问:
Can we define the generating function for all $x$ and all $t$
The Chebyshev polynomials of second kind are defined for any $x \in \Bbb R$ (or
even $x \in \Bbb C$), e.g. via the recurrence relation
$$
  U_0(x) = 1 \\
  U_1(x) = 2x \\
  U_{n+1}(x) = 2x U_n(x) - U_{n-1}(x)
$$

The ordinary generating function is

$$∑_{n=0}^{∞}U_{n}(x)t^n=1/(1-2xt+t^2)$$

defined for all $|x|<1$ and $|t|<1$. (https://mathworld.wolfram.com/ChebyshevPolynomialoftheSecondKind.html)

My question: Since $U_n(x)$ is defined for all reals $x$, then why we need $|t|<1$. Is it related only on the condition $|x|<1$. Can we define the generating function for all $x$ and all $t$.

答:
The generating function can be defined formally. But in that case $x$ and $t$ doesn't represent numbers, they are only formal variables that doesn't take any values, it makes no sense saying "for all $x$ and all $t$".

If one want to consider the generating functions as actual functions with variables that are real (or complex) numbers, then the convergence of the series must be taken into account. $|x|<1$ and $|t|<1$ are sufficient conditions for the series to converge.


问:
Complex representation and Dual representation notation
Let's say we have a representation $\rho$ of $G$ on a vector space $V$. Wikipedia refers to the dual representation as $V^*$, but the dual vector space as $\overline{V}$. It does the opposite for the conjugate representation and the conjugate vector space. What is normal accepted notation for these, and is the discrepancy just an error by someone who edited Wikipedia?

Thank you.

答: The standard notation for the dual vector space to a vector space $V$ is $V^*$.  If $V$ is a $G$-module for some group $G$ (equivalently, one has a representation $\rho \colon G \to \operatorname{End} V$), then one often also writes $V^*$ for the dual module.  The notion of a conjugate vector space (and representation) is less common.</p>

ContentLicense: CC BY-SA 3.0


问:
Bipartite Graph.
Scientists have identified 20 traits for humans.
Each person has 8 of these 20 traits before birth. In addition, human beings are unique and no one set of traits is the same.
Show that a human being may acquire a new trait during his or her lifetime so that the human being remains unique.</p>

答:
Hint: For simplicity of argument, let's suppose that every possible set of $8$ traits corresponds to one of the humans so that there are the maximal number $\binom {20}{8}$ humans; if there aren't enough humans, then it is only easier to ensure that the humans remain unique after gaining a trait.

For each set $A$ of $8$ traits and $B$ of $9$ traits, connect $A$ to $B$ if and only if $A \subset B$. Each such set $A$ is connected to $20 - 8 = 12$ sets. A pair of distinct sets $A_1,A_2$ connect to a common set $B$ of $9$ traits if and only if $|A_1 \setminus A_2| = 1$, and in this case they both connect to $B = A_1 \cup A_2$.

Using this result: verify that the degree of each subset with size $8$ (each "human") has degree $20 - 8 = 12$, while each subset of size $9$ has degree $9$. The conclusion follows from the fact that $12 \geq 9$.

ContentLicense: CC BY-SA 4.0

Title: Bipartite Graph.


问:
Find convex closure of certain set
I'm having trouble solving this problem:

Given the set $X\subset\mathbb{R}^2$ defined as
$$X=\{(x,y)\in\mathbb{Z}\times\mathbb{R} : y\geq 0,x^2\leq b^2+y\},$$
being $b\in\mathbb{R}$. Is $X$ covex? Study the convex closure of $X$ in $\mathbb{R}^2$ in function of $b$. Define this closure in an analogous way to the set $X$, adding if necessary one new inequality. Note: consider absolute value.

I know how to prove $X$ is not convex, but I don't know how to answer the second question. The first thing I need to do is obviously change the condition $(x,y)\in\mathbb{Z}\times\mathbb{R}$ to $(x,y)\in\mathbb{R}^2$. But after this, I don't know how to express it. I clearly see the shape in my head, but I don't understand how to expess it with just one more inequality. I understand that
$$\{(x,y)\in\mathbb{R}^2 : y\geq 0,x^2\leq b^2+y\}$$
is NOT the convex closure of $X$, since you don't need the full parabola $x^2=b^2+y$ as the set's edge (instead, the edge of the convex closure will be an infinte number of segments joining each consecutive intersection points between $x\in\mathbb{Z}$ and $x^2=b^2+y$). I hope I have explained my point correctly

How can I solve this? Any help or hint will be appreciated, thanks in advance

答:
This is just a formal writing following @EDX's advice.

The required set is $$Y = \{(x,y) \in \mathbb{R}^2 \mid y \ge \lfloor x \rfloor^2 - b^2  + \underbrace{(x - \lfloor x \rfloor) (2 \lfloor x \rfloor + 1)}_{=0 \text { if } x \in \mathbb{Z}}\}.$$  It's easy to see that $X \subseteq Y$.
The key to prove the convexity of $Y$ is to observe that it's a piecewise linear function.  Choose an arbitrary point $x \in \mathbb{R}$ belonging to the interval $[k, k+1)$ with $k \in \mathbb{Z}$, then choose arbitrary $s \in [k,k+1)$.  We have to show that $f(s) = \lfloor s \rfloor^2 - b^2  + (s - \lfloor s \rfloor) (2 \lfloor s \rfloor + 1)$ is linear on the closed interval $[k,k+1]$.

We start with some basic observations.

The points $P_0 = (k,k^2 - b^2), P_1 = (k+1, k^2 - b^2 + 2k + 1)$ belongs to $X$.

The floor function is a step function, and $\lfloor s \rfloor = k$.

Compute the slope of the line segment joining $(k,f(k))$ and $(s,f(s))$, and observe that it's the same as that joining $(k,f(k))$ and $(k+1,f(k+1))$.

$$\frac{f(s)-f(k)}{s - k} = \frac{k^2 - b^2 + (s-k)(2k+1)-(k^2-b^2)}{s - k} = 2k + 1 = f(k+1)-f(k).$$

It's clear that $Y$ is the intersection of upper half-planes whose boundary joins a pair of "adjacent (in the sense of $\mathbb{Z}$ of the $x$-component) points".

$$Y = \bigcap_{k \in \mathbb{Z}} \{(x,y) \in \mathbb{R}^2 \mid y \ge k^2 - b + (2k + 1)x \}$$

As each half-plane above is closed, so does its countable intersection.

I hope you would be convinced that $Y$ is the convex envelop of $X$, even though I've only showed that the boundary points of $Y$ are a convex combination of $X$. (i.e. $\{ (x,y) \in \mathbb{R}^2 \mid y = f(x) \} \in \mathrm{conv}(X)$).

 For each $\epsilon > 0$ and $x \notin \mathbb{Z}$ such that $(x,f(x) + \epsilon) \in Y$, write it as a convex combination of $(\lfloor x \rfloor, f(\lfloor x \rfloor) + \epsilon)$ and $(\lfloor x+1 \rfloor, f(\lfloor x+1 \rfloor) + \epsilon)$, which belong to $X$.  In other words, if there's an uplift $\epsilon$, just do the same on your chosen pair of adjacent points in $X$.

Title: Find convex closure of certain set


问:
How do I solve this equation over integers: $1=91p+74q$
Like in title. I need to solve this equation: $$1=91p+74q$$ with $p, q$ being integers. I brute forced the solution to be $(p,q)=(-13,16)$, but I'd love to know how to solve it without just mashing keys on the calculator, hoping for the best.

Body:
Use the extended Euclidean algorithm.

$91=1\cdot74+17$

$74=4\cdot17+6$

$17=2\cdot6+5$

$6=1\cdot5+1$.

Therefore,

$1=6-1\cdot5=6-1(17-2\cdot6)=3\cdot6-17=3(74-4\cdot17)-17=3\cdot74-13\cdot17$

$=3\cdot74-13(91-74)=16\cdot74-13\cdot91$.


问:
Counting number of distinct regions with intersecting circles
Given $n$ circles of possibly different radii, how many distinct regions can there be?
For small $n$, I can work it out with pictures. (I'm pretty sure $n=4$ can yield 13 distinct regions, but not positive.)

Just curious what an approach to the problem could be.

E.g. I have considered that is could be a quadratic sequence based on skimpy evidence, but don't know why that would be, or how to prove it.

I would also be interested to know if there was a way of doing this problem with regular $n$-gons. (I am assuming that there would be more regions, but am not sure.)

答:
The source quoted by Ross Millikan (OEIS) gives different formula: </p>

<p>$$n^2+n+2$$</p>

Body:
The source quoted by Ross Millikan (OEIS) gives different formula:

$$n^2+n+2$$


问:
Ring Inside an Algebraic Field Extension
Let $E|F$ be an algebraic field extension and a ring $K$ such that $F\subseteq K\subseteq E$. It is true that $K$ is a field?</p>

答:
Yes. Let $a \in E$ with $a\ne0$.

Since $a \in E$ is algebraic over $F$, we have that $F[a]$ is a finite-dimensional $F$-algebra because $F[a]$ is by definition the $F$-vector space generated by the powers of $a$ and powers having exponent larger than the degree of $a$ can be replaced by linear combinations of lower powers by using a monic polynomial equation satisfied by $a$.

Since $F[a]$ is a finite-dimensional $F$-algebra, the map $x \mapsto ax$ on $F[a]$ is $F$-linear and injective and hence surjective. Therefore, $1$ is in the image and $a$ is invertible in $F[a]$.

Since $F[a]$ is by definition the smallest ring containing $F$ and $a$, we have that $a$ is invertible in every subring $K$ of $E$ that contains $a$.
In particular, every nonzero element of $K$ is invertible, and $K$ is a field.

Actually, the following are equivalent (proof left as an exercise):

$a$ is algebraic over $F$

$F[a]$ is a finite-dimensional $F$-algebra

$F[a]$ is a field

$F[a]=F(a)$


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

This is what I have thus far.

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

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

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

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

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

答: If the circles of radii $b,\,a$ have respective centres $(a,\,0,\,0),\,(0,\,0,\,0)$, their intersection is confined to $|z|\le b<a$. For such $z$, the intersection satisfies$$0\le(x-a)^2+y^2\le b^2-z^2,\,0\le x^2+y^2\le a^2-z^2.$$Rewriting the $xy$-plane with polar coordinates, the first double inequality is$$\begin{align}(r-a\cos\theta)^2&\le b^2-z^2-a^2\sin^2\theta\\\iff r&\in\left[a\cos\theta-\sqrt{b^2-z^2-a^2\sin^2\theta},\,a\cos\theta+\sqrt{b^2-z^2-a^2\sin^2\theta}\right],\end{align}$$which is satisfiable iff $\sin^2\theta\le\frac{b^2-z^2}{a^2}$. The second sphere's double inequality is equivalent to $0\le r\le\sqrt{a^2-z^2}$. So the intersection is$$\begin{align}|\theta|&\le\arcsin\frac{\sqrt{b^2-z^2}}{a},\\r&\in\left[\max\left\{0,\,a\cos\theta-\sqrt{b^2-z^2-a^2\sin^2\theta}\right\},\,\min\left\{\sqrt{a^2-z^2},\,a\cos\theta+\sqrt{b^2-z^2-a^2\sin^2\theta}\right\}\right].\end{align}$$If we right this more concisely as $-f(z)\le\theta\le f(z),\,g_-(\theta)\le r\le g_+(\theta)$, the volume of the intersection is$$\int_{-b}^bdz\int_{-f}^fd\theta\int_{-g}^grdr.$$


问:
Divide $n > 10$ distinct elements $\{1,2, \dots, n\}$ to $3$ distinct sets $A,B,C$ where $|A|=5$ and $|B|>|C|$
I am trying to use the inclusion-exclusion principle here but having a problem calculating the cases.

Assume $D_1$ - the possible ways to choose a number not equal to $5$ members for $A$

and $D_2$ - the possible ways to divide the elements where $|B| \leq |A|$

and $U$ - all possible divisions of $n$ elements to $A,B,C$

And I want to calculate $\bar{D_1} \cap \bar{D_2}=|U| - |D_1|-|D_2|+|D_1 \cap D_2|$

Now, $|U|$ is the number of functions from a set of the cardinality of $n$ to a set of the cardinality of $3$ which given $|U|=3^n$

But the other calculations seem too confusing. Should we solve it this way? how to calculate the rest of the expression if so? or how to solve it in a better way?

答:
You can use a greatly simplified inclusion-exclusion idea.

Once you have chosen the 5 elements for set A you must split the remaining $n-5$ elements into two sets. When you do this you either have or do not have two sets of equal size.

If $n-5$ is odd

The sets have to be unequal in size and so half the possibilities have B larger than C. The number of possibilities is $2^{n-6}\binom{n}{5}$.

If $n-5=2k$ is even

We have to exclude the cases when the sets are equal. The number of possibilities is $\frac{1}{2}(2^{n-5}-\binom{n-5}{k})\binom{n}{5}$.


问:
sample space in Hewitt-Savage 0-1 law
I'm reading Durrett's Probability: Theory and Examples (5th ed).

He state Hewitt-Savage 0-1 law as followings

If $X_1, X_2,...$ are independent and identically distributed(i.i.d) and $ A \in \varepsilon $ then $P(A) \in \{0,1\}$

where $ \varepsilon $ is exchangeable $\sigma$-field.

In proof, He claimed that

Let $A_n \in \sigma(X_1, ... X_n)$ so that $P(A_n \Delta A) \rightarrow 0$

Then $A_n$ can be written as $\{\omega : (\omega_1, ... , \omega_n) \in B_n\}$ where $B_n \in S^{n} $

I understand that such $A_n$ exists in meausure theoretical sense.

But I can't understand what "$A_n$ can be written as~" means.

I think $\omega = (\omega_1 , \omega_2 , ...) $ if $X_i(\omega) = \omega_i$ for given random variables. But, Is it well-defined?

Since random variable may not be injective, this expression using coordinates is not well-defined I think. (different element of sample space can be expressed as same coordinates)

How can I understand this expression?

Thanks.

答:
Suppose $X : (\Omega, \mu) \to \mathbb{R}$ is a random variable. You can then define a new random variable $X' : (\mathbb{R}, \mu') \to \mathbb{R}$ by $X'(\omega)=\omega$ where the measure on the domain is $$\mu'(E)=P(X \in E) = \mu(X^{-1}(E))$$ for any measurable $E \subseteq \mathbb{R}$. So in some sense you are throwing away the old probability space $(\Omega, \mu)$. [$\mu'$ is called the pushforward measure.]

So if you have $X_1, \ldots, X_n$ all represented as maps $(\mathbb{R}, \mu_n) \to \mathbb{R}$ you can take the product of the measures $\nu = \mu_1 \otimes \cdots \otimes \mu_n$ on $\mathbb{R}^n$ and reformulate each $X_i$ as a map $(\mathbb{R}^n, \nu) \to \mathbb{R}$ where $X_i((\omega_1,\ldots,\omega_n)) = \omega_i$.

Title: sample space in Hewitt-Savage 0-1 law


问:
Covariance of Wiener Process with nonzero $x(0)$
Let $x(t)$ be a Wiener process and let $\sigma_1^2 = var(x(1))$, $\sigma_0^2 = var(x(0))$. If $x(0) = 0$, then we know that $C(t,s) = \sigma_1^2 min(t,s)$ (please see Mean and covariance of Wiener process). I want to prove that if $x(0)$ is not zero and instead a r.v then $C(t,s) = \sigma_0^2 + min(t,s)(\sigma_1^2 - \sigma_0^2)$. Any ideas to start with?

答: Write $x(t)=B(t)+x(0)$ where $B$ is a Wiener process started at $0$ with Var$[B(1)]=\sigma_1^2-\sigma_0^2$ and $x(0)$ is independent of $B$. Now use bilinearity of covariance to obtain the claim.


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

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

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

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

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

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

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

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

Thank you :)

答:
In the context of the chain rule, it is understood as you said (and as I said in the comments), in the way that you described as "magically".

Let me give an example to show a slightly different perspective, which hopefully will be the more "natural" interpretation you are looking for.

Consider a circle, with radius $r$. Its area is $A = \pi r^2$. Its circumference is $C = 2\pi r$. We can write $A$ in terms of $C$. While this might seem a little unnatural at first, just bear with me. Take the equation for $C$ and square both sides to get $C^2 = 4\pi^2 r^2$. Now just factor out $4\pi$ to get $C^2 = 4\pi \cdot \pi r^2 = 4\pi A$. Solving for $A$, you get
$$ A = \frac{C^2}{4\pi} $$

Now, we have $A$ is a function of $C$, and $C$ is a function of $r$, so we are in the situation where we could use the chain rule, and it would tell us: $\frac{dA}{dr} = \frac{dA}{dC} \frac{dC}{dr}$. This of course gives $2\pi r$ either way you compute it, but that's not the point. My point is that because of the equation $A = \frac{C^2}{4\pi}$ above, we can make sense of the expression $\frac{dA}{dC}$ without even mentioning $r$! If you just "forget" for a moment that $C$ depends on $r$, and think of $C$ itself as being the independent variable (taking the equation $A = \frac{C^2}{4\pi}$ out of context), then $\frac{dA}{dC} = \frac{C}{2\pi}$ makes perfect sense, as just the ordinary derivative (no magical trick of notation).

My point here was that you can interpret $\frac{df}{dg}$ as just an ordinary derivative, thinking of $g$ itself as the variable (forgetting that $g$ is actually a function). The example above was supposed to illustrate that in many "real-life" situations where the functions have intuitive geometrical meanings, they can be either thought of as variables in their own right, or as functions of another variable. In my circle example, $C$ could either be interprepeted as a function of $r$, or we could just think of $C$ itself as a variable (and ignore the relation between $C$ and $r$).

Title: How to use Leibniz notation properly


答:
I see that others have mentioned the Euclidean Algorithm. That would surely work. I would like nominate an approach I constantly use.

So we first look at $91$ and $74$. They are coprime, which means that such $p,q$ exist. What we want to do is to write $1$ as a linear combination of $91$ and $74$ with integer coefficients. We have $91-74=17$, so $17$ can be written as a linear combination of $91$ and $74$ with integer coefficients (which are $1$ and $-1$). Next we want to find a even smaller number that can also be written as a linear combination. We have $91-5\cdot 17=6$. Now we have $91-15\cdot 6=1$, i.e.,

$$1=91-15\cdot 6=91-15\cdot(91-5\cdot 17)=91-15\cdot(91-5\cdot(91-74))\\=91-15\cdot(5\cdot 74-4\cdot 91)=61\cdot 91-75\cdot 74$$

Title: How do I solve this equation over integers: $1=91p+74q$

Body: Like in title. I need to solve this equation: $$1=91p+74q$$ with $p, q$ being integers. I brute forced the solution to be $(p,q)=(-13,16)$, but I'd love to know how to solve it without just mashing keys on the calculator, hoping for the best.


答: I am going to tell you a very beautiful approach to solve these types of problems note that general solution is x=-13+74t  and y=16-91t where t is a nonnegative integer now I tell you how I get that I assume that you know a little bit about continued fractions since I will apply it here now just  write  91/74 as a continued fraction now look at the second last convergent say p(n-1) /q(n-1) now use cassini identity which says that p(n) q(n-1) -q(n) p(n-1) =${-1}^n$ by using this I got general solution I hope it helps


问:
What does the output of a derivative actually say in real life?
Perhaps the wording of the question will need to be adjusted, but I do not know how else to convey what I am looking for in one sentence.

Here is what I understand:

The derivative of a function in essence shows the "rate of change" of a function at any given point. In practical application, determining the current speed of a car at a given time along it's route, as opposed to the average speed of a car over it's entire route.

The value (output) given by the derivative function expressed mathematically is the "slope of a line",the tangent, at any given point (output) of the function as displayed on a graph, which represents the how much the output of the function has changed with the input.

But what exactly is the output of the derivative telling me about this change? If the output of the derivative of some 'x' is '2', what is this '2'? It is a slope, yes, but what is this slope telling me in real life?

Here is an example of what I am trying to understand. We will use the following function... $$f(x)=x^2$$ and its derivative... $$f'(x)=2x$$

Let us imagine I have an extremely unique and completely absurd job that lasts for a '1' hour shift each day, and that has a pay rate of f(x)=x^2 per minute, where 'x' is the number of minutes worked in the shift so far.

If we look at just the first five minutes of the shift, both the 'pay rate per minute' and the 'rate of change (slope) of pay per minute', we get...
$$f(0\text{min})=$0;f'(0\text{min})=0$$ $$f(1\text{min})=$1;f'(1\text{min})=2$$ $$f(2\text{min})=$4;f'(2\text{min})=4$$ $$f(3\text{min})=$9;f'(3\text{min})=6$$ $$f(4\text{min})=$16;f'(4\text{min})=8$$ $$f(5\text{min})=$25;f'(5\text{min})=10$$

Now, again, the output for the derivative, if graphed, is the slope of the line at any given point along the main function.

But what does that mean in practice?

If I asked someone, "How much more money am I earning from minute '4' of my shift?", and that someone answered, "according to the derivative of the money earned, 8." , what is that? '8' what? Dollars? Percent? No unit? Just '8", or any output given by the derivative, doesn't give me any useful understanding of what is changing.

The question-answer bit I gave at the end may be worded incorrectly. (let me know what I should change it to, if so), but I hope the point of this question accross.

Also, I am not looking for proofs in the answer. Proofs are not hard to find. I am looking for answers that are not written in proofs.

答:
The derivative of $f(x)$ with respect to $x$ always has units of $$\frac{\text{units of $f$}}{\text{units of x}}.$$  So, for your function, if $$f(x) = x^2 \text{ dollars per minute}$$ then the derivative will be $$f'(x) = 2x \text{ dollars per minute per minute}.$$  In practice, that means that tells you how fast your pay per minute in increasing.  If, for example, you say $f'(4) = 8$ what this really means is that a minute $4$ your pay per minute is increasing by $\$8$ per minute per minute.

Let's say that we don't actually know what $f(x)$ is, just that $f(4) = \$16$ per minute and $f'(4) = \$8$ per minute per minute.  That would not only tell you that your pay per minute was increasing, but that at minute $5$ you would be making approximately an additional $\$8$ per minute than you were at minute $4$.


问:
Inverse of function of projective space $\mathbb{R}P^1$ in $\mathbb{S}^1$?
Is it possible to explicitly define the inverse function of this function of the projective space $\mathbb{R}P^1$ in $\mathbb{S}^1$?

$$f([x,y]) = \left( \dfrac{2xy}{x^2+y^2}, \dfrac{x^2-y^2}{x^2+y^2}\right).$$

答:
I guess you define $\mathbb RP^1 = (\mathbb C \setminus \{0\})/ \sim$, where $z \sim z'$ iff $z' = tz$ for some $t \in \mathbb R \setminus \{0\}$. Note that I identified $\mathbb R^2$ with $\mathbb C$. Then your map $f$ is essentially given by $f([z]) = \dfrac{z^2}{\lvert z \rvert^2}$ (note that I exchanged the first and second coordinate of your original $f([z])$ which does not play a role conceptually because the "flip-map" $\tau(x+iy) = i \overline z = y + ix$ is a homeomorphism on $S^1$).

This map is a well-defined continuous bijection, hence a homeomorphism. You ask for its inverse $f^{-1} : S^1 \to \mathbb RP^1$. It is well-known that for each complex number $w \ne 0$ we can choose a square root $\sqrt w$ (there are two possible choices which differ by a factor $-1$, but we shall not be specific how to choose $\sqrt w$). Note that $\sqrt w \in S^1$ for $w \in S^1$. We claim that
$$f^{-1}(w) = [\sqrt w] .$$
Note that the RHS does not depend on the choice of $\sqrt w$ because $\sqrt w \sim -\sqrt w$. But now we have $f([\sqrt w]) = \dfrac{\sqrt w ^2}{\lvert \sqrt w \rvert^2} = w$ which proves our claim.

Concerning square roots it may be interesting to have a look at my answer to Analytic Functions - Entire Function .


问:
Proving $6n^4>(n+2)^4$ for $n\geq 4$, by induction

I need to prove that:
$$6n^4>(n+2)^4$$
for $n\geq4$ is true using proof by induction.

I've started by showing that when $P(4)$ is true since $\text{LHS}=1536$ and $\text{RHS}=1296$.

I then let $k≥4$ and assumed $P(k)$ is true, which is $6k^4>(k+2)^4$.

So we need to show that P(k+1) is true, so,
$$6(k+1)^4>(k+3)^4$$

This is where I get stuck, so would be very grateful if anybody could help finish the proof.

Thanks
Kara :-)

答:
Starting from your induction step of $P(k)$, you want to somehow "change" the inequality so or one side or the other becomes that of $P(k+1)$ while also showing the inequality still holds. With the left side, to get from $6k^4$ to $6(k+1)^4$, since it's multiplicative, one way to do this is to divide by $k^4$ and then multiply by $(k+1)^4$, i.e., multiply both sides by $\left(\frac{k+1}{k}\right)^4$. Doing this gives

$$\begin{equation}\begin{aligned}
6k^4 & \gt (k + 2)^4 \\
6k^4\left(\frac{k+1}{k}\right)^4 & \gt (k + 2)^4\left(\frac{k+1}{k}\right)^4 \\
6(k+1)^4 & \gt \left(\frac{(k+2)(k+1)}{k}\right)^4 \\
& = \left(\frac{k^2 + 3k + 2}{k}\right)^4 \\
& = \left(k + 3 + \frac{2}{k}\right)^4 \\
& \gt (k + 3)^4
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$

This proves $P(k + 1)$ is true.


问:
Find the equation of a plane containing line $L$ and equidistant from two points
I want to find the equation for a plane that is equidistant from the points $P_0 = (1, 2, 3)$ and $P_1=(3, 2, 1)$ and contains the line $\ell=(1, 1, 1) + t(0, 1, -1).$

I know that the midpoint of the two points $P_0$ and $P_1$ must be on the plane, which is $MP=(2, 2, 2).$ However, I notice this point is on the line $\ell$. I need another vector in the plane to preform the cross product and find a normal to the plane, however I can't think of a way to find such vector. I know that the length of the orthogonal complements of the two points on the normal must be the same for all points on the plane, but setting up such an equation doesn't seem like the right way to go about it. Any help would be appreciated.

答:
The set of points that is equidistant from two points in $\mathbb R^3$ is a plane. The normal vector of this plane is $\overrightarrow{P_0P_1}$ and contains the midpoint $(2,2,2)$.

Therefore the place you require is

$$2x-2z = 0$$

or

$$x=z$$

However this plane does not contain the line $l$; their intersection is the point $(1,1,1)$, so I suspect something is wrong with the question: the line $l$ itself is not equidistant from the two points.


问:
Conditional Probability, Lack of Dependence on a Parameter
I am trying to understand why the following is true:
$$
p(f(Y) = f(y) \mid Y = y) = p(f(Y) = f(y) \mid X = x, Y = y) \qquad \ldots \text{(Eq. 1)}
$$
where $Y$ and $X$ are random variables, and $f(Y)$ is a function of a random variable. $p(Y = y)$ represents the probability density function of random variable $Y$, evaluated at the point $y$.

Equation 1 states the event $f(Y) = f(y)$ conditioned on the value of the random variable $Y$, has no dependence, or is unaffected by the value of random variable $X$. Is there a formal name for Equation 1, or any additional information to help me understand further the process behind Equation 1?

The context for my question is Markov Chains in Information Theory. $X$, $Y$, and $f(Y)$ form a Markov Chain in That Order. Equation 1 must be true to imply the Markov Chain of the three random variables.

Thanks.

答: $f(Y)$ depends on $Y$, and $X$ depends on $Y$, and $f(Y)$ is conditionally independent of $X$ given $Y$ due to your Markov Chain assumption. Therefore, $p(f(Y)|X,Y) = p(f(Y)|Y)$. In general this is not true, rather $p(f(Y)|X,Y) = \int p(f(Y)|X,Y) p(X|Y) \mathrm{d}X$


问:
20 Gifts and four stacks, How many Gifts in each pile?
There are twenty gifts stacked up into $4$ piles.
The first pile has $3$ less than the second pile.
The second pile has $2$ more than the third pile.
The fourth pile has twice as many as the second pile.
How many gifts are in each pile?

This is for my $8$ year old son... Embarassing that I can't figure it out.

答:
Let
$$\begin{align*}
A & =\text{number of gifts in the first pile}\\
B & =\text{number of gifts in the second pile}\\
C & =\text{number of gifts in the third pile}\\
D & =\text{number of gifts in the fourth pile}
\end{align*}$$
Then the information given is that
$$A=B-3,\quad B=C+2,\quad D=2B,\quad A+B+C+D=20.$$
We can re-express $B=C+2$ as $C=B-2$. Replacing term by term:
$$\underbrace{A}_{B-3}+B+\underbrace{C}_{B-2}+\underbrace{D}_{2B}=20$$
we see that
$$(B-3)+B+(B-2)+2B=5B-5=20$$
which implies $B=5$, and then $A=2$, $C=3$, and $D=10$.</p>


问:
Exponential parametrization for diagonal SL(2,R)
I already saw this question (and the linked questions) but mine is slightly different. I am asked to prove the fact that the exponential parameterization isn't a good one for $SL(2,\mathbb{R})$ using some diagonal matrices as examples. I'm aware that one can find some (upper triangular) matrices that don't have a bijective exponential parameterization, but I'm pretty sure that any $SL$ matrices of the form $diag(x,x^{-1})$ can be written as $diag(e^a,e^{-a})$. Am I wrong, or is the question just not answerable using diagonal matrices?

edit: thanks to Moishe Kohan, I noticed that there are some $x$ so that $diag(x,x^{-1})$ can't be written as $diag(e^a,e^{-a})$ - the negatives. But I also noticed that a diagonal matrix written as an exponential of another matrix $diag(x, x^{-1})=e^A$ doesn't imply that A is diagonal! So this still doesn't answer the question: can all diagonal $SL$ matrices be written with an exponential parameterization?

答:
Hints:

Suppose that $D$ is a diagonal matrix and $D=\exp(A)$. Relate the eigenvalues of $D$ and of $A$. (If this is too  difficult, assume that $D$ has distinct diagonal entries.)

Consider what happens when $D$ has negative diagonal entries.


问:
Degree of the map
What is a degree of a map $f:T^2\rightarrow T^2$ given by $f(z,w)=(z^2, w^3)$? Is it 5?

答: Use the Künneth formula to obtain an explicit form of the fundamental class of $T$ in terms of the two circle factors. Then, use the explicit form of $f$ acting on the two circle factors to conclude that the degree is 6.

Title: Degree of the map


答:
Use the forward version of the Extended Euclidean Algorithm:

$$\begin{array}{ccl} & 91 & 74\\
--- & --- & --- & ----\\
 91 & 1 & 0 & (1) \\
74 & 0 & 1 & (2) \\
17 & 1 &  -1 & (3) = (1)-(2)\\
6 & -4 & 5 & (4) = (2)-4\times(3)\\
-1 & 13 & - 16 & (5) = (3)-3\times(4) \end{array}$$

Therefore $-1 = 13(91)-16(74) \implies 1=(-13)\times 91+16\times (74)$

Note: when I did regular EEA it's very easy to make calculation mistake because of its backward substitution nature. With this forward method it almost never happened. And when it does, you can track each line easily to find where you made a mistake. For example,

Line $(1): 91 = 1\times 91 + 0 \times 74$

Line $(4): 6 = -4 \times 91 + 5 \times 74$

etc.


问:
Having number of walks between two vertices how to calculate paths
I know that we can calculate number of walks with n length between two vertices of an undirected graph by calculating the nth power of the adjancy matrix but what if you want to avoid repetition in edges and vertices can you still use the nth power of the adjancy matrix and divise it by sth or perform some operation to remove the repetition?
Any help is appreciated</p>

答:
I would like to formalize your question first.

Let $G=(V,E)$ be an undirected graph with $n$ vertices and $m$ edges. Without the loss of generality we can assume that $G$ is connected (because if $G$ is not connected, we could simply apply our algorithm to all components of $G$ and sum up the results).

Def. 1: A walk of length $k$ is a sequence of vertices $v_0,\cdots,v_k \in V$ such that $v_i$ and $v_{i+1}$ are connected by an edge of $E$, for any $0 \leq i < k$.

Def. 2: A path $P$ of length $k$ is a walk $v_0,\cdots,v_k$ such that any two vertices $v_i$ and $v_j$ are distinct, for $0 \leq i < j \leq k$. We call a path from vertex $v_i$ to vertex $v_j$ a $v_iv_j$-path.

Let us call a path from a vertex $v_i$ to a vertex $v_j$ simply $v_iv_j$ and denote its length $|v_iv_j|$.

Let us denote the set of $v_iv_j$-paths in a graph $G$ by $\mathcal{P}_G(v_iv_j)$ for any given $v_i,v_j \in V$.

Problem Instance: Two vertices $v_i,v_j$ of a graph $G=(V,E)$ and an integer $k$.

Question: What is the cardinality of $\mathcal{P}_G(v_iv_j)$ for $|v_iv_j|=k$?

First notice that this cannot be done in general using the adjacency matrix of $G$ alone, at least as far as I am concerned. Actually, I don't know of any nor have I found a method that generally solve this problem efficiently. What is generally known is that this problem is hard, in fact it is $\#P$-complete.

I think that you could generate all possible solutions using the Depth-First-Search (DFS) and then backtracking for listing all possible paths, then filter for your desired length. Keep in mind that this has a running time complexity of $O(n!)$ for $n$ being the number of vertices, which is insane. The algorithm is described here.

However, there are a few papers that I would like to propose. If your paths are short, there exists a general enumeration formula. Take a look at

"On the number of paths of lengths 3 and 4 in a graph" by Movarraei and Shikare,

"On the number of paths of length 5 in a graph" by Movarraei and Boxwala.

both in the International Journal of Applied Mathematical Research.

Not answering your question directly, but still closely related are the following paper.

"A General Purpose Algorithm for Counting Simple Cycles and Simple Paths of Any Length" by Giscard et. al, Springer,

"Optimal Listing of Cycles and st-Paths in Undirected Graphs" by Birmelé et. al in Proceedings of the twenty-fourth annual ACM-SIAM symposium on Discrete algorithms.

This is not new, in fact there is a publication from 1966 by Ponstein called "Self-avoiding paths and the adjcency matrix of a graph".

There is a method for counting the number of $v_iv_j$-paths using a stochastical approach. It is called

"Estimating the Number of s-t Paths in a Graph" by Roberts and Kroese in Journal of Graph Algorithms and Applications.


问:
Combinatorial proof for $ \frac{1}{2} \binom{2n}{n} = \binom{2n-1}{n}$
I have to find a combinatorial proof for the following identity:

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

I am really not sure how to formulate the proof because I have a scalar in front of binomial, in this case, a fraction.

答:
There is a identity which says that $ \binom{n}{k} =  \frac{n}{k}
 \binom{n-1}{k-1}$ now just compare left and right hand side combinatorial proofs will be like this first we have to choose n persons from 2n persons which is same thing as if one person is fixed then total ways are apparant from here I suppose you can complete


问:
Orthogonal Projection onto a Variation of the Unit Simplex

Given the set $T_{\alpha}=\{x\in\mathbb{R}^n:\sum x_i=1,0\leq x_i\leq \alpha\}$

For which $\alpha$ the set is non-empty?

Find a dual problem with one dual decision variable to the problem of finding
the orthogonal projection of a given vector $y ∈\mathbb{R}^n$ onto $T_α$.

I found that $\alpha\geq1/n$ and when i tried to calculate the dual function  I did the following.

$\min$ $||x-y||^2$

s.t

$\sum x_i=1$

$0\leq x_i\leq\alpha\to X=\{x:0\leq x_i\leq\alpha\}$

Setting the lagrangian to be $L(x,\lambda)=||x-y||^2+2\lambda(\sum x_i-1)$,we would like to minimize the lagrangian w.r.t x then :

$$\underset{x\in X}{\min}L(x,\lambda)=\sum x_i^2+2(\lambda-y_i)x_i+||y||^2-2\lambda$$

$$\frac{\partial L}{\partial x_i}=2x_i+2(\lambda-y_i)=0\to x_i^*=\left\{\begin{array}{rcl} y_i-\lambda&0\leq y_i-\lambda\leq\alpha\\
\alpha&y_i-\lambda>\alpha\\
0&y_i-\lambda<0\end{array}\right.$$

and the dual function is $q(\lambda)=-2\lambda+||y||^2+\left\{\begin{array}{rcl}n\alpha^2+2n\alpha\lambda-2\alpha\sum y_i&y_i-\lambda>\alpha\\
-\sum [y_i-\lambda]_+^2&\mbox{else}\end{array}\right.$
and the derivative of $q$ is:

$$q'(\lambda)=-2+\left\{\begin{array}{rcl}2n\alpha &y_i-\lambda>\alpha\\
2\sum[y_i-\lambda]_+^2&\mbox{else}\end{array}\right.$$

I'm not sure if I've found the right dual function and if I've found the right derivative and write matlab code that solves the dual and finds the projection of $y$

答:
Adding to @Royi 's answer,

The Lagrangian is:
$$\mathcal L(\mu) = \sum_{i=1}^n ({\left( {y}_{i}-\mu \right)}_{0 \leq \cdot \leq \alpha}^2 -2(y_i-\mu){\left( {y}_{i}-\mu \right)}_{0 \leq \cdot \leq \alpha}) + ||y||_2^2-2\mu
   $$

From duality, we know this function is concave. It is also $\to -\infty$ as $\mu \to \infty$ (sum becomes zero, and $-2\mu\to -\infty$) and as $\mu \to -\infty$ (square in the sum is finite;$||y||_2^2$ is finite; we end up with $\mu(2\alpha n-2)$ but since $\alpha > \frac{1}{n}$ then $2\alpha n-2>0$). It is continuous. The question is - is it differentiable?

We note that while $[x]_{0 \leq \cdot \leq \alpha}^2$ is not differentiable, $[x]_{0 \leq \cdot \leq \alpha}^2 - 2x[x]_{0 \leq \cdot \leq \alpha}$ surprisingly is! For $x < 0$ the term is $=0$ and so the derivative is $0$. For $x>\alpha$ we get $\alpha^2-2\alpha x$, whose derivative is $-2\alpha$. For $0 < x < \alpha$ we get $x^2-2x^2=-x^2$, whose derivative is $-2x$. This goes to $0$ as $x\to 0$ and goes to $-2\alpha$ as $x\to \alpha$. Hence the Lagrangian is also differentiable! And the derivative is $- 2[x]_{0 \leq \cdot \leq \alpha}$.

So the dual problem is to find the maximum of the Lagrangian, under $\mu\in\mathbb R$ (as it is an equality constraint). From all of the above, if we differentiate and equate to zero we will find the maximum of this dual problem. Differentiating and equating to zero, we get:
$$\mathcal L' = \sum_{i=1}^n -2{\left( {y}_{i}-\mu \right)}_{0 \leq \cdot \leq \alpha}(-1) -2 = 0 \Rightarrow \sum_{i=1}^n {\left( {y}_{i}-\mu \right)}_{0 \leq \cdot \leq \alpha} -1 = 0
   $$

Which is the h function in Royi's answer.


答:
First transform the identity to
$$\binom{2n}{n} = 2 \binom{2n - 1}{n}\ .$$
The left hand side is the number of ways of choosing an $n$ elements subset out of a $2n$ element set $S$. Now specify one element $r\in S$, it can be either included in the chosen subset or not. If it is not included, then we need to choose $n$ elements out of the remaining $2n-1$ elements to form the subset, of which there are $\binom{2n-1}{n}$ ways. If it is included, then we need to choose $n-1$ elements out of $S\setminus \{r\}$, or equivalently, we can choose $n$ elements out of the remaining $2n-1$ elements to exclude, of which there are also $\binom{2n-1}{n}$ ways.

Title: Combinatorial proof for $ \frac{1}{2} \binom{2n}{n} = \binom{2n-1}{n}$

Body:
I have to find a combinatorial proof for the following identity:

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

I am really not sure how to formulate the proof because I have a scalar in front of binomial, in this case, a fraction.


问:
Simple partial derivatives chain rule proof
Let $z=f(x,y)$, suppose $x=s+t$ and $y=s-t$ show that

$(\frac{\partial z}{\partial x})^2 - (\frac{\partial z}{\partial y})^2=\frac{\partial z}{\partial s}\frac{\partial z}{\partial t}$

I have rearranged to get them both in terms of s but this hasn't yielded any success. Any help with this would be greatly appreciated.

答:
We have:

$$\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}=\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}$$

(as $\frac{\partial x}{\partial s}=1, \frac{\partial y}{\partial s}=1$) and

$$\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}=\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}$$

(as $\frac{\partial x}{\partial t}=1, \frac{\partial y}{\partial t}=-1$).

Multiply those two equalities above to reach the desired equality.


问:
Uniform distribution,probability
Fix a positive integer $ b>0 $.

Let $ U\sim U[0,1] $ be uniformly distributed random variable, over the segment $[0,1]$.

Let $ (A_n)_n $ be a sequence of random variables, such that

$$ \mathbb{P}\left(U=\sum_{n=1}^\infty \frac{A_n}{b^n}\right)=1 .$$

Show that $ (A_n)_n $ are independent random variables, all uniform on the set $\{ 0,1,\ldots,b-1\}  $.

I tried to think of a solution, but couldnt find any idea.

I'd appreciate any help, thanks in advance.

答: In order for $U$ to be uniform, each $A_k$ has to have a distribution $P(A_k=j)=\frac{1}{b}$ for $0\le j\le b-1$.  For any $n$ coefficient $A_k$, this means $P((A_{k1}=j1)\cap( A_{k2}=j2)....\cap (A_{kn}=jn))$$ =P(A_{k1}=j1)\times P( A_{k2}=j2)....\times P (A_{kn}=jn)=\frac{1}{b^n}$ which shows independence.

Title: Uniform distribution,probability


答:
At that age they are pretty clever, but usually innocent of algebra. So the expected approach is trial and error, in the schools often called "guess and check." But one might as well be reasonably smart about it.

Let's visualize the piles of gifts. The second pile is bigger than its two neighbours, and the fourth is twice as big as the second. So the fourth is quite big, and has an even number of gifts. Let's guess that the fourth has $12$ gifts. Then the second has $6$, the third has $4$, and the first has $3$. Add up, we get $25$. Too bad, not right.

So $12$ for the fourth is too big. Let's try for $10$. Then the second has $5$, the third has $3$, the first has $2$. Add up. Bingo!

One can work in this way from any of the piles. For example, the first is clearly the smallest. If we guess it is of size $1$, we get $1, 4, 2, 8$, too small a sum. But $2, 5, 3, 10$ get us to the right place.

Title: 20 Gifts and four stacks, How many Gifts in each pile?

Body:
There are twenty gifts stacked up into $4$ piles.
The first pile has $3$ less than the second pile.
The second pile has $2$ more than the third pile.
The fourth pile has twice as many as the second pile.
How many gifts are in each pile?

This is for my $8$ year old son... Embarassing that I can't figure it out.


问:
Evaluate $\int \frac{dt}{(t^2-1)^2}$
I have this problem of evaluating the indefinite integral of a rational function

$$\int \frac{dt}{(t^2-1)^2}$$

and I'm a bit unsure about how to proceed. I could use partial fractions:

$$\frac{1}{(t+1)(t-1)(t+1)(t-1)}$$

$$\frac{1}{(t+1)^2(t-1)^2}$$

$$\frac{A}{t+1} + \frac{B}{t-1} + \frac{C}{t+1} + \frac{D}{(t+1)^2}$$

and go from there... but that seems long and complicated. Is there a trick I'm missing?

答:
Integrate directly by parts

$$\int \frac{dt}{(t^2-1)^2}=\int \frac1{2t}d\left( \frac{-t^2}{t^2-1}\right)= -\frac12 \frac{t}{t^2-1}-\frac12\int \frac{dt}{t^2-1}
$$

Title: Evaluate $\int \frac{dt}{(t^2-1)^2}$


问:
Solution of $Ax\leq b$ implies the existence of solution of $AA^Ty\leq b$.
Let we are given with a matrix A and a vector $b$. Given that there exists a solution of $Ax\leq b$. Show that the solution of $$AA^Ty\leq b$$ also exists.

My Effort: I tried to prove it by contradiction. Suppose that the system $AA^Ty\leq b$  has no solution. This means that for all y, we have that $$AA^Ty>b.$$
This means $$Ax>b\ \forall\ x=A^Ty.$$ How to arrive at a contradiction? Or otherwise how to approach this problem.

答:
1.)  in general you have a very useful inequality

$\text{rank}\big(AB\big)\leq \text{rank}\big(A\big)$

this holds because we always have  $\text{column space}\big(AB\big) \subseteq \text{column space}\big(A\big)$.

The inequality is met with equality iff  $\text{column space}\big(A\big) \subseteq \text{column space}\big(AB\big)$ i.e. iff $\text{column space}\big(A\big) = \text{column space}\big(AB\big)$

2.) set $B:=A^T$ and

$\text{rank}\big(AA^T\big)= \text{rank}\big(A\big)\implies \text{column space}\big(A\big) = \text{column space}\big(AA^T\big)$

(use inner product argument or SVD for LHS)

note: I infer/assume we are working over $\mathbb R$ here

3.) Thus the existence of some $\mathbf x$ such that

$A\mathbf x \leq \mathbf b$
reads: there is a vector in the column space / image of $A$ that is component-wise $\leq \mathbf b$

but since $A$ and $AA^T$ have the same column space / image, then there is a vector in the the column space / image of $AA^T$ that is component-wise $\leq \mathbf b$  (for avoidance of doubt: since said vector exists in the image of $AA^T$, then there is at least one vector in its pre-image, and call this $\mathbf y$.)


问:
Lagrange multipliers with 3 constrains
So I have this problem with the following task.

Find the points that satisfy necessary condition for existance of minimas:

$f(x,y)=-(x^2+y^2)$ constrains $\begin{cases} x + 2y\leq3\\ x\geq0 \\ y\geq0 \end{cases}$

The problem is that after creating system of equations for gradients I compare them to 0 and end up with one more unknown variable.

答:
Introducing three slack variables $\{s_k\}$ we transform the inequalities into equations and the Lagrangian formulation gives

$$
L(x,y,\lambda_1,\lambda_2,\lambda_3,s_1,s_2,s_3) = x^2+y^2+\lambda_1(x+2y-3+s_1^2)+\lambda_2(x-s_2^2)+\lambda_3(y-s_3^2)
$$

so the stationary points are given by the solutions to

$$
\nabla L = 0 = \cases{ \lambda_1+\lambda_2+2 x \\
 2 \lambda_1+\lambda_3+2 y \\
 s_1^2+x+2 y-3 \\
 x-s_2^2 \\
 y-s_3^2 \\
 \lambda_1 s_1 \\
 \lambda_2 s_2 \\
 \lambda_3 s_3 \\}
$$

with $f(x,y) = x^2+y^2$ the feasible results are

$f$
$x$
$y$
$s_1^2$
$s_2^2$
$s_3^2$

$0$
$0$
$0$
$3$
$0$
$0$

$\dfrac{9}{5}$
$\dfrac{3}{5}$
$\dfrac{6}{5}$
$0$
$\dfrac{3}{5}$
$\dfrac{6}{5}$

$\dfrac{9}{4}$
$0$
$\dfrac{3}{2}$
$0$
$0$
$\dfrac{3}{2}$

$9$
$3$
$0$
$0$
$3$
$0$

Here $s_k = 0$ indicates that the corresponding restriction is actuating.

Title: Lagrange multipliers with 3 constrains


问:
Definition of immersion and globally injective immersion.
Let $f:N \to M$ be a smooth constant rank map between smooth manifolds.

The rank theorem says that such a map can be injective only if its rank is equal to the dimension of $N,$ which in turn has to be smaller or equal to that of $M.$

Say that some smooth map $f$ is an immersion when the rank of $f$ at a point $p$ is equal to the dimension of the domain for every point $p.$

From the above claim, we have that any injective smooth map of constant rank must be an immersion.

On the other hand, I find the definition of globally injective immersion as a smooth map that is an immersion which is also bijective onto its image.

But any constant rank injective smooth map must automatically be both bijective onto its image and also an immersion.

So isn't the definition of globally inctive immersion just saying that $f$ is a constant rank smooth injective map? If yes, what's the use of this definition?

答:
Yes, you gave a correct interpretation of an injective immersion. What your definition is good for, I do not know: Personally, I do not find it particularly useful. However, with immersions, the point is that they do not have to be injective. Furthermore, it is often easier to check that a map is an immersion than to verify injectivity: Verifying that a map is an immersion boils down to understanding the derivative of a map and then injectivity of linear maps.

Just think of functions of one variable. Suppose that somebody asked you to show that a smooth function $f$ is injective, say,
$$
f(x)=x^5+ 2x+10.
$$
This sounds like a hard task. But you can use the 1st derivative test
$$
f'(x)=5x^4+ 2
$$
and notice that $f'(x)> 0$ for all $x$, implying that $f$ is strictly monotonic (increasing). Hence,  $f$ is also 1-1. In terms of immersion, what you prove is that $f$ is an immersion since $f'(x)\ne 0$ for all $x$. Hence, $f$ is strictly monotonic (assuming it is defined and is differentiable on an interval). This method of proving injectivity fails in higher dimensions, so one settles for immersions rather than injective maps.


问:
We have an integer n. We have n boxes where each box contains a non-negative amount of balls. Find all the permutations which satisfy some criteria
Let $n$ be a positive integer. We have $n$ boxes where each box contains a non-negative number of pebbles. In each move we are allowed to take two pebbles from a box we choose, throwaway one of the pebbles and put the other pebble in another box we choose. An initial configuration of pebbles is called solvable if it is possible to reach a configuration with no empty box, in a finite(possibly zero) number of moves. Determine all initial configurations of pebbles which are not solvable, but become solvable when an additional pebble is added to a box, no matter which box is chosen.

I state that for a permutation $(m_1, m_2, \dots, m_n)$, $S=\left\lceil{\frac{m_1}{2}}\right\rceil+\dots+\left\lceil{\frac{m_n}{2}}\right\rceil$

After a lot of trial and error I have worked out that it is not solvable if and only if $S\le n-1$ however I can't prove it. This obviously leads to the solution of the question, if it were proved. Could you please explain to me how to solve the question, using my method or any other method you find best suited?

答: You can prove that the quantity $\lceil m_1/2\rceil + \dots + \lceil m_n/2\rceil$ does not increase as you make valid moves (prove this). Furthermore, once all the boxes are nonempty, the quantity is at least $n$. Therefore, is a configuration is solvable, the value of the quantity must initially be at least $n$. Conversely, if the quantity is at least $n$, then a simple greedy algorithm works to fill all the boxes: simply always take balls from a box with at least $3$ balls, and put the result into an empty box.


问:
Product forcing of symmetric systems
Given a family of forcing notions $(P_i)_{i\in I}$ we can take the product $P:=\prod_{i\in I}P_i$ as a forcing notion to create a generic filter of the form $G=(G_i)_{i\in I}$ such that for each $i\in I$ the projection $G_i$ corresponds to the generic filter created when forcing with $P_i$. This is called product forcing and allows us to adjoin several different types of generic objects at once. (For a more detailed discussion of the subject see Product forcing and generic objects)

Now my question is if and how product forcing can be combined with symmetric forcing. Assume we have a family of forcing notions as above and a family of groups $(\mathcal{G}_i)_{i\in I}$ as well as $(\mathcal{F}_i)_{i\in I}$ such that $\mathcal{G}_i$ is a subgroup of $Aut(P_i)$ and $\mathcal{F}_i$ is a normal filter on $\mathcal{G}_i$ for all $i\in I$. Can we just define $P$ as above with $\mathcal{G}:=\prod_{i\in I}\mathcal{G}_i$ acting on $P$ componentwise and $\mathcal{F}\simeq\prod_{i\in I}\mathcal{F}_i$ as a normal filter on $\mathcal{G}$ ?

E.g. consider Cohen's original symmetric model of $ZF+\neg AC$ where he adjoins countably many generic reals and then proceeds to construct an infinite subset $A\subset \mathbb{R}$ without any countably  infinite subsets. Then the construction described above should allow us to adjoin $I$ many such sets $(A_i)_{i\in I}$ at once.

Are there any complications one might encounter of with this type construction (i.e. symmetric product forcing)? Is there any literature on the subject?

答:
Yes, there's a lot of this in the literature. Although very little in the ways of "abstract framework". This is something that was done essentially from the very early days of forcing, and you can find evidence of that in early papers.

In my works

Karagila, Asaf, Iterating symmetric extensions, J. Symb. Log. 84, No. 1, 123-159 (2019). ZBL1448.03038.

Karagila, Asaf, The Morris model, Proc. Am. Math. Soc. 148, No. 3, 1311-1323 (2020). ZBL07159661.

You can find a more general treatment. Products are a particular case of an iteration, and the first paper deals with the case where the support is finite. In the case of a product, however, we can dispense with some of the difficulties in generalising iterations to arbitrary supports, and some of the work is done in the second paper.

In addition to that you can see products defined "by hand" in many places, it's easy to see that the definitions hold for any kind of symmetric systems (but the products are normally used with Cohen-style forcings). Here are some recent examples, mainly from my work that revolved this topic quite often, and older examples.

Hayut, Yair; Karagila, Asaf, Spectra of uniformity.,
Commentat. Math. Univ. Carol. 60, No. 2, 287-300 (2019). ZBL07144894.

Karagila, Asaf, Embedding orders into the cardinals with (\mathsf {DC}_{\kappa} ), Fundam. Math. 226, No. 2, 143-156 (2014). ZBL1341.03068.

Karagila, A., Fodor’s lemma can fail everywhere, Acta Math. Hung. 154, No. 1, 231-242 (2018). ZBL1413.03012.

Monro, G. P., Independence results concerning Dedekind-finite sets, J. Aust. Math. Soc., Ser. A 19, 35-46 (1975). ZBL0298.02066.

Roguski, Stanisław, A proper class of pairwise incomparable cardinals, Colloq. Math. 58, No. 2, 163-166 (1990). ZBL0706.03038.

Between all of these you'll see finite supports, countable (or $\kappa$-)supports, Easton supports, and you'll see that leaping towards anything else (which is now just other-kind-of-mixed support is really just the same).

In fact, we even have more power now since we can talk about changing the support in the product of the filters and the groups. You'd think that this means that we can say a whole lot more, but in fact, it's usually irrelevant.

In my paper about iterations I described a concept called "tenacity". Towards the end of my Ph.D. in one of the many discussions I had with Yair Hayut we decided to try and figure what really lies beneath that concept. And it turned out that every symmetric system is equivalent to a tenacious one. And that means that playing with different supports (i.e. finite support on the filters while using Easton on the forcing) is usually just equivalent to whatever smallest support you're using. Not necessarily always, but usually.

As for the Cohen model, that's a bit tricky. Each generic is a real, and we not only care about those, we also care about the set of all generics. So this is actually not a product, but rather an iteration of adding each real, violating choice by not adding the set of all reals, and then forcing to add the set of generics without its well-ordering. All this makes the approach of just thinking about it as a single extension a lot simpler.

Title: Product forcing of symmetric systems


答:
As in my comment above, you can solve to get $s = \frac{x+y}{2}$ and $t = \frac{x-y}{2}$. Now use the chain rule to evaluate $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$. You can see from the equations above that $\frac{\partial s}{\partial x} = \frac{1}{2}$, $\frac{\partial s}{\partial y} = \frac{1}{2}$, $\frac{\partial t}{\partial x} = \frac{1}{2}$, and $\frac{\partial t}{\partial y} = -\frac{1}{2}$.

$$ \frac{\partial z}{\partial x} = \frac{\partial z}{\partial s}\frac{\partial s}{\partial x} + \frac{\partial z}{\partial t}\frac{\partial t}{\partial x} = \frac{1}{2} \left( \frac{\partial z}{\partial s} + \frac{\partial z}{\partial t} \right) $$

$$ \frac{\partial z}{\partial y} = \frac{\partial z}{\partial s}\frac{\partial s}{\partial y} + \frac{\partial z}{\partial t}\frac{\partial t}{\partial y} = \frac{1}{2} \left( \frac{\partial z}{\partial s} - \frac{\partial z}{\partial t} \right) $$

Now plug these expressions into the left-hand side of your equation. From here on out, I'm going to use the notations $z_s$ and $z_t$ instead of $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}$, to make typing easier.

$$ \begin {align*} z_x^2 - z_y^2 &= \frac{1}{4} (z_s+z_t)^2 - \frac{1}{4}(z_s-z_t)^2 \\ &= \frac{1}{4} (z_s^2 + 2z_sz_t + z_t^2) - \frac{1}{4} (z_s^2 - 2z_sz_t + z_t^2) \\ &= \frac{1}{4} \left( z_s^2 + 2z_sz_t + z_t^2 - z_s^2 + 2z_sz_t - z_t^2 \right) \\ &= \frac{1}{4} \left( 4z_sz_t \right) \\ &= z_s z_t \end{align*} $$

Title: Simple partial derivatives chain rule proof

Body:
Let $z=f(x,y)$, suppose $x=s+t$ and $y=s-t$ show that

$(\frac{\partial z}{\partial x})^2 - (\frac{\partial z}{\partial y})^2=\frac{\partial z}{\partial s}\frac{\partial z}{\partial t}$

I have rearranged to get them both in terms of s but this hasn't yielded any success. Any help with this would be greatly appreciated.


问:
Continuity, differentiability and double-differentiability of $\sum_{k=1}^{\infty} \frac{\sin(x/k)}{k}$ and $\sum_{k=1}^{\infty}\frac{\sin(kx)}{k^3}$
I have solved the following exercise except for the last point and I would like to know if I have made any mistakes and I would appreciate any hint about how to complete the last question in part (b) about $g$ being twice-differentiable; thanks.

Consider $f(x)=\sum_{k=1}^{\infty} \frac{\sin(x/k)}{k}$ and $g(x)=\sum_{k=1}^{\infty}\frac{\sin(kx)}{k^3}$.

(a) Where is $f$ defined? Continuous? Differentiable? Twice-differentiable?

(b) Show that $g(x)$ is differentiable and that $g'(x)$ is continuous

(c) Can we determine if $g(x)$ is twice-differentiable?

My solution:

(a) f defined Let $x\in\mathbb{R}$: then $|\frac{1}{k}\sin(\frac{x}{k})|\leq\frac{1}{k}|\sin(\frac{x}{k})|\leq\frac{1}{k}\frac{|x|}{k}=\frac{|x|}{k^2}$ (where we have used the fact that $|\sin(x)|\leq |x|$ for all $x\in\mathbb{R}$) and since $\sum_{k=1}^{\infty}\frac{|x|}{k^2}=|x|\sum_{k=1}^{\infty}\frac{1}{k^2}<\infty$ by Comparison Test it must be $\sum_{k=1}^{\infty}|\frac{1}{k}\sin(\frac{x}{k})|<\infty$ which implies $\sum_{k=1}^{\infty}\frac{1}{k}\sin(\frac{x}{k})<\infty$ by Absolute Convergence Test. $f$ is thus well-defined for all $x\in\mathbb{R}$.

f continuous and differentiable $|f'_k(x)|=|\frac{1}{k^2}\cos(\frac{x}{k})|\leq\frac{1}{k^2}$ and $\sum_{k=1}^{\infty}\frac{1}{k^2}<\infty$ so $\sum_{k=1}^{\infty} f'_k(x)$ converges uniformly on $\mathbb{R}$ by Weierstrass M-test and since $\sum_{n=1}^{\infty}f_n(0)=0$ by Term-by-Term Differentiability Theorem we have that $f$ si differentiable (and hence continuous) on $\mathbb{R}$ and $f'(x)=\sum_{k=1}^{\infty}f'_k(x)=\sum_{k=1}^{\infty}\frac{1}{k^2}\cos(\frac{x}{k})$.

f twice-differentiable $|f''_k(x)|=|-\sin(\frac{x}{k})\frac{1}{k^3}|\leq\frac{1}{k^3}$ and $\sum_{k=1}^{\infty}\frac{1}{k^3}<\infty$ so $\sum_{k=1}^{\infty}f''_k(x)$ converges uniformly on $\mathbb{R}$ by Weierstrass M-Test and since $\sum_{k=1}^{\infty} f'_k(x)=\sum_{k=1}^{\infty}\frac{1}{k}\cos(\frac{0}{k})=\sum_{k=1}^{\infty}\frac{1}{k^2}<\infty$ by Term-by-Term Differentiability Theorem we have that $f'(x)$ is differentiable on $\mathbb{R}$  and $f''(x)=\sum_{k=1}^{\infty}f''_k(x)=\sum_{k=1}^{\infty}-\frac{1}{k^3}\sin(\frac{x}{k})$.

(b) g differentiable with g' continuous $|g'_k(x)|=|\frac{\cos(kx)}{k^2}|\leq\frac{1}{k^2}$ and $\sum_{k=1}^{\infty}\frac{1}{k^2}<\infty$ so $\sum_{k=1}^{\infty}g'_k(x)$ converges uniformly on $\mathbb{R}$ by Weierstrass M-Test and since $\sum_{k=1}^{\infty}\frac{\sin(k\cdot 0)}{k^3}=0$ by Term-by-Term Differentiability Theorem $\sum_{k=1}^{\infty}g_k(x)$ converges uniformly to a differentiable function $g(x)=\sum_{k=1}^{\infty}g_k(x)=\sum_{k=1}^{\infty}\frac{\sin(kx)}{k^3}$ on $\mathbb{R}$ and $g'(x)=\sum_{k=1}^{\infty}g'_k(x)=\frac{\cos(kx)}{k^2}$ and since each $g'_k$ is continuous by Term-by-Term Continuity Theorem $g'$ is continuous too.

g twice-differentiable? Here I have tried the estimates $|g''_k(x)|=|-\frac{\sin(kx)}{k}|\leq\frac{1}{k}$ which tells me nothing and neither does the estimate  $|g''_k(x)|=|-\frac{\sin(kx)}{k}|\leq\frac{k|x|}{k}=|x|$ since in both of these cases I can't use Weierstrass M-Test so I don't see how I could prove that $g$ is twice differentiable.

答:
You need to use a partial summation/Dirichlet test to show that $h(x)=\sum_k \frac{\sin(kx)}{k}$ converges (locally uniformly, to a continuous function) away from $2\pi \Bbb{Z}$ (ie. using that $\sum_{k< K} \sin(kx)= \Im(\frac{e^{i Kx}-1}{e^{ix}-1})$ is bounded away from $2\pi \Bbb{Z}$)

Integrating $h$ two times (integrating termwise, allowed since the convergence is locally uniform) gives that $g$ is $C^2(\Bbb{R-2\pi Z})$.


问:
Only 3 $n$ where $q=\left\lfloor (2 p_n+p_{n+2}) (p_{n}+p_{n+1}+p_{n+2})\over p_{n}\right\rfloor,\;\text{isPrime}(q)$?
Consider: $$q=\left\lfloor (2 p_n+p_{n+2}) (p_{n}+p_{n+1}+p_{n+2})\over p_{n}\right\rfloor,\;\text{isPrime}(q)$$

where $p_n$ denotes the $n$th prime.

Other than: $$n=6\quad\text{or}\quad n=8\quad\text{or}\quad n=9$$
$$p_6=17\quad\text{or}\quad p_8=23\quad\text{or}\quad p_9=29$$

Are there any $n$ such that $\text{isPrime(q)}=\text{true}$?

I have tested all $p_n < 10^7$, so if someone could explain why this is the case(if it is the case), or present an unconditional argument; It would be greatly appreciated.

Cheers!

Edit:

Observation:

Most $q,\;q>(p_9p_{10})$ is of form such that $q=3k$

答:
Let $p_{n+1}=p_n+G_n$ and $p_{n+2}=p_n+H_n$ then
$$
\begin{align}
q &= \left\lfloor \frac{1}{p_n}(3p_n+H_n)(3p_n+G_n+H_n)\right\rfloor \\
& = 9p_n+3G_n+6H_n+\left\lfloor\frac{H_n(G_n+H_n)}{p_n}\right\rfloor \\
& = 3p_{n+1}+6p_{n+2}+\left\lfloor\frac{H_n(G_n+H_n)}{p_n}\right\rfloor
\end{align}
$$

If $H_n<\sqrt{p_n}$ then the last term is either zero or 1 and $q$ is divisible by 3 or 2 respectively. This seems to be the case for $p_n>1327$, and would follow as @IvanLoh said from Cramér's conjecture for $n$ large enough.

I'll hypothesize that to establish that no other $q$ is prime would require showing that there cannot be gaps large enough that $2p_n<H_n(G_n+H_n)$, since there are plenty of $n$ for which $3p_{n+1}+6p_{n+2}+2$ is prime.  This condition, though weaker than Cramér's conjecture, is still tighter than the best currently available bounds on prime gaps.


问:
Can the "evaluate" linear function be represented?
Let $CB(\mathbb{R})$ be the set of Bounded Continuous functions from the real line onto itself. Think of it as a linear subspace of $L^\infty(\mathbb{R})$. Define $A\colon CB(\mathbb{R}) \to \mathbb{R}$ by $A(f) = f(0)$.  This is a continuous linear function and $\|A\|=1$. By the Hahn-Banach Theorem, $A$ can be extended to $A'\colon L^\infty(\mathbb{R})\to \mathbb{R} $ preserving the norm, i.e. $\|A'\|=1$. The question is: Can $A'$ be represented by some $g \in L^1(\mathbb{R})$ ? That is: Can we find  $g \in L^1(\mathbb{R})$ such that for all $f \in L^\infty(\mathbb{R})$ the following equality holds:
$$
A'(f) = \int_\mathbb{R} f\cdot g \, d\lambda
$$

All i know so far is that if that $g$ exists then it must satisfy $\|g\|_1\geq 1 $ beacuse $T(f) \leq \|g\|_1\cdot\|f\|_\infty$.
Any help would be appreciated.

答:
There is no $g \in L^1(\mathbb{R})$ such that for all $f \in L^\infty(\mathbb{R})$ the following equality holds:
$$
A'(f) = \int_\mathbb{R} f\cdot g \, d\lambda
$$

Proof:

Suppose such $g$ exists.

For $n \in \Bbb N$, let $f_n$ be a continuous bounded function such that:

for all $x \in \Bbb R $, $0 \leqslant f_n(x)\leqslant 1$;

$ \textrm{supp} f_n \subseteq [-\frac{1}{n+1}, \frac{1}{n+1}] $;

$f_n(0)=1$.

Then, for all $n \in \Bbb N$, we have
$$1=|f_n(0)| = |A(f_n)| = |A'(f_n)|= \left | \int_{\Bbb R} f \cdot g \,d\lambda \right | \leqslant  \int_{\Bbb R}| f |\cdot | g |\,d\lambda  \leqslant  \int_{\Bbb R} \chi_{[-\frac{1}{n+1}, \frac{1}{n+1}]}\cdot | g |\,d\lambda $$

So, for all $n \in \Bbb N$, we have

$$1  \leqslant  \int_{\Bbb R} \chi_{[-\frac{1}{n+1}, \frac{1}{n+1}]}\cdot | g |\,d\lambda \tag{1}$$

Now, note that $\chi_{[-\frac{1}{n+1}, \frac{1}{n+1}]}\cdot | g |$ converge pointwise to $0$ a.e.. Since  $g \in L^1(\mathbb{R})$ and, for all $n \in \Bbb N$, $\chi_{[-\frac{1}{n+1}, \frac{1}{n+1}]}\cdot | g | \leqslant |g|$, we have, by Lebesgue Dominated Convergence Theorem, that:
$$ \lim_n \int_{\Bbb R} \chi_{[-\frac{1}{n+1}, \frac{1}{n+1}]}\cdot | g |\,d\lambda = \int_{\Bbb R} \lim_n  \chi_{[-\frac{1}{n+1}, \frac{1}{n+1}]}\cdot | g |\,d\lambda =  \int_{\Bbb R} 0 \, d\lambda =0$$
Contradiction to $(1)$.


问:
How to numerically plot the function $f(x)=\cos(x)\int^x_0\frac{\sin t}{\cos ^2(t) \sqrt{2+\sin^{2}\left(t\right)}}dt $
I'd like to numerically plot the function

$$f(x)=\cos(x)\int^x_0\frac{\sin t}{\cos ^2(t) \sqrt{2+\sin^{2}\left(t\right)}}dt \tag{1}$$

I can plot this analytically by evaluating the integral in $(1)$;

$$f(x)=\frac{\sqrt{2+\sin^{2}\left(x\right)}}{3}-\frac{\sqrt{2}\cos\left(x\right)}{3} \tag{2}$$

which exists on the range $(0,\pi)$. The problem is when I want to numerically plot this function from equation $(1)$ without explicitly integrating the integral in $(1)$.

I ask this because I would like to plot similar functions for powers other than $2$ and $1/2$, for example I would like to plot

$$f(x)=\cos(x)\int^x_0\frac{\sin t}{\cos ^2(t) \sqrt[m]{2+\sin^{m}\left(t\right)}}dt \tag{3}$$

but for larger powers the integral doesn't exist in closed form and so a nice formula like eq $(2)$ cannot be used and the plotting needs to be done numerically.

When I try and plot this numerically, for example by plotting $(1)$ with Desmos, the function is only defined up to $x=\frac{\pi}{2}$.

See the image below where I compare the numerical plot of $(1)$ in blue with the analytic plot of $(2)$ in red.

The integral in $(1)$ diverges as $x\to\frac{\pi}{2}$ however because $\cos x \to 0$ in the limit $f(\frac{\pi}{2})$ is finite,  I think because of the diverging integral my plotter registers the value of $(1)$ as undefined and stops plotting once the plotter reaches this point. Is there a way to plot this function in its entirety numerically?

Any suggestions would be welcome.

答:
So i found a way to plot it by using integration by parts to pull the divergent part of the integrand out of the integral. By taking

$$u=\frac{\sin(x)}{\sqrt{2+\sin^2\left(x\right)}} \qquad \frac{dv}{dx}=\sec^2(x) \implies \frac{du}{dx}=\frac{2\cos(x)}{(2+\sin^2x)^{\frac{3}{2}}} \qquad v=\tan(x)$$

We can perform integration by parts to find

$$f(x):=\cos(x)\int^x_0\frac{\sin x}{\cos^{2}\left(x\right)\sqrt{2+\sin^2x}}dx=\frac{\sin^2x}{\sqrt{2-\sin^2x}}-\cos(x)\int^x_0\frac{2\sin x}{(2+\sin^2x)^{\frac{3}{2}}}dx$$

The divergent part of the integral has now been removed and cancelled by the $\cos (x)$ term out front of the initial integral. Since the remaining integral is finite over $(0,\pi)$ my plotter can now plot it fine.


问:
Differential equations of third order.
I'm asked to solve this equation $y''-xy'''+(y''')^3=0$ by the reduction of order method. I tried by the change of varaible $p=y''$ and get $p=xp'- (p')^3$, then if I differentiate $p'=p'+xp''-3(p')^2p' \iff xp''+3(p')^2p'=0$, but then I have no idea of how to solve.

Please help me, I think what I've done is incorrect.

答:
As was already mentioned, you missed noting one derivation. However, just before that you already found the form
$$
p=p'x-(p')^3
$$
which is a Clairaut differential equation with a linear solution family
$$
y''=p=Cx-C^3\implies y=A+Bx+\frac{C}6x^3-\frac{C^3}{2}x^2
$$
and a singular solution
$$
x=3p'^2\implies y''=p=\frac23p'x=\pm\frac2{3^{3/2}}x^{3/2}
$$
which also can be now easily further integrated.

Title: Differential equations of third order.


问:
Find the equation of a plane containing line $L$ and equidistant from two points
I want to find the equation for a plane that is equidistant from the points $P_0 = (1, 2, 3)$ and $P_1=(3, 2, 1)$ and contains the line $\ell=(1, 1, 1) + t(0, 1, -1).$

I know that the midpoint of the two points $P_0$ and $P_1$ must be on the plane, which is $MP=(2, 2, 2).$ However, I notice this point is on the line $\ell$. I need another vector in the plane to preform the cross product and find a normal to the plane, however I can't think of a way to find such vector. I know that the length of the orthogonal complements of the two points on the normal must be the same for all points on the plane, but setting up such an equation doesn't seem like the right way to go about it. Any help would be appreciated.

答:
The plane that is equidistant from the points $P_0=(1,2,3);\;P_1=(3,2,1)$ is the symmetry plane of the segment $P_0P_1$. Its equation can be found calling $P=(x,y,z)$ a generic point of the space and solving
$$PP_0^2=PP_1^2$$
$$(x-1)^2+(z-3)^2=(x-3)^2+(z-1)^2$$
That simplifies in $$x-z=0$$
which does not contain the line $\ell=(1,1+t,1-t)$ because in the parametric equation of $\ell$ we have $x\ne z$.


问:
When are the roots of a Lie algebra the differences of the weights?
In books about group theory written for physicists, there's a strange procedure used to find the roots of a Lie algebra. The steps are:

Write down the fundamental representation. This is the 'physicist's' fundamental representation, i.e. the one that just takes the matrix representatives for a matrix Lie algebra.

Extract the Cartan generators in this representation and hence the weights of the fundamental.

Take all possible differences of pairs of these weights to find candidate roots.

Sometimes this gives too many roots. In this case, we throw out roots we know can't work, such as the ones that are integer multiples of other roots. If there are still too many, which happens rarely, we toss out more roots until we get the right number, while keeping the root diagram symmetrical.

This procedure is done in Georgi's Lie Algebras in Particle Physics and Zee's Group Theory in a Nutshell for Physicists. In both cases, the physicist can arrive at the roots for every semi-simple Lie algebra with almost no effort.

But why should this procedure work at all? Why must the roots be a subset of the differences of weights of the fundamental, and what's the rigorous way to 'throw out' unneeded roots? Is there a reason that this works when it does?

As requested in the comments, here's an example with $\mathfrak{so}(5)$. The fundamental representation contains Hermitian $5 \times 5$ matrices. Thus a Cartan sublagebra is spanned by
$$H^1 = \begin{pmatrix} & -i & & & \\ i & & & & \\ & & & & \\ & & & & \\ & & & & \end{pmatrix}, \quad H^2 = \begin{pmatrix} & & & & \\ & & & & \\ & & & -i & \\ & & i & & \\ & & & & \end{pmatrix}$$
which may be diagonalized to
$$H^1 = \text{diag}(-1, 1, 0, 0, 0), \quad H^2 = \text{diag}(0, 0, -1, 1, 0).$$
We thus read off the five weights
$$w^1 = (-1, 0), \quad w^2 = (1, 0), \quad w^3 = (0, -1), \quad w^4 = (0, 1), \quad w^5 = (0, 0).$$
There are ten distinct nonzero differences of weights. We throw out the candidate roots $(2, 0)$ and $(0, 2)$ since they are twice $(1, 0)$ and $(0, 1)$, giving the eight roots of $\mathfrak{so}(5)$.

答:
I think the reasoning behind this procedure is as follows:
We know that the differences between weights in any representation are linear combinations of roots (though not all these combinations need be roots). In fact, we can prove that the set of these differences contain all the roots of the simple lie algebra. Because if any root were not included in this set, then this would imply that the creation or annihilation operator corresponding to that root acts as a null matrix on the representation space, and the lie algebra would not be simple. Therefore, for simple lie algebras, all the roots must be part of this set of differences.</p>


问:
How do I find the series' formula knowing its generating function: $f(x) = \frac{x}{6x^2-5x+1}$
$f(x) = \frac{x}{6x^2-5x+1}$ is the generating function of a series. How do I go about finding the formula of this series?

I started by finding the partial fractions: $$f(x) = \frac{1}{2x-1} + \frac{-1}{3x-1}$$ How do I proceed from here?

Body:
hint

As recomended, by Leon,

If $ |X|<1 $, then

$$\frac {1}{1-X}=1+X+X^2+...=\sum_{k=0}^{+\infty}X^k$$

So, if $ |3x|<1 $,

$$f(x)=\frac{1}{1-3x}-\frac{1}{1-2x}=$$

$$\sum_{k=0}^{+\infty}((3x)^k-(2x)^k)$$


问:
A simple issue with contour integration
This is probably something quite simple, but I can't get my head around this.
Consider the following holomophic funtion : $$f(z)=1-|z|^2$$
This function is zero all along the unit circle C.

So I would say that  for every complex number $a$ within the open unit disc, $$\int_{C}\frac{f(z)}{z-a}dz=0$$

But according to Cauchy's Integral formula, that would mean that $f$ is $0$ everywhere inside the open unit disc, which is obviously false.

Am I missing something obvious here ?

答:
Your function is not holomorphic. Try to think what it series might be.

More formally, zeros of a holomorphic function are isolated, which is not the case with your function.

Title: A simple issue with contour integration


问:
Deterministic equivalence test of multilinear polynomials with coefficients 0 or 1
I know that there's no known poly-time equivalence testing algorithm for polynomials of $n$ variables(also called identity test) when the polynomials are given as circuit. Notice that when the polynomial is given in the canonical form this problem is trivial, but when it's given as a circuit it's there's no known poly-time algorithm to solve it.

I wanted to know if the same problem in the case of multilinear polynomials ($n$ variables) with coefficients in $\{0,1\}$ can be solved in poly-time? A multilinear polynomial is one polynomial of degree $\leq n$ and in each product of the polynomial, each variable can be multiplied at most one time.

For example, $x_1 +x_2x_3 +x_1x_3$ is a multilinear polynomial with coef. in $\{0,1\}$, but $2x_1$ is just multilinear. Furthermore, $x_1^2$ is not multilinear.

EDIT: Polynomial equivalence is not known to be "difficult" (as I previously stated) but neither is known to be poly-time solvable.

答:
The decision problem of polynomial identity testing is closely related to the reconstruction of a polynomial from its evaluations.  In particular an algorithm that determines equality or not of any two polynomials must in the worst case (that the polynomials are equal) use evaluations at a set of assignments sufficient to exactly reconstruct that polynomial (by a standard adversary argument, although the work to do the reconstruction can be omitted).

The restrictions on the polynomials considered here (multilinear, monic terms) make the problem equivalent to one of "hidden hypergraphs", namely searching for a family of subsets of the given variables $x_1,\ldots,x_n$ (corresponding to those that are multiplied to form each individual term of such polynomials).

The literature contains as of 2010 a survey that the reconstruction of hidden hypergraphs from edge-counting queries with deterministic polynomial complexity is an open problem.  However edge-counting queries amount to using binary (0/1) assignments for the variables, and the computational queries considered here are stronger (assignments of arbitrary integer arguments, if I understand correctly).

We can show that if arbitrarily large integer arguments are used, a single assignment suffices to distinguish (test the identity of) any two polynomials.  While deterministic, such an algorithm cannot be called "polynomial complexity" as the evaluations will have bit-lengths $2^n$ for $n$ variables.  Hence the final comparison of those will require exponential time.

One can also show that the family of all $2^n$ assignments of binary $x_i \in \{0,1\}$ suffices to distinguish any two nonidentical polynomials here.  However that in the worst case requires exponentially many such queries.

Between those extremes there remains a lot of room for investigation.  I'm interested to summarize the required queries for smallish numbers of variables, to see if a trend can be discerned.  I will have to post more later on this.


问:
Combinatorial proof for $ \frac{1}{2} \binom{2n}{n} = \binom{2n-1}{n}$
I have to find a combinatorial proof for the following identity:

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

I am really not sure how to formulate the proof because I have a scalar in front of binomial, in this case, a fraction.

答:
Divide a class of $2n$ students into two teams of $n$ members each. The oldest student is present in only one of them. Of all such team formations, $\binom{2n}{n}$ in total, he is present in only half of them and absent in other half.

All possible $n$-strong teams that include the oldest student can also be generated in $\binom{2n-1}{n}$ ways by selecting $n$ of the other $2n{-}1$ students to exclude from joining a team with him.


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

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

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

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

答:
Recall that
$$
\bigcap_{x \in J} A_j = \{x \ | \ \forall j \in J, x \in A_j \}.
$$
We want to show that $N \triangleleft G$ ie.
$$
\forall g \in G, gN = Ng
$$
and $N$ is a subgroup of $G$ (we see that it's an intersection of subgroups of $G$ so it's a subgroup of $G$).
Let $g \in G, n \in N$. So
$$
\forall a' \in G, n \in a'^{-1}Ha'.
$$
Take $a'=ag$. We get
$$
\forall a \in G, n \in g^{-1}a^{-1}Hag
$$
So
$$\forall a \in G, gng^{-1} \in a^{-1}Ha.
$$
This shows that
$$\forall g \in G,gNg^{-1} \subset N.$$
Thus
$$\forall g \in G, gN \subset Ng.$$
Using $g^{-1}$ we get
$$
\forall g \in G, g^{-1}N \subset Ng^{-1}
$$ so
$$\forall g \in G, Ng \subset gN.$$
Thus
$$
\forall g \in G, gN=Ng.
$$


问:
Convergence of Fourier series in $[-\pi,\pi]$
Suppose $f$ equals $-1$ on $[-\pi, 0]$ and $+1$ on $(0, \pi)$. Prove that $S_{N} f(x) \rightarrow f(x)$ if $x \in(-\pi, 0) \cup(0, \pi)$ and that
$$
\lim _{N \rightarrow \infty} S_{N} f(0)=0
$$
where we define $S_{N}$ :
$$
\begin{array}{c}
S_{N} f(x)=\frac{1}{\sqrt{2 \pi}} \sum_{|n| \leq N} c_{n} e^{i n x} \\
c_{n}=\frac{1}{\sqrt{2 \pi}} \int f(y) e^{-i n y} d y
\end{array}
$$
I think I was able to show the second part and I find $S_{N} f(0)=0$ for every $N$!

But for the first part I have some problems to show the pointwise convergence. I know the theorem in the link pointwise convergence But I dont know that we can use it for this problem or not.

答: Hint: If $f$ is differentiable at $x,$ then $S_N(f,x)\to f(x).$ You can derive this by using the Dirichlet kernel representation of $S_N(f,x).$


问:
Theorem 9.8 Atiyah and Macdonald

I don't understand why $\mathfrak b=\mathfrak a_{\mathfrak p}$. Since $\mathfrak b$ is a fractional ideal of $A$, we have that $\mathfrak b\subset Q(A)$ (where $Q(A)$ is the quotient field of $A$), but if we contract $\mathfrak b$ to $\mathfrak b\cap A$, we get an ideal in $A$. Its extension $\mathfrak a_{\mathfrak p}$ can't be equal to $\mathfrak b$ (if $\mathfrak b$ wasn't an ideal of $A$ to begin with), because the extension is going to be an ideal in $A$ (and not just a fractional ideal).

Where am I making a mistake?

答:
Note that $\mathfrak a_\mathfrak p=\{a/s:a\in\mathfrak a,s\notin\mathfrak p\}$. I will show both inclusions.

Why $\mathfrak b\subseteq\mathfrak a_\mathfrak p$:

Let $a/s\in\mathfrak b$ where $a\in A$ and $s\in A\setminus\mathfrak p$ be arbitrary. Then clearly $a=s\cdot a/s\in\mathfrak b\cap A=\mathfrak a$, so that $a/s\in\mathfrak a_\mathfrak p$.

Why $\mathfrak b\supseteq\mathfrak a_\mathfrak p$:

Let $a/s\in\mathfrak a_\mathfrak p$ be arbitrary. Here $a\in\mathfrak a\subseteq\mathfrak b$, so $a/s=1/s\cdot a\subseteq \mathfrak b$.

Title: Theorem 9.8 Atiyah and Macdonald


问:
Derivative equal to original function
How many different functions $f(x)$ exist such that $f'(x) = f(x)$?

The ones I know of right now are $f(x) = 0$ and $f(x) = ne^x$, for any real number $n$. What other functions satisfy this property?

答:
This is a simple Ordinary differential equation. And all solutions are of the form
$$f(x)=Ce^x$$
for some real constant $C$. Note that $f(x)=0$ as soon as you set $C=0$.

proof.
\begin{align}
f'(x)& =f(x), \forall x\\
f'(x)e^{-x}- f(x)e^{-x}&=0, \forall x\\
f'(x)e^{-x}+ f(x)(e^{-x})'&=0, \forall x\\
(f(x)e^{-x})'& =0, \forall x\\
f(x)e^{-x}&=C, \forall x\\
f(x)&=Ce^x, \forall x
\end{align}

Title: Derivative equal to original function


问:
Galois extension of $\mathbb{Q}_2$ with Galois group $(\mathbb{Z}/2\mathbb{Z})^3$
I am trying to solve this exercise:
Prove that $\mathbb{Q}_2$ has a unique Galois extension F with Galois group $(\mathbb{Z}/2\mathbb{Z})^3$. Compute it's ramification groups.

Here is what I have done so far:

Im will use the following lemma: If F has residue class degree f, then it contains $K=\mathbb{Q}_2(\zeta_{q^f-1})$ and F/K is totally ramified, while $K/\mathbb{Q}_2$ is unramified. The extension is of degree 8 and so it has residue class degree f = 1,2,4 or 8.

f=8: This can't happen since unramified extensions of $\mathbb{Q}_2$ have cyclic Galois group.

f=4: This can't happen either, since the extension has an unramified part of order 4, which is $\mathbb{Q}_2(\zeta_{15})/\mathbb{Q}_2$ and the Galois group of this extension is cyclic of order 4. (F has no element of order 4)

f=2: This works for example for $F=\mathbb{Q}_2(\zeta_3,\zeta_8)$, so this must be the only solution... Im not sure how to prove it though. By the lemma I've been using it must contain $\zeta_3$ and the other extension is an Eisenstein polynomial of order 4.

f=1: F is a totally ramified extension, so it comes from adjoining a root of an Eisensteinpolynomial. How do I get a contradiction here?

答:
You can prove that $\mathbb{Q}_2$ has exactly three quadratic extensions such that any one of them is disjoint from the compositum of the other two. More precisely, you can prove that $\mathbb{Q}_2^\times/(\mathbb{Q}_2^\times)^2\cong\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$. That will be enough because, as David said in his answer, coset representatives of $\mathbb{Q}_2^\times/(\mathbb{Q}_2^\times)^2$ are in bijection with quadratic extensions of $\mathbb{Q}_2$.

To prove the claimed isomorphism, write down everything you know about the group structure of $\mathbb{Q}_2^\times$.

As an exercise, you might want to repeat for $\mathbb{Q}_p$, $p$ odd.


问:
If $f(z)$ is analytic on the unit disk and $\left| f \right| \leq \frac{1}{1- \left| z\right|}$ then bound the derivative
Let $f(z)$ is analytic on $D = \{ z \in \mathbb{C}: \left| z\right| < 1\}$ and $\left| f \right| \leq \frac{1}{1- \left| z\right|}$ then show that $\left| f '(z) \right| \leq \frac{4}{(1 - \left|z\right|)^2}$.

I don't know what to do here, we know that the function is bounded by the series $\sum_k \left|z\right|^k$ and that it looks like we are taking the derivative at both sides of the inequality (I know that this is not correct). I tried to use Cauchy's integral formula without success.

答:
Cauchy works applying it for a point $|z|<1$ in a radius $r<1-|z|$ so getting:

$f'(z)=\frac{1}{2\pi i}\oint_{|w-z|=r}\frac{f(w)dw}{(w-z)^2}$ and then estimating to ($|dw|=rdt$):

$|f'(z)| \le \frac{1}{(1-(|z|+r))r}$ since $|w-z| \le r$ means $|w| \le |z|+r$ so $|f(w)|\le \frac{1}{1-|w|} \le \frac{1}{1-(|z|+r)}$

But now choosing $2r=1-|z|$ (to maximimize the denominator under the condition $r+|z| <1$) one gets $1-(|z|+r)=\frac{1-|z|}{2}$ hence

$|f'(z)|  \le \frac{4}{(1-|z|)^2)}$ so we are done


问:
$X\thicksim$ Cauchy$(\theta, a=1) \Longrightarrow \theta \equiv \mbox{ mode } \equiv \mbox{ median }$
Let $X$ be a random variable, $X\thicksim$ Cauchy$(\theta, a=1)$ where $\theta$ is not known. $X$'s PDF is:
$$f(x; \theta)=\frac{1}{\pi}\cdot \frac{1}{1+(x-\theta)^2}$$
I want to prove that $\theta$ is the same as the mode and the median of $X$. How can I do this?

答: By inspection, $f$ is maximal at $\theta$ (so $\theta$ is the mode) and even in $x-\theta$ (so $\theta$ is the median).


问:
Study the convergence of this sequence in $L^{2}(\mathbb{R})$
Let $x_{n}$ be a seuquence in $\mathbb{R}$, define the sequence of functions:
$$u_{n}=\chi_{B(x_{n},1)}$$
where $\chi$ denotes the characteristic function. Study the convergence of $u_{n}\in L^{2}(\mathbb{R})$ in the strong topology and weak topology in the following cases cases:

a) $x_{n}\longrightarrow0$

b)$|x_{n}|\longrightarrow+\infty$

The first case should be the easier, since we expect the limit to be $0$, if it exists. So we may compute:
$$\int_{\mathbb{R}}(\chi_{B(x_{n},1)})^{2}dx=\int_{x_{n}-1}^{x_{n}+1}dx=2$$
Thus there can't be strong convergence. Regarding weak convergence, we know
$$u_{n}\rightharpoonup0\iff\int_{\mathbb{R}}u_{n}\varphi dx\longrightarrow0\quad\forall\varphi\in L^{2}(\mathbb{R})
$$
so why not $\varphi=u_{n}$ again. Then we would have $u_{n}$ doesn't converge weakly in this case.
Does this idea work? Am I missing something? Does case $b)$ follow similarly?

答: No, in the first case where $x_n\to 0,$ you should expect strong convergence to $\chi_{B(0,1)}.$ In the second case, how about weak convergence to $0$ and no strong convergence?


问:
For which positive integer $n$, is $\frac{2^n-1}{3}$ a factor of $4m^2+1$ for some integer $m$? (How to prove that one answer group is the only one?)
Edit:
Dietrich Burde kindly gave me a hint but I, in my inexperience, am not able to understand the logic. I've tried but failed. So, right now my problem is understanding him. So, if you can explain the logic to me, I would be grateful.

For which positive integer $n$, is $\frac{2^n-1}{3}$ a factor of $4m^2+1$ for some integer $m$?
So, I've gotten that $n=2^q,\ q\in\mathbb{Z}_+$ is one solution, and I'm pretty certain that it is the only solution, but I don't know how to prove this (correct or wrong).

If you're interested on how I got to this point, well here's briefly what I've done:
$n$ must be an even integer, as $3\mid 2^n-1$ as $2^{2k}-1=4^k-1\equiv 1^k-1\equiv 0\ (mod\ 3)$

$n=2$ is a quite trivial solution.

$\frac{2^n-1}{3}$ must be a factor of $4m^2+1$, which gives us this (Diophantine) equation: $$4m^2+1=\frac{d(2^n-1)}{3}$$
For $d=1$, we get $4m^2=\frac{2^n-4}{3}$ and because $n$ must be even, we get:
$$4m^2=\frac{4^k-4}{3}$$
$$m^2=\frac{4^{k-1}-1}{3}$$
$$3m^2=(2^{k-1})^2-1^2$$
$$3m^2=(2^{k-1}-1)(2^{k-1}+1)$$
From this you can quite easily get $k=1, m=\pm 1$ as a solution so $n=4$.

But what about for other values of $d$?

Well, we have $n=2,4=2^1, 2^2$. Maybe it works for all $n=2^q$.
I proved this with induction using the fact that $2^{(2^{l+1})}=(2^2)^{2^l}$.

But is there other solutions? I don't know. I feel like there is no other ones because I haven't been able to find anything for the integers in between.

Body:
$$\frac{4m^2+1}{\frac{2^n-1}{3}}=\frac{3 (4 m^2 + 1)}{(2^n - 1)}$$

This answer does not address the theoretical solutions to this problem but, $-1500\le n \le 1500$ reveals:

$$n=\in\{\pm1,\pm2\}\implies m\in\{\mathbb{Z}-0\}$$
$$
n=\pm 4\implies m\in\{\pm1,\pm4,\pm6,\pm9,\pm11,\pm14,\pm16,\pm19,\cdots\}\\ \text{(OEIS A047209)}$$

$$n=\pm 8
\implies |m|\in\{19,36,49,66,104,121,134,151,\cdots\}$$
$$n=\pm 16
\implies |m|\in\{3606,5646,7461,9501,\cdots\}$$

For $n\in\{32,64,128,\cdots\}$ no solutions were found using $15$-digit precision.


问:
The normal order of $\nu(n)$ and its normal distribution
I understand that for $\nu(n)$, the number of divisors of $n$, let me denote $\log_2 x=\log\log x$. Then $$\sum_{n\le x}\Big(\nu(n)-\log_2x\Big)^2=O(x\log_2x)$$

The book told me about the theorem (due to Erdos and Kac), for $S(x;\alpha,\beta)=\#\left\{ n\le x : \alpha\le\dfrac{\nu(n)-\log_2n}{\sqrt{\log_2(n)}}\le\beta\right\}:=\#S$, then $$\lim_{x\rightarrow\infty}\dfrac{S(x;\alpha,\beta)}{x}=\dfrac{1}{\sqrt{2\pi}}\int_{\alpha}^\beta e^{-t^2/2} dt$$

I want to prove that $\displaystyle\sum_{n\le x}\Big(\nu(n)-\log_2n\Big)^2\ll x\log_2 x$ by using the theorem stated above.

What I am confused is that if $n\in S$ then $\displaystyle \delta(n):=\left|\dfrac{\nu(n)-\log_2 n}{\sqrt{\log_2 (n)}}\right|\le \max(|\alpha|,|\beta|)$ and if we choose $|\alpha|,|\beta|\le \sqrt x$ then the statement I want follows for $n\in S$, but what about $n\not\in S, n\le x$? I confuse about the phrase "normally distributed". What does it mean exactly in this sense?

As I guess, if we choose $\alpha,\beta$ that $\beta-\alpha$ is wide enough then $\delta(n)$ that satisfies the condition in $S$ would be wide enough, too. Therefore, the number of terms left from the distribution (the term on the left and right part of normal distribution) would be small.

How can I add the left terms that I mentioned to cover all $n\le x$?

答: It is not possible in general to show that weak convergence implies convergence of moments. One needs some extra assumption, which in your example just happens to be the bound that you want to prove.</p>


问:
$R, L$ are left and right adjoint in an equivalence of categories.
In my category theory course, we've defined an equivalence of categories to be a pair of adjoint functors $(L, R)$, i.e. $\text{Hom}_{\mathcal D}(LC, D) \cong \text{Hom}_{\mathcal C}(C, RD)$ in a natural way, such that $L$ and $R$ are fully faithfull. An equivalent definition is that $(L, R)$ is an adjunction and the unit and the counit $\eta, \varepsilon$ are natural isomorphisms.

It follows from the second definition, by reversing $\eta$ and $\varepsilon$, that $(R, L)$ is still an adjunction therefore $R, L$ are both left and right adjoint. Indeed
$$\text{Hom}_{\mathcal D}(D, LC) \overset{R}{\longrightarrow} \text{Hom}_{\mathcal C}(RD, RLC)\overset{\eta^{-1}_C \circ -}{\longrightarrow}\text{Hom}_{\mathcal C}(RD, C)$$
and we can go the other way around by using $L$ and $\varepsilon_D^{-1}$. Now I have some troubles to show the fact that $(R, L)$ is still an adjunction but only by using the first definition of equivalence of category, i.e. the fully faithfullness of $R, L$. Does anyone know how to do that ?

答:
Here's a proof:

Observe that $L$ and $R$ being full and faithful is equivalent to there being natural bijections:
$${\rm Hom}_{\mathcal{C}}(C,C') \cong {\rm Hom}_{\mathcal{D}}(LC,LC') \ \ (1), \ \ \ \ {\rm Hom}_{\mathcal{D}}(D,D') \cong {\rm Hom}_{\mathcal{C}}(RD,RD') \ \ (2).$$
Then we have the following natural bijection:
\begin{equation}
\begin{split}
{\rm Hom}_{\mathcal{D}}(D,LC) & \cong {\rm Hom}_{\mathcal{C}}(RD,RLC) \ \text{ by (1),}\\
&\cong {\rm Hom}_\mathcal{D}(LRD,LC) \ \text{ since $L \dashv R$,}\\
&\cong {\rm Hom}_\mathcal{C}(RD,C) \ \text{ by (2).}
\end{split}
\end{equation}
Thus $R \dashv L$.


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

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

Is then true that:

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

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

答:
Suppose $\alpha\notin g(\mathbb R\setminus\{\beta\})$.

Let $\epsilon>0$ be arbitrary. Then there exists a $\delta>0$ such that $0<|x-\alpha|<\delta\Rightarrow|f(x)-L|<\epsilon$.

Furthermore, there exists a $\eta>0$ such that $0<|y-\beta|<\eta\Rightarrow|g(y)-\alpha|<\delta$. Since $g(y)\ne\alpha$ by our hypothesis, we have $0<|g(y)-\alpha|<\delta$. Then, by my statement above, this implies that $|f(g(y))-L|<\epsilon$, which shows that $\lim_{y\to\beta}f(g(y))=L$.

P.S.: You can get away with assuming that for a small enough $\epsilon>0$ there are no $x\in\mathbb R\setminus\{\beta\}$ with $|x-\beta|<\epsilon$ such that $\alpha=g(x)$.

When this is not the case, there is a sequence of points $y_n\in\mathbb R\setminus\{\beta\}$ with $y_n\to\beta \ (n\to\infty)$ such that $g(y_n)=\alpha$. Then, $\lim_{n\to\infty}f(g(y_n))=f(\alpha)$, so that $\lim_{y\to\beta}f(g(y))=f(\alpha)$ if this limit exists.


问:
Is the Banach Algebra generated by a separable subset separable?
Let $A$ be a Banach algebra and $S \subset A$ a subset. Suppose $B$ is the Banach algebra generated by $S$ and that $S$ is a separable subset of $A$.

Does this imply that $B$ is separable?

My intuition says the answer is yes, but I can't quite prove it. Here's my progress. For any  $T \subset A$, I define
$$
W(T):=\{ t_1\cdots t_m : m \in \mathbb{N}, t_j \in T\} \subset A
$$
By definition of Banach algebra generated by a set we have that
$$
B_1:=\left\{ \sum_{k=1}^n \lambda_k w_k  : n \in \mathbb{N}, \lambda_k \in \mathbb{C} , w_k \in W(S) \right\}
$$
is a dense subset of $B$. Let $D$ be a countable dense subset of $S$. I claim that
$$
B_2:=\left\{ \sum_{k=1}^n \lambda_k w_k : n \in \mathbb{N}, \lambda_k \in \mathbb{Q}+i\mathbb{Q} , w_k \in W(D) \right\}
$$
is a countable dense subset of $B$. That $B_2$ is countable is clear. I can't quite prove that is dense because I am having issues showing that an any element of $W(S)$ is arbitrarily close to an element in $W(D)$.

I would appreciate any help with this. It might as well be that my intuition is wrong or that I am trying to prove it in a complicated way.

Thanks in advance.

答:
It's known that

$\mathrm{cl}(B_1)=B$.

$B$ is the smallest Banach algebra containing the set $S$.

On the one hand, $B_2 \subseteq B_1$ and therefore $\mathrm{cl}(B_2) \subseteq B$, by 1. above.

On the other hand, $\mathrm{cl}(B_2)$ is already a Banach algebra that contains $S$ (because $D$ is dense in $S$) and therefore $B \subseteq \mathrm{cl}(B_2)$, by 2. above.

Therefore, $\mathrm{cl}(B_2)=B$, so indeed $B$ is separable.


问:
Finding the volume bounded to the area of $z = x + y+ 2, z=0$ and cylinder $x^2 + y^2 = 16$
I want to find the volume bounded to the area of $z = x  + y+ 2, z=0$ and cylinder $x^2 + y^2 = 16$ with triple Integral.

I appreciate someone to give me the approach or a hint on what to do?

Thank you.

答:
As I mentioned in my comment, the plane $z = x + y + 2$ cuts the cylinder below and above $XY$ plane. So you have volume bound between the plane, cylinder and plane $z = 0$ both above $XY$ plane and below. So if you are doing by hand and not an online calculator, it is a bit tricky to get the right bounds and a workable integral without change of variable.

So the approach I suggest is to rotate the plane by $\frac{\pi}{4}$ such that it is parallel to $y-$axis. So use the below change of variable,

$u = \frac{x + y}{\sqrt2}, v = \frac{x - y}{\sqrt2} \,$ $\big($or in cylindrical coordinates, $u = r \cos (\theta - \frac{\pi}{4}), v = r \sin (\theta - \frac{\pi}{4})\big)$

So the plane $z = x + y +2 \,$ will become $z = u \sqrt2 + 2$

The equation of the cylinder will be $u^2 + v^2 = 16$ (as it is rotation, it would remain the same. If you plug in $x, y$ in terms of $u, v$ in $x^2 + y^2 = 16$, you can check).

Now you need to find the volume of cylinder $u^2 + v^2 = 16$ between planes $z = u\sqrt2 + 2$ and $z = 0$.

This is a lot more workable in both cartesian as well as cylindrical coordinates.

By the way if you use online calculator, using absolute function or min, max function gives you the volume easily. The above approach is when you have to do it by hand.


答:
This is true, and the $\varepsilon-\delta$-approach isn't that tedious, actually: For every $\varepsilon>0$ there is a $\lambda>0$ such that $\vert f(x)-L\vert<\varepsilon$ for all $x$ with $\vert x-\alpha\vert<\lambda$. Also, for each $\lambda>0$, there is a $\delta>0$ such that $\vert g(y)-\alpha\vert<\lambda$ for all $y$ with $\vert y-\beta\vert<\delta$. Put both facts together to get that for all $\varepsilon>0$ there is a $\delta>0$ such that $\vert f(g(y))-L\vert<\varepsilon$ for all $y$ with $\vert y-\beta\vert<\delta$.

Edit: To be clear, this assumes the following definition of the limit of a function. Let $X\subseteq\mathbb R$, $f:X\to \mathbb R$, $x_0\in X$. Then $\lim_{x\to x_0}f(x)=y$ if for all $\varepsilon>0$ there exists a $\delta>0$ such that for all $x\in X$ with $\vert x-x_0\vert<\delta$ we have $\vert f(x)-y\vert<\varepsilon$. In a more topological language, for every open neighborhood $V$ of $y$ in $X$ there is an open neighborhood $U$ of $x_0$ such that $f(U)\subseteq V$. This is an contrast to a slightly different definition where the condition need only hold for all $x\in X\backslash\{x_0\}$, or in a topological language, for all punctured neighborhoods of $x_0$. The two are not equivalent, and my argument only holds with the former. To work with the latter, $g$ must also be continuous.

Title: Does this property of limits hold when we compose functions?

Body:
Suppose we have two functions $f,g:\mathbb{R}\to{\mathbb{R}}$ such that:

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

Is then true that:

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

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


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

答:
It is false. Let $f(z)=0$ for all $z\ne0$, and $f(0)=1$.

Now, if you require $f$ to be analytic, then the assertion is true: it's Liouville's Theorem, as mentioned by julien.