问:
A basic question on the definition of $E[X]$
My question is regarding the definition of $E[X]$ in a probability book. It starts with the definition in case of a simple random variable (a random variable which takes only finite number of values) where $E[X]$ is defined by :

$$E[X]=\sum_{i=1}^{n}x_i P(X=x_i)$$

Now a 3-dimensional figure is drawn (I am confused about how to describe the figure in words) and it is said that from that figure it can be understood that the previous sum can be viewed as an integral over $\Omega$. With this motivation the following equation is written :

$$E[X] = \int _{\Omega}X(\omega)P(d\omega)$$

I am confused about the whole method of defining $E[X]$.

Actually, my question was why a "3 dimensional figure" is needed to describe expectation ? The finite sum I have mentioned can be thought of an uncountable sum $X(\omega)P(d\omega)$ over all $\omega$. Now, if our probability measure is Lebesgue measure, then the  uncountable sum is equivalent to $X(\omega)d\omega$ over all $\omega$ which is nothing but $\int_{\Omega}X(\omega)d\omega$ which is equal to $\int_{\Omega}X(\omega)P(d\omega)$

答:
The expectation value is an average. For example, if I roll a pair of six sided dice a number of times and record the values, what is the average value I will get?

To see why the formula is what it is, let's look at the average value of a set of numbers. To calculate the average of a set of values, we sum the values and divide by the number of values, i.e. $$\bar x=\frac{1}{n}\sum_{i=1}^{n}x_i$$ Now suppose the values of $x_i$ cannot be any number but are restricted to some finite set of values as in the dice example. We can rewrite the above formula as $$\bar x=\frac{1}{n}\sum_{i=1}^{k}n_ix_i$$ where $n_i$ is the number of times we get the value $x_i$. Note that $\sum_{i=i}^{k}n_i = n$ and $k$ is the number of values the $x_i$ can be. In our dice example, $x_i\in \{2,3,..., 12\}$ so  $k=11$.

If we move the $\frac{1}{n}$ inside the sum we get $$\bar x=\sum_{i=1}^{k}\frac{n_i}{n}x_i$$ The fraction $\frac{n_i}{n}$ is the number of times we got $x_i$ divided by the total number of times we recorded a value, i.e. $\frac{n_i}{n}$ is the probability of getting a value of $x_i$. Therefore, we can rewrite this average as $$\bar x=\sum_{i=1}^{k}P(x_i)x_i$$ which is the expectation value.


问:
Dirichlet form of 2 Complex Functions is a conformal invariant
Let complex functions $g,h$ be defined on a bounded domain $U$. The Dirichlet form of $g$ and $h$ is:
$$
D_U(g,h)=\int\int_U \left [ \frac{\partial g}{\partial x}\frac{\partial \bar{h}}{\partial x} +\frac{\partial g}{\partial y}\frac{\partial \bar{h}}{\partial y} \right ] dxdy
$$

I am trying to show that the Dirichlet form is conformal invariant: Let $z = f(\xi)$ be a 1-1 analytic function from a bounded domain $V$ to $U$, then:
$$
D_U(g,h)=D_V(g \circ f, h \circ f)
$$

I can see quite clearly the image of this theorem. However, I couldn't find a way to transform the integral from in $V$ to in $U$.

答:
Expressing the partial derivatives as column vectors, we have $D_U(g,h)=\int\int_U\left[\left(\frac{\partial\bar{h}}{\partial (x, y)}\right)^T  \frac{\partial g}{\partial (x, y)}\right]dxdy.$

The Jacobian matrix of $f$ can be written as $f'=\frac{\partial(x,y)}{\partial(s,t)}=\rm{Re}(f') \left(\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right) + \rm{Im}(f') \left(\begin{array}{cc}
0 & -1\\
1 & 0
\end{array}\right).$

Now $\left[\left(\frac{\partial\bar{h}}{\partial (s, t)}\right)^T  \frac{\partial g}{\partial (s, t)}\right]dsdt=\left[\left({f}'\frac{\partial\bar{h}}{\partial (x, y)}\right)^T  f' \frac{\partial g}{\partial (x, y)}\right]dsdt=\left[\left(\frac{\partial\bar{h}}{\partial (x, y)}\right)^T   \frac{\partial g}{\partial (x, y)}\right][(\rm{Re}(f')^2-\rm{Im}(f')^2)dsdt]=\left[\left(\frac{\partial\bar{h}}{\partial (x, y)}\right)^T \frac{\partial g}{\partial (x, y)}\right]dxdy.$

ContentLicense: CC BY-SA 4.0


问:
How to show this inequality?

Show that $$-2 \le \cos \theta(\sin \theta+\sqrt{\sin^2 \theta +3})\le2$$

Trial: I know that $-\dfrac 1 2 \le \cos \theta\cdot\sin \theta \le \dfrac 1 2$ and $\sqrt 3\le\sqrt{\sin^2 \theta +3}\le2$. The problem looks simple to me but I am stuck to solve this. Please help. Thanks in advance.

答: $ \left[\cos \theta\sin \theta+\cos\theta\sqrt{\sin^2 \theta +3})\right]^2\le \left(\cos^2\theta+\bigg(\sqrt{\sin^2\theta+3}\bigg)^2\right)\left(\cos^2\theta+\sin^2\theta\right)=4$ by the Cauchy-Schwarz inequality which  gives you the desired result.</p>

ContentLicense: CC BY-SA 3.0

Title: How to show this inequality?


问:
If $A=(0,1)$ in a topological space, how would you find its boundary?
Let $\partial A=\overline{A}\cap\overline{X\setminus A}$ be the boundary of $A$ in $X$.

Given $A=(0,1)$, $\overline{A}=[0,1]$ and $\overline{X\setminus A}=(-\infty,0]\cap[0,+\infty)$, then
$$\partial A=[0,1] \cap \big((-\infty, 0] \cap [0,+\infty)\big)=\{0,1\}\text.$$
So, $\partial A=\{0,1\}$. Is this correct?

答:
Yes, the procedure is right (you just apply the topological definition of boundary, actually). Assuming that $X=\mathbb R$ with the usual (Euclidean) topology, of course.

Only one note: the closure of the complementary set $\overline{X\smallsetminus A}$ is $(-\infty,0]\cup[1,+\infty)$, not $(-\infty,0]\cap[0,+\infty)$.

But I assumed it was just a typo, since you got the correct final result.


问:
Question from Gagilardo-Nirenberg Sobolev inequality

Why the constant C in this inequality depends on $q$ but not depend on $u$.

In the proof $C= (\int_{\mathbb{R}^n} |u|^{(\gamma -1)\frac{p}{p-1}}dx )^\frac{p}{p-1}$.

Can anyone explain why it does not depend on $|u|$?

答:
That is not what the proof tells you. It is derived
$$\left(\int |u|^{\frac{\gamma n}{n-1}}\right)^\frac{n-1}{n} \leq \gamma \left(
 \int |u|^{(\gamma-1)\frac{p}{p-1}} \right)^\frac{p-1}{p} \|Du\|_{L^p}^p.$$
But the proof does not stop here. Now, $\gamma>1$ is chosen s.t. $\frac{\gamma n}{n-1}=(\gamma-1)\frac{p}{p-1}=p^*$, hence it implies
$$\left(\int |u|^{p^*}\right)^\frac{n-1}{n} \leq \gamma \left(
 \int |u|^{p^*} \right)^\frac{p-1}{p} \|Du\|_{L^p}^p.$$
Hence
$$\left(\int |u|^{p^*}\right)^{\frac{n-1}{n}-\frac{p-1}{p}} \leq \gamma \|Du\|_{L^p}^p.$$
Notice $\frac{n-1}{n}-\frac{p-1}{p}=\frac{1}{p^*}$, i.e., $C=\gamma$.


问:
Create a function such that:
$f(3) = 0$, $f$ has a horizontal asymptote at $y = 2$, $f$ has vertical asymptotes at $x = 4$ and $x = -4$ and $f(0) = 1$. Basically, write a function for a graph that will match these criteria.

答:
Starting with the asymptotes, a good candidate is any function of the form
$$f(x) = 2 + \frac{ax+b}{(x-4)(x+4)}.$$
Notice this has division by zero when $x=\pm4$, which gives you the desired vertical asymptotes, and also when $x\to\pm\infty$, the value of $f(x)$ gets closer and closer to $2$; this is what it means to have a horizontal asymptote at $y=2$. (More precisely, $f(x) = 2 + O(\frac1x) \sim 2$.)

Now for the other two requirements, you can substitute $x=3$ and $x=0$ into $f(x)$ so that $f(3)=0$ and $f(0)=1$ give you simultaneous equations in $a$ and $b$, where you should find that $a=-\frac23$ and $b=16$. Thus the function
$$f(x) = 2 + \frac{-\frac23x+16}{(x-4)(x+4)}$$
which simplifies to
$$f(x)=\frac{2 (x-3) (3 x+8)}{3 \left(x^2-16\right)}$$
satisfies all the requirements.

Title: Create a function such that:


问:
Injective representation of nonabelian group
Let $H$ be a nonabelian group of order $pq$ where $q<p$ are prime. Then show that $H$ has an injective representation $\rho:H \to GL_2(\mathbb{F}_p)$.

What I have shown so far is that if $H$ is a group of order $pq$, then it is either $C_{pq}$ or it is $C_p \rtimes_{\phi} C_q$ where $\phi:C_q \to Aut(C_p)$ is injective, so because $H$ is nonabelian it must be the latter. However I am not sure where to go from here.

答:
There exists only  one $H_p$ Sylow subgroup of order $p$, otherwise you have at least $(p+1)(p-1)=p^2-1>pq$. This imples that $H_p$ is normal. Let $H_q$ be a Sylow subgroup of order $q$, $H_q$ acts non trivially on $H_p$ by conjugation since $G$ is not commutative, since $H_q$ acts by linear maps of the vector space $\mathbb{F_q}=H_p$, we deduce that $H_q\subset \mathbb{F}^*_p =\mathbb{F}_p-\{0\}$.

This implies that you can embed $G$ in $Gl_2(\mathbb{F}_p)$ has a subgroup of
$\pmatrix{a&c\cr 0&1}$ where $H_p$ is the subgroup $\pmatrix{1&c\cr 0&1}$ and $H_q$ is a subgroup of $\pmatrix{a&0\cr 0&1}$.


问:
An open set containg all irrationals
Let $A$ be an open set in $\mathbb{R}$ with usual topology and $A \cup\mathbb Q=\mathbb{R}$. Does it imply $A=\mathbb{R}$ ?</p>

答:
No, I can remove any set of rationals that contains all its limit points.

So I could remove $\mathbb Z$ for example or $\{0\}\cup\{\frac 1n : n\in\mathbb N\}$ for example, or many more complex ones.

Title: An open set containg all irrationals


问:
Formula to work out likely number of unique items
There is an unknown quantity of unique prizes. Each time you play the game, you get a prize. The likelihood of getting any type of prize X is equal to getting the likelihood of any other type, and winning a prize does not decrease the likelihood that that type will be won in future.

I won 14 prizes. Where '1' represents a type of prize I have never won before, and '0' represents winning a type of prize I have won before, I got the following ordered list:

(1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0)

I know that each time I play the game, the probability of getting a prize that I have not won before is (N-a)/N where a is the number of unique prizes I have already won.

What formula can I use to determine what the most likely number of prizes is?

答:
There is more than one possible approach.

One is to say that if there are $n$ types of prizes then the probability of the pattern you have seen (i.e. the likelihood) would have been $\frac{n}{n}\frac{n-1}{n}\frac{n-2}{n}\frac{n-3}{n}\frac{n-4}{n}\frac{n-5}{n}\frac{n-6}{n}\frac{n-7}{n}\frac{8}{n}\frac{8}{n}\frac{n-8}{n}\frac{n-9}{n}\frac{10}{n}\frac{10}{n}$.  If you try to find the $n$ which maximises this, you will find the maximum likelihood estimate of $n$.  I think it may give  $\hat n \approx 17.9741$ or $18$ if you restrict $n$ to an integer

An alternative might be to say that if there are $n$ types, then the expected number won after $14$ prizes is $n\left(1 - \left(1-\frac1n\right)^{14}\right)$ and if you set this equal to $10$ then you would get the method of moments estimate which I think is $\hat n \approx 18.4865$

A third method might be to assign a (possibly impropor) prior distribution to $n$, use the likelihood to update this to a posterior distribution for $n$, and then chose one of the central statistics of the posterior distribution such as the mean, median or mode as your estimator.

In any case, $18$ looks a plausible estimate, but it is quite possible that $n$ is some other value and the third method would give you a reasonable sense of that uncertainty.


问:
What's the intuition behind $T^*T$?
Let $T : V \to W$ be a linear transformation across inner product spaces over some field $\mathbb{F}$ and $T^* : W \to V$ be its adjoint. When considering $T$ composed with $T^*$, we have some unique properties:

If $T^*T = TT^*$, $V = W$, and $\mathbb{F} = \mathbb{C}$, then $V$ has an orthonormal eigenbasis with respect to $T$.

The eigenvalues of $T^*T$ are the singular values of $T$, which allows us to express $T$ as a diagonal matrix with respect to two orthonormal bases.

All of this seems pretty cool, but I don't really have any intuition for what $T^*T$ (or $TT^*$) means. Does anyone have an intuitive way of thinking about the transformation $T^*T$?

答:
This transformation is called the Gramian matrix. It serves a multitude of roles (such as SVD/PCA). It is a "sort of squaring" for matrices (that can be rectangular, unlike the normal squaring). The best analogy for this "sort of squaring" I know of is in the complex numbers. Where $T^2$ is like $z^2$, you can think of $T^* T$ to be like $z \bar{z} = |z|^2$. For complex-valued matrices of size $1*1$, these two notions are even precisely equivalent. Both the (conjugate) transpose and conjugation are involutions; and this corresponds to the product of a value with its involution.

Also note, if your product is non-commutative (or only defined one way around), then $T^*T$ and $TT^*$ might not be equal.

Title: What's the intuition behind $T^*T$?


问:
Nearly Consistent, is the following true?
Continuing my previous question:

Let A be a set of axioms, we say that A is nearly consistent is there
is a finite sub set of A (A') for which A\A' is consistent.

Is it true to say that:

A is nearly consistent iff each finite sub set of A is  nearly consistent?

Which side is correct and which one isn't?

答:
The answer is essentially contained in the answers/comments to your previous question:

Using the fact that a subset of a consistent set is consistent, we can easily prove that any subset of a nearly consistent set is nearly consistent. (This is basically the content of J.G.'s answer.) So the left-to-right direction holds.

In the other direction, in the comments below my answer I mentioned that every finite set of sentences is nearly consistent (do you see why?). So no matter what $A$ is, every finite subset of $A$ is nearly consistent. From this it's easy to get a counterexample to the right-to-left direction:

 Just take $A$ to be any set which is not nearly consistent.

(Note that the fact that every finite set is nearly consistent makes my argument in the first bulletpoint unnecessary, but it's still worth understanding.)


问:
The smallest number that can be discribed as a sum of three squares in two different ways with equal multiple.

What is the smallest number $k$ that is a sum of three squares in two different ways(there does not exist a pair of numbers from different triple which are the same)
$$a_0^2+b_0^2+c_0^2=a_1^2+b_1^2+c_1^2=k$$
Such that $a_0b_0c_0=a_1b_1c_1$?

Also, does there exist a solution in two numbers?

Note: All the numbers mentioned are non-zero natural numbers

My Attempt:

I’ve started with the two variable case as it seemed easier, and I got to the equation:
$$a_0^2k-a_0^4=a_1^2k-a_1^4$$
Needing to have two solutions where $a$ and $b$ are distinct for some fixed $k$.

It looks like it implies no solutions but I had not managed to prove it.

I’ve also tried parity arguments to no success.

I've found a pair of triples, $1^2+12^2+10^2=2^2+4^2+15^2 = 245$ and $1\cdot 12 \cdot 10 = 2\cdot 4\cdot 15=120$ but I'm unsure of it's minimality.

Body:
Triples under $99$ (excluding equal product considerations) include

$$1^2 + 4^2 + 4^2 = 2^2 + 2^2 + 5^2 = 33$$
$$1^2 + 1^2 + 6^2 = 2^2 + 3^2 + 5^2 = 38$$
$$1^2 + 2^2 + 6^2 = 3^2 + 4^2 + 4^2 = 41$$
$$1^2 + 2^2 + 7^2 = 2^2 + 5^2 + 5^2 = 3^2 + 3^2 + 6^2 = 54  $$
$$1^2 + 5^2 + 6^2 = 2^2 + 3^2 + 7^2 = 62 $$
$$1^2 + 1^2 + 8^2 = 1^2 + 4^2 + 7^2 = 66$$
$$1^2 + 2^2 + 8^2 = 2^2 + 4^2 + 7^2 = 69 $$
$$1^2 + 3^2 + 8^2 = 3^2 + 4^2 + 7^2 = 74 $$
$$2^2 + 3^2 + 8^2 = 4^2 + 5^2 + 6^2 = 77 $$
$$1^2 + 4^2 + 8^2 = 3^2 + 6^2 + 6^2 = 81 $$
$$1^2 + 1^2 + 9^2 = 3^2 + 5^2 + 7^2 = 83 $$
$$1^2 + 2^2 + 9^2 = 1^2 + 6^2 + 7^2 = 5^2 + 5^2 + 6^2 = 86 $$
$$2^2 + 2^2 + 9^2 = 2^2 + 7^2 + 6^2 = 3^2 + 4^2 + 8^2 = 89 $$
$$1^2 + 5^2 + 8^2 = 4^2 + 5^2 + 7^2 = 90$$
$$2^2 + 3^2 + 9^2 = 3^2 + 2^2 + 9^2 = 3^2 + 6^2 + 7^2 = 94$$
$$1^2 + 4^2 + 9^2 = 3^2 + 5^2 + 8^2 = 98 $$
$$1^2 + 7^2 + 7^2 = 3^2 + 3^2 + 9^2 = 5^2 + 5^2 + 7^2 = 99$$
I put these and all others up to 245 in a spreadsheet with product formulas for each triple. Only those for the sum 245 had more than one triple with the same product.


问:
Gradient of distance-squared on Riemannian manifold
Problem:

Let $M$ be a Riemannian manifold. Consider the function $f: M \rightarrow \mathbb{R}$ where $f(x)=\text{dist}_M^2(p,x)$, and $p \in M$ is fixed. Show that $\text{grad}(f)=-2\exp^{-1}_x(p)$ as vectors in $T_xM$. (Assuming that $\exp^{-1}$ exists and is smooth etc.)

My attempt at a proof:

We must show that $\langle -2\exp^{-1}_x(p), \cdot \rangle_x = df(\cdot)$ as 1-forms at $x$.

Let $e_1,\dots,e_n$ be an orthonormal basis of $T_pM$ and introduce normal coordinates at $p$: $x=(x^1,\dots,x^n) \leftrightarrow x=\exp_p(x^i e_i)$. If $q=\exp_p(v)$ then $\text{dist}_M (p,q)=||v||_p$ and so $f(x)=||x^i e_i||^2_p=\sum (x^i)^2$, where the second equality follows from the fact that the metric is the identity at $p$ (in these coordinates).

So $df=\sum 2x^i dx^i$. $(*)$

On the other hand, the geodesic "from x to p" is the same as the geodesic "from p to x", except that the direction is reversed. It should therefore be true that
$p=\exp_x(-S_{p \rightarrow x}x^i e_i )=\exp_x(-x^i S_{p \rightarrow x} e_i)$ where $S_{p \rightarrow x}$ is the parallel transport from $p$ to $x$. Hence $\exp^{-1}_x(p)=-\sum x^i S_{p \rightarrow x} e_i $. Hence
$\langle -2\exp^{-1}_x(p), \cdot \rangle_x=\langle 2x^i S_{p \rightarrow x}e_i, \cdot \rangle = 2x^i \langle S_{p \rightarrow x} e_i, \cdot \rangle = 2x^i dx_i$ which gives the result on comparison with $(*)$. $\square$

My question:

When writing that out, I felt like I was manipulating symbols without really understanding what they mean. For example, when I write $x^i$ I'm not sure whether I mean the coordinate function $x^i$ or its particular value at the point $x$. I was also concerned that when I talk about parallel transport I should really have been talking about the derivative of $\exp$. There's a lemma in Do Carmo which states $\langle (D\exp_p)_v(v),D\exp_p)_v(w)\rangle=\langle v,w \rangle$ for $v \in T_pM$ and $w \in T_v(T_pM) \simeq T_pM$. I can see this might be useful, though I'm not sure precisely how.

I would appreciate it very much if you checked to see if this proof is correct and "expand it out", particularly the second half) including more details to make clear what is going on.

答:
Here is another proof that uses the Gauss lemma which I believe is what you are referring to in do Carmo.

Let $\gamma_t(s)$ be a geodesic variation through with $\gamma_0(0) = p, \gamma_0(1) = x$ where we note that $\frac{d}{dt}\bigg\vert_{t = 0}\gamma_0(t) = \dot{\gamma}(0) = \log_p(x)$ by definition of $\exp_p$. By reversing the direction of this geodesic we find that $\frac{d}{dt}\bigg\vert_{t = 0}\gamma_0(1-t) = -\dot{\gamma}(1) = \log_x(p)$. This is exactly the statement that you wanted regarding parallel transport, the parallel transport of $\dot{\gamma}(0)$ along $\gamma_0(t)$ gives $\dot{\gamma}(1)$ at $x$ (as the velocity of a geodesic is parallel) which is equal to $-\log_x(p)$.

Take $\frac{d}{ds}\gamma_s(1) = v$. The squared distance function can be written as $f(x) = d(x,p)^2 = \vert \log_p(x) \vert^2$ so that $$df(v) = \frac{d}{ds}\bigg\vert_{s = 0}\vert \log_p(\gamma_s(1))\vert^2  = 2\langle \log_{p*}(v),\log_p(\gamma_0(1))\rangle_p$$
Note that as a function of $s$, $\log_p(\gamma_s(1))$ always lies in $T_pM$ and thus the  inner product in this expression is the inner product of $T_pM$ and so it is as if we are working in $\mathbb{R}^n$ (points in $T_pM$ and tangent vectors can be canonically identified, justifying the last equality above). Under this canonical identification of a vector space and its tangent spaces we have $\log_p(\gamma_0(1)) = \log_{p*}(\dot{\gamma}_0(1))$ which can be seen by finding the coordinate expression for $\gamma_0(t)$ in normal coordinates.

Now we use the Gauss lemma, as $\dot{\gamma}_0(1)$ is a radial the Gauss lemma implies that

$$df(v) =  2\langle \log_{p*}(v),\log_p(\gamma_0(1))\rangle_p = 2\langle \log_{p*}(v),\log_{p*}(\dot{\gamma}_0(1)) \rangle_p = 2\langle v,\dot{\gamma}_0(1) \rangle_x$$
Thus $\text{grad}(f) = 2\dot{\gamma}_0(1) = -2\log_x(p)$ as desired.

Your proof looks to be more or less correct other than the last step which I believe requires some extra justification like the Gauss lemma. The reason being that the normal coordinate vector fields $\partial_i$ are not parallel in general (only the radial ones are!). We thus cannot conclude that the parallel transport of $\partial_{i,p}$ to $x$ is equal to $\partial_{i,x}$.


问:
How do we know if two objects in a category are the same or not?
I am wondering whether the category of sets, $\mathbf{Set}$, and the category of groups, $\mathbf{Grp}$, are well defined or not.

Suppose we choose a singleton $\{a\}$ from $\mathbf{Set}$. And suppose we choose a singleton $\{b\}$ from $\mathbf{Set}$.  How could we possibly know if we have chosen the same singleton or not?

Similarly, if we refer to a singleton $\{a\}$, then how could we possibly refer to it a second time and be convinced that it is indeed the same singleton?

How can I know if $\{\text{“horse”}\}$ and $\{\text{“caballo”}\}$ are different singletons, or simply two expressions of the same singleton?

The same questions come up with trivial groups.  Suppose I choose two trivial groups. How do I know if I have chosen the same one twice or not?

And how can I tell whether or not the category $\mathbf{Set}$ contains multiple instances of the same singleton?

It seems that there is no possible language that could make sense of the objects of $\mathbf{Set}$.  In that case, it seems fair to admit that $\mathbf{Set}$ is, on some level, ill defined. Please correct me! or please affirm me.

A similar question was asked earlier, How to treat isomorphic objects in a category?, but I don't think the answers address my concerns.

Personally, it would make sense to me if the objects in $\mathbf{Set}$ that are equivalent were defined to be equal, as sets.  But that's apparently not the way it works in category theory. You can have many distinct sets that are all equivalent, from what I understand.

答:
When you denote the sets with the symbolic names "{a}" and "{b}," you are also intrinsically utilizing the labels "a" and "b" as names referring to certain objects. When two symbolic names X and Y, for example, are simply two names with the same referent, then we write "X=Y," which is supposed to mean that the two names refer to the same object. By the definition of equality for sets, the set named "{a}" and the set named "{b}" are the same set when the object named "a" and the object named "b" are the same object. So your question defers to that of when objects are numerically identitical in general, or at least when the objects that are elements of sets are numerically identical. The answer will be context dependent, i.e. it will depend on the nature of the objects or the type of object to which your labels are referring, in so much as the type or nature of the object will determine the properties of that object that constitute its numerical identity.

Language in general, however, is a social construct. You have to agree with your audience about what your symbols mean or else you can't even get off the ground. For example, I can read your question and understand what it means, and try to answer it. But to do that I have to parse your language and assign meanings to the sentences you have written in accordance with the English language, otherwise successful communication with you is impossible. You likewise must be able to comprehend my message here for that communication to occur. Do you understand what I mean?

You have to understand that the letter "a," for example, is the same symbol with the same context-dependent functions as any other letter "a," independent of font or position or time. The same question that you ask could be asked of the symbols you use to refer to an object. For example, how do I know that the symbol you used in referring to the set {a} is even the same symbol you used when referring to a set {a} the second time, let alone whether it is referring to the same object?

The answer is that we both understand each other, we both speak English, and we have social agreements about how we practice communication.


问:
Is it guaranteed to be true that a subspace can be the empty set?
Just like the title, but can a subspace be the empty vector set? Is it GUARANTEED to be true?</p>

答:
No. One of the axioms in the definition of a vector space postulates the existence of the zero vector.

Therefore, every vector space has at least one element.


问:
Unfamiliarity with complex geometry
With some drawings, the answer is likely a line perpendicular to $AO$.

In the plane are given a circle with center $O$ and radius $r$ and a point $A$ outside the circle. For any point $M$ on the circle, let $N$ be the diametrically opposite point. Find the locus of the circumcenter of triangle $AMN$ when $M$ describes the circle.

If we let $O'$ be the circumcenter of $\triangle AMN$ and the projection of $O'$ unto $AO$ to be $P$, then the plan is to prove that $P$ is not dependent upon $\angle AON$. There is probably a cozy synthetic method, but I'd like to get some practice with complex numbers.

Let $\angle AON=\alpha,|AO|=L.$ Let $m=0,n=2,a=Le^{i\alpha}+1,$ such that the diameter of the circle is $2$. I know of a formula $\frac{xy(\overline{x}-\overline{y})}{\overline{x}y-x\overline{y}}$ to calculate the circumcenter of a triangle $XYZ$ when $Z=0.$ I'm not sure if my usage here is correct:

Substituting our values with $x=n=2$ and $y=a=Le^{i\alpha}+1$, we get $$O'=\frac{2(Le^{i\alpha}+1)(2-(Le^{-i\alpha}+1))}{2(Le^{i\alpha}+1)-2(Le^{-i\alpha}+1)}=\frac{1-L^2e^{2i\alpha}}{2L\sin\alpha},$$

where we get that second step from a denominator of $Le^{i\alpha}-Le^{-i\alpha}.$ That simplifies to $$O'=\frac{1-L^2(\cos2\alpha + i\sin2\alpha)}{2L\sin\alpha}.$$

To continue, we find $p$ as the foot of the altitude from $O'$ to $AO$. However, it turned out for me that I didn't get to anything meaningful because $p$ was dependent upon $\alpha$. I would write it here, but I'm worried that my result above is already incorrect. Is there anything I can fix, or does that look fine and that using the projection formula and some bash would get me to the result of $p$ being independent from $\angle AON$, and hence that $O'$ would be along the line $AO$?

答:
It is hard to understand what happens in the computations. (Starting from $m=0$ and $n=2$. If these are the afixes of $M,N$, then i expect them to move around a center.)

Here are some words and a complex numbers approach to the story.
For a point $Q$ in the plane i will denote by $z_Q\in\Bbb C$ the afix of $Q$.

Here is a picture for the convenience of the reader.

We will place $A$ in the origin, since we will use a formula which is nice to use when $A$ is in the origin.
The center $C$ of the circle, where the opposite points $M,N$ move on is taken on the real axis to simplify calculations, so let $c\in\Bbb R$ be the afix of $C$.
(I am not using the letter $O$ from the OP to avoid confusions.)
Let $O'$ be the circumcenter of $AMN$. Then we have:
$$
\begin{aligned}
z_A &= 0\ ,\\
z_C &= c\in\Bbb R\ ,\\
z_M &= c+re^{it}\text{ for some suitable }t\in\Bbb R\ ,\\
z_N &= c-re^{it}\ ,\\
z_{O'} &=\frac{z_Mz_N(\bar z_M-\bar z_N)}{\bar z_Mz_N-\bar z_N z_M}\\
&=
\frac
{(c^2-r^2e^{2it})\Big(\ (c+re^{-it})-(c-re^{-it})\ \Big)}
{(c+re^{-it})(c-re^{it})-(c-re^{-it})(c+re^{it})}
\\
&=
\frac 1
{-2cr(e^{it}-e^{-it})}
(c^2-r^2e^{2it})\cdot 2re^{-it}
\\
&=\frac 1{c\cdot 2i\sin t}\color{blue}{(r^2e^{it}-c^2e^{-it})}\ .
\end{aligned}
$$
The quantity in the blue bracket is explicitly
$$ (r^2-c^2)\cos t + (r^2+c^2)i\sin t\ .$$
This shows that the real part of $z_{O'}$ is constant,
$$
\Re z_{O'}=\frac{(r^2+c^2)i\sin t}{c\cdot 2i\sin t}
=
\frac 1{2c}(r^2+c^2)\ .
\ .
$$

This corresponds to the "plan" from the OP, the projection of $O'$ on $AC$ (the real axis) is a fixed point, the point $P$ with $z_P=\frac 1{2c}(r^2+c^2)\in\Bbb C$.

$\square$

(The imaginary part involves the cotangent function, so that the whole perpendicular in $P$ on $AC$ is taken, not only a part of it.)

Title: Unfamiliarity with complex geometry


答:
Here's some guidance for answering this for yourself: try coming up with some functions that satisfy each requirement individually, then see if we can combine them in a way that doesn't break anything:

$f(3) = 0$. We could take the zero function, but that would likely cause problems with the combining, so instead, we'll look for something that's only zero at $3$.

A vertical asymptote at $x = 4$. The obvious function here has no problems.

A vertical asymptote at $x = -4$. Again, choose the obvious function.

At this point, you can combine these in a fairly obvious way to get a function that satisfies all three properties. I'll call that function $g$.

You can then modify $g$ to get some new function $h$ satisfying your first four requirements by just scaling it in $y$, assuming that $g$ has neither a zero nor an asymptote at $x = 0$. Otherwise, fix that first.

A horizontal asymptote at $y = 2$. First, make sure $h$ has a horizontal asymptote somewhere. If it's at $2$ already, you're done. If it's at some $c \neq 2$, construct a function $k$ which (a) is zero at $0$, (b) is zero at $3$, and (c) has a horizontal asymptote at $2 - c$, then take $f = h + k$. Points (a) and (b) are fairly easy: proceed as in (1) above. For (c), first make it have a horizontal asymptote somewhere other than $y = 0$, then scale it in $y$ to put the asymptote in the right place.

Title: Create a function such that:

Body: $f(3) = 0$, $f$ has a horizontal asymptote at $y = 2$, $f$ has vertical asymptotes at $x = 4$ and $x = -4$ and $f(0) = 1$. Basically, write a function for a graph that will match these criteria.


问:
Finding four square roots of $\small\begin{bmatrix}3 & -4\\4 & 3\end{bmatrix}$
This question is about finding the 4(!!) square roots of a 2 by 2 matrix. The method I use is through Diagonalization (or actually doing the "opposite"). I have no problem finding the four roots of a 2 by 2 matrix that has real eigenvalues (and thus real eigen vectors). Heck, I found 8 roots of a 3 by 3 matrix that has real eigen values. The problem I am running into is if the to be square rooted matrix has complex eigenvalues. In my view, the method should work the same, but somehow the algebra is letting me down. Example: Find the four square roots of
$ \begin{bmatrix}3 & -4\\4 & 3\end{bmatrix}$. Note, I found this matrix by squaring $ \begin{bmatrix}2 & -1\\1 & 2\end{bmatrix}$, so obviously this is one root as well as taking opposite of its entries. So here is my work: The characteristic equation of $ \begin{bmatrix}3 & -4\\4 & 3\end{bmatrix}$ is $(3-\lambda)^2+16=0$ from which we find $\lambda=3+4i$ and $\lambda=3-4i$. An eigenvector for $\lambda=3+4i$ is $ \begin{bmatrix}1 \\-i  \end{bmatrix}$ and an eigenvector for $\lambda=3-4i$ is $ \begin{bmatrix}1 \\i  \end{bmatrix}$. So the given matrix can now be diagonalized: $ \begin{bmatrix}3 & -4\\4 & 3\end{bmatrix}$=$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix}$$ \begin{bmatrix}3+4i & 0\\0 & 3-4i\end{bmatrix}$$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix}^{-1}$. I found$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix}^{-1}$=$\frac{1}{2i} \begin{bmatrix}i & -1\\i & 1\end{bmatrix}$ which is $ \begin{bmatrix}0.5 & 0.5i\\0.5 & -0.5i\end{bmatrix}$. It's easy to find out that the square roots of $3+4i$ are $2+i,-2-i$, and of $3-4i$ are $2-i,-2+i$. So now I would arrive at my 4 solutions:$$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix} \begin{bmatrix}2+i & 0\\0 & 2-i\end{bmatrix} \begin{bmatrix}0.5 & 0.5i\\0.5 & -0.5i\end{bmatrix}$$
$$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix} \begin{bmatrix}2+i & 0\\0 & -2+i\end{bmatrix} \begin{bmatrix}0.5 & 0.5i\\0.5 & -0.5i\end{bmatrix}$$
$$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix} \begin{bmatrix}-2-i & 0\\0 & 2-i\end{bmatrix} \begin{bmatrix}0.5 & 0.5i\\0.5 & -0.5i\end{bmatrix}$$
$$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix} \begin{bmatrix}-2-i & 0\\0 & -2+i\end{bmatrix} \begin{bmatrix}0.5 & 0.5i\\0.5 & -0.5i\end{bmatrix}$$
When I work out my first candidate, I get a matrix, but when squared it doesn't give me $ \begin{bmatrix}3 & -4\\4 & 3\end{bmatrix}$ and so something isn't right. I don't know what I did wrong and ANY help from the math community is greatly appreciated. I feel that my line of thinking is correct, where did I go wrong? And does anyone happen to know online software that can do matrix multiplications with complex entries, because the TI doesn't work

答:
Let $A = [\begin{smallmatrix}3 & -4\\4 & 3\end{smallmatrix}]$ and $\chi_A(\lambda)$ be its characteristic polynomial. Since
$$\chi_{A}(\lambda) = \det(\lambda I_2 - A) = \lambda^2 - 6\lambda + 25$$
By Cayley Hamiltonian, we have
$$A^2 - 6A + 25 I_2 = 0
\iff (A+5I_2)^2 = 16 A \iff (A - 5I_2)^2 = -4A
$$
Taking "square root" on both sides and rearrange, one find $A$ has following four square roots.
$$\pm \frac14(A + 5I_2) = \pm \begin{bmatrix}2 & - 1\\1 & 2\end{bmatrix},\quad \pm \frac{i}{2}(A - 5 I_2) = \pm i\begin{bmatrix}-1 & -2\\2 & -1\end{bmatrix}$$


问:
Find the norm of a bounded compact operator
In space $C[0,1]$ is defined bounded and compact operator $A$:
$$ Ax(t)= 2x(0) - tx(1),\enspace  t\in [0,1] $$
Find the norm of operator $A$. ($\|A\|=$ ?)

Body:
The norm of $A$ is $3$.
$$
|Ax(t)|\leq 2|x(0)|+t\,|x(1)|\leq 2\|x\|+\|x\|=3\|x\|.
$$
So $\|A\|\leq 3$. Now let $x(t)=1-2t$. The range of $x$ is $[-1,1]$, so $\|x\|=1$. And
$$
|Ax(1)|=|2x(0)-x(1)|=2-(-1)=3.
$$
Thus $\|A\|\geq3$, showing the equality.


问:
What are good ways of understanding a permutation group from a set of generators?
I'm trying to understand the structure of a Rubik's Cube-style puzzle I'm playing with; I have an expression of the solutions as the permutation group generated by four elements of $S_{16}$, each a simple cycle. My first inclination is to check all words of a given length (I think I can check all 8-element words easily, and maybe all 9- or 10-element words with a little effort) to see which ones move as few pieces as possible and then try and use that to understand more of the structure (for instance, I already know that there are elements of odd sign, so it's not a subgroup of $A_{16}$); above and beyond that, though, what tricks and techniques should I be aware of?  In particular, are there any good ways of finding useful/meaningful expressions of my group as some sort of wreath product, or of understanding its quotient in $S_{16}$ (assuming it's even normal!)?  I'm presuming that understanding $S_{16}/G$ will provide a sense of the constraints that any valid position has to satisfy...

EDIT: For anyone who's curious enough and has better tools to play with it than I do, the four generators are the cycles: $\alpha = (1\quad 2\quad 3\quad 7\quad 11\quad 10\quad 9\quad 5)$ , $\beta = (2\quad 3\quad 4\quad 8\quad 12\quad 11\quad 10\quad 6)$, $\gamma = (5\quad 6\quad 7\quad 11\quad 15\quad 14\quad 13\quad 9)$, and $\delta = (6\quad 7\quad 8\quad 12\quad 16\quad 13\quad 14\quad 10)$.  I wouldn't be surprised to learn it's all of $S_{16}$, but that still leaves a lot of obviously-open questions from a puzzle perspective (minimal-length words to swap two elements and the like), and any good suggestions on ways of finding particular elements of the group (where to look for interesting commutators and conjugates, etc.) would be welcome; I have a little background from doing some basic Rubik's-Cube stuff (for instance, my first look was at the element $(\alpha^2\beta^2)^2$ ), but not much.

Body:
Yes, the group generated by the permutations is all of $S_{16}$.

Here's my GAP:

gap> g:= (1,2,3,7,11,10,9,5);

(1,2,3,7,11,10,9,5)

gap> h:= (2,3,4,8,12,11,10,6);

(2,3,4,8,12,11,10,6)

gap> k:=(5,6,7,11,15,14,13,9);

(5,6,7,11,15,14,13,9)

gap> m:=(6,7,8,12,16,13,14,10);

(6,7,8,12,16,13,14,10)

gap> G:=Group( g,h,k,m);

Group([ (1,2,3,7,11,10,9,5), (2,3,4,8,12,11,10,6), (5,6,7,11,15,14,13,9), >(6,7,8,12,16,13,14,10) ])

gap> Size(G);

20922789888000

gap> Size(SymmetricGroup(16));

20922789888000


问:
Find the variance of $W=5X+2Y$.
The problem:

Toss a die 10 times. Let $X$ be the number of ones and let $Y$ be the number of sixes.

Find $P\left(X+Y=2\right)$ and then find the variance of $W=5X+2Y$.

I found $P\left(X+Y=2\right)$ using the binomial distribution.

I am attempting to find $W=5X+2Y$ using $E[aX+bY]\ =v^{T}Mv$, where

$v=
\begin{bmatrix}
    5\\2\\
\end{bmatrix}$, $M=\begin{bmatrix}
                           Var(X)&Cov(X,Y)\\Cov(X,Y)&Var(Y)\end{bmatrix}$

So now I have,

$\begin{bmatrix}
    5&2
\end{bmatrix}$ $\begin{bmatrix}
                           Var(X)&Cov(X,Y)\\Cov(X,Y)&Var(Y)\end{bmatrix}$
$\begin{bmatrix}
    5\\2\\
\end{bmatrix}$

The problem I am having is that I can't figure out how to find the covariance of $Cov(X,Y)$.

From the binomial distribution, $\mu=n\theta$ and $\sigma^2=n\theta\left(1-\theta\right)$, but I don't know if either of those relationships will help me find $Cov(X,Y)$. Any feedback would be greatly appreciated.

答:
A slick way to get the covariance:

Let $X_1, X_2, \ldots, X_6$ be the number of ones, twos, etc. respectively. So $X=X_1$ and $Y=X_6$.

Note that $X_1 + \cdots + X_6 = 10$, and that the sequence $X_1, \ldots, X_6$ is exchangeable since no face of the die is intrinsically different from any other face.
Then $\text{Var}(X_i)= \text{Var}(X_1)$ for any $i$, and $\text{Cov}(X_i, X_j) = \text{Cov}(X_1, X_2)$ for any $i \ne j$. Thus,
$$0 = \text{Var}(X_1 + \cdots + X_6) = \sum_{i=1}^6 \text{Var}(X_i) + 2 \sum_{1 \le i < j \le 6}\text{Cov}(X_i, X_j) = 6 \text{Var}(X_1) + 6 \cdot 5 \cdot \text{Cov}(X_1, X_2).$$
Thus $\text{Cov}(X_1, X_2) = - \frac{1}{5} \text{Var}(X_1) = - \frac{1}{5} \cdot 10 \cdot \frac{1}{6} \cdot \frac{5}{6} = -\frac{5}{18}$.

Title: Find the variance of $W=5X+2Y$.


问:
$EX = 0$ and $\textrm{MGF}(t)= \infty$ for all $t>0$
This is a Durret exercise 2.7.5:

Suppose that $X_1,X_2,\ldots$ are iid with $EX_1 = 0$ and $E[e^{tX_1}]
 = \infty$ for all $t>0$. Let $S_n = X_1+\cdots+X_n$. Prove that for all $a>0$: $$\frac{1}{n} \log P(S_n \geq na) \xrightarrow{n\to\infty}
 0.$$

I have the intuition for this exercise: $P(X\geq a)$ decays slow enough so  that the moment generating function infinite. But i don't know how to start with this. Can you someone please help?

答:
Let $g\geq 0$ be a borel function, then for all $a> 0$, we have that:
\begin{align*}
E[g(X)] &= g(0) + \int_{0}^{\infty} g(t) P(X > t)dt \\
&= g(0) + \int_{0}^{\infty} g(t) P(X > t)dt \\
&= g(0) + \sum_{n\geq 0} \int_{n}^{n+1} g(t) P(X > t)dt \\
&\leq g(0) + \sum_{n\geq 0} \left(\sup_{n\leq t\leq n+1}g(t)\right) P(X>n)
\end{align*}
For $e^{\theta X}$, we get that:
$$E[e^\theta X] = 1 + \sum_{n\geq 0}e^{\theta(n+1)}P(X>n) = \infty \implies \sum_{n\geq 0} e^{\theta n} P(X>n) = \infty.$$
Let $\epsilon>0$, then we must have $P(X_n>an) > e^{-(\theta + \epsilon)n}$ infinitely often since $$\sum_{n\geq 1} e^{-\epsilon n} < \infty.$$
Therefore,
$$\limsup\limits_{n\to\infty} \frac{1}{n} \log P(X>n) \geq -(\theta + \epsilon).$$
Taking $\epsilon \to 0$ and then $\theta \to 0$ gives us that $\limsup\frac{1}{n} \log P(X>n) = 0$. This in turn gives us that $\lim\frac{1}{n} \log P(S_n>an) = 0$ for all $a\in\mathbb{R}$ by a lemma we proved in class.


问:
How to show this inequality?

Show that $$-2 \le \cos \theta(\sin \theta+\sqrt{\sin^2 \theta +3})\le2$$

Trial: I know that $-\dfrac 1 2 \le \cos \theta\cdot\sin \theta \le \dfrac 1 2$ and $\sqrt 3\le\sqrt{\sin^2 \theta +3}\le2$. The problem looks simple to me but I am stuck to solve this. Please help. Thanks in advance.

Title: How to show this inequality?

Body:
Essentially, we want the extremum of
$$f(a) = a\sqrt{1-a^2} + a\sqrt{4-a^2}$$
We have
$$f'(a) = \sqrt{1-a^2} + \sqrt{4-a^2} - \dfrac{a^2}{\sqrt{1-a^2}} - \dfrac{a^2}{\sqrt{4-a^2}} = \dfrac{1-2a^2}{\sqrt{1-a^2}} + \dfrac{4-2a^2}{\sqrt{4-a^2}}$$
Let $a^2 = x$, we then want to solve for
\begin{align}\dfrac{1-2x}{\sqrt{1-x}} + \dfrac{4-2x}{\sqrt{4-x}} = 0 \implies (1-2x)^2(4-x) = 4(2-x)^2(1-x)\\
(-4x^3+20x^2-17x+4)-(-4x^3+20x^2-32x+16) = 0 \implies15x-12=0 \implies x = \dfrac45
\end{align}
Hence, $a^*= \pm \dfrac2{\sqrt5}$, which gives us $f(a^*) = \pm 2$.


问:
The set of all points $x\in\mathbb{R}$ that must belong to an infinite number of sets.
Let $\{E_k\}_{k=1}^{\infty}$ be a collection of subsets of $\mathbb{R}$. Why does the set

$$A:=\bigcap_{k=1}^{\infty}\left(\bigcup_{n=k}^{\infty} E_n\right),$$

consist of the points $x\in\mathbb{R}$ that are in infinitely many of the $E_k$'s?

答:
Suppose $x \in \mathbb{R}$ belongs to only finitely many of those sets. Then, there is a $k \in \mathbb{N}$ that is the largest for which $x \in E_k$. Then, for all $n \geq k+1, x \notin E_n$, so $x \notin \cup_{n=k+1}^\infty E_n$.

Conversely, if $x \in E_n$ for infinitely many $n$, then for all $k \in \mathbb{N}$, there is a $k' > k$ for which $x \in E_{k'}$. Therefore, $x \in \cup_{n  = k}^\infty E_n$ for all $k$, so $x$ is in the intersection of all such sets.


问:
Is $f$ continuous if $f_n \to f$ point-wise (with additional info in post)
I have a question relating to a previous example.

We have:

Let $(f_n)$ be a sequence of differentiable functions on the interval $[0, 2]$

Assume that $(f_n)$ converges pointwise to a function $f$ on $[0, 2]$

Assume there exists positive real $M$ st $\forall n,$ $ \sup_{x\in[0,2]}\{|f'_n(x)|\}=M$

Is the limit $f$ of $(f_n)$ continuous?

I know that there is a theorem that states that if $f_n$ converges uniformly to $f$ and each$f_n$ is continuous then f is continuous but I am unsure whether that is true here. Or should I be looking at this another way?

答:
The condition on the derivatives implies that the sequence is equicontinuous because, by the Mean Value Theorem,

$$ |g(a) - g(b)| \le \sup_{c \in (a,b)}|g'(c)| |a - b|. $$

So then we can apply the Arzelà–Ascoli theorem to conclude that there is a uniformly convergent subsequence.

Therefore, because $f$ is the limit of a uniformly convergent sequence, it must be continuous. This wouldn't necessarily be true if we didn't have equicontinuity.


问:
Prove if given L|M and gcd(M, D) = 1, then gdc (D, L) = 1 is true
I want to prove if given $L|M$ and $gcd(M, D) = 1,$ then $gdc (D, L) = 1$ is true (As stated in the title)

This seems very easy, yet I am having trouble on where to really start?

I feel like Bezout's Identity should be used here, however I don't know how to really apply it.

Any suggestions, or beginning steps would be gladly appreciated

答:
Let $M = nL$.

Since $GCD(M,D) = 1$, there exist integers $a, b$ such that

$$\begin{align*}
aM+bD &= 1\\
anL + bD &= 1
\end{align*}$$

So $1$ is a multiple of $GCD(D, L)$.

From transitivity of divisibilities $GCD(D,L) \mid L$ and $L\mid M$, then $GCD(D,L) \mid M$.

Then $GCD(D,L)$ is a common divisor of both $M$ and $D$, and so $GCD(D,L)\mid GCD(M,D)$.

From given that $GCD(M,D) = 1$, then $GCD(D,L)$ is also $1$.


问:
If $r>0$ and $r\notin \mathbb{N}$, is there a simple method to evaluate $ \sum_{n=\lceil r \rceil}^{\infty} {\binom{n}{r}^{-1}}?$
Let $r>0,r\in \mathbb{R}\setminus\mathbb{N}$. Empirically, I have noticed the following relation:
$$
\sum_{n=0}^{\lfloor r \rfloor} \frac{1}{\binom{n}{r}} = - \sum_{n=\lceil r \rceil}^{\infty} \frac{1}{\binom{n}{r}};
$$in particular, $\displaystyle{\sum_{n=0}^{\infty} \frac{1}{\binom{n}{r}} =0}$. Note that if $r$ is an integer, the finite sum is not well-defined, although we do have
$$
\sum_{j=0}^{k-1} \operatorname{Res}  \left(\frac{1}{\binom{z}{k}},z=j\right)= k\cdot\sum_{m=0}^{k-1}\binom{k-1}{m}(-1)^m=0,
$$ so in this sense the sum 'cancels'. Mathematica returns the closed-form of
$$
\sum_{n=\lceil r \rceil}^{\infty} \frac{1}{\binom{n}{r}}=  \frac{\lceil r\rceil }{(r-1) \binom{\lceil r\rceil }{r}},
$$which when $r\in\mathbb{N}$ reduces to this question, but I don't know how to derive that myself. Maybe I'm not fully understanding the answers there but I don't think the same tricks apply when the sum doesn't telescope. So in summary, my questions are:

Can someone explain the closed-form?

Is there a simple, conceptual reason the finite sum is the negative of the infinite sum?

答:
Here's a computation of the total sum from $n=0$ to $\infty$, which may lead a way to compute the finite sum. Since
$$\binom xy\binom yx=\operatorname{sinc}(\pi(x-y)),$$
where $\operatorname{sinc}(x)=\sin(x)/x$, we have
$$\frac{1}{\binom xy}=\binom yx\frac{\pi(x-y)}{\sin(\pi(x-y))};$$
in particular,
$$\frac{1}{\binom nr}=\binom rn\frac{\pi(r-n)}{\sin(\pi(r-n))}=\pi(r-n)\binom rn\frac{(-1)^n}{\sin \pi r}.$$
So, we want to evaluate
$$\frac{\pi}{\sin \pi r}\sum_{n=0}^\infty (-1)^n(r-n)\binom rn.$$
Consider
$$f(x)=(1+x)^r=\sum_{n=0}^\infty \binom rnx^n.$$
We have
$$\frac{d}{dx}(x^{-r}f(x))=\sum_{n=0}^\infty \binom rn\frac{dx^{n-r}}{x}=\sum_{n=0}^\infty \binom rn(n-r)x^{n-r-1};$$
also,
$$\frac{d}{dx}(x^{-r}f(x))=\frac{d}{dx}\left(\frac{1+x}{x}\right)^r=-\frac{r\left(\frac{1+x}{x}\right)^{r-1}}{x^2},$$
and so we have the identity
$$\sum_{n=0}^\infty \binom rn(-1)^n(r-n)\binom rn=(-1)^{r+1}\frac{d}{dx}\left(x^{-r}f(x)\right)\bigg|_{x=-1}=0$$
whenever $r>1$.


答: In fact, as noted, it follows that the convergence is uniform. But if all you want is to show that $f$ is continuous, that's completely trivial: $$|f(x)-f(y)|=\lim|f_n(x)-f_n(y)|\le M|x-y|.$$

Title: Is $f$ continuous if $f_n \to f$ point-wise (with additional info in post)

Body:
I have a question relating to a previous example.

We have:

Let $(f_n)$ be a sequence of differentiable functions on the interval $[0, 2]$

Assume that $(f_n)$ converges pointwise to a function $f$ on $[0, 2]$

Assume there exists positive real $M$ st $\forall n,$ $ \sup_{x\in[0,2]}\{|f'_n(x)|\}=M$

Is the limit $f$ of $(f_n)$ continuous?

I know that there is a theorem that states that if $f_n$ converges uniformly to $f$ and each$f_n$ is continuous then f is continuous but I am unsure whether that is true here. Or should I be looking at this another way?


答:
Without using Arzelà–Ascoli:

Let $\varepsilon>0$, and by compactness, cover $[0,2]$ by finitely many $\varepsilon$-balls centered at $\{x_1,\ldots,x_n\}$. Fix $x\in[0,2]$; then there exists $x_j$, $1\leq j\leq n$ such that $x\in B_\varepsilon(x_j)$. By the triangle inequality:
$$|f_n(x)-f(x)|\leq|f_n(x)-f_n(x_j)|+|f_n(x_j)-f(x_j)|+|f(x_j)-f(x)|.$$
By the mean value theorem, $|f_n(x)-f_n(x_j)|\leq M|x-x_j|<M\varepsilon$. Choose $N_j\in\mathbb{N}$ such that $|f_n(x_j)-f(x_j)|<\varepsilon$ for all $n\geq N_j$ by pointwise convergence.

For the last term, we want to show that $|f(x_j)-f(x)|\leq M|x-x_j|$ as well. Note that:
$$\big||f(x_j)-f(x)|-|f_n(x_j)-f_n(x)|\big|\leq|f_n(x_j)-f(x_j)|+|f_n(x)-f(x)|<\varepsilon$$
by the triangle inequality and for large enough $n$ by pointwise convergence. Hence:
$$|f(x_j)-f(x)|<\varepsilon+|f_n(x_j)-f_n(x)|\leq\varepsilon+M|x-x_j|,$$
and since $\varepsilon$ was arbitrary we have $|f(x_j)-f(x)|\leq M|x-x_j|<M\varepsilon$. It follows that:
$$|f_n(x)-f(x)|\leq\varepsilon(1+2M).$$
Note that $N_j$ depends only on $x_j$, so choose $N=\max\{N_1,\ldots,N_n\}$. Then this $N$ works for each $x\in[0,2]$, so the convergence is uniform and a fortiori $f$ is continuous.


问:
Why don't we consider the zero subspace (which is readily $T$-invariant) in the definition of direct sum of linear operators?

Why don't we consider the zero subspace (which is readily $T$-invariant) in the definition of direct sum of linear operators?

REF: Schaum's Outline of Linear Algebra

答:
Technically, you could.

A decomposition of an operator is much like a factorization of a natural number. We can consider factorizations of numbers with '1' as one of the factor, but then we really are interested in is non-trivial factorizations.

I would prefer your definition, allowing for zero subspaces, but we don't lose much either way. I suspect that is why (nonzero) is in parentheses - if they felt it was a hard rule, they wouldn't need to parenthesize that, would they?

Essentially, we are interested in whether an operator is "decomposable." Non-decomposable operators are, in some sense, primes. But all operators are decomposables if we allow for the null space, so we would then need to define non-decomposable as "only has trivial decompositions."


问:
Multiplying Taylor series for $e^x$ with terms from converging sequence
Let $(a_n)$ be a sequence that converges to $a\in \mathbb{R}.$

For which $x \in \Bbb{R}$ does $\sum_{k=0}^{\infty} \frac{a_k x^k}{k!}$ converge?

Here is my attempt:

Let $\epsilon =1.$ Then $\exists N \in \mathbb{N}$ st $\forall n\geq N, |a_n -a| < 1$. Also $a_n < 1 +a.$

Then $\sum_{k=0}^{\infty} \frac{a_k x^k}{k!} = \sum_{k=0}^{N-1} \frac{a_k x^k}{k!}+ \sum_{k=N}^{\infty} \frac{a_k x^k}{k!}.$ The first sum is finite (say it equals Y) so $ \leq Y +   \sum_{k=N}^{\infty} \frac{(a+1) x^k}{k!} \leq Y + (a+1)\sum_{k=N}^{\infty}  \frac{x^k}{k!}.$ This is the serie for $e^x$ but minus first $N-1$ terms so it must converge. Then multiplying by $(a+1)$ will still make it converge so $\sum_{k=0}^{\infty} \frac{a_k x^k}{k!}$ must converge for all R.

Does this make sense?

答:
Apply the ratio test:

We need:

$\lim_\limits{k\to\infty}\frac {a_{k+1}}{a_k} \frac {x^{k+1}}{x^k} \frac {k!}{(k+1)!} < 1$

Simplifying:

$\lim_\limits{k\to\infty}\frac {a_{k+1}}{a_k} \frac {x}{k+1} < 1$

Since $a_k$ is convergent $\lim_\limits{k\to\infty}\frac {a_{k+1}}{a_k} = 1$

$\lim_\limits{k\to\infty}\frac {x}{k+1} < 1$

The series will converge for all $x.$


问:
Two goats on a sliding leash in a square garden
I have two goat problems I am trying to figure out.  Here are my problems.

You have a square with sides of length 100 m. There are two goats and each goat is on their own diagonal on a 10 m leash that they are able to slide along their diagonal (ignore that they might tangle up where the diagonals intersect). A cabbage is thrown into the square and it could land at any point with equal probability.

(a) Find the probability that at least one goat can reach the cabbage.

(b) Find the probability that both goats can reach the cabbage.

This is what I have done for each.  I think I am correct but if anyone sees a mistake or if I am doing something wrong please let me know.  Thank you for any help.

Area of Triangle EML = $1/2*80*40 = 1600*4$ triangles $= 6400$.  Then I took the area of the whole square $(100*100 = 10000)$ and subtract $10000-6400 = 3600$ to get the area of the region the goats can cover.  So, my probability for (a) is $3600/10000 =.36$

For (b), the probability that both goats can reach the cabbage is in that square in the middle of the figure, so $20*20=400$.  That gives me a probability of $400/10000=.04$

答:
[NB: your diagram is inaccurate: you have used this flawed diagram in your answer to (a), which has thrown its answer out, but reasoned from the correct version of the diagram in (b)]

To verify your solution, I'll proceed by a different method, and see if we get to the same conclusion.

First, note that this problem is scale invariant, so we'll rescale such that the big square has side length $1$, so that our probabilities are exactly the areas of the relevant shapes. Now, it's also symmetrical under quarter-rotations around the central point, so we'll restrict to the lower-left quarter, then quadruple our areas.

Now, (b) is asking for exactly four times the area of the small triangle in what is now the top-right. This triangle is right-isosceles with hypotenuse $0.2$. Pythagoras then gives that the two shorter sides are each $\sqrt{0.2^2/2} = \sqrt{0.02}$, and so the area of this triangle is $\frac{1}{2}(\sqrt{0.02})^2 = 0.01$, so our probability is $4 \times 0.01 = 0.04$, agreeing with your answer.

On the other hand, (a) is asking for exactly four times the area of the central strip. That rectangle differs from the rectangle of the same width extended to the corners of our small square by four small triangles ("fold in the corners"), each of which has half of the area of the triangle in (b) (so those four small triangles have combined area $0.02$). That rectangle has one side of length $0.2$, and the other is the diagonal of this quarter-square, so has length $\sqrt{2}/2$. Thus, the area of this rectangle is $0.2\sqrt{2}/2 = \sqrt{2}/10$. Subtracting those four small triangles gives the area for the central strip as $\frac{\sqrt{2}}{10} - 0.02$. The desired probability in (a) is hence $4\left(\frac{\sqrt{2}}{10} - 0.02\right) = \frac{10\sqrt{2}-2}{25} \approx 0.485685425$.

Your error in (a) is that triangle $EML$ does not actually have hypotenuse $80$ or altitude $40$, contra your diagram. Indeed, length $CL$ is actually $10\sqrt{2}\mathrm{m}$, so length $EL$ is $(100 - 20\sqrt{2})\mathrm{m}$, and length $OM$ is actually $20\sqrt{2}\mathrm{m}$, so your altitude is $\frac{1}{2}(100 - 20\sqrt{2})\mathrm{m} = (50 - 10\sqrt{2})\mathrm{m}$. Using those values in your method will give the correct answers.


问:
How do you integrate over the directional derivative?
Question: How do you integrate the directional derivative of a function over a rectangle?

Let's say $K$ is a rectangle in $\mathbb{R}^2$, and let's say that $\beta$ is a 2D vector that specifies a direction. Lastly, say $u(x,y)$ is a multivariable scalar function. How do I solve the following
\begin{equation}
\int _K \beta \,\,\cdot \nabla u
\end{equation}
where $\beta \,\,\cdot\nabla$ denotes the directional derivative in the direction of $\beta$.

I know that in 1D integrating over the derivative of a function gives back the original function, but I'm not sure how this works in 2D.

Thanks!

答:
I'm not sure what kind of answer you're looking for, but here is one way to change this integral into another, possibly simpler, integral.

The idea is to use Green's Theorem (or Stokes' Theorem). If $\beta = (a,b)$, then your integrand is just $a \, u_x + b \, u_y$. Consider the vector field
$$ (-b \, u, \; a \, u) $$

Then if we write this vector field as $(P,Q)$, then $Q_x = au_x$ and $P_y = -bu_y$, so we get
$$ Q_x - P_y = au_x + bu_y $$
So by Green's Theorem, if $C$ is the boundary of the rectangle $K$ (meaning the four line segments),
$$ \int_K (a \, u_x + b \, u_y) \, dx \, dy = \int_C a \, u \, dy - b \, u \, dx $$

Again, I'm not sure if this is the kind of answer you were looking for, but in some sense it has been simplified because we've traded an integral over a $2$-dimensional region for an integral over a $1$-dimensional region.


问:
Does there exist an epimorphism from the ring $\mathbb{Z_{24}}$ onto the ring $\mathbb{Z_{7}}$?
Does there exist an epimorphism from the ring $\mathbb{Z_{24}}$ onto the ring $\mathbb{Z_{7}}$?

an epimorphism is a surjective homomorphism

Could someone give me an idea how to know this? to be honest I have no idea

答:
Given any homomorphism $f:\Bbb Z_{24}\to\Bbb Z_7$, by definition of homomorphism we must have
$$
f(0)=f(24)=f(1+1+\cdots+1)\\
=f(1)+\cdots+f(1)
$$
where there are $24$ terms in each of the two long additions. Now, what could $f(1)$ possibly be? Which elements of $\Bbb Z_7$ are such that adding together $24$ copies of it gives $0$ (since $f(0)=0$)?

This argument actually doesn't use all of the ring structure, only the additive group. So the answer would be the same if we were asking about the groups $\Bbb Z_{24}$ and $\Bbb Z_7$.


问:
(Proof Verification) : ACT for Integrals
I would like to kindly request your assistance in looking at my work and giving me a feedback on my work if its missing something or wrong.

Question :  that if $\int_{a}^{\infty}|f(x)| d x$ converges then $\int_{a}^{\infty} f(x) d x$ converges moreover
$$
\left|\int_{a}^{\infty} f(x) d x\right| \leq \int_{a}^{\infty}|f(x)| d x
$$

My proof : First, we note that $f(x)\in\mathcal{R}\implies|f(x)|\in\mathcal{R}$. Moreover, $f(x)\leq|f(x)|$ and thus $0\leq\int_{a}^{\infty}f(x)\;\text{d}x\leq\int_{a}^{\infty}|f(x)|\;\text{d}x$, since $\int_{a}^{\infty}|f(x)|\;\text{d}x$ converges then by comparison test we have $\int_{a}^{\infty}f(x)\;\text{d}x$ converges. Since $f(x)\in\mathcal{R}$, then we define $\mathcal{P}$ to be a partition of $[a,b]$, $\mathcal{P}=\{\underbrace{x_{1}}_{a},x_{2},\cdots,\underbrace{x_{3}}_{b}\}$ where $b>a$ such that $\lim_{b\to\infty}\int_{a}^{b}f(x)\;\text{d}x$ exists due to convergence of the integral. Thus, by definition we have that $L(\mathcal{P},f)=U(\mathcal{P},f)=\sum_{i=1}^{n}M_{i}x$. Therefore :
\begin{align*}
&\left|\int_{a}^{b} f(x) d x\right|=\left|\sum_{i=1}^{n}M_{i}x\right|\leq\sum_{i=1}^{n}|M_{i}|x=\int_{a}^{b} |f(x)| d x \\
\implies&\lim_{b\to\infty}\left|\int_{a}^{b} f(x) d x\right|=\lim_{n\to\infty}\left|\sum_{i=1}^{n}M_{i}x\right|\leq\lim_{n\to\infty}\sum_{i=1}^{n}|M_{i}|x=\lim_{b\to\infty}\int_{a}^{b} |f(x)| d x \\
\implies&\left|\int_{a}^{\infty} f(x) d x\right| \leq \int_{a}^{\infty}|f(x)| d x
\end{align*}

答:
You are misusing the  comparison test which applies when $0 \leqslant f(x) \leqslant g(x)$. Then convergence of $\int_a^\infty g(x) \, dx $ implies convergence of $\int_a^\infty f(x) \, dx $ .  You don't have $f$ nonnegative as an assumption here. You write $0 \leqslant \int_a^\infty f(x) \, dx$.  Even if the integral converges (which you have not yet established)  it might be negative.

(1) To prove convergence of the integral of $f$ we can apply the Cauchy criterion since
$$\left|\int_{c_1}^{c_2} f(x) \, dx \right| \leqslant \int_{c_1}^{c_2} |f(x)| \, dx < \epsilon$$

(2) To prove the inequality note that

$$-\int_a^b |f(x)| \, dx \leqslant \int_a^b f(x) \, dx \leqslant \int_a^b |f(x)| \, dx,$$

and, hence,

$$-\int_a^\infty |f(x)| \, dx \leqslant \lim_{b \to \infty}\int_a^b f(x) \, dx  = \int_a^\infty f(x) \, dx \leqslant\int_a^\infty |f(x)| \, dx,$$

which implies

$$\left|\int_a^\infty f(x) \, dx\right| \leqslant \int_a^\infty |f(x)| \, dx$$


问:
Cauchy’s functional equation for non-negative arguments
Function $f:[0,+\infty)\rightarrow\mathbb{R}$ satisfies $f(x+y)=f(x)+f(y)$ for every non-negative $x$ and $y$. It’s bounded from below with some non-positive constant $m$. Does it imply that $f$ has the form $f(x)=cx$ or is there another function satisfying these conditions?</p>

答:
Yes.

It follows easily from induction that $f(nx)=nf(x) \, \forall n \in \mathbb{Z}^+$. Thus $f(\frac{m}{n})=\frac{1}{n}f(m)=\frac{m}{n}f(1)$.

If $f(y)<0$ for some $y \in[0, +\infty)$, then $f(ny)=nf(y)<m$ for sufficiently large $n$. Thus $f(y) \geq 0 \forall y \in [0, +\infty)$. Now if $a \geq b$ then $f(b+(a-b))=f(b)+f(a-b) \geq f(b)$, so $f(x)$ is monotonic. This is sufficient to imply that $f(x)=xf(1)$ (where here we require $f(1) \geq 0$).


答: For another perspective, note that ring homomorphisms are always homomorphisms of the underlying additive groups. So if $f\colon \Bbb Z_{24}\to\Bbb Z_7$ is surjective, the first isomorphism theorem implies $7$ divides $24$. If you think about it, you can see how this is equivalent to Arthur's answer.

Body:
Does there exist an epimorphism from the ring $\mathbb{Z_{24}}$ onto the ring $\mathbb{Z_{7}}$?

an epimorphism is a surjective homomorphism

Could someone give me an idea how to know this? to be honest I have no idea


问:
Showing that an arithmetic function is multiplicative
Let $f$ be an arithmetic function that counts the number of consecutive integers between $1$ and $n$ (inclusive) such that both integers are coprime to $n$. More formally,

$$ f(n) = \sum_{\substack{1 \leq t \leq n \\ (t,n)=1 \\ (t+1,n)=1}}1. $$

An immediate observation is that $f(n)=0$ when $n$ is even. After playing around with this function, I suspect that it is multiplicative. That is, if $(a,b)=1$, then $f(ab)=f(a)f(b)$. However, I am unable to prove this.

Is this function really multiplicative or are there counterexamples?

Also, what happens when we modify $f$ so that it counts the number of consecutive triplets that are all coprime to $n$?

答:
Suppose $\gcd(m,n)=1$

Then let $(i,i+1)$  be a good pair for $m$ and let $(j,j+1)$ be a good pair for $n$.  The Chinese Remainder theorem gives us a unique  $k$ with $1≤k≤mn-1$ such that $$k\equiv i\pmod m\quad \&\quad k\equiv j \pmod n$$

Clearly we have
$$k+1\equiv i+1\pmod m\quad \&\quad k+1\equiv j+1 \pmod n$$

So $(k,k+1)$ is a good pair for $mn$.

It remains to be shown that every good pair for $mn$ arises this way.  But if $(k,k+1)$ is  a good pair for $mn$ then remark that $tm<k<\left((t+1)m-1\right)$ for some $t$ so $(k\pmod m,(k+1)\pmod m)$ is a good pair for $m$ (and similarly for $n$) and we are done.


问:
How do I prove that if a function mapping from $ C \times C \to \Bbb R$ is continuous on $ C \times C $, then it is uniformly continuous there?
Let C be the Cantor Middle Third set in [0,1]. Suppose $ f: C\times C \to \Bbb R$ is continuous on $C\times C$. Show f is uniformly continuous there.

I'm not really sure what the link is here. Somehow, C being a Cantor set means continuity implies uniform continuity, but I don't really know how to argue that.

答: $C\times C$ is compact so any continuous function from it to $\mathbb{R}$ will be uniformly continuous.


问:
Which topics of mathematics should I study?
I'm a first year econometrics student with a great interest in mathematics. I very much enjoy my study, but still I am interested to learn about more topics in mathematics which are not part of my study. Some of the main topics which I am already familar with or which will soon be covered in my study are: calculus, linear algebra, optimization, statistics/probability/combinatorics.

Which topics in mathematics would you advise me to study still (that are not part of this list)? I have a general interest in mathematics, so any advice on interesting and or essential topics in mathematics that are worth studying is appreciated.

If possible, could you also give me advice on books/references which I should study from?

Thanks in advance.

Edit: To be more specific, I am looking for topics on which the general consensus is that they are essential to know of for any mathematician or very interesting to study.

答:
Study what you enjoy!

There is no way I or anyone else can tell you what you might find interesting. Read around on wikipedia and try to get an overview of what some subjects are about and see which one appeals to you. Try to gain mathematical maturity by studying many different branches of mathematics. Some combinatorics, abstract algebra and real analysis are fun subjects and after taking them maybe you will find something that appeals to you more than others.

Try to take classes with friends too, if possible. It will make the reading a lot more fun and you can discuss with eachother.


答:
Let me give the argument for a function $f\colon C\to \Bbb R$, as this extends analogously to $C\times C$. Let $\varepsilon>0$, and, for every $p\in C$, choose $\delta_p>0$ such that $|x-p|<\delta_p$ implies $|f(x)-f(p)|<\varepsilon$. The sets $(p-\delta_p,p+\delta_p)$ form an open cover of $C$ as $p$ ranges over all points in $C$, and by compactness there is a finite subcover corresponding to points $p_1,\ldots p_n$. Simply take $\delta=\min\{\delta_{p_1},\ldots, \delta_{p_n}\}$ and you're done.

Now repeat this using open balls in $C\times C$, etc. Or prove once and for all that continuous functions on compact sets are uniformly continuous (or cite this fact, depending on the class).

Title: How do I prove that if a function mapping from $ C \times C \to \Bbb R$ is continuous on $ C \times C $, then it is uniformly continuous there?

Body:
Let C be the Cantor Middle Third set in [0,1]. Suppose $ f: C\times C \to \Bbb R$ is continuous on $C\times C$. Show f is uniformly continuous there.

I'm not really sure what the link is here. Somehow, C being a Cantor set means continuity implies uniform continuity, but I don't really know how to argue that.


答:
Let $m$ and $n$ be coprime positive integers. Chinese reminder theorem gives us a ring isomorphism

$$
g: \mathbb{Z}/mn\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z} \\
x\pmod{mn} \mapsto (x\pmod m, x\pmod n).
$$

For integer $k$, define $A_k \subset \mathbb{Z}/k\mathbb{Z}$ as

$$
A_k = \{t \in \mathbb{Z}/k\mathbb{Z} \, | \, \gcd(t, k) = 1 \land \gcd(t + 1, k) = 1\}.
$$

Note that $x \in A_k$ if and only if both $x$ and $x + 1$ are units in the ring $\mathbb{Z}/k\mathbb{Z}$ i.e. $x \in A_k$ if and only if there exist $y \in \mathbb{Z}/k\mathbb{Z}$ and $z \in \mathbb{Z}/k\mathbb{Z}$ such that

$$
xy = 1 \pmod k \\
(x + 1)z = 1 \pmod k.
$$

Since $g$ preserves inverses, we see that  $t \in A_{mn}$ if and only if $g(t) = (t_1, t_2)$ with $t_1 \in A_m$ and $t_2 \in A_n$. Thus, $g$ restricted to $A_{mn}$ is a bijection between $A_{mn}$ and $A_m \times A_n$. Therefore, $|A_{mn}| = |A_m| |A_n|$.

Multiplicativity of $f$ follows since $f(k) = |A_k|$.

Using the result above and the observation that $f(p^k) = p^{k-1}(p - 2)$ for prime $p$, one can derive a formula for $f(n)$ where $n = p_1^{k_1}p_2^{k_2}\dots p_r^{k_r}$

$$
f(n) = \prod_{i=1}^r p_i^{k_i-1} (p_i - 2) = n \prod_{i=1}^r \left(1 - \frac{2}{p_i}\right)
$$

which is reminiscent of a similar formula for the Euler totient function.

Title: Showing that an arithmetic function is multiplicative

Body:
Let $f$ be an arithmetic function that counts the number of consecutive integers between $1$ and $n$ (inclusive) such that both integers are coprime to $n$. More formally,

$$ f(n) = \sum_{\substack{1 \leq t \leq n \\ (t,n)=1 \\ (t+1,n)=1}}1. $$

An immediate observation is that $f(n)=0$ when $n$ is even. After playing around with this function, I suspect that it is multiplicative. That is, if $(a,b)=1$, then $f(ab)=f(a)f(b)$. However, I am unable to prove this.

Is this function really multiplicative or are there counterexamples?

Also, what happens when we modify $f$ so that it counts the number of consecutive triplets that are all coprime to $n$?


答:
Yes.

We can restrict to values $(a,b)$ such that $a,b>2$ are odd and coprime.
Now we use the following observations:

$f(n)$ counts the number of points $(x,y)$ with coordinates in the ring $\Bbb Z/n$ such that
$$
\tag{$C(n)$}
x(x+1)y=1\ .
$$

If $u,v$ are invertible modulo $n$, then we can use alternatively (set e.g. $x=ux'$, etc.)
$$
\tag{$C(n;u,v)$}
x(x+u)y=v\ .
$$

$\Bbb Z/ab\cong (\Bbb Z/a)\times(\Bbb Z/b)$, $x\cong(x',x'')$ in the category of rings. (Chinese Remainder Theorem, CRT.)

We need an explicit isomorphism. So we write $1=as+bt$ in $\Bbb Z/ab$, where $(a,b)=(s,b)=(t,a)=(s,t)=1$.

The explicit ring isomorphism from the CRT is $(x',x'')\to x:=btx'+asx''$. (So $(1',1'')\to 1$.) The converse morphism is the canonical projection.

Let $(x,y)$ be a solution modulo $ab$. Then with $x=btx'+asx''$ we have
$$
1=x(x+1)y=(btx'+asx'')(btx'+asx''+as+bt)y\ ,
$$
and reading this modulo $a$, respectively modulo $b$ we obtain solutions in $\Bbb Z/a$ and $\Bbb Z/b$ for the equations $C(a;1,1/(bt)^2=1)$ and $C(b; 1, 1/(as)^2=1)$.

Conversely, let us start with a solution $(x', ?')$ modulo $a$, and a solution $(x'',?'')$ modulo $b$ for the equations above, the corresponding $x$ leads to a solution modulo $ab$.

$\square$

Note: Since for a prime $p$ we have $f(p^k)=(p-2)p^{k-1}$, we have in general for a number $N$ with prime factor decomposition
$$
N=p_1^{k_1}p_2^{k_2}\dots p_r^{k_r}
$$
the formula
$$
f(N)
=N
\left(1-\frac 2{p_1}\right)
\left(1-\frac 2{p_2}\right)
\cdots
\left(1-\frac 2{p_r}\right)
\ .
$$


问:
Two countries are trying to form a cartel. To keep the price of gold high, they agree to limit their production to 4 tons and 1 ton, respectively...
Two countries are trying to form a cartel. To keep the price of gold high, they agree to limit their production to 4 tons and 1 ton, respectively. For each country, cheating is equivalent to producing 1 more ton  per day. Depending on your decisions, total production would be 5, 6, or 7 tons per day, with corresponding profit margins of $ 16, $ 12, and $ 8,000 per ton. Describe this situation as a non-zero sum game, writing the game's matrices.

My attempt

\begin{array}{c|lcr}
I/II & \text{1} & \text{2} \\
\hline
1 & (4,1) & (4,2)  \\
2 & (5,1) & (5,2)  \\
\end{array}

Is correct? Should I use the profit figures?

Thanks in advance for any comments.

答:
A rational firm would likely want to maximize its profit regardless of the number of units sold, so you should use the profit margin times the number of units sold. This gives the game, with units in thousands of dollars,

\begin{array}{c|lcr}
I/II & \text{1} & \text{2} \\
\hline
1 & (64,16) & (48,24)  \\
2 & (60,12) & (40,16)  \\
\end{array}

This game only has a single pure Nash equilibrium.


问:
Show $\varepsilon > 0$ and strictly increasing seq of indices $\{n_k\}$ s.t. $k$, $\vert f(s_{n_k}) - f(t_{n_k}) \vert\ge \varepsilon$
This is a question from Advanced Calculus, Fitzpatrick. Section 3.4 Q12

Suppose that the function $f : D \rightarrow \mathbb{R}$ is not uniformly continuous. Then by definition there are sequences $\{s_n\}$ and $\{t_n\}$ in $D$ such that $$\lim_\limits{n \to \infty}[s_n - t_n] = 0 \text{, but } \lim_\limits{n \to \infty}[f(s_n) - f(t_n)] \neq 0$$
A. Show that there is an $\epsilon > 0$ and a strictly increasing sequence of indices $\{n_k\}$ such that for each index $k$, $\vert f(s_{n_k}) - f(t_{n_k}) \vert \geq \varepsilon$

My attempt
\begin{align}
\lim_\limits{n \to \infty}[f(s_n) - f(t_n)] \neq 0 && \tag{1}\\
\neg (\forall \varepsilon > 0, \exists N \text{ s.t. } \forall n \geq N \; \vert f(s_n)-f(t_n)\vert < \varepsilon) && \text{logical definition}\tag{2}\\
\exists \epsilon \gt 0, \forall N, \exists k \geq N \; \vert f(s_k)-f(t_k)\vert \geq \varepsilon && \text{apply negation throughout}\tag{3}
\end{align}

I'm stuck here. It seems like this would define a subsequence $\{n_k\}$ but I don't think it's necessarily strictly increasing.

For example, we could have a sequence $N = [1, 2, 3, 4, 5, 6]$

For $N=1, k=3$ --> $N_k = [3]$
For $N=2, k=2$ --> $N_k = [3, 2]$
For $N=3, k=5$ --> $N_k = [3, 2, 5]$ .... etc

Question
Is this a reasonable way to approach this problem? Are there any errors in my first 3 arguments?

答:
You can do this directly. Set $c_n=|f(s_n) - f(t_n)|.$ Then you know that $\lim_\limits{n \to \infty}c_n \neq 0.$ This means that $\overline {\lim}c_n$ is different from zero. (why?). If it is infinite, the result follows trivially. If $\overline {\lim}c_n=c>0$ then write $\overline {\lim}c_n=\underset{k\to \infty}\lim \sup_{k\le n}c_n=\underset{k\to \infty}\lim g_k=c.$

Each $g_k\ge c.$ Let $K$ be large enough so that $k\ge K\Rightarrow |g_k-c|<c/4.$ Thus, this inequality is satisfied for the subsequence $\{g_K,\ g_{K+1},\cdots ,\}$, which we relabel $\{g_1,\ g_{2},\cdots ,\}$ for convenience.

Using the definition of the supremum, we find an integer $k_1$ such that $g_1-c_{k_1}<c/4;\ k_1\ge 1.$ And an integer $k_2$ such that $g_{k_1+1}-c_{k_2}<c/4;\ k_2>k_1$ (by construction). Inductively, we obtain a sequence $(c_{k_j})$ that satifies

$|c_{k_j}-c|\le |c_{k_j}-g_{k_j+1}|+|g_{k_j+1}-c|\le c/4+c/4=c/2\Rightarrow c_{k_j}>c/2$ so the claim follows with $\epsilon=c/2.$


问:
Does the notation $x \mapsto(x+2 \mathbb{Z}, x+3 \mathbb{Z})$ refers to $x \mapsto(x+2 k, x+3 k) , k \in Z$
I found a notation in a problem and which was not used in the lecture and I didn't cross before:

It asks to show that the application is a ring homomorphism:
$$
\begin{aligned}
\varphi: \mathbb{Z} & \longrightarrow \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z} \\
& x \quad \mapsto(x+2 \mathbb{Z}, x+3 \mathbb{Z})
\end{aligned}
$$

I looked up wikipedia pages on "ring" and modular arithmetic but I don't find use of such notation, I am guessing it is some convention of notation and it refers to $x \mapsto(x+2 k, x+3 k) , k \in Z$ am I correct?

答:
No, your interpretation is not correct.

$x + 2\mathbb{Z}$ is the coset of $\mathbb{Z}/2\mathbb{Z}$ that contains $x$. So for example if you write the elements of $\mathbb{Z}_n$ as $\{0, 1, \ldots, n-1\}$ then in this example the image of $8$ is the pair $(0,2)$.


问:
How to prove that $11^{n+2} + 12^{2n+1}$ can be divided by $133$ where $n\in\mathbb N$
I know that I have to get to a point where an summation is multiplied by $133$ but I don't know how to get there.
I tried formating the equation like $11^2\times11^n + 12 \times 12^ {2n}$ but I got stuck there and I don't really know how to prove it with induction. I really need to solve this.

Thank you for your help and sorry if I misspelled anything. Also sorry that you have to edit it, I don't really know how to make superscripts.

Body:
If you know modular arithmetic, it's easy:

$$11^{n+2}+12^{2n+1}=121\cdot11^{n}+12\cdot144^{n}\equiv121\cdot11^{n}+12\cdot11^{n}=133\cdot 11^{n}\equiv0\bmod133.$$


问:
Can I show a limit exists if it exists in a sum?
I know that if $$ \lim_{x\rightarrow\infty}(f(x)-a)=0 $$ for some a in the reals can I say $$\lim_{x\rightarrow\infty}f(x) =a$$

I know that the sum of limits existing does not imply that each limit must exist but in this case, since a is just a number I do not see how the limit of f(x) could not exist.

If so, how might I prove this better?

答:
If $\lim_{x\to\infty}(f(x)-a)$ exists and is equal to $0$, then by the limit sum formula

$$\lim_{x\to\infty}f(x) = \lim_{x\to\infty}(f(x)-a) + \lim_{x\to\infty}a = a$$


答:
This is part of a somewhat broader bit of notation for quotient rings such as $\mathbb Z/2\mathbb Z$. In particular, $2\mathbb Z$ refers literally to the set of even numbers $\{2k : k\in \mathbb Z\}$ and $1+2\mathbb Z$ refers to the set of odd numbers $\{1+2k:k\in \mathbb Z\}$. It's fairly common to see arithmetic done on sets this way, where we use a set like $\mathbb Z$ in an expression to stand in for "the set of all possible evaluations of this expression" - for instance, one sometimes sees equations even like $4\mathbb Z + 6 \mathbb Z = 2\mathbb Z$, which is literally true of sets. This is related to various operations on ideals in ring theory (although one should be really careful that it is far less common to use this notation with sets being multiplied together - for the good reason that multiplication in a quotient ring is not necessarily multiplication of each element of a coset).

The point of the notation is that the ring $\mathbb Z/2\mathbb Z$ consists of the additive cosets of $2\mathbb Z$ - it is a ring whose two elements are "the set of even numbers" and "the set of odd numbers" and the usual projection that reduces mod $2$ happens to be given by
$$f:\mathbb Z\rightarrow\mathbb Z/2\mathbb Z$$
$$f(x)=x+2\mathbb Z.$$
It's common to represent the elements of $\mathbb Z/2\mathbb Z$ by the numerals $\{0,1\}$ - or sometimes with various ornaments like $\bar 0$ and $\bar 1$ which tend to be defined as "the set of even numbers $2\mathbb Z$" or "the set of odd numbers $1+2\mathbb Z$", but the literal meaning of a quotient involves these cosets - which are given by shifting the set $2\mathbb Z$ by addition. Note that the notation of cosets is rather flexible, since it lets you write $0+2\mathbb Z$ and $2+2\mathbb Z$ and $4+2\mathbb Z$ and so on as equally valid (and equal) elements of $\mathbb Z/2\mathbb Z$. Mod $3$ works similarly.

So the notation
$$g(x)=(x+2\mathbb Z,x+3\mathbb Z)$$
is really sending each $x\in \mathbb Z$ to a pair consisting of a coset of $2\mathbb Z$ (in $\mathbb Z/2\mathbb Z$) and a coset of $3\mathbb Z$ - as is required by the literal definition of these things. That is, it takes an integer to a pair of sets, the first set being in $\mathbb Z/2\mathbb Z$, the second in $\mathbb Z/3\mathbb Z$.

Title: Does the notation $x \mapsto(x+2 \mathbb{Z}, x+3 \mathbb{Z})$ refers to $x \mapsto(x+2 k, x+3 k) , k \in Z$

Body:
I found a notation in a problem and which was not used in the lecture and I didn't cross before:

It asks to show that the application is a ring homomorphism:
$$
\begin{aligned}
\varphi: \mathbb{Z} & \longrightarrow \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 3 \mathbb{Z} \\
& x \quad \mapsto(x+2 \mathbb{Z}, x+3 \mathbb{Z})
\end{aligned}
$$

I looked up wikipedia pages on "ring" and modular arithmetic but I don't find use of such notation, I am guessing it is some convention of notation and it refers to $x \mapsto(x+2 k, x+3 k) , k \in Z$ am I correct?


问:
Complete elliptic integrals
Is there a simple way to relate complete elliptic integrals of the first and second kind of the form

$$
K(1/m)\,,\quad K(-m)\,,\quad E(1/m)\,,\quad\text{and}\quad E(-m)
$$

to $E(m)$ or $K(m)$. Here, $K(m)$ and $E(m)$ are complete elliptic integrals of the first and second kind, respectively and $m$ is a real number.

答:
DLMF 19.7.3 gives
$$K(1/m)=\sqrt m(K(m)-iK(1-m))$$
$$E(1/m)=\frac1{\sqrt m}(E(m)+iE(1-m)-(1-m)K(m)-imK(1-m))$$
DLMF 19.7.5 gives
$$K(-m)=\frac1{\sqrt{1+m}}K\left(\frac m{1+m}\right)$$
$$E(-m)=\sqrt{1+m}E\left(\frac m{1+m}\right)$$
Here $m$ is the parameter.

Title: Complete elliptic integrals


问:
How can I prove that $\sum _{n=1}^{\infty }\:\:\sin\left(\frac{n}{n^2+1}\right)$ diverges?
I have tried all of the series tests that I know of, but I can't seem to find an answer. According to wolfram alpha, it is possible using the comparison test, however I can't figure out what I would compare the series with to show that it diverges.</p>

答: WolframAlpha in fact says that, by the Comparison Test, this series diverges.


问:
Prove that $\frac{n-1}{\sigma^2}S^2 \sim \chi^2_{n-1}$.
Let $\Bbb Y = (Y_1, \cdots, Y_n)' \sim N_n(\Bbb \theta J_n, \sigma^2I_n)$ be a random vector and $S^2=\frac{1}{n-1}\sum\limits_{i=1}^n(Y_i-\bar Y)^2$. Prove, that $\frac{n-1}{\sigma^2}S^2 \sim \chi^2_{n-1}$.

So I could start with writing that $({n-1})S^2=\sum\limits_{i=1}^n(Y_i-\bar Y)^2=\sum\limits_{i=1}^nY_i^2-n\bar Y^2=Y'Y-\frac{1}{n}Y'J_nJ_n'Y=Y'HY$ where $H$ is a symetric idempotent matrix, so it gives me $(HY)'HY$ and I would be much happier if $HY \sim N(0,I_n)$. If $H$ was orthogonal I could then write that $HY \sim N(H\theta J_n, \sigma^2I_n)$ and dividing it by $\sigma^2$ would almost give me a desired distribution.

The other idea is to somehow standarize my random variable $Y_i \sim N(\theta_i,\sigma^2)$, but I don't really know how it is distributed if I subtract the sample mean instead of $\mu$.

答:
Let $Z \sim N_n(0 J_n,  I_n)$ so that $Y = \theta J_n + \sigma Z$.

Note that $Y'HY = \sigma^2 Z' H Z$ because $H J_n = 0$. It then suffices to show $Z'H Z \sim \chi^2_{n-1}$.

Cryo has given a great hint: because $H$ is symmetric and idempotent with trace $n-1$, it is diagonalizable [with respect to an orthonormal basis] with eigenvalues $1$ (multiplicity $n-1$) and $0$ (multiplicity $1$). Let $H = UDU'$ be that diagonalization, where $U$ is orthogonal.

Then $Z' H Z = Z'UDU'Z = (U'Z)' D (U'Z)$. Use rotational invariance of the $N(0, I_n)$ distribution to note that $\tilde{Z} := U'Z$ also follows the $N(0, I_n)$ distribution as well. Then note that $\tilde{Z}' D \tilde{Z}$ is the sum of squares of $n-1$ components of $\tilde{Z}$, so it follows a $\chi^2_{n-1}$ distribution.


问:
A convergent series for the Trigamma function $\psi_1(n) =\sum_{k=n}^{\infty} \frac1{k^2} $
I just came up with
the following
convergent series
for the Trigamma function
defined by

$\psi_1(n)
=\sum_{k=n}^{\infty} \frac1{k^2}
$.

\begin{align*}
\psi_1(n)
&=\lim_{m \to \infty} \sum_{j=1}^m \frac{(j-1)!}{j\prod_{i=0}^{j-1}(n+i)}\\
&=\frac1{n}+\frac{1}{2n(n+1)}+\frac{2}{3n(n+1)(n+2)}+\ldots\\
\end{align*}

This contrasts with
the usual asymptotic series
for $\psi_1(n)$
which is asymptotic,
does not converge,
and involves the
Bernoulli numbers.

I'm sure this isn't new,
but I could not find it here.

So,
my questions are,
as is often the case,

(1) Is this new?

(2) Is there a reasonably simple proof?
(Mine is moderately messy.)

I'll post my proof
in a few days
if anyone is interested.

Thanks

答:
This is not a proof, not an answer but it is too long for comments.

Let
$$a_j=\frac{(j-1)!}{j\prod_{i=0}^{j-1}(n+i)}=\frac{(j-1)!}{j n (n+1)_{j-1}}\qquad \text{and} \qquad S_m=\sum_{j=1}^m a_j$$

$$\frac{(m+1) \Gamma (n+1) \Gamma (m+n+1)}{\Gamma (n) }S_m=$$
$$(m+1) n \psi ^{(1)}(n) \Gamma (m+n+1)-$$ $$\Gamma (m+1) \Gamma (n+1) \,
   _3F_2(1,m+1,m+1;m+2,m+n+1;1)$$

Simplifying
$$S_m=\psi ^{(1)}(n)-\frac{\Gamma (m+1) \Gamma (n)}{(m+1)
   \Gamma (m+n+1)} \, _3F_2(1,m+1,m+1;m+2,m+n+1;1)$$

What seems interesting is to make a contour plot of the second term; it shows how small it is.

We can also notice that for large values of $m$ $(m\gg n)$, the coefficient
$$\frac{\Gamma (m+1) \Gamma (n)}{(m+1) \Gamma (m+n+1)}\sim \frac{ \Gamma (n) }{m^{n+1}}$$

Unfortunately, I have not been to find the asymptotics of the hypergeometric function.


问:
A Countable Basis and Countably Locally Finite Collections

Show that if $X$ has a countable basis, a collection $\mathcal{A}$ of subsets of $X$ is countably locally finite if and only if it is countable.

I am going to give my proof of the $\impliedby$ implication first. Suppose that $\mathcal{A}$ is a countable collection of subsets of $X$. Then we can enumerate them by $\Bbb{N}$ likeso: $\mathcal{A} = \{A_n \mid n \in \Bbb{N}\}$. But then $\mathcal{A} = \bigcup_{n=1}^\infty \{A_n\}$, where $\{A_n\}$ is locally finite since any given open set can intersect at most one set in $\{A_n\}$, and hence, intersect finitely many sets in $\{A_n\}$. This means that $\mathcal{A}$ is countably locally finite.

<s>For the other direction, I am having a little trouble. A countably locally finite set can be written as a countable union of locally finite sets, so it seems that it suffices to show any locally finite collection is countable. Let $\mathcal{A}$ be any such collection. As with most arguments about cardinality, it seems that my best option is to construct a function $f : \mathcal{B} \to \mathcal{A}$ that is surjective. Here's an attempt. Assume that $A \in \mathcal{A}$ is not empty. Then there exists $a \in A$, and since $\mathcal{B}$ is a basis $B \in \mathcal{B}$ such that $a \in B$, so that $A \cap B \neq \emptyset$. So we might define $f(B) = A$ such that $A \cap B \neq \emptyset$ (note: there could be may sets in $\mathcal{A}$ satisfying this, so perhaps this won't work...?) Is it well-defined? Well, if $B=B'$, then $\emptyset \neq A \cap B = A \cap B'$, which I guess means that $f(B) = f(B')$ (?). But by definition, this is surjective, so $|\Bbb{N}| = |\mathcal{B}| \ge |\mathcal{A}|$, implying that $\mathcal{A}$ is countable.

As the periodic question marks in the above paragraph indicate, I don't feel terribly comfortable with some parts of the proof. Another thing is, I don't seem to use $\mathcal{A}$'s local finiteness anywheree. Is there a way of clarifying the proof? If it's incorrect, is there a way of fixing it? If not, I could use a hint on how to prove the theorem.</s>

EDIT: Hold on! I may have a proof. Just give me a few minutes to type it up.

Okay. My proof is slightly inspired by Henno's answer given here, although I didn't read the entire proof until I came up with my own, because I wanted to try to solve the problem own my own as much as possible (indeed, the only thing I saw in the link was his definition of $I(n)$). Here is my second attempt at proving the "$\implies$" implication, which is I believe is slightly different from Henno's since I don't do a proof by contradiction.

First, let $\mathcal{A} = \{A_i \mid i \in I\}$ be some locally finite collection. We want to show that $I$ is countable. For each $n \in \Bbb{N}$, define $I(n) = \{i \in I \mid A_i \cap B_n \neq \emptyset \}$. Clearly the union of these sets is contained in $I$. Suppose that $i \in I$. Then given $A_i \neq \emptyset$, there exists $a_i \in A_i$; and since $\mathcal{B} = \{B_n \mid n \in \Bbb{N}\}$ is a basis, there exists an $n \in \Bbb{N}$ such that $a_i \in B_n$, which implies that $i \in I(n)$. Hence $I = \bigcup_{n=1}^\infty I(n)$, so that we need only prove that $I(n)$ is countable. To this end, let $n \in \Bbb{N}$ and let $x \in B_n$. Since $\mathcal{A}$ is locally finite, there exists a basis element $B_{x_k}$ containing $x$ that intersects only finitely many elements in $\mathcal{A}$. Hence, $\{B_{x_k}\}$ is a cover of $B_n$, each of which intersects only finitely many sets in $\mathcal{A}$; moreover, since $\mathcal{B}$ is countable, $\{B_{k_x}\}$ must also be countable and hence can be enumerated likeso: $\{B_{\alpha_k} \mid k \in \Bbb{N} \}$. Now, if $A_i \cap B_n \neq \emptyset$, then it must also intersect a set in the cover of $B_n$, which means that $I(n) \subseteq \bigcup_{k=1}^\infty I(\alpha_k)$. Seeing as the latter set is a countable union of finite sets, which is makes it countable, $I(n)$ is als countable.

How does this sound?

答:
I'm going to fill in some of the steps left out above for others if they didn't understand it, cause I had a hard time.

A = $\bigcup{C_{N}}$. Where $C_{N}$ is a locally finite collection. Note that if $C_{N}$ is countable collection, then by simple analysis A is a countable collection since then A would be a countable union of a countable amount of sets. Thus the problem is that $C_{N}$ may not be a countable collection.

Since X is second countable, X = $\bigcup{B_{i}}$; where a countable set of indices i can be used.

Not every basis element may intersect $C_{N}$ in only finitely many places. Thus we need some sort of method to remove the bad basis elements for my proof to work.

Since X can be written as the union of non-empty basis elements, every $x \in X$ is contained within a basis element $B_{i}$. The local finiteness of $C_{N}$ implies some neighborhood U of x must intersect $C_{N}$ in finitely many sets. If the neighborhood U is contained within $B_{i}$, then pick a basis element $B_{j} \subset U \subset B_{i}$ that contains x and throw out the original basis element $B_{i}$. Otherwise relabel $B_{i} = B_{j}$. Then for sure $B_{j}$ intersects $C_{N}$ in finitely many sets or zero. Since we still have $x \in B_{j}$, and x was arbitrary in X, the space X = $\bigcup{B_{j}}$ where the basis elements $B_{j}$ intersects $C_{N}$ in finitely many sets or zero sets.

Define $A_{j} = B_{j}\bigcap{C_{N}} = \bigcup_{M=1}^{<\infty}{\{C_{M}\}}$. The $C_{M}$ are the finite sets that $B_{j}$ intersects $C_{N}$ in. If its zero sets that they intersect in, then just leave $A_{j}$ empty for now.

Every element $y \in C \in C_{N}$ lies in X, obviously because C are subsets of X. Then we know at least one basis element $B_{j}$ in our refined collection intersects any arbitrary C of $C_{N}$; otherwise C is empty because we could just pick x to be in C, and find the $B_{j}$ that contains x and only intersects $C_{N}$ finitely. Thus taking the union over basis elements: $\bigcup{A_{j}} \supset C_{N}$.

Hence we have that $C_{N}$ is at most a countable number of sets because $C_{N} \subset \bigcup{A_{j}} = \bigcup{\{\bigcup_{M=1}^{<\infty}{\{C_{M}}\}\}_{j}}$ which is a countable union of a finite numbers of sets.

Knowing this, we must have A = $\bigcup{C_{N}}$ is a countable union of a at most countable number of sets.


问:
Find the matrix representing the relations S o R, where the matrices representing R and S.
I am having difficulty understanding the following:

Find the matrix representing the relations $S\circ R$, where the matrices representing $R$ and $S$:

$$M_R = \begin{bmatrix}1 \ 0 \ 1 \\ 1 \ 1 \ 0 \\ 0 \ 0 \ 0\end{bmatrix} \text{ and }M_C = \begin{bmatrix}0 \ 1 \ 0 \\ 0 \ 0 \ 1 \\ 1 \ 0 \ 1\end{bmatrix} \implies M_R \odot M_S \implies \begin{bmatrix}1 \ 1 \ 1 \\ 0 \ 1 \ 1 \\ 0 \ 0 \ 0\end{bmatrix}$$

I know that if the composite matrix has $t_{ij} = 1$ if and only if $r_{ik} = s_{kj} = 1$ for some $k$. I am unsure how to arrive at the answer to the right.

Body:
Using standard matrix multiplication,

$$\pmatrix{1&0&1\\1&1&0\\0&0&0}\pmatrix{0&1&0\\0&0&1\\1&0&1}=\pmatrix{1&1&1\\0&1&1\\0&0&0}.$$


问:
Is the $\mathbb{C}$-vector space $C_{\text{per}}^k(\mathbb{R}^n)$ with the map $\Vert \cdot \Vert _{C^k}$ a $\mathbb{C}$-Banach space?
Below $(l^\infty_\mathbb{C}(\mathbb{R}^n),\Vert \cdot \Vert _\infty)$ denote the $\mathbb{C}$-Banach space of bounded functions $f:\mathbb{R}^n\to\mathbb{C}$ where $\Vert f\Vert _\infty:=\sup_{x\in\mathbb{R}^n}|f(x)|$.

Let $k\in \mathbb{N}_0\cup\{\infty\}$ and consider the set:

$$C_{\text{per}}^k(\mathbb{R}^n):=\big\{f\in C^k_\mathbb{C}(\mathbb{R}^n):f\text{ is }2\pi\text{-periodic}\big\}.$$

Considering the multi-index notation, we can prove that if $f\in C_{\text{per}}^k(\mathbb{R}^n)$, then $\partial ^\alpha (f)\in l^\infty _\mathbb{C}(\mathbb{R}^n)$ for all $\alpha\in\mathbb{N}_0^n$ satisfying $|\alpha|\leq k$. Knowing that we can conclude that the map $\Vert \cdot \Vert _{C^k}:C^k_\text{per}(\mathbb{R}^n)\to \mathbb{R}$ given by $\Vert f\Vert _{C^k}:=\sum_{|\alpha |\leq k}\Vert \partial ^\alpha (f )\Vert _\infty $ is well-defined.

My question is: is the $\mathbb{C}$-vector space $C_{\text{per}}^k(\mathbb{R}^n)$ with the map $\Vert \cdot \Vert _{C^k}$ a $\mathbb{C}$-Banach space? If so, how do I prove it?

Thank you for your attention!

答:
The definition of the norm in $C_{\text{per}}^k(\mathbb{R}^n)$ seems to require that $k<\infty$, so I will make that assumption.

The answer is yes.  Let us begin with an auxiliary result:

Lemma.  Let $\{f_n\}_n$ be a sequence in $C^1(\mathbb{R}^n)$ converging pointwise to some function $f$, and such
that, for a given $k$, the sequence $\{\partial f_n/\partial x_k\}_n$ converges uniformly over compact sets to a
function $g$.  Then $\partial f/\partial x_k=g$.

Proof.  For every $x\in \mathbb{R}^n$, and every $t\in \mathbb R$, we have
$$
  f(x+te_k) -   f(x) =
  \lim_{n\to\infty} f_n(x+te_k) -   f_n(x) = $$$$ =
  \lim_{n\to\infty}\int_0^t \frac{\partial f_n}{\partial x_k}(x+se_k)\, ds =
  \int_0^t g(x+se_k)\, ds.
  $$
The conclusion follows by taking the derivative relative to $t$ of the above.  QED

This said, suppose that $\{f_n\}_n$ is a Cauchy sequence in $C_{\text{per}}^k(\mathbb{R}^n)$.  Then, for every
$\alpha\in\mathbb{N}_0^n$ with $|\alpha|\leq k$, we have that
$$
  \|\partial^\alpha f_n -  \partial^\alpha f_m\|_\infty  \leq    \|f_n - f_m\|_{C^k},
  $$
so the sequence $\{\partial^\alpha f_n\}_n$ is  Cauchy  in the Banach space
$l^\infty(\mathbb{R}^n)$, and hence  it converges uniformly to some $g_\alpha $ in that space.

Setting $f=g_{(0,0,\ldots ,0)}$, we may use the Lemma and  induction on $|\alpha |$ to prove that $\partial^\alpha f=g_\alpha $, whenever $|\alpha|\leq k$, so we conclude
that $f_n\to f$ relative to $\Vert \cdot \Vert_{C^k}$.


答:
Going off of Matthew Pilling's comment, we have that
\begin{align*}
\lim_{n\to \infty} \frac{\sin\left(\frac{n}{n^2+1}\right)}{1/n} &\stackrel{L.H.}{=}
\lim_{n\to\infty} \frac{\left(\frac{1-n^2}{(n^2+1)^2}\right)\cos\left(\frac{n}{n^2+1}\right)}{-1/n^2} \\
&= \lim_{n\to \infty} \frac{n^4-n^2}{(n^2+1)^2}\cos\left(\frac{n}{n^2+1}\right) \\
&= 1 >0
\end{align*}
Thus, by the limit comparison test, since the harmonic series diverges, we have that
$\sum_{n=1}^\infty\sin\left(\frac{n}{n^2+1}\right)$ diverges.

Title: How can I prove that $\sum _{n=1}^{\infty }\:\:\sin\left(\frac{n}{n^2+1}\right)$ diverges?

Body: I have tried all of the series tests that I know of, but I can't seem to find an answer. According to wolfram alpha, it is possible using the comparison test, however I can't figure out what I would compare the series with to show that it diverges.</p>


答:
Note that for any $n\in\mathbb N$:
$$\frac{1}{2n}=\frac{n}{n^2+n^2}\le \frac{n}{n^2+1} \le 1$$
Furthermore, $\sin x\ge\frac{x}{2}$ for $0\le x\le 1$. Therefore:
$$\sum_{n=1}^N \sin \frac{n}{n^2+1} \ge \frac{1}{4} \sum_{n=1}^N \frac{1}{n} \xrightarrow[N\to\infty]{} \infty$$


问:
Value of integration $\int_0^1 \sin^{-1}(3x-4x^3)dx$
I have stumbled across an definite integration problem.
The question is to evaluate $$I=\int_0^1 \sin^{-1}(3x-4x^3)dx$$
I carried out the integration as this. Let $x = \sin A$ so that A = $\sin^{-1}(x)$ then
$$\begin {align} I &=\int \sin^{-1}(3\sin A-4\sin^3A)dx\\ &=\int \sin^{-1}(\sin3A)dx\\&=\int3Adx\\&=3\int \sin^{-1}(x)\\&=3[x\sin^{-1}x+\sqrt{1-x^2}] + c\end{align}$$

Then,
$$\begin{align}I&= \biggl.3[x \sin^{-1}x+\sqrt{1-x^2}]\biggl|_0^1\\&=3[(\sin^{-1}1+\sqrt{1-1^2})-(0\times \sin^{-1}0+\sqrt{1-0})]\\&=3(\frac\pi2-1)\\&=1.71238...\end{align}$$

But the correct answer is different as given by mathematica and many other ways,
$$I=\int_0^1 \sin^{-1}(3x-4x^3)dx = 0.62535...$$

I couldn't find the mistake in my process. And I don't even know which answer is correct actually? Can you please help me here?

Body:
Note

\begin{align}
I&=\int_0^1 \sin^{-1}(3x-4x^3)dx \>\>\>\>\>\>\>(x=\sin A)\\
&= \int_0^{\frac\pi2} \sin^{-1}(\sin3A)\cos AdA \\
 &= \int_0^{\frac\pi6} \sin^{-1}(\sin3A)\cos AdA
 +  \int_{\frac\pi6}^{\frac\pi2} \sin^{-1}(\sin(\pi-3A))\cos AdA \\
 &=  \int_0^{\frac\pi6} 3A\cos AdA
 +  \int_{\frac\pi6}^{\frac\pi2} (\pi-3A)\cos AdA \\
 &=\left( \frac{3\sqrt3}2-3 +\frac\pi4 \right)+ \left(\frac{3\sqrt3}2-\frac{3\pi}4\right)\\
&=3\sqrt3-3-\frac\pi2
\end{align}


问:
How to simplify boolean expressions.
I'm struggling to understand what rules to apply when simplifying boolean expression.

For example:

$$ B+(A\cdot(C+B) \overline C) $$

I'm not sure how to simplify this expression. Here is my attempt.

$$
= B+AB\cdot(C+\overline C)
\\
= B+AB\cdot(1)
\\
= B(A+B)
\\
= B(B+A)
\\
= B
$$

答:
Two really useful principles are:

Absorption

$P + PQ = P$

Reduction

$P+ P'Q = P + Q$

With these:

$ B+A(C+B)C' \overset{Reduction}{=}$

$B+ ABC' \overset{Absorption}{=}$

$B$

Done!

Title: How to simplify boolean expressions.


问:
If $R$ is not integral domain, and $P$ is prime in $R$, is $P^e$ not equal to $PR_P$?
In Dummit & Foote, Proposition $46$(2) on page $718$ says

If $R$ is an integral domain, then $R_P$ is an integral domain. The ring $R$ injects into the local ring $R_P$, and, identifying $R$ with its image in $R_P$, the unique maximal ideal of $R_P$ is $PR_P$.

Let $\varphi: R \to R_P$ be the canonical map, where $R$ is not an integral domain. Is it not true in general that $P^e (= \varphi(P)R_P)$ is not equal to $PR_P$?

答:
If $\varphi: R \to S$ is a ring homomorphism of commutative rings, then the extension of an ideal $I \subset R$ is the ideal $\varphi(I)S$, and the contraction of an ideal $J \subset S$ is the ideal $\varphi^{-1}(J)$.

In the special case where $R$ is a subring of $S$ and $\varphi$ is the natural injection, the extension of $I \subset R$ is written as $IS$, and the contraction of $J \subset S$ is written as $J \cap R$.

If $R$ is an integral domain and $P$ is prime, then the natural map $R \to R_P$ is injective and we can identify $R$ as a subring of $R_P$. Then the maximal ideal of $R_P$ is $PR_P$ by the special case above.

In general, since $P^e$ is the ideal of $R_P$ generated by all products $\frac{a}{1}\frac{r}{u}$ where $a \in P$, $r \in R$, $u \notin P$, then this is the same as the ideal $PR_P$. But it seems Dummit & Foote reserve the notation $PR_P$ for when $R$ injects into $R_P$.


问:
$a\mid b,c\mid d\Rightarrow\,\gcd(a,c)\mid \gcd(b,d)$
So far I have that $a|b$ implies $b=ax$ for some x in the integers and $c|d$ implies that $d=cy$ for some y in the integers.  From here I can see that $gcd(a,c)|gcd(b,d)$ is logically equivalent to $gcd(a,c)|gcd(ax,cy)$ but I am not sure where to go from there.</p>

答:
By  gcd Universal Property $\  x\mid y,z\iff x\mid (y,z)\, \ $ [where $\,(y,z):=\gcd(y,z)$], $ $ so

$$\begin{align}
a\mid b \,\Rightarrow\, \color{#0a0}{(a,c)}\mid a\mid \color{#c00}b\\
c\mid d \,\Rightarrow\, \color{#0a0}{(a,c)}\mid c\mid\color{#c00} d\end{align}\,\Rightarrow\, \color{#0a0}{(a,c)}\mid (\color{#c00}{b,d})\qquad\!\!$$

Or more generally notice  $\ \begin{align}  a\mid b&\,\Rightarrow\, b = a\bar a\\ c\mid d&\,\Rightarrow\, d = c\bar c\end{align}\ \ $ so it follows by below.

Lemma $\,\ \color{#c00}{(a,c)}(\bar a,\bar c)\,\mid\, (\color{#c00}a\bar a, \color{#c00}c\bar c)$

Proof $\ \ \ \ \ (a,c)(\bar a,\bar c) = (a\bar a,c\bar c,a\bar c,c\bar a )\mid a\bar a,c\bar c\,$ so it $\mid\, (a\bar a,c\bar c)\,$ by above universal property.

Or we can use $\,(w,x,y,z) = ((w,x),(y,z))\mid (w,x)\,$ by the gcd associative law.

The proof uses "gcd polynomial arithmetic" (associate, commutative, distributive) to expand the product, i.e. a gcd analog of $(a\!+\!c)(\bar a\!+\!\bar c) = a\bar a\! +\! c\bar c\! +\! a\bar c\! +\! c\bar a,\,$ as explained in the prior link.


问:
A $\zeta$-like sum of the positive roots of $\csc x=x$.
From this answer, I was very interested to learn the following.

Let $x_n$ be the $n$-th positive root of the equation $\csc x=x$. Then $$\sum_{n\ge1}\frac{1}{x_n^2}=1,$$
and, setting $s(k)=\sum_{n\ge1}x_n^{-k}$, we have $$\sum_{k\ge1}s(2k)x^{2k}=\frac{x}{2}\cdot\frac{1+x\cot x}{\csc x-x}.\tag 1$$

This result was very surprising and I've never seen anything like it before. The proof was discussed rather briefly in the comments and apparently it can be shown via a contour integral, but I've never done anything like that before so I have no idea how. Could I have some help proving $(1)$? Thanks! :)

答:
The $x_n$ are the zero of $z \sin z-1$ (all simple). The sequence is almost periodic (because the large zeros of $\sin z - 1/z$ are close to $\pi n$). So
$$f(z)=\frac{(z\sin z -1)'}{z\sin z -1}-\sum_n (\frac1{z-x_n}+\frac{1}{x_n})$$
converges, it is entire and $f'$ is bounded and vanishes at $\infty$ so $f'=0$ and $f$ is constant, since $f$ is odd then $f=0$.

For $|z|<|x_1|$ the smallest zero
$$g(z)=\frac{(z\sin z -1)'}{z\sin z -1}=\sum_n (\frac1{z-x_n}+\frac{1}{x_n})$$ is analytic and $$g^{(2k)}(0)=0, \qquad g^{(2k+1)}(0)=- (2k+1)!\sum_{n} \frac1{x_n^{2k+2}}$$
whence
$$g(z)=-\sum_{k=0}^\infty z^{2k+1}\sum_{n} \frac1{x_n^{2k+2}}$$


问:
Is there a chain rule for integration?
I know the chain rule for derivatives. The way as I apply it, is to get rid of specific 'bits' of a complex equation in stages, i.e I will derive the $5$th root first in the equation $(2x+3)^5$ and continue with the rest.

I wonder if there is something similar with integration. I tried to integrate that way $(2x+3)^5$ but it doesn't seem to work. Well, it works in the first stage, i.e it's fine to raise in the power of $6$ and divide with $6$ to get rid of the power $5$, but afterwards, if we would apply the chain rule, we should multiply by the integral of $2x+3$!, But it doesn't work like that, we just need to multiply by $1/2$ and that's it.

So my question is, is there chain rule for integrals? I want to be able to calculate integrals of complex equations as easy as I do with chain rule for derivatives.

答:
Consider the functions z(y) and y(x).  I am showing an example of a chain rule style formula to calculate
$I(x) = \int dx z(y(x)),$
where z(y) can be triply integrated over dy, and where
$y(x)=\sqrt{x}$ or $y(x)=x.$

To construct a formula for I(x), first define F(y) as the triple integration of z(y) over dy, that is
$F(y) = \iiint dy dy dy z(y).$
Next evaluate F(y) for y(x), that is define
$G(x) = F(y(x)).$
Differentiate G(x) twice over dx and then divide by $(dy/dx)^3,$ yielding
$I(x) = \int dx z(y(x)) = G''(x) / y'^3.$

Consider an example calculation of I(x) where $z = y^3.$

First let $y(x)=\sqrt{x},$ so $dy/dx = (1/2) {x^{-1/2}}.$

Then
$F(y) = y^6 / 120 + ay^2/2 + by + c,$ which yields
$G(x) = F(y(x)) = x^3 /120  + ax/2 + bx^{1/2} + c,$ and then
$G''(x) = x/20 - (1/4)bx^{-3/2},$ so that
$I(x) = y'^{-3} G''(x) = 8 x^{3/2} [ x/20 - (1/4)bx^{-3/2} ]= (2/5)x^{5/2}  - 2b.$
Directly integrating for $y = x^{1/2}$ and $z = y^3$ yields
$I(x) = \int dx z(y(x)) = \int dx y^3  = \int dx x^{3/2} = (2/5) [x^{5/2}] + constant.$

Next for this same example z = $y^3$ let y = x.
Then
$F(y) = y^6/120 + ay^2/2 + by + c,$ which yields

$G(x) = F(y(x)) = x^6/120 + ax^2/2 + bx + c,$  so that

$I(x) = y'^{-3}G''(x)  = 1^{-3} [x^4/4 + a]  = x^4/4 + a.$

Directly integrating yields
$I(x) = \int dx z(y(x)) =  \int dx y^3  =  \int dx x^3   = x^4/4 + constant.$

This demonstrates that the direct and chain rule methods agree with each other to within a constant for $y(x)=x$ and $y(x)=\sqrt{x}$ for the specific function $z(y) = {y^3}.$  This agreement should work for any function z(y) where $y(x)=x$ or $y(x)=\sqrt{x}.$

Title: Is there a chain rule for integration?