + \iint_{\text{plane}}F\cdot dS = \iiint_V \nabla \cdot F dV$$ $$ = \int_0^{2\pi}\int_{\frac{1}{\sqrt{3}}}^1 \int_{\cos^{-1}\left(\frac{1}{\sqrt{3}\rho}\right)}^\pi\rho^2\sin\phi\:d\phi d\rho d\theta + \frac{4}{3}\pi\left(\frac{1}{\sqrt{3}}\right)^3$$

Since $\nabla\cdot F = 1$, we only want the volume, thus we have implicitly taken advantage of the rotation from option 2 to set up our bounds. The remaining surface integral can be calculated as follows

$$\iint_{\text{plane}}F\cdot dS = \iint_{\text{plane}}(x,0,-x)\cdot\frac{(1,1,1)}{\sqrt{3}}dS = 0$$

therefore the answer will only be the volume integral.

Title: Surface integral on $S=\{(x,y,z)|x^2+y^2+z^2=1,x+y+z\leq 1\}$

Body:
Let $S=\{(x,y,z)|x^2+y^2+z^2=1,x+y+z\leq 1\}$, $F(x,y,z)=(x,0,-x)$ and $n(x,y,z)$ be the unit normal vector of $S$ such that $n(0,0,-1)=(0,0,-1)$.
I want to evaluate $\displaystyle \iint_{S}F(x,y,z)\cdot n(x,y,z)dS$.

My Attempt

Let $f(x,y,z)=x^2+y^2+z^2-1$. Then $n$ can be calculated by $n=\frac{\nabla f}{|\nabla f|}=(x,y,z)$. This satisfies the condition stated in the problem.

Therefore we have $\displaystyle \iint_{S}F(x,y,z)\cdot n(x,y,z)dS=\iint_{S}(x^2-zx)dS$. Now we need to calculate this surface integral, but I'm encountering issues.

According to this website, I have two options. One option is to find an orthogonal projection of $S$. The other option is to find a parameterization of $S$. However, I couldn't do either of them. Is there a simple expression for them? Any help is appreciated.


问:
Calculating $\sqrt{28\cdot 29 \cdot 30\cdot 31+1}$
Is it possible to calculate $\sqrt{28 \cdot 29 \cdot 30 \cdot 31 +1}$ without any kind of electronic aid?

I tried to factor it using equations like $(x+y)^2=x^2+2xy+y^2$ but it didn't work.

答:
One notes in base 28, that the numbers are 1.0, 1.1, 1.2, 1.3.  The product of the inner and outer pairs are 1.3.0 and 1.3.2.  Adding one to this product gives $1.3.1^2$, whence the square root is 1.3.1.

Evaluating this in decimal gives 869.

When this series starts with 35, the result is the cube 1331.


问:
A proof for $\nabla |u|^p = p\ \text{sgn}(u)|u|^{p-1}\nabla u$.
I believe the formula
$$\nabla |u|^p = p\ \text{sgn}(u)|u|^{p-1}\nabla u
$$
to be true for $u\in W^{1,p}(\Omega)$, where $\Omega$ is an open domain with nice boundary, but failed to find a good reference for this.  I am sorry if this has been asked already.

The usual assumption that I usually see regarding the chain rule for Sobolev functions, i.e.
$$
\nabla F(u) = F'(u)\nabla u,
$$
is that $F\in C^1$ with bounded derivative (or perhaps a Lipschitz function). I want $F(t)=|t|^p$, which doesn't satisfy that assumption.

My idea of the proof goes as follow: I'd like to mimic the proof of the case $p=1$ so first we show a similar result for
$$
F_\varepsilon(t) = (\varepsilon^2 + t^2)^{p/2}
$$
by approximation with $F_\varepsilon(u_n)$, where $u_n\to u$ in $W^{1,p}$ and $u_n\in C_c^\infty(\Omega)$. Then we take $\varepsilon \to 0$ to approximate $|\cdot|^p$. While the usual argument using the Dominated Convergence Theorem wouldn't work, I think Vitali Convergence Theorem might do the trick.

Does anyone know a reference to this result? Also, if the formula doesn't hold in general spaces (i.e. $u\notin W^{1,p}$) then does anyone have a counterexample?

I have seen the proofs of the cases $p=1,2$. Finding a reference for a general $p\in [1,\infty)$ turns out to be a lot harder than I anticipated.

答: A nice proof in Lieb and Loss's Analysis Chapter 6 (Distributions)  page 152.


问:
Regularizing a solenoidal vector field $u\in L^p(\Omega)^N$.
Let $\Omega\subset\mathbb{R}^N$ be a bounded smooth domain and suppose that $u\in L^p(\Omega)^N$, $p\in (1,\infty)$. Assume that in the sense of distributions, $\operatorname{div}u=0$ where $\operatorname{div}u=\sum_{i=1}^N\frac{\partial u_i}{\partial x_i}$.

Given $\epsilon>0$, can I find $v\in C^1(\overline{\Omega})$ such that $\|v-u\|_p<\epsilon$ and $\operatorname{div}v=0$?

Thank you

答:
The paper Extension and Representation of Divergence-free Vector Fields
on Bounded Domains may be helpful here.

I also add a couple of remarks.

Remark 1. In the Banach space of $L^p$-vector fields, the subspaces of conservative fields and of divergence-free fields are closed. Moreover, in the reflexive range the space of divergence-free $L^p$ fields is precisely the annihilator of conservative fields in $L^q$, the dual space.

Remark 2. For $1<p<\infty$, for smooth domains, the above subspaces are complemented in $L^p$: that is, they admit bounded projections. This comes from the boundedness of singular integrals on $L^p$. The key term is $L^p$ Hodge decomposition.

Having a bounded projection allows you to mollify a divergence-free field and then project it back onto the subspace, without  going too far from the original field.

As a starting point, look at this paper and references given there.

From an older answer:

Remark 3, on reflecting a divergence-free field in the hyperplane $x_0=0$. Given $x=(x_1,\dots,x_n)$, denote $x^*=(x_1,x_2,\dots,-x_n)$. Here is how the reflection of divergence-free vector field $u$ should be defined:

Write $u$ in components: $u=(u_1,\dots,u_n)$.

When $j<n$, use odd reflection $u_j(x^*)=-u_j(x)$.

For $j=n$ use even reflection: $u_n(x^*)=u_n(x)$.

I claim that $\int u\cdot \nabla \varphi=0$ for any test function $\varphi$. Indeed, any $\varphi$ can be decomposed into even part
$\frac12(\varphi(x)+\varphi(x^*))$ and odd part $\frac12(\varphi(x)-\varphi(x^*))$.

If $\varphi$ is even, then $u\cdot \nabla \varphi$ is odd by construction, and therefore integrates to zero.

If $\varphi$ is odd, then $u\cdot \nabla \varphi$ is even, therefore it suffices to integrate it over one half-space. And since $\varphi$ is odd, it vanishes on the boundary. By the divergence-free property  of the original field $u$, the integral of  $u\cdot \nabla \varphi$ is zero.


问:
Expected number of unpecked chicks - NYT article
In this article, the winner of the math competition answered this question correctly:

In a barn, 100 chicks sit peacefully in a circle. Suddenly, each chick randomly pecks the chick immediately to its left or right. What is the expected number of unpecked chicks?

The answer was given as 25. I am interested to know the correct method for solving this problem as well as a general way to find this out for N chicks. I was thinking of trying to solve by figuring out the number of occurrences of (something?) inside a binary string of length 100 (or N), but I don't know if that is the right way to approach it.

答: Represent chickens as variables ($a_{1}, a_{2}, \dots, a_{100})$. Assign it a $1$ if unpecked, $0$ otherwise. We want to find the expected value of the sum $a_{1} + \dots + a_{100}$. Notice that the expected value of each of these is $\frac{1}{4}$, so by linearity of expectation, the expected value of the sum is $100 \cdot \frac{1}{4} = 25$


答:
Follows immediately by CCRT for congruences in A.P. (Arithmetic Progression), i.e.

$\!\!\bmod \color{#c00}{77\!+\!k}\!:\ x \equiv 15\!+\!\color{#c00}k \equiv 15\color{#c00}{-77} \equiv -62\ \ $ [for $\,k=0,1$]

so $\,x\equiv -62\pmod{\!77\ \&\ 78}\!\iff\! x\equiv -62\pmod{77\cdot 78}\ $ by CCRT = Constant-case CRT

so $\,x\equiv -62\equiv 8\pmod{\!\color{#0a0}{14}}\,$ by $\,\color{#0a0}{2(7)}\mid 78(77)\,$ & congruences persist mod factors of modulus.

Generally, $ $ as above: $\,\ \begin{align}&x\equiv a\ \ \ \pmod{\!n}\\ &x\equiv a\!+\!1\!\!\!\!\pmod{n\!+\!1}\end{align}\! \iff x\equiv a\!-\!n\pmod{\!\!n(n\!+\!1)}$

therefore again by persistence we infer$\ d\mid n(n\!+\!1)\,\Rightarrow\, x\equiv a\!-\!n\pmod{\!d}$

Remark $ $ Replying to a comment, note the above persistence follows from,$\,\color{#0a0}2\mid 78\,$ and $\,\color{#0a0}7\mid 77\,$ so alternatively we could use such persistence before applying CRT to get $\,x\equiv 16\equiv 0\pmod{\!2}\,$ and $\,x\equiv 15\equiv \color{#c00}8\pmod{\!7}$ so $\,x\equiv 8\pmod{\!14}\,$ by CRT, as in J.W.T's answer (alternatively $\,x\bmod{14}= 2(\color{#c00}{x/2} \bmod 7) = 2(\color{#c00}4)\,$ by $\,2\mid x\,$ and the mod Distributive Law).

The A.P. method has the advantage that we also get the (unique) solution $\bmod 77\cdot 78\,$ just as easily (and the techniques used are instructional). It will always be easy when the A.P. has very small step-size (like $\,s = 1\,$ above) since the linked formula involves inverting only $\,s\ $ (not the possibly larger moduli as when rotely applying the common CRT formula).

But it helps to be familiar with both methods, i.e. reducing before and after CRT. Interesting reults may arise by performing operations in different order, so it may well be worthwhile to investigate the ramifications of such a reordering.

Title: Find $r$ from three moduli

Body:
Determine $r$ in $$x\equiv  15 \pmod{77}\\
x\equiv 16 \pmod{78}\\
x\equiv r \pmod{14}$$

With lists in Excel, I have discovered that $x = 5944$.

$$5944\equiv  15 \pmod{77}\\
5944\equiv 16 \pmod{78}\\
5944\equiv r \pmod{14}\implies r=8$$

This is not an elegant solution. How can I take on the task in a more mathematical way?


问:
the notation $d^2y$ has nothing to do with "squared"?
I have looked up lots of posts asking about why it's $d^2/dx^2$ and how $dx$ itself is a variable, but I can't find answers explaining how the notation, specifically "$d^2y$" is derived apart from $dx^2$.

I am confused if $d^2y$ means that $d$ is squared or it merely is an operator just like $d/dx$ itself is (so it has nothing to do with being "squared"). I think the latter explanation is right but I haven't seen any clear explanation how it's derived that way.

答:
No, it has nothing to do with the squaring of anything. It is basically interpreted as the second derivative of the evaluated function.

A common mistake for beginners to calculus.


问:
Graph Theory: Prove Ramsey Number R(3, 4)=9
Most of the proofs I've seen prove this by showing that R(3, 4)≤9 and at the same time ≥9. The former one is natural and easy to understand. To prove R(3, 4)≥9, a lot of proofs construct a graph with 8 vertices which contains no K4 and whose complement contains no K3. The counterexample is often like this
the counterexample

As far as I know, R(m, n)=R(n, m), so if I want to show that R(3, 4)≥9, the graph I construct should not contain any monochromatic K3 or K4. Go back to the counterexample. Though there is no blue K4 and red K3, it can be easily found that there are blue K3s in the first graph. Is the counterexample wrong? Or is there any problem with my understanding about Ramsey Number? Thanks a lot for the help.

答:
The counterexample is correct. Note it's not the graph itself not containing $K_n$ and $K_m$:

If we say $R(n,m) = k$, then this means every graph $G$ with $k$ vertices, either $G$ contains a $K_n$ or the complement $G'$ contains a $K_m$. If we want to show $R(n,m) > k$, then we show that there exists some graph $G$ with $k$ vertices that fails that condition, namely $G$ does not contain $K_n$ and $G'$ does not contain $K_m$. We just need to find one such example.

Since in the counterexample you found, the 8-vertex graph $G$ (blue) has no $K_4$ and its complement $G'$ (red) contains no $K_3$, we see that $R(4,3)>8$. Now it is true that $R(4,3)=R(3,4)$: To show $R(3,4)>8$ as well, notice the graph $G'$ (red) has no $K_3$ and its complement $G''=G$ (blue) has no $K_4$.


问:
Are terminating decimals dense in the reals?
I have read that rationals are dense in the reals, but I've been curious if the terminating decimals are as well. Since, in applied math, all effectively "real" numbers are approximated with terminating decimals, I would expect the terminating decimals to have this property, and if not I'm curious why. Here I'm defining the terminating decimals like this:

$$D = \{\frac{m}{10^n} \in \mathbb{Q} : m \in \mathbb{Z} \wedge n \in \mathbb{N}_0\}$$.

答:
We have

$\quad D \subset \Bbb Q \subset \Bbb R$

We know $\Bbb Q$ is dense in $\Bbb R$.

We'll now show that $D$ is dense in $\Bbb Q$, leaving it to the OP to

complete the argument (i.e. denseness is transitive) that $D$ is dense in $\Bbb R$.

Let $s \in \Bbb Q$ so that we can write

$\quad s = \Large \frac{u}{v} \normalsize \quad \text{with } u,v \in \Bbb Z \text{ and } v \ge 1$

If $n \in \Bbb N$ consider the division of $u\cdot 10^n$ by $v$,

$\quad u\cdot 10^n = vq + r \text{ with } 0 \le r \lt v$

But then

$\quad \Large \frac{u}{v} - \large q \cdot 10^{-n} = \Large \frac{r}{v} \large 10^{-n}$

Since $q\cdot 10^{-n} \in D$ and

$\quad \large 0  \le \Large \frac{r}{v} \large 10^{-n} \lt 10^{-n}$

we've demonstrated that $s \in \Bbb Q$ can be approximated

to any specified precision with a terminating decimal.

Using the above theory a simple algorithm can be specified to find these expansions.

Example: Approximate $\large \frac{23}{7}$ up to $6$ (fractional) decimals of precision;

i.e. repeat $6$ times: multiply the numerator (the residue) by $10$ and then divide by $7$.

Divide
Approximate (append quotient)

$23\cdot 10 = 7 \cdot 32 + 6$
$\large \frac{23}{7} \approx 3.2$

$6\cdot 10 = 7 \cdot 8 + 4$
$\large \frac{23}{7} \approx 3.28$

$4\cdot 10 = 7 \cdot 5 + 5$
$\large \frac{23}{7} \approx 3.285$

$5\cdot 10 = 7 \cdot 7 + 1$
$\large \frac{23}{7} \approx 3.2857$

$1\cdot 10 = 7 \cdot 1 + 3$
$\large \frac{23}{7} \approx 3.28571$

$3\cdot 10 = 7 \cdot 4 + 2$
$\large \frac{23}{7} \approx 3.285714$

Since the decimal expansions of fractions eventually keep repeating a block of digits, this algorithm is useful. For the above, the algorithm can be designed to determine that

$\large \frac{23}{7} = 3.\overline{285714}$

Note that the above theory and algorithm design does not depend on the base being equal to $10$;

it follows the same logic path for any base $b \ge 2$.


问:
Statement about divisors of polynomials and their roots
I am a 10th grade student and there is a statement in my math book

If $a$ is a root of the polynomial $f(x)$ then $(x-a)$ is a divisor of $f(x)$

Why is $(x-a)$ a divisor of $f(x)$? Can you please tell me?

答:
Lemma $:$ For a field $F, \alpha \in F$ is a root of $a(x)$ if and only if $x-\alpha$ divides $a(x)$

Proof. $(\Longrightarrow)$ Assume that $\alpha$ is a root, i.e., $a(\alpha)=0 .$ Then we can write $a(x)$ as
$$
a(x)=(x-\alpha) q(x)+r(x)
$$
where $\operatorname{deg}(r(x))<\operatorname{deg}(x-\alpha)=1,$ i.e., $r(x)$ is a constant $r,$ where
$$
r=a(x)-(x-\alpha) q(x)
$$
Setting $x=\alpha$ in the above equation gives
$$
r=a(\alpha)-(\alpha-\alpha) q(\alpha)=0-0 \cdot q(\alpha)=0
$$
Hence $x-\alpha$ divides $a(x)$

($\Longleftarrow$) To prove the other direction, assume that $x-\alpha$ divides $a(x),$ i.e., $a(x)=$ $(x-\alpha) q(x)$ for some $q(x) .$ Then $a(\alpha)=(\alpha-\alpha) q(\alpha)=0,$ i.e., $\alpha$ is a root of $a(x)$


答:
$dy/dx$ is not a ratio of two quantities—$dy$ and $dx$. However, sometimes we can 'pretend' that it is. The notation for the second derivative comes from
$$
\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d^2y}{(dx)^2} \, .
$$
Because mathematicians are lazy, we often omit the parentheses on the bottom and write
$$
\frac{d^2y}{dx^2} \, .
$$
But to reiterate, this is just notation. The fact that we 'multiply' the $d$'s together to obtain this notation does not mean that $dy$  and $dx$ are independent quantities, and nor does it mean that $dx$ is the product of two variables $d$ and $x$.

Title: the notation $d^2y$ has nothing to do with "squared"?

Body:
I have looked up lots of posts asking about why it's $d^2/dx^2$ and how $dx$ itself is a variable, but I can't find answers explaining how the notation, specifically "$d^2y$" is derived apart from $dx^2$.

I am confused if $d^2y$ means that $d$ is squared or it merely is an operator just like $d/dx$ itself is (so it has nothing to do with being "squared"). I think the latter explanation is right but I haven't seen any clear explanation how it's derived that way.


问:
Sum of all $4$-digit numbers that can be formed with the digits $0,1,2,3$

What is the sum of all 4-digit numbers that can be formed with the digits 0,1,2,3 when repetition not allowed?

All I have been able to do is figure out that there will be $18$ such $4$ digit numbers ..but I am stuck and would like to know how we would proceed with this.

Also how would we approach this problem if repetition was allowed ?

答:
It would be easy to write down four blanks side by side, and then try to see how many possible choices could be filled in each block that would satisfy the condition.

```<code>      _____   ______  _____ _____
</code>```

Check the different cases, for which repetition is allowed, and for repetition not being allowed.

For case 1, repetition being allowed, there are 3 choices for the first blank(0 not incuded), 4 for the next three.
So total number of choices, as per counting principle, total= 3 * 4 * 4 * 4=192

For case 2, repetition not being allowed, there are 3 choices for the first blank, 3 for the next, 2 for the third, and 1 for the last.
so total number of choices, total=3 * 3 * 2 * 1=18.

Body:
It would be easy to write down four blanks side by side, and then try to see how many possible choices could be filled in each block that would satisfy the condition.

      _____   ______  _____ _____

Check the different cases, for which repetition is allowed, and for repetition not being allowed.

For case 1, repetition being allowed, there are 3 choices for the first blank(0 not incuded), 4 for the next three.
So total number of choices, as per counting principle, total= 3 * 4 * 4 * 4=192

For case 2, repetition not being allowed, there are 3 choices for the first blank, 3 for the next, 2 for the third, and 1 for the last.
so total number of choices, total=3 * 3 * 2 * 1=18.


答:
For the no-repetition case:

In the units place, each of $1,2,3$ appear the same number of times as the number of $3$-digit numbers formed from $0$ and two other digits from ${1,2,3}$, to contribute to a sum of $$(3!-2)(1+2+3)$$ The same thing goes for the tens and hundreds place : $$(3!-2)(10+20+30)+(3!-2)(100+200+300)$$

For the thousands place, fixing one digit there lends us $3!$ choices to arrange the other digits, as there is no restriction for $0$ now. This contributes a sum of $$3!(1000+2000+3000)$$ and adding the three expressions will give the answer.

For the repetition case:

Fix one of $1,2,3$ in the units place. We have $4$ choices for every other place except the thousands place, in which we have $3$ choices. This gives $$4^2\cdot 3(1+2+3)$$ Similarly for the tens and hundreds: $$4^2\cdot 3(10+20+30+100+200+300)$$

For the thousands place, fixing one digit lends us $4^3$ such numbers: $$4^3(1000+2000+3000)$$

Title: Sum of all $4$-digit numbers that can be formed with the digits $0,1,2,3$

Body:
What is the sum of all 4-digit numbers that can be formed with the digits 0,1,2,3 when repetition not allowed?

All I have been able to do is figure out that there will be $18$ such $4$ digit numbers ..but I am stuck and would like to know how we would proceed with this.

Also how would we approach this problem if repetition was allowed ?


问:
Prove that, if L is regular, $P(L) = \{ w \in L\text{ } |\text{for any prefix } w' \text{ of w, } w' \in L\}$ is regular
I have found one specific problem about regular languages which I am not sure how to solve:

Prove that the following language $$P(L) = \{ w \in L\text{ } |\text{for any prefix } w' \text{ of w, } w' \in L\}$$ is regular, given that L is regular too.

I thought of using closure properties, but I can't think of any other language to use for that. Any tips?

答:
It is possible to construct a non-deterministic automaton for $P(L)$:

Consider a DFA of $L$, and construct an NFA by adding an accepting state with no outgoing edges, and an $\epsilon$ edge from each of the states that have a path to an accepting state in the DFA, to the new accepting state.

This way, if a word is a prefix, then there is a path from the end of its run on the DFA to an accepting state, and there is an $\epsilon$ transition to the new accepting state.

If there is a run which reaches an accepting state, then it is either a run of the DFA (and then the word is in $L\subset P(L)$) or a run reaching the new accepting state, and therefore must be a word that reached a state that has a path to an accepting state in the DFA, therefore a prefix.


问:
Check if an arithmetic function is (completely) multiplicative
In Paul J. McCarthy's book about arithmetic functions, he constructs a completely multiplicative function $g$ by setting $g(p)$ equal to a root of the equation
$$X^2+f^{-1}(p)X+f^{-1}(p^2)=0, $$
where $f$ is a multiplicative arithmetic function and $p$ is prime. He does not further explain why this function is multiplicative or even completely multiplicative. I have tried to come up with an explanation for hours but I couldn't. I got to the point where I questioned if I do understand the assertion correctly. I tried showing that for two primes $p,q$ fulfilling
$$g(p)^2+f^{-1}(p)g(p)+f^{-1}(p^2)=0,\quad g(q)^2+f^{-1}(q)g(q)+f^{-1}(q^2)=0 $$
it is also true that
$$(g(p)g(q))^2+f^{-1}(pq)(g(p)g(q))+f^{-1}((pq)^2)=0. $$
Which would show that $g(pq)=g(p)g(q)$ for primes $p$ and $q$ and thus imply at least multiplicativity. One can use the facts that for multiplicative arithmetic functions
$$f^{-1}(p)=-f(p),\quad f^{-1}(p^2)=f(p)^2-f(p^2).$$
I would greatly appreciate any help. I guess that I miss something very obvious.

答:
You can construct any completely multiplicative function $g$ by first defining $g(p)$ for primes $p$ and then the multiplicativity property forces $g(n)$ for general $n$ to be

$$ n=\prod p^e \quad\implies\quad g(n)=\prod f(p)^e. $$

You can also define any (not necessarily completely) multiplicative function $g$ by defining $g(p^e)$ for all prime powers $p^e$ and then $g(n)=\prod f(p^e)$. (Note $g(1)=1$ is assumed.)


问:
Number of elements with a certain property in $S_5$: what does $S_5$ mean?
I am working on the following problem:

Determine number of elements in $\{
\rho \in S_5\vdots \ \rho (2)=5
\}$

I have difficulty to understand what is asked for. Specifically:

Is $S_5=\{1,2,3,4,5\}$?

What does $\rho (2)=5$ mean?

答: $\rho$ is a permutation. After fixing $\rho(2)=5$ there are $4!$ ways to assign the results of applying $\rho$ to $1,3,4,5$, so the answer is $24$.


问:
Condition For No Existence Of Real Root
$2x^{4}+5ax^{3}-2bx^{2}+1=0$ has no real root in $(5,2014)$

Find the conditions for $a$ and $b$

I am suspicious of even the existence of its solution and at a loss.

答:
The region(s) in the $ab$-plane where $(a,b)$ leads to $p_{(a,b)}(x)=2x^4+5ax^3-2bx^2+1$ having a root inside $(5,2013)$ is bordered by curves (possibly straight lines) where one of the following happens:

$p_{(a,b)}$ has a root at 5

$p_{(a,b)}$ has a root at 2013

$p_{(a,b)}$ has a double root inside $(5,2013)$

$p_{(a,b)}$ has a quadruple root inside $(5,2013)$

This is because if there is a root inside $(5,2013)$ and we perturb $(a,b)$ by $\epsilon$, these are the only ways that the root could suddenly disappear from the interval $(5,2013)$.

So I suggest mapping out the curves that define these boundaries and then examining the regions they contain. The first two boundary curves are quickly seen to be straight lines defined by setting $p(5)=0$ and $p(2013)=0$.  A quadruple root is not possible since $p$'s linear term is $0$, and the only way for that to happen with a fourth degree polynomial that has a quadruple root is if the polynomial is $cx^4$.

So the difficult curve is the one that defines where $p_{(a,b)}$ has a double root inside $(5,2013)$. For $p$ to have a double root means $p'$ has that root with multiplicity one. $p'$ in this problem has been made rather easy to analyze: $p'(x)=x(8x^2+15ax-4b)$. Clearly the common root between $p$ and $p'$ would not be $0$, so a double root would be one of the numbers $\frac{-15a\pm\sqrt{225a^2+128b}}{16}$. Plugging this back into $p(x)=0$, we have a quartic relation in $a$ and $b$: $$2\left(\frac{-15a\pm\sqrt{225a^2+128b}}{16}\right)^4+5a\left(\frac{-15a\pm\sqrt{225a^2+128b}}{16}\right)^3-2b\left(\frac{-15a\pm\sqrt{225a^2+128b}}{16}\right)^2+1=0$$

Title: Condition For No Existence Of Real Root


问:
Characterization of cross product (proof that only exists in dimension 3)
In order to show that the cross product exists only in $\mathbb R^3$, our teacher puts this remark, "Exercise left to the reader". I have no idea how to solve it. Here's the problem:

"Let $B:\mathbb R^n \times \mathbb R^n \to \mathbb R^n$ be a surjective bilinear map and antisymmetric so that $B(u,v)=0 \iff \text{u and v are linearly dependent}$. Then $n = 3$."

答:
That's wrong. There is also a 7D binary cross product that satisfies the given properties.

There are exactly two cross products on a 3D inner product space, one for both possible orientations of space. But there are infinitely many 7D cross products - in fact, the space of all cross products on a 7D inner product space is topologically the same as  $\mathrm{SO}(7)/G_2\simeq S^7$.

In general, an $n$-dimensional $k$-ary cross product $X$ is defined to be an antisymmetric multilinear function of $k$ vectors in $\mathbb{R}^n$ which also satisfies a volume constraint: namely, $\|X(v_1,\cdots,v_k)\|$ ought to be the volume of the parallelepiped formed from $v_1,\cdots,v_k$ - which may be calculated using the Gramian determinant, $\mathrm{vol}^2=\det(\langle v_i,v_j\rangle)$.

The full classification of $k$-ary $n$-dimensional cross products is as follows:

Nullary cross product exist in any dimension.

Unary cross products exist in even dimensions.

Binary cross products exist in dimensions $0,1,3,7$.

Ternary cross products exist only in dimensions $4,8$.

Co-unary cross products exist whenever $k=n-1$.

Over-ary cross products exist whenever $k>n$.

Some commentary on these:

The nullary and over-ary cross products are both identically $0$.

The unary cross product is essentially multiplication-by-$i$ on $\mathbb{C}^n$ as a $2n$-dimensional real inner product space (note that taking the real part of the sesquilinear complex inner product yields a real inner product).

The binary cross products are given by the imaginary parts $\mathrm{Im}(ab)=a\times b$ in the real normed division algebras $\mathbb{R},\mathbb{C},\mathbb{H},\mathbb{O}$ respectively. Here, $\mathbb{H}$ is the quaternions and $\mathbb{O}$ is the octonions. The $0$D and $1$D binary cross products are both over-ary.

The 7D binary and 8D ternary cross product both relate to the octonions (see here).

The co-unary cross product $X(v_1,\cdots,v_{n-1})$ is the unique vector perpendicular to the span of $\{v_1,\cdots,v_{n-1}\}$, oriented appropriately, of the correct magnitude. The nullary 1D, unary 2D, binary 3D, and ternary 4D cross products are all co-unary.


问:
Check whether the equation $3x^3+5y^5+z^3+4z=13$ defines the function $z(x,y)$ at a point $A(1,1,1)$

Consider the equation
$$
3x^3+5y^5+z^3+4z=13
$$
Check whether the equation defines the function $z(x,y)$ at a point $A(1,1,1)$

I don't understand what am I supposed to in this problem. Can someone give me a clue as to how to start?

答:
We have
$$
z^3+4z=f(z)=g(x,y) = 13-3x^3-5y^5
$$

but $f(z)$ has inverse because $f'(z) = 3z^2+4 > 0\ \  \forall z\in\mathbb{R}$ so there exists

$$
z = f^{-1}\left(g(x,y)\right)\ \ \forall \{x,y\}\in\mathbb{R}^2
$$

Attached a plot representation for $z = f^{-1}\left(g(x,y)\right)$. In black the point $(1,1,1)$


答:
Hi hope you're having a jolly christmas!

$p$ is just a function of the form(just an example)$f(x)=(2,5)(1,3,4)$ were $f(2)=5$ and $f(3)=1$, $f(4)=1$. And $g(x)=(1,2,3,4,5)$ with $g(2)=3$ is another example. It's the permutation of $S_5$.(Sorry my English is really bad)

Title: Number of elements with a certain property in $S_5$: what does $S_5$ mean?

Body:
I am working on the following problem:

Determine number of elements in $\{
\rho \in S_5\vdots \ \rho (2)=5
\}$

I have difficulty to understand what is asked for. Specifically:

Is $S_5=\{1,2,3,4,5\}$?

What does $\rho (2)=5$ mean?


问:
independent sub spaces
Let V be a vector space with dimV=4.

$$S=\{v_1,v_2,v_3\} , R=\{u_1,u_2,u_3\}$$

both are INDEPENDED.

I need to prove that there exists a vector $$v\in V$$
such that $$R\cup \{v\} $$ $$S\cup \{v\} $$
will both be a basis of V.

答: This is false. For example, take $S$ to be a linearly dependent set.

Title: independent sub spaces


问:
The smallest nontrivial conjugacy class in $S_n$

Find the smallest nontrivial conjugacy class in $S_n$.

For small $n$, the answer can be found by counting the permutations of each possible cycle type. The result is:
$$\begin{array}{ccc}
n & \text{smallest nontrivial class(es)} & \text{size} \\
\hline
1 & \text{none} & \\
2 & \text{transpositions} & 1 \\
3 & 3\text{-cycles} & 2 \\
4 & \text{double transpositions} & 3 \\
5 & \text{transpositions} & 10 \\
6 & \text{transpositions, triple transpositions} & 15
\end{array}$$

For $n\geq 7$, I think the unique smallest nontrivial conjugacy class in $S_n$ is given by the transpositions.

The conjugacy classes in $S_n$ are given by the sets of permutations of the same cycle type. For the cycle type $\alpha = 1^{a_1} 2^{a_2} \ldots$, the size of the respective conjugacy class is known to be
$$N_\alpha = \frac{n!}{\prod_i (i^{a_i} a_i!)}.$$
In particular, the number of transpositions in $S_n$ is $\binom{n}{2}$.

In this way, it remains to show the following:

Let $n \geq 7$ and $\alpha$ be a partition of $n$ distinct from $1^n$ and $2^1 1^{n-2}$. Show that $N_\alpha > \binom{n}{2}$.

So far I can only come up with a lengthy and technical case-by-case study.
I'm looking for a more elegant proof of the statement.

答:
I don't think there is a very conceptual approach giving a meaning to the comparison of conjugacy class sizes, but I think there is a way to make the argument quite short. The first thing to do is to cancel the term $a_1!$ for $i=1$ (for the points remaining fixed) in the denominator against a part of the numerator $n!$, which leaves the numerator equal to $n(n-1)\ldots(a_1+1)$ (the number of factors is the sum of the sizes of the cycles for the conjugacy class, in the group-theoretic sense where cycles cannot have length$~1$).

A first easy simplification rids us the case of mixed cycle structures, classes with is more than one different cycle length${}\geq2$. For such cycle types we can map permutations in the class to another non-trivial cycle type by simply omitting the cycle(s) of the longest length$~l$ from the decomposition. This map commutes with any conjugation by a permutations (which simple relabels elements in the cycles) and is therefore surjective to a single new conjugacy class; moreover it is not injective since there is always more than one way to reconstruct the length$~l$ cycle(s). Therefore such classes are never the smallest non-trivial ones.

The next step is to show that we need only deal with the case of transpositions, possibly flying in formation. More precisely (and with three exceptions that do not occur for $n\geq7$), the class with $m$ disjoint $k$-cycles with $k>2$ is larger than the one with $m$ disjoint $2$-cycles. We have $a_k=m$ and factor our formula as
$$
  \frac{n(n-1)\ldots(n-mk+1)}{k^m}\times\frac1{m!},
$$
where the second factor is identical to the one for $k=2$. In the first factor replacing $k$ by $2$ means dropping the last $m(k-2)$ factors from the numerator and dividing the denominator by $(k/2)^m$. Using that a product of $k-2$ distinct positive integers can only be${}\leq k/2$ if $k\leq4$, we see that this decreases the first factor except when $a_1=0$, and $(m,k)\in\{(1,3),(1,4),(2,3)\}$, which happens only for $n=3,4,6$ (it explains that $3$-cycles are least numerous for $n=3$, that for $n=4$ there are as many $4$-cycles as $2$-cycles, namely $6$, and that for $n=6$ there are fewer ($40$) pairs of $3$-cycles then pairs of transpositions $(45)$).

We have reduced to the case of $m$ disjoint transpositions, which we shall compare with isolated transpositions ($m=1$). For fixed $m$, increasing $n$ leaves the denominator unchanged and increases all factors in the numerator by$~1$. This gives more relative increase for $m>1$ than for $m=1$, so if the class with $m$ disjoint transpositions is to be less numerous than that with one transposition, it must already be the case for the minimal possible value of $n$, which is $n=2m$. The formula for the number of products of  $m$ disjoint transpositions when $n=2m$ is the product $1\times 3\times\cdots\times(2m-1)$ (which some people, defying the ambiguity, write as $(2m-1)!!$) which is${}\leq\binom{2m}2$ only for $m=2,3$; this explains these cases for $n=4,6$. For $n\geq7$ this does not happen any more.


答: Write $B_i=\sum_{jk}\varepsilon_{ijk}u_jv_k$. Hereafter, we'll drop the $\sum$. There are constraints on $\varepsilon_{ijk}$, e.g. antisymmetry ($\varepsilon_{ijk}=-\varepsilon_{jik}$) and orthogonality ($\varepsilon_{ijk}=-\varepsilon_{ikj}$). But beyond three dimensions, these leave some degrees of freedom, whereas in $3$ dimensions the symbol $\varepsilon_{ijk}$ is determined up to scaling. The former fact is why the product @runway44 mentioned isn't unique, even up to scaling.

Title: Characterization of cross product (proof that only exists in dimension 3)

Body:
In order to show that the cross product exists only in $\mathbb R^3$, our teacher puts this remark, "Exercise left to the reader". I have no idea how to solve it. Here's the problem:

"Let $B:\mathbb R^n \times \mathbb R^n \to \mathbb R^n$ be a surjective bilinear map and antisymmetric so that $B(u,v)=0 \iff \text{u and v are linearly dependent}$. Then $n = 3$."


问:
Triple summation verification $\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}$
I am attempting the following triple summation. It would be great if someone would verify whether what I've done is correct.

$$\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}$$

$$\begin{aligned}\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}&=\sum_{i=1}^{\infty}\sum_{j=i+1}^{\infty}\sum_{k=j+1}^{\infty}\frac{1}{2^i3^j5^k}\\&^= \sum_{i=1}^{\infty}\sum_{j=i+1}^{\infty}\frac{1}{2^i3^j}\frac{1/5^{j+1}}{4/5}\\&=\frac{1}{4}\sum_{i=1}^{\infty}\frac{1}{2^i}\frac{1/15^{i+1}}{14/15}\\&=\frac{1}{56}\sum_{i=1}^{\infty}\frac{1}{30^{i}}=\frac{1}{56}\cdot\frac{1}{29}\end{aligned}$$

答:
The index region of the sum
\begin{align*}
\sum_{\color{blue}{1\leq i <j<k}}\frac{1}{5^i3^j2^k}\tag{1}
\end{align*}
is specified by the inequality chain
\begin{align*}
1\leq i <j<k
\end{align*}
which has $1$ as lower limit and $k-1$ as upper limit. We have two indices $i$ and $j$, which means we can write it as double sum as shown in the evaluation below.

We obtain
\begin{align*}
\color{blue}{\sum_{1\leq i<j<k}\frac{1}{5^{i}3^j2^k}}
&=\frac{1}{2^k}\sum_{i= 1}^{k-2}\frac{1}{5^{i}}\sum_{j=i+1}^{k-1}\frac{1}{3^j}\tag{2}\\
&=\frac{1}{2^k}\sum_{i= 1}^{k-2}\frac{1}{5^{i}}\left(\frac{\left(\frac{1}{3}\right)^{i+1}-\left(\frac{1}{3}\right)^k}{1-\frac{1}{3}}\right)\tag{3}\\
&=\frac{1}{2^k}\sum_{i = 1}^{k-2}\frac{1}{5^i}\,\frac{1}{2}\left(\frac{1}{3^i}-\frac{1}{3^{k-1}}\right)\tag{4}\\
&= \frac{1}{2^{k+1}}\sum_{i=1}^{k-2}\frac{1}{15^i}- \frac{1}{2^{k+1}\,3^{k-1}}\sum_{i=1}^{k-2}\frac{1}{5^i}\\
&=\frac{1}{2^{k+1}}\left(\frac{\frac{1}{15}-\left(\frac{1}{15}\right)^{k-1}}{1-\frac{1}{15}}\right)-\frac{1}{2^{k+1}\,3^{k-1}}\left(\frac{\frac{1}{5}-\left(\frac{1}{5}\right)^{k-1}}{1-\frac{1}{5}}\right)\tag{5}\\
&=\frac{1}{2^{k+2}\cdot7}\left(1-\frac{1}{15^{k-2}}\right)-\frac{1}{2^{k+3 }\,3^{k-1}}\left(1-\frac{1}{5^{k-2}}\right)\\
&\,\,\color{blue}{=\frac{1}{2^{k+2}\cdot7}-\frac{1}{2^{k+3}\,3^{k-1}}+\frac{1}{2^{k+3}\,3^{k-1}\,5^{k-2}\cdot7}}\tag{6}
\end{align*}

Comment:

In (2) we factor out $\frac{1}{2^k}$ and reorder the double sum using another common style.

In (3) we evaluate the inner sum using the finite geometric summation formula.

In (4) we do a simplification and multiply out in the next line.

In (5) we apply the finite geometric summation formula twice and do a simplification in the following lines.

Notes:

Varying $k$ is not admissible since $k-1$ is an upper limit. Here $k$ is a free variable whereas the indices $i$ and $j$ are bound variables. This is different to the situation
\begin{align*}
\sum_{1\leq i <j<k\color{blue}{<\infty}}\frac{1}{5^{i}3^j2^k}
\end{align*}
where $k$ is an index bound by the upper limit $\infty$ and where $k$ varies between $j$ and $\infty$.

The result (6) is also given by Wolfram Alpha with input

```<code> sum(sum 1/(5^i*3^j*2^k), j=i+1..k-1),i=1..k-2
</code>```

Hint: You might find chapter 2: Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik helpful. It provides a thorough introduction in the usage of sums.

Body:
The index region of the sum
\begin{align*}
\sum_{\color{blue}{1\leq i <j<k}}\frac{1}{5^i3^j2^k}\tag{1}
\end{align*}
is specified by the inequality chain
\begin{align*}
1\leq i <j<k
\end{align*}
which has $1$ as lower limit and $k-1$ as upper limit. We have two indices $i$ and $j$, which means we can write it as double sum as shown in the evaluation below.

We obtain
\begin{align*}
\color{blue}{\sum_{1\leq i<j<k}\frac{1}{5^{i}3^j2^k}}
&=\frac{1}{2^k}\sum_{i= 1}^{k-2}\frac{1}{5^{i}}\sum_{j=i+1}^{k-1}\frac{1}{3^j}\tag{2}\\
&=\frac{1}{2^k}\sum_{i= 1}^{k-2}\frac{1}{5^{i}}\left(\frac{\left(\frac{1}{3}\right)^{i+1}-\left(\frac{1}{3}\right)^k}{1-\frac{1}{3}}\right)\tag{3}\\
&=\frac{1}{2^k}\sum_{i = 1}^{k-2}\frac{1}{5^i}\,\frac{1}{2}\left(\frac{1}{3^i}-\frac{1}{3^{k-1}}\right)\tag{4}\\
&= \frac{1}{2^{k+1}}\sum_{i=1}^{k-2}\frac{1}{15^i}- \frac{1}{2^{k+1}\,3^{k-1}}\sum_{i=1}^{k-2}\frac{1}{5^i}\\
&=\frac{1}{2^{k+1}}\left(\frac{\frac{1}{15}-\left(\frac{1}{15}\right)^{k-1}}{1-\frac{1}{15}}\right)-\frac{1}{2^{k+1}\,3^{k-1}}\left(\frac{\frac{1}{5}-\left(\frac{1}{5}\right)^{k-1}}{1-\frac{1}{5}}\right)\tag{5}\\
&=\frac{1}{2^{k+2}\cdot7}\left(1-\frac{1}{15^{k-2}}\right)-\frac{1}{2^{k+3 }\,3^{k-1}}\left(1-\frac{1}{5^{k-2}}\right)\\
&\,\,\color{blue}{=\frac{1}{2^{k+2}\cdot7}-\frac{1}{2^{k+3}\,3^{k-1}}+\frac{1}{2^{k+3}\,3^{k-1}\,5^{k-2}\cdot7}}\tag{6}
\end{align*}

Comment:

In (2) we factor out $\frac{1}{2^k}$ and reorder the double sum using another common style.

In (3) we evaluate the inner sum using the finite geometric summation formula.

In (4) we do a simplification and multiply out in the next line.

In (5) we apply the finite geometric summation formula twice and do a simplification in the following lines.

Notes:

Varying $k$ is not admissible since $k-1$ is an upper limit. Here $k$ is a free variable whereas the indices $i$ and $j$ are bound variables. This is different to the situation
\begin{align*}
\sum_{1\leq i <j<k\color{blue}{<\infty}}\frac{1}{5^{i}3^j2^k}
\end{align*}
where $k$ is an index bound by the upper limit $\infty$ and where $k$ varies between $j$ and $\infty$.

The result (6) is also given by Wolfram Alpha with input

 sum(sum 1/(5^i*3^j*2^k), j=i+1..k-1),i=1..k-2

Hint: You might find chapter 2: Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik helpful. It provides a thorough introduction in the usage of sums.


问:
Show that $\int_0^a \frac{\sin x}{x} dx \leq 1+\log a$ for all $a \geq 1$
Show that $\int_0^a \frac{\sin x}{x} dx \leq 1+\log a$ for all $a \geq 1$.

My attempt: define $f(a)=1+\log a-\int_0^a \frac{\sin x}{x}dx$. Then the inequality is equivalent to show that $f(a) \geq 0$ for all $a \geq 1$.

$f$ is differentiable for all $a \geq 1$, since it a sum of differentiable functions (a constant, a logarithm and an integral of a continuous function that is differentiable in the interval $0 < a \leq 1$ for the fundamental theorem of calculus) in the interval $a\in[1,\infty)$.

SO it has sense to evaluate $f'$, and it is
$$f'(a)=\frac{1}{a}-\frac{\sin a}{a}=\frac{1}{a}\left(1-\sin a\right) \geq 0 \ \ \ \ \ \forall a \geq1$$
So $f$ is increasing for all $a \geq 1$, this means that if $f(1) \geq 0$ then $f(a) \geq 0$ for all $a \geq 1$ for monotonicity. It is
$$f(1)=1-\int_0^1 \frac{\sin x}{x} dx$$
So it is
$$f(1) \geq 0 \iff \int_0^1 \frac{\sin x}{x} dx \leq 1$$
Since $\sin t \leq t$ for all $t \geq 0$ and since $x\in[0,1]$ it is $\sin x \leq x \iff \frac{\sin x}{x} \leq 1$, so for the monotonicity of the integral it is $\int_0^1 \frac{\sin x}{x} dx \leq \int_0^1 1 dx=1$ as well; this shows that the inequality is true for all $a \geq 1$.

Is my approach correct? If there is mistakes or imprecisions let me know please.

If it is correct I would be glad to see other way to prove this. Thanks.

答:
It looks good to me, one other way to approach this problem is
$$ \int_0^a \frac{\sin(x)}{x}dx=\int_0^1\frac{\sin(x)}{x}dx+\int_1^a\frac{\sin(x)}{x}dx $$
But as you said $\displaystyle\int_0^1 \frac{\sin(x)}{x}dx\leqslant 1$ because $\sin(x)\leqslant x$, and on the other hand
$$ \int_1^a\frac{\sin(x)}{x}dx\leqslant \int_1^a\frac{dx}{x}=\log(a)$$
Therefore $$ \int_0^a \frac{\sin(x)}{x}dx\leqslant 1+\log(a) $$


问:
Find $r$ from three moduli
Determine $r$ in $$x\equiv  15 \pmod{77}\\
x\equiv 16 \pmod{78}\\
x\equiv r \pmod{14}$$

With lists in Excel, I have discovered that $x = 5944$.

$$5944\equiv  15 \pmod{77}\\
5944\equiv 16 \pmod{78}\\
5944\equiv r \pmod{14}\implies r=8$$

This is not an elegant solution. How can I take on the task in a more mathematical way?

答:
$x\equiv15\bmod77\implies x\equiv 1\bmod 7$.

$x\equiv16\bmod78\implies x\equiv 0\bmod 2$.

Therefore, by the Chinese remainder theorem,

$x\equiv15\bmod77$ and $x\equiv16\bmod78\implies x\equiv\color{blue}8\bmod14$.

Title: Find $r$ from three moduli


问:
How can two extensive-form games be made equivalent with different player orders?
Consider a game like the following linked one.

Neither player is aware of what the other player has played or will play.

Suppose then I model a slightly different game. In this game, player two moves first and can tell the future. She knows the move that player one will make and can, therefore, be represented to have moved after player one, thus:

While this conveys the strict rules of the game, it is misleading to the actual process of play because it is shown as if player two moves after player one. Is there a better way to show player two's extra knowledge while still having them play higher up in the tree?

I thought of using a baysian game where the one branch is the true game and the other is a contrived game to force player one to choose certain things with a probability of 1:0. But then I didn't know how to show that player one believes the probability to be 0:1.

Please ask if this last paragraph was not clear. Though it was not paramount to the question, it indicates what I have tried.

答:
This was too long to be a comment:

There’s an issue here with talking about “knowledge”. Nash-style equilibrium concepts all revolve around players knowing other players’ strategies because they’re all flawless objective Bayesian rationalists. l think what you’re suggesting is that 2 is clairvoyant and knows 1’s actions in advance. Then 2 really has two information sets, one in which they know 1 plays U and one in which they know 1 plays D. 2 then chooses U’ or D’ and 1 is forced to choose what 2 foresaw, as 2 is correct. This isn’t a game because there’s nothing that determines which information set 2 winds up in. You could make nature decide, but that wouldn’t make it equivalent to the second game pictured. 2 can know that 1 will do something irrational forcing 1 to do it. You could insist that 1 can only play rational strategies, but then depicting the game in this way would go part of the way to solving it.

Addendum: You could abandon game trees altogether by thinking of 1 causing 2 to know 1’s action in the past. With retrocausality the linearity of a game trees falls apart. If moves represent causation and we want to keep the axioms of game trees except for perfect recall and the game tree having a root then there’s one way to do this. Let $\mathbb{Z}_z$ be the set of all integers less than or equal to $z$. Nodes are functions $f:\mathbb{Z}_z\to\{U,D,U’,D’\}$ such that even integers maps into $\{U’,D’\}$ and odds into $\{U,D\}$ (the choice of even vs odd is arbitrary). We have four information sets consisting of all nodes $f$ with domains $\mathbb{Z}_z$ (allowing $z$ to vary with $f$) with the same value of $f(z)$. We assign $f$ with $z$ even to $1$ and $f$ with $z$ odd to $2$. The information a player has at a given information set is represented by $f(z)$, which is what the other player has/will played/play. The player may then choose from their respective set of moves. After player $i$ players move $m$ at node $f:\mathbb{Z}_z\to\{U,D,U’,D’\}$, we move to node $f_m:\mathbb{Z}_{z+1}\to\{U,D,U’,D’\}$ such that $f_m(z+1)=m$ and for all $y\in\mathbb{Z}_z$, $f_m(y)=f(y)$. Note that most of the nodes only exists so you can keep the idea of both players making choices, all but eight of them could clearly never be reached since they’d involve different moves being played at the same information set. You’d then need to define solving this game “tree”. You could count some of the nodes as identical, for instance compressing both of 2’s information sets into one node and the members of particular information sets of 1 that follow members of particular information sets of 2. This is a lot more workable, but now there are loops of moves. Of course, there’s retrocausality, so perhaps that’s not so concerning. However, if you want to keep the idea of 1 freely taking an action and this causing something for $2$ in the past and have a normal game tree, well you can’t.


问:
If $a,b,c$ are three non coplanar vectors, then prove that the vector equation $r=(1-p-q)\vec a+p\vec b +q\vec c$ represents a plane
Let $r=x\vec a + y\vec b+z\vec c$

Comparing with the given equation, we obtain
$$x+y+z=1$$ which is a plane

What I don’t understand is how does this say $r$ is a plane, since r is actually $x \vec a+y\vec b+z\vec c$

答: Rearrange to $$\vec r = \vec a + p(\vec b-\vec a) +q (\vec c-\vec a)$$ And this is of the form of the plane equation $$\vec r= \vec p + \lambda \vec s +\mu \vec q $$ More precisely, this is the plane that passes through the point with position vector $\vec a$ and with normal vector $(\vec b-\vec a)\times (\vec c-\vec a)$.


问:
Does a derivative exist if it is equal to infinity
I recently was reading through my James Stewart calculus book and learnt  the limit laws . Take for example the sum rule if the limit of 2 functions exist then the sum of their limits is equal to the limit if their sums.this we can use to prove the sum rule for derivatives.
However when they say exists what exactly do they mean for example does the limit need to be a real number.for example if the limit of some function is
$\lim_{x \to a} f(x)$=$+\infty$
Does the limit not exist and hence we cannot use the sum rule ?

答: As some people in the comments have said, there really lacks a consensus about whether a limit exists if it is equal to $\infty$. Some will say "it diverges" if it is unbounded, others will say "it converges to $\infty$", saving "diverges" for when it oscillates or something. As far as the sum rule goes, if you are adding $\infty$ to any constant then it works, but if you are adding $\infty$ to $-\infty$, then you have indeterminate and should try calculating the derivative another way.


问:
Complex analysis. Cauchy's integral formula
I want to find $f(1)$ and $f'(3)$, where
$$f(z) = \dfrac{1}{2\pi i} \int_{|\zeta| = 2} \dfrac {e^{1/({\zeta - 5})}}{\zeta - z} d \zeta$$
where
the integral is taken over a positively oriented circle. I think, that $f'(3) = 0$ because of Cauchy theorem. But I have some problems with $f(1)$.

答:
$$f(z)=\frac1{2\pi i}\oint_{|\zeta|=2}\frac{e^{1/(\zeta-5)}}{\zeta-z}d\zeta\implies f'(z)=\frac1{2\pi i}\oint_{|\zeta|=2}\frac d{dz}\left(\frac{e^{1/(\zeta-5)}}{\zeta-z}\right)d\zeta=$$

$$=\frac1{2\pi i}\oint_{|\zeta|=2}\frac{e^{1/(\zeta-5)}}{(\zeta-z)^2}d\zeta=\frac{2\pi i}{1!\cdot2\pi i}\frac d{dz}\left(e^{1/(z-5)}\right)=\left(-\frac1{(z-5)^2}\,e^{1/(z-5)}\right)\implies$$

$$f'(3)=-\frac1{(-2)^2}\,e^{-1/2}=-\frac1{4\sqrt e}$$

Observe that the last expression in the first line and the first expression in the second line is just the integral theorem by Cauchy


问:
Prove that $\mathbb{R}^2 \setminus \{0\}$ is connected?
Prove that  $\mathbb{R}^2 \setminus \{0\}$ is connected ?

My attempt : take $A= (-\infty, 0)$ and $B= (0,\infty)$

$\mathbb{R} \setminus \{0\}=A\cup B = (-\infty, 0)\cup (0,\infty)$

squaring both side we have

$\mathbb{R}^2 \setminus \{0\}=(A\cup B)^2 = ((-\infty, 0)\cup (0,\infty))^2=(0,\infty) \cup (0, \infty)$

where$(0,\infty) \cap (0, \infty)\neq \emptyset$ since $(-\infty,0)^2= (0,\infty)$ and $(-\infty)^2=\infty$

Therefore $\mathbb{R^2} \setminus \{0\}$ is connected

Is its correct ?

答:
In this context $X^2$ does not denote the set of all squares formed from elements of $X$, as you seem to be interpreting it.

For a set $X$, the notation $X^2$ denotes the cartesian product
$$
X^2 := X\times X := \{\, (a,b) \,|\, a,b\in X \,\}
$$
consisting of all pairs with both entries elements of $X$.

Hence $\mathbb R^2$ is the two-dimensional Euclidean space (the plane) and $\mathbb R^2\setminus\{0\}$ is the plane with the origin removed. Here $0$ is just a short notation for $(0,0)$.


答:
Your answer is correct but there is a easier (to remember) way to derive the result.

To simplify the sum, the first thing one should do is get rid of the "$<$" constraint among the summation indices. If you introduce index $\ell, m$ such that $j = i + \ell$ and $k = j + m$, then
the constraint $1 \le i < j < k$ simplifies to $1 \le i,\ell,m$. This will
allow you to rewrite your iterated sum as a product of geometric series:
$$\begin{align}\sum_{\substack{(i,j,k)\\1 \le i < j < k}}\frac{1}{2^i 3^j 5^k}
&= \sum_{\substack{(i,j,k)\\1 \le i, \ell, m }}\frac{1}{2^i 3^{i+\ell} 5^{i+\ell + m}}
=\sum_{i=1}^\infty \frac{1}{30^i}\sum_{\ell=1}^\infty\frac{1}{15^\ell}\sum_{m=1}^\infty\frac{1}{5^m}\\
&= \frac{\frac{1}{30}}{1-\frac1{30}}\frac{\frac{1}{15}}{1-\frac{1}{15}} \frac{\frac15}{1-\frac15} = \frac{1}{29}\frac{1}{14}\frac{1}{4} = \frac{1}{1624}
\end{align}
$$

Title: Triple summation verification $\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}$

Body:
I am attempting the following triple summation. It would be great if someone would verify whether what I've done is correct.

$$\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}$$

$$\begin{aligned}\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}&=\sum_{i=1}^{\infty}\sum_{j=i+1}^{\infty}\sum_{k=j+1}^{\infty}\frac{1}{2^i3^j5^k}\\&^= \sum_{i=1}^{\infty}\sum_{j=i+1}^{\infty}\frac{1}{2^i3^j}\frac{1/5^{j+1}}{4/5}\\&=\frac{1}{4}\sum_{i=1}^{\infty}\frac{1}{2^i}\frac{1/15^{i+1}}{14/15}\\&=\frac{1}{56}\sum_{i=1}^{\infty}\frac{1}{30^{i}}=\frac{1}{56}\cdot\frac{1}{29}\end{aligned}$$


问:
Writing Vectors As A Linear Combination Of Other Vectors
I've been tangled with a section from a question :

In the vector space of  $P_3(R)$ given a group of vectors :

$$S= \{u_1=1+t+t^2 , u_2=2+3t+3t^2 , u_3=2+6t-t^2 , u_4=1-2t+2t^2+5t^3\}$$

I have managed to check whether they are linearly independent or not and found the basis and the dimension of $S$

$$B_S=\{(1,0,0,1),(0,1,0,1),(0,0,1,-1)\}$$
$${\operatorname{dim}_S=3}$$

and what is left is to type all the vectors that belong to S as a combination for the basis i found , kind of stuck there , any suggestions ?

答:
Given a polynomial $p \in P_3(R)$, where $p(x) =a +bx + cx^2 + dx^3$.

From the coefficient of $1, x, x^2$, we can conclude that $$p(x) = a(1+x^3) + b(x+x^3) + c(x^2-x^3)$$


问:
understanding someone proof :Characterize those power series that converge uniformly on (−∞, ∞).
Characterize those power series that converge uniformly on ($-\infty$, $\infty$).

I can't understand why he wrote, "Now this is only possible if an+1=0."
for any epsilon, there is another N and another an+1 it seems he only proves that an+1 converges to 0 but not equal to 0.
could someone assist?

答:
Consider the function
$$\begin{array}{rcl}
f:\Bbb R &\longrightarrow& \Bbb R\\
x & \mapsto & \vert a(x-c)^n\vert
\end{array}$$
for some constants $a,c\in \Bbb R$ and some $n\in \Bbb N_0$.

If $a,n \neq 0$ we find that $f(x_n)\rightarrow \infty$ for $x_n \rightarrow \pm \infty$. So if $f$ is bounded ie. satisfies $\forall x: f(x) < C$ for some $C\in \Bbb R$, then either $a$ or $n$ has to be 0. If we know that $n\geq 1$, $a$ has to be 0.


问:
In what base system a given decimal-based number with no terminating decimal expansion can be represented in a finite string of digits?
I was reading about precision floating numbers in the mpmath python module documentation and came across this https://mpmath.org/doc/current/contexts.html#arbitrary-precision-interval-arithmetic-iv:

The fact that '0.1' results in an interval of nonzero width indicates
that 1/10 cannot be represented using binary floating-point numbers at
this precision level (in fact, it cannot be represented exactly at any
precision).

This made me wonder if there's a way to determine in what base system a given decimal-based number with no terminating decimal expansion can be transferred in order to be represented in a finite string of digits.

答:
It depends what you mean by 'base system'. In base $\pi$, for instance, the units column is $1$, meaning that the 'tens' column is $\pi$, and so on. $\pi$ can be represented as $10$ in base $\pi$. However, if you require that the base is an integer, then no, there isn't an irrational number that can be written as a finite string. We can prove this by contraposition. If a number $n$ has a finite number of digits, then we can write the digits that come after the decimal point as
$$
.d_1d_2d_3\ldots d_k \, .
$$
This means that $$n=\text{integer part + }(b^{-1} \cdot d_1)+(b^{-2}\cdot d_2)+(b^{-3}\cdot d_3)+\ldots+(b^{-k}\cdot d_k) \, .$$The sum of finitely many rationals is rational. Hence, finite number of digits $\implies$ the number is rational. This means that irrational numbers can't have finitely many digits, and so they must have infinitely many digits.


问:
Torus interior homeomorphic to torus exterior
Let $T^2 \subset \mathbb{R^3}$, then $X_i$ be its interior and $X_e$ its exterior. By computing homotopy groups of $X_i \cup T^2$ and $X_e \cup T^2$ and corresponding isomorphisms between, one could show with Whitehead's theorem that $X_i$ is homotopy equivalent to $X_e$.

But how could I show that $X_e$ is homeomorphic to $X_i$?

答:
Here's how I think of it. The one point compactification of $\mathbb R^3$ is $S^3$. If you dig out a standard ball from $S^3$ it's believable that the complement is a ball, but you can also prove this directly using stereographic projection. If you remove a standard ball, the complement is a union of rays emanating from a sphere out to infinity. That is, it is a cone on a sphere, which is a ball.

So now I have a $S^3$ written as the union of $2$ balls, one of which contains $\infty$. Drill a straight hole through the center of the bounded ball. This has the effect of creating a bounded solid torus. Now what happens to the unbounded ball is that a solid tube gets added to it along its boundary. Abstractly, if you add a solid tube to a $3$-ball to get an orientable $3$-manifold, the result is a solid torus.


问:
Derivative of $f(x) = \lfloor x\rfloor \sin ^2 (\pi x) $
To help with the problem I first plotted the graph.

Hence for $ x \in (0,1): f'(x) = 0 $

For $x \in (n,n+1), n \in \mathbb{N}$

$f(x) = n \sin^2 (\pi x) $ and thus $f'(x) = 2 \pi n x \sin (\pi x ) \cos (\pi x )$

Finally for $x \in (-n, -n+1) , \ f(x) = -n \sin^2 (\pi x) $ so that $f'(x) = -2 \pi n x \sin (\pi x ) \cos (\pi x )$

Is it correct to assume that the function is not differentiable at $x \in \mathbb{Z} $ because the left and right-hand derivatives would differ in signs and hence the function is not differentiable there.

答: As @player3236 has noted, when you actually evaluate each of those derivatives in the limit at the integer $n$, even though the left and right derivatives are different functions of $n$, both of those functions take the value of $0$ at integers, since $\sin(\pi x)$ is $0$ for integer $x$. Thus, the derivatives are the same, and the derivative exists and is equal to $0$ at integers, since the left and right side limits of it match up and are equal to $0$.


问:
Integral of $1 / \sqrt {(a x + b) (p x + q)}$
I am working on this:

$\displaystyle \int \dfrac 1 {\sqrt {(a x + b) (p x + q)} } \ \mathrm d x$

valid where $(a x + b)(p x + q) > 0$.

I am specifically interested in the case where $\dfrac {b p - a q} p < 0$.

I do this:

Let $u = \sqrt {a x + b}$

$\leadsto x = \dfrac {u^2 - b} a$

$\leadsto \sqrt {p x + q} = \sqrt {p \left({\dfrac {u^2 - b} a}\right) + q}$

$= \sqrt {\dfrac {p \left({u^2 - b}\right) + a q} a}$

$= \sqrt {\dfrac {p u^2 - b p + a q} a}$

$= \sqrt {\dfrac p a} \sqrt {u^2 - \left({\dfrac {b p - a q} p}\right) }$

Then by a standard result:

$\displaystyle \int \frac {\mathrm d x} {\sqrt {(a x + b) (p x + q) } } = \int \frac {2 u \mathrm d u} {a \sqrt {\frac p a} \sqrt {u^2 - \left({\frac {b p - a q} p}\right) } u}$

$= \dfrac 2 {\sqrt {a p} } \int \dfrac {\mathrm d u} {\sqrt {u^2 - \left({\dfrac {b p - a q} p}\right) } }$

Now let $\dfrac {b p - a q} p < 0$.

Let $c^2 = -\dfrac {b p - a q} p$.

This gives us:

$\displaystyle \int \dfrac 1 {\sqrt {(a x + b) (p x + q)} } \ \mathrm d x = \frac 2 {\sqrt {a p} } \int \frac {\mathrm d u} {\sqrt {u^2 + c^2} }$

$= \dfrac 2 {\sqrt {a p} } \sinh^{-1} {\dfrac u c} + C$ (standard result)

$= \dfrac 2 {\sqrt {a p} } \sinh^{-1} \dfrac {\sqrt {a x + b} } {\sqrt {\dfrac {a q - b p} p} } + C$

(because $\dfrac {b p - a q} p = -\dfrac {a q - b p} p$)

$= \dfrac 2 {\sqrt {a p} } \sinh^{-1} \sqrt {\dfrac {p (a x + b) } {a q - b p} } + C$

But when I look in the book (in this case Murray Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968), I see this as result $14.120$:

$\displaystyle \int \dfrac 1 {\sqrt {(a x + b) (p x + q)} } \ \mathrm d x = \dfrac 2 {\sqrt {-a p} } \arctan \sqrt {\dfrac {-p (a x + b) } {b p - a q} } + C$

Can't see where I'm doing it wrong.

答:
Amplified comment: "Convert an $\arcsin$ to an $\arctan$."

I learned this long ago because certain computer languages had only $\arctan$ as built-in function, and you had to use formulas if you wanted $\arcsin, \arccos$, etc.   (BASIC had ATN arctangent, but none of the other "arc" trig functions)

So here it is:

$$
\arcsin t = 2 \arctan\left(\frac{t}{1+\sqrt{1-t^2}}\right)
$$

Here are the graphs.

Identical


答: The set $S$ is equal to$$\{a+bt+ct^2+dt^3\mid a+b=c+d\}.$$This is a $3$-dimensional vector subspace of, which contains $1+t^3$, $t+t^3$, and $t^2-t^3$, which are the elements of your basis.

Title: Writing Vectors As A Linear Combination Of Other Vectors

Body:
I've been tangled with a section from a question :

In the vector space of  $P_3(R)$ given a group of vectors :

$$S= \{u_1=1+t+t^2 , u_2=2+3t+3t^2 , u_3=2+6t-t^2 , u_4=1-2t+2t^2+5t^3\}$$

I have managed to check whether they are linearly independent or not and found the basis and the dimension of $S$

$$B_S=\{(1,0,0,1),(0,1,0,1),(0,0,1,-1)\}$$
$${\operatorname{dim}_S=3}$$

and what is left is to type all the vectors that belong to S as a combination for the basis i found , kind of stuck there , any suggestions ?


答:
It helps to understand why a certain number is terminating or non-terminating in a given base. Let us take base $b$. If you express a number $n$ in base $b$ as:
$$n=d_kd_{k-1}\cdots d_0 \space . \space d_{-1}d_{-2}\cdots d_{-l}$$
then this is equivalent to saying:
$$n=\sum_{i=-l}^k d_i \cdot b^i \implies b^ln=\sum_{i=0}^{k+l} d_{i-l} b^i$$
Now, it is clear that it is necessary $b^ln$ to be an integer for some $l \in \mathbb{N}$ for $n$ to terminate in base $b$. This is clearly a sufficient condition as well since if $b^ln$ is an integer, it clearly terminates in base $b$ (since all integers terminate in all bases), and thus, when dividing $b^ln$ by $b^l$, since it is base $b$, only digits move across the decimal point.

Hence, we have proven that a number $n$ terminates in base $b$ if and only if there exists $l \in \mathbb{N}$ such that $b^ln$ is an integer. This clearly cannot hold for irrational numbers $n$ since:
$$b^ln \in \mathbb{Z} \implies b^ln=t \implies n=\frac{t}{b^l}$$
where both $t$ and $b^l$ are non-zero integers, contradicting the fact that $n$ is irrational. Hence, irrational numbers are non-terminating in all bases.

Although, when you consider rational numbers, some rational numbers terminate in some bases but not in others. Consider the rational number $n=\frac{x}{y}$ in its simplest form. If $y$ has a prime factor $p$ that does not divide $b$, then no matter what $l$ you take, $b^ln$ will still have the prime factor $p$ in its denominator, and hence $b^ln$ cannot be an integer, which means that $n$ is non-terminating in base $b$.

If $y$ does not contain any such prime factor, then all prime factors of $y$ divide $b$. Define $l$ to be the largest power of a prime in the prime factorization of $y$. Then, it is clear that $b^ln$ is an integer since all the primes get cleared from the denominator. Hence, $n$ will terminate in base $b$.

Thus, a rational number is terminating in base $b$, if and only if for the denominator of $n$ in its simplest form, all its prime factors are also prime factors of $b$.

Title: In what base system a given decimal-based number with no terminating decimal expansion can be represented in a finite string of digits?

Body:
I was reading about precision floating numbers in the mpmath python module documentation and came across this https://mpmath.org/doc/current/contexts.html#arbitrary-precision-interval-arithmetic-iv:

The fact that '0.1' results in an interval of nonzero width indicates
that 1/10 cannot be represented using binary floating-point numbers at
this precision level (in fact, it cannot be represented exactly at any
precision).

This made me wonder if there's a way to determine in what base system a given decimal-based number with no terminating decimal expansion can be transferred in order to be represented in a finite string of digits.


问:
Why is $\frac{987654321}{123456789} = 8.0000000729?!$
Many years ago,
I noticed that $987654321/123456789 = 8.0000000729\ldots$.

I sent it in to Martin Gardner at Scientific American
and he published it in his column!!!

My life has gone downhill since then:)

My questions are:

Why is this so?

What happens beyond the "$729$"?

What happens in bases other than $10$?

Body:
$$729=9^3$$
$$66339=9^3\cdot 91$$
$$6036849=9^3\cdot 91^2$$
$$...$$
$$987654321/123456789=8+9^3\cdot 10^{-10}\cdot\displaystyle\sum_{n=0}^{\infty}(91\cdot 10^{-10})^n$$


答:
Hint:

Just  show that $\mathbf R^2$ is path-connected, noting this it is obvious for two points which are at the same distance from the origin.

Title: Prove that $\mathbb{R}^2 \setminus \{0\}$ is connected?

Body:
Prove that  $\mathbb{R}^2 \setminus \{0\}$ is connected ?

My attempt : take $A= (-\infty, 0)$ and $B= (0,\infty)$

$\mathbb{R} \setminus \{0\}=A\cup B = (-\infty, 0)\cup (0,\infty)$

squaring both side we have

$\mathbb{R}^2 \setminus \{0\}=(A\cup B)^2 = ((-\infty, 0)\cup (0,\infty))^2=(0,\infty) \cup (0, \infty)$

where$(0,\infty) \cap (0, \infty)\neq \emptyset$ since $(-\infty,0)^2= (0,\infty)$ and $(-\infty)^2=\infty$

Therefore $\mathbb{R^2} \setminus \{0\}$ is connected

Is its correct ?


问:
Limit of sin function
I need some help solving the following limit:

Here is how I started:

$$\lim_{x \to 0^+}\frac{\sin(6x)}{\sqrt{\sin(2x)}} = \lim_{x \to 0^+}\frac{\sin(6x)\sqrt{\sin(2x)}}{{\sin(2x)}}=\lim_{x \to 0^+}\frac{\frac{6 \sin(6x)}{6x}\frac{\sqrt{\sin(2x)}}{x}}{{\frac{2 \sin(2x)}{2x}}}$$

I don't know how to deal with this sqrt...

Thanks! :)

答:
The limit equals:

$$\lim_{x \to 0^+} \frac{\sin(6x)}{6x} \frac{1}{\sqrt{\frac{\sin(2x)}{2x}}} \cdot 6x \cdot \frac{1}{\sqrt{2x}} = \lim_{x \to 0^+} 1 \cdot 1 \cdot 6x \cdot \frac{1}{\sqrt{2x}} = \lim_{x \to 0^+} \frac{6}{\sqrt{2}} \sqrt{x} = 0.$$

Title: Limit of sin function


问:
Combinatorics problem with a deck of cards

I don't seem to get this question.. what exactly is going on?
Do we start with an empty sequence, pick a card from the deck (let's say 2 Heart), put it inside the sequence, so our sequence is now (2 Heart), put it back into the deck, shuffle and pick another card (let's say Queen Spade) so our sequence is now (2 Heart, Queen Spade) and so on n times?
and the question is in how many of the possible sequences all the Jacks appear at least once?
How would I approach this question?

答:
Apply Principle of Inclusion Exclusion.

Number of all possible sequences $ = 52^n \,$ as each element in the sequence can be one of the $52$. $(n \ge 4)$

Now there are $51^n$ sequences out of these where a specific jack is missing (but it will also have sequences where other jacks are missing).

Similarly $50^n$ sequences where two specific jacks are missing, $49^n$ where three jacks are missing and $48^n$ where all $4$ jacks are missing.

So applying P.I.E to take out overcounting, number of sequences where you have one or more jacks missing

$ = {4 \choose 1} \times 51^n - {4 \choose 2} \times 50^n + {4 \choose 3} 49^n - {4 \choose 4}48^n$

Now if we subtract these from $52^n$, you get all possible sequences with all jacks at least appearing once in the sequence.


Title: Limit of sin function

Body:
I need some help solving the following limit:

Here is how I started:

$$\lim_{x \to 0^+}\frac{\sin(6x)}{\sqrt{\sin(2x)}} = \lim_{x \to 0^+}\frac{\sin(6x)\sqrt{\sin(2x)}}{{\sin(2x)}}=\lim_{x \to 0^+}\frac{\frac{6 \sin(6x)}{6x}\frac{\sqrt{\sin(2x)}}{x}}{{\frac{2 \sin(2x)}{2x}}}$$

I don't know how to deal with this sqrt...

Thanks! :)

Body:
Substitute $y=2x\to 0^+$:

$$\lim_{y\to 0^+} \frac{\sin 3y}{\sqrt{\sin y}}=\lim_{y\to 0^+} \frac{3\sin y-4\sin^3 y}{\sqrt{\sin y}} =\lim_{y\to 0^+} 3\sqrt{\sin y}-4(\sin y)^{5/2} =0 $$


问:
Power of group (order 6)
Determine all powers to $x$, the result should be a group of
order 6. Is this group isomorphic with $S_3$?

$$x=\begin{pmatrix}
1 &2  &3  &4  &5 \\2
 &4  &5  &1  &3
\end{pmatrix}$$

How do i start?

答: Hint: All elements in the (sub)group that you get will commute with each other. Is that the case in $S_3$?

Title: Power of group (order 6)


问:
Why does the ODE $xy''+y'+xy=0$ require only one initial condition?
In order to solve second order differential equations with constant coefficients of the form
$$ay''+by'+cy=0$$
we need two initial conditions, for example $y(0)=1$ and $y'(0)=2$.

However, I just crossed the following differential equation with varying coefficients:
$$xy''+y'+xy=0 \\y(0)=1$$ known as the Bessel's Equation which can be solved using power series to give the solution $$y(x)=\sum_{n=0}^{+\infty}\frac{(-1)^n}{4^n(n!)^2}x^{2n}.$$
Why did this equation only require one initial condition? is it because it is not with constant coefficients?

答: Plugging in $x=0$ into the equation gives $y'(0)=0$, so the second initial condition follows from the equation.


问:
Sum of all $4$-digit numbers that can be formed with the digits $0,1,2,3$

What is the sum of all 4-digit numbers that can be formed with the digits 0,1,2,3 when repetition not allowed?

All I have been able to do is figure out that there will be $18$ such $4$ digit numbers ..but I am stuck and would like to know how we would proceed with this.

Also how would we approach this problem if repetition was allowed ?

答:
We'll find sums at each place (units, tens, etc.) and add them.

At thousands place each of $1,2,3$ occurs $6$ times. Sum is  $6(1+2+3)\cdot 1000=36000$.

For hundreds place, we first look at numbers with thousands digit $1$. These are six numbers in which each of $2,3,0$ occurs twice. Hence in the $18$ nos, each of $1,2,3$ occurs $4$ times while $0$ occurs $18-3\cdot 4=6$ times as expected. Sum at hundreds place is $4(1+2+3)\cdot 100=2400$.

Sum at tens and units places are similar - $240$ and $24$ respectively.

Desired sum is $36000+2400+240+24=38664$.

The repetition case is exactly similar and probably easier.

We have $3\cdot 4^3$ nos in all, with $4^3$ starting with $1,2,3$ each. Sum at thousands place is $4^3(1+2+3)\cdot 1000$

For hundreds place, lets fix any digit as $$\square \, 0 \,  \square \, \square$$ There'll be $3\cdot 4^2=48$ occurences of the fixed digit. Thus sum at hundreds place is $48(0+1+2+3)\cdot 100$.

Similarly at tens and units place - $48(0+1+2+3)\cdot 10$ and $48(0+1+2+3)\cdot 1$ resp.

Desired sum is $$(1+2+3)(64000 + 48\cdot 111) = 415968$$


答:
HINT: We can write $x = (124)(35)$ in cycle notation. To find the powers of $x$, we can use the fact that disjoint cycles commute.

Now, a group $G$ is cyclic if and only if $G = \langle x \rangle$ for some $x \in G$, i.e. $G$ can be generated by only one element. Is $S_3$ a cyclic group?

Title: Power of group (order 6)

Body:
Determine all powers to $x$, the result should be a group of
order 6. Is this group isomorphic with $S_3$?

$$x=\begin{pmatrix}
1 &2  &3  &4  &5 \\2
 &4  &5  &1  &3
\end{pmatrix}$$

How do i start?


问:
Show that $\int_0^a \frac{\sin x}{x} dx \leq 1+\log a$ for all $a \geq 1$
Show that $\int_0^a \frac{\sin x}{x} dx \leq 1+\log a$ for all $a \geq 1$.

My attempt: define $f(a)=1+\log a-\int_0^a \frac{\sin x}{x}dx$. Then the inequality is equivalent to show that $f(a) \geq 0$ for all $a \geq 1$.

$f$ is differentiable for all $a \geq 1$, since it a sum of differentiable functions (a constant, a logarithm and an integral of a continuous function that is differentiable in the interval $0 < a \leq 1$ for the fundamental theorem of calculus) in the interval $a\in[1,\infty)$.

SO it has sense to evaluate $f'$, and it is
$$f'(a)=\frac{1}{a}-\frac{\sin a}{a}=\frac{1}{a}\left(1-\sin a\right) \geq 0 \ \ \ \ \ \forall a \geq1$$
So $f$ is increasing for all $a \geq 1$, this means that if $f(1) \geq 0$ then $f(a) \geq 0$ for all $a \geq 1$ for monotonicity. It is
$$f(1)=1-\int_0^1 \frac{\sin x}{x} dx$$
So it is
$$f(1) \geq 0 \iff \int_0^1 \frac{\sin x}{x} dx \leq 1$$
Since $\sin t \leq t$ for all $t \geq 0$ and since $x\in[0,1]$ it is $\sin x \leq x \iff \frac{\sin x}{x} \leq 1$, so for the monotonicity of the integral it is $\int_0^1 \frac{\sin x}{x} dx \leq \int_0^1 1 dx=1$ as well; this shows that the inequality is true for all $a \geq 1$.

Is my approach correct? If there is mistakes or imprecisions let me know please.

If it is correct I would be glad to see other way to prove this. Thanks.

答:
By the self-adjointness of the Laplace transform

$$ \int_{0}^{a}\frac{\sin x}{x}\,dx = \int_{0}^{+\infty}\frac{1-(s\sin(a)+\cos(a))e^{-as}}{1+s^2}\,ds$$
hence by the Cauchy-Schwarz inequality
$$ \left|\int_{0}^{a}\frac{\sin(x)}{x}\right|\leq \frac{\pi}{2}+\int_{0}^{+\infty}\frac{ds}{e^{as}\sqrt{s^2+1}}\,ds\leq \frac{\pi}{2}+\sqrt{\frac{\pi}{4a}}. $$


答:
In order to take a permutation to a power, you just iterate. So, to get $x^2$, remember that $x$ is $$ x=\begin{pmatrix} 1 &2  &3  &4  &5 \\ 2 & 4  &5  &1  &3 \end{pmatrix}$$ So if I apply that permutation twice, $1$ will get sent to $2$ the first time, and the second time that 2 will get sent to 4. So 1 gets sent to 4. Likewise, 2 gets sent first to 4 and then to 1, 3 gets sent to 5 and then back to 3, 4 gets sent 1 and then to 2, and 5 gets sent to 3 and back to 5. So $$ x^2=\begin{pmatrix} 1 &2  &3  &4  &5 \\ 4 & 1  & 3  &2  &5 \end{pmatrix}$$ In the same way, if do the transformation again, 3 times, 1 will get sent to 2 and then to 4 and then ultimately back to 1. And so on, you can compute all of the powers of $x$. When you do it 6 times, you'll get back to the identity, so $x^0, x^1, x^2, x^3, x^4, x^5$ will form a group under composition with 6 elements.

To see why it's not isomorphic to $S_3$, as @Aryaman has noted, notice that $x^a \cdot x^b = x^{a+b}$. So $x^b \cdot x^a = x^{b + a} = x^{a + b} = x^a \cdot x^b$. So elements in this group always commute with each other, this group is abelian. The same is not true of $S_3$ (can you see why?), so this group cannot be isomorphic to $S_3$.

Title: Power of group (order 6)