问:
A question related to cyclotomic polynomials in -x
Consider the following problem from Hungerford Algebra :

If n is odd, then $g_{2n}(x) = g_n(-x)$ where $g_n(x) $ are cyclotomic polynomials over $\mathbb{Q}$.

So, $g_{2n} (x) = \frac {x^{2n}-1} {\prod_{d|2n, d<2n}  g_d(x) }$ and $g_n(-x) = \frac{(-1)(-x)^n -1}{x+1 ...}  $ can be written similarly . The problem I am facing is that in denominator of g_n(-x)  , the $g_d(x)$ terms will be have difference due to $(-x)^n$ in the $g_d(x)$, so I am not able to get an idea on how should I proceed.

So, Can you please help me?

答:
Some hints:

Irreducible polynomials over $\Bbb Q$ are separable. In particular, division by such a polynomial amounts to evaluating at its roots and checking they are zero.

 If $\xi$ is a root of $g_{n}(-x)$ then $(-\xi)^{n} = (-1)^n\xi^n = 1$ and so $\xi^{2n} = 1$. Hence $g_n(-x) \mid g_{2n}(x)$. But why are these two equal?

 Since $n$ is odd we have $\deg g_{2n} = \varphi(2n) = \varphi(n)$, and both are monic, so they must coincide.


问:
Conditions for a bipartite graph
Is a 8 vertices graph with degree $3$ and $12$ edges a bipartite graph? For instance, $24$ is the sum of degree and it is twice the number of edges and theorem 1 it says any graph is bipartite if the sum of the degree of all vertices is equal to twice the number of edges. The problem is that when I give the partition some of the vertices are adjacent to each other and even if they are in the same set. Like $V1 = \{a,c,f,h\}$ and $V2 = \{b,d,e,g\}$ and $a$ is adjacent to $h$ or $e$ is adjacent to $d$. Does it matter if they are adjacent in the same set?

答:
This graph can be both bipartite and unbipartite and the info you gave isn't enough to decide whether it is or it isn't. The only theorem about bipartite graphs based on their properties is that the graph $G$ is bipartite iff it doesn't have any odd cycles and clearly your graph can be of both types.

For a example of a bipartite graph of this type, draw four vertices $A_1,A_2,A_3,A_4$ in the first set and $B_1,B_2,B_3,B_4$ in the second set. Then connect $A_1$ to $B_2,B_3,B_4$ , connect $A_2$ to $B_1,B_3,B_4$ , etc. and the graph is bipartite. For the opposite, make a $C_4$ with $A_1,A_2,B_1,B_2$ and another $C_4$ with the other vertices and it will be unbipartite because it has several cycles of length 3.

ContentLicense: CC BY-SA 4.0

Title: Conditions for a bipartite graph


问:
What does area represent?
Since any two Euclidean shapes have an infinite number of points inside of them, shapes with different area have the same infinite number of points in them (and any object has the same number of points in it as are inside the plane it is inside). So area isn't a measure of the amount of points in an object, right? And if this is true, what does the area of a shape actually represent? In other words beyond the abstract idea of the amount of space in an object, what does an area of, for example, $4$, mean?</p>

答:
Yes, area isn't a measure of the amount of points of an object.  A positive measure P of an object J implies that J can get further subdivided into parts, since P consists of a positive real number and for all positive numbers, there exists a real number X such (X+X)=P.  A point T consists of something which has no part, making subdivision into parts impossible and thus T has measure of zero.

The area of a shape represents how much space a 2-dimensional object would take up in a 2-dimensional plane.

ContentLicense: CC BY-SA 3.0

Title: What does area represent?


问:
$f:\mathbb{N}\to\mathbb{Z}$ which is surjective but not injective
I'm looking for a really simple example of $f:\mathbb{N}\to\mathbb{Z}$ which is surjective but not injective. I thought about:
$$
f(x)=\begin{cases}
0 & x=0\\
\frac{x+1}{2} & x\text{ is odd}\\
-\frac{x}{2} & x\text{ is even}\wedge x\neq0
\end{cases}
$$
Is there a better and more simple  example I can use?

答:
$0 \notin \mathbb{N}$.

Let $f:\mathbb{N} \to \mathbb{Z}$ be a surjective and injective function. Define:

$$g(n) = \cases {f(1),\quad n =1\\
                 f(1),\quad n =2\\
                 f(n-1),\quad n\geq3}$$

Then $g:\mathbb{N} \to \mathbb{Z}$ is surjective but not injective.


问:
What is a slower decay rate than hyperbolic?
Hyperbolic decay

$$f_\alpha(x)=\frac{1}{\alpha x + 1}
$$

is slower than exponential decay

$$f(x) = e^{-x}$$

where $\alpha > 0$ is a scaling factor. The larger that $\alpha$ is, the steeper the descent towards $0$.

What would be a slower decay function than hyperbolic?

答: Any reciprocal of a function that grows slower than linear. E. g. a much slower decay rate than hyperbolic would be $$\frac{1}{\ln(x)}$$ or even $$\frac{1}{\ln(\ln(x))}$$


问:
Components of $X$ are connected disjoint subspaces whose union is $X$ s.t each nonempty connected subspace intersects only one of them
Hi i am reading Topology by Munkres and have one doubt in the proof of theorem 25.1. There it says

By the result just proved $A_x\subset C$

I am unable to understand the above quoted statement. The theorem states that :The components of $X$ are connected disjoint subspaces of $X$, such that each nonempty connected subspace of $X$ intersects only one of them.

I don't know how the result just proved above gives us $A_x\subset C$
? For reference i am attaching the screenshot of the theorem's proof where i have highlighted the part in which i have problem. Btw only the highlighted part i am unable to understand everything else is fine.

答:
As Munkres says in the previous paragraph, a connected subspace only intersects exactly one component. And $A_x$ surely intersects $C$, since they both contain the point $x \in C$. Hence $A_x$ does only intersect $C$, and since the equivalence classes $\mathcal C$ partition $X$, we have $A_x \subset C$.

In symbols

$$
A_x = A_x \cap X = A_x \cap \bigcup_{S \in \mathcal C}S = \bigcup_{S \in \mathcal C} A_x \cap S = A_x \cap C,
$$

as $A_x \cap S = \varnothing$ for all $S \neq C$ in $\mathcal S$. Hence $A_x= A_x \cap C$ and so $A_x \subset C$.


问:
Prove that $\overline{e^{ix}}=e^{-ix}$
I am trying to construct properly the imaginary exponential as an extension of the real exponential. For this, I need to show that  $\overline{e^{ix}}=e^{-ix}$.

I know that there is a very simple proof using an infinite sum of powers. But I am really looking for something purely algebraic using mainly if not only the properties of the real exponential extended to complex numbers.

One can easily show that $1=e^{ix}\cdot e^{-ix}$ and therefore that if $z=e^{ix}$, then $e^{-ix}=\lambda\cdot \overline z$ with $\lambda\in\mathbb R$, but how can we show that $\lambda=1$ ?

Once again, I am really looking for a proof that doesn't use any infinite sum.

答:
Write $e^{ix} = a + bi$ where $a, b \in \mathbb{R}$. Clearly $\overline{e^{ix}} = a-bi$. Notice now the following:
$$e^{-ix} = \frac{1}{e^{ix}} = \frac{1}{a+bi} = \frac{a-bi}{a^2 + b^2} = \frac{\overline{e^{ix}}}{a^2 + b^2}$$
Hence $(a^2 +b^2)e^{-ix} = \overline{e^{ix}}$. Since $a^2 + b^2 = 1$, the result follows.

Edit: the fact that $|e^{ix}| = a^2 + b^2 = 1$ can be proven without using infinite sums or Euler's formula, as is done here.

Title: Prove that $\overline{e^{ix}}=e^{-ix}$


问:
Probability Joint PDF
Every night Joe goes to the casino and takes with him an amount of money in dollars, X, that is distributed according to the pdf:

f(x) = Ax^2 for 0 < x < 10

where A is a constant that you need to determine. He returns with an amount of money Y that is uniformly distributed between 0 and 3X. Find

E[Y] w/out finding the distribution for Y (i.e. using E[E(Y|X)])

The pdf for Y

E[Y] using the distribution for Y. Compare with your result of part
a.

The prob he returns w/ at least 5 dollars

If he returns w/ 5 dollars, what is the prob he left with at least 7
dollars

If he return w/ 5 dollars, what is the expected value of the amount
of money with which he left?

I am having a hard time finding the joint pdf for this situation. I know f(x) = 3x^2/1000 for 0 < x < 10, but I keep getting 1/3X for the distribution of Y, which doesn't seem right. Any hints on how to form the joint pdf here?

答:
Given that $X$ has value $x_0, 0 < x_0 < 10$, $Y$ is (conditionally) uniformly distributed on $[0, 3x_0]$ and hence $E[Y\mid X=x_0] = 3x_0/2$. (If you can't do this last step by inspection, work out the value of $\int_0^{3x_0} y \cdot \frac{1}{3x_0} \,\mathrm dy$).
Since $E[Y\mid X=x_0]$ depends on the value of $X$, it can be  regarded as a random variable_ that is a function of the random variable $X$). We denote
this random variable by $E[Y\mid X]$ and note that it happens to be the
function $3X/2$ in this instance.  Can you now
find $E[Y] = E[E[Y\mid X]] = E[3X/2]$ without needing to find the pdf of $Y$?

For the pdf of $Y$, note that
$$f_{Y\mid X}(y\mid x)
= \begin{cases}\frac{1}{3x}, &0 < y < 3x,\\\quad\\0, &\text{otherwise,}\end{cases}$$
and so,
$$f_{X,Y}(x,y) = f_{Y\mid X}(y\mid x)f_X(x)
= \begin{cases}\frac{1}{3x}\cdot\frac{3x^2}{1000}, &0 < y < 3x, 0 < x < 10,\\\quad\\0, &\text{otherwise.}\end{cases}$$
You can find the pdf of $Y$ from this.  Remember to sketch the region where the
joint pdf is nonzero; it will help you set the limits of integration correctly.

Title: Probability Joint PDF


问:
Is it possible to solve $\frac{x(x-1)}{(x+y)(x+y-1)} =\frac12$, with $x>y$ and $100 \leq x+y\leq500$, without writing some program?
I am reading a puzzle that ends up based on the given info to require the solution of the following equation:

$$\frac{x(x-1)}{(x + y)(x + y -1)} =\frac12$$

also knowing that:

$$x \gt y \qquad\text{and}\qquad 100 \leq x + y \leq 500$$  and $x >= 0 \land y >= 0$

The solution states that it requires a computer to figure out the values of $x$ and $y$ (which are $x=85$ and $y = 35$).

I was wondering is this only possible to be solved with a computer program or is there a way to progress the equation even further?

Trying out I could reach nowhere to be honest:

$$\begin{align}
2x(x-1) &= (x + y) (x + y -1) \\[4pt]
2x^2 - 2x &= x^2 + xy - x + yx + y^2 - y \\[4pt]
x^2 - x &= y^2 +2xy -y
\end{align}$$

but I don't see how this helps at all.

Is there any way to do some progression?

答:
Let $z=x+y$, then $$2(2x-1)^2=(2z-1)^2+1$$
This is a Pellian equation, which you can look up.


答:
$$f(x) = \frac{1}{x}.$$

Reciprical square root,

$$g(x)=\frac{1}{\sqrt{x}}
$$

or $x^{-p}$ for any $0<p<1$:

$$h_p(x)=\frac{1}{x^p}.
$$

These are much slower than $\frac{1}{x}$ as $x \to \infty$, since

$\large{\frac{f(x)}{h_p(x)} = \frac{\frac{1}{x}}{\frac{1}{x^p}} = \frac{x^p}{x} =}$ $x^{p-1}\to 0 $ as $ x \to \infty\ $ because $p-1<0. $

Title: What is a slower decay rate than hyperbolic?

Body:
Hyperbolic decay

$$f_\alpha(x)=\frac{1}{\alpha x + 1}
$$

is slower than exponential decay

$$f(x) = e^{-x}$$

where $\alpha > 0$ is a scaling factor. The larger that $\alpha$ is, the steeper the descent towards $0$.

What would be a slower decay function than hyperbolic?


问:
Fourier Series of a $2\pi$-Periodic Function

Assume that $f$ is a $2\pi$−periodic and integrable function defined on $\mathbb{R}$, and suppose that $f(\theta + \pi) = f(\theta)$ for all $\theta \in \mathbb{R}$. Show that $\hat{f}(n) = 0$ for all odd $n$.

Using here that $$\hat{f}(n) = \frac{1}{2\pi} \int_{-\pi}^\pi f(x) e^{-inx}dx$$

I tried expanding this to $$\hat{f}(n) = \frac{1}{2\pi} \int_{-\pi}^\pi f(x) \big(\cos(nx) + i\sin(nx)\big)dx$$ but I don't see how the expression changes for odd $n$.

答:
Note that $$\int_{0}^\pi f(x)e^{-inx}dx=\int_{-\pi}^0 f(x+\pi)e^{-in(x+\pi)}dx=-\int_{-\pi}^0f(x)e^{-inx}dx$$
since $e^{-in(x+\pi)}=e^{-inx}e^{-in\pi}=-e^{-inx}$ as $n$ is odd. Hence $$\int_{-\pi}^\pi f(x)e^{-inx}dx=0$$


问:
Prove that $x^2 + 1$ divides $f(x) - \frac{(f(i) - f(-i))}{2i} x - \frac{(f(i) + f(-i))}{2}$ in the ring $\mathbb{Q}(i)[x]$.
Let $\mathbb{Q}(i) = \{a + bi ~|~ a, b \in \mathbb{Q}\}$. With addition and multiplication as in $\mathbb{C}$, $\mathbb{Q}(i)$ is a field.

(Note: You are not being asked to prove that $\mathbb{Q}(i)$ is a field.)

Let $f(x) \in \mathbb{Q}(i)[x]$. Prove that $x^2 + 1$ divides
\begin{align*}
    f(x) - \frac{(f(i) - f(-i))}{2i} x - \frac{(f(i) + f(-i))}{2}
\end{align*}
in the ring $\mathbb{Q}(i)[x]$.

$\textbf{My Attempt:}$

Since, $i, i^2, i^3, i^4 \in \mathbb{Q}(i)$. Also, $i^5 = i \in \mathbb{Q}(i)$.

So, for some $n \in \mathbb{N}$, $i^n \in \mathbb{Q}(i)$.

Since, we let $f(x) \in \mathbb{Q}(i)[x]$.

Let $f(x)$ has degree of $n$ where $n \in \mathbb{N} \cup \{ 0 \}$.

Then, $f(i) = a_n i^n + \cdots + a_0$ and $f(-i) = a_n (-i)^n + \cdots + a_0$. for some $a_j\in \mathbb{Q}(i)$.

Since, $\mathbb{Q}(i)$ is a field with addition and multiplication as in $\mathbb{C}$.

Then, $f(i) \in \mathbb{Q}(i)$ and $f(-i) \in \mathbb{Q}(i)$.

Also, $(f(i) - f(-i)) \in \mathbb{Q}(i)$ and $(f(i) + f(-i)) \in \mathbb{Q}(i)$.

Since, notice that $-\frac{1}{2i} = \frac{1}{2}i \in \mathbb{Q}(i)$.

Then, $-\frac{(f(i) - f(-i))}{2i} = (\frac{1}{2}i)(f(i) - f(-i)) \in \mathbb{Q}(i)$ and $-\frac{(f(i) + f(-i))}{2} = (-\frac{1}{2})(f(i) + f(-i)) \in \mathbb{Q}(i)$.

Let $\alpha = -\frac{(f(i) - f(-i))}{2i}$ and $\beta = -\frac{(f(i) + f(-i))}{2}$. Where $\alpha, \beta \in \mathbb{Q}(i)$.

Then, $f(x) - \frac{(f(i) - f(-i))}{2i}x - \frac{(f(i) + f(-i))}{2} = f(x) + \alpha x + \beta$.

Since, we let $f(x) \in \mathbb{Q}(i)[x]$ and we set $f(x)$ has a degree of $n$.

So, $f(x) = b_n x^n + \cdots + b_1 x + b_0$ where $b_j \in \mathbb{Q}(i)$.

Then, $f(x) + \alpha x + \beta = (b_n x^n + \cdots + b_1 x + b_0) + \alpha x + \beta = b_n x^n + \cdots + (b_1 + \alpha)x + (b_0 + \beta)$.

Since, $b_1, b_0, \alpha, \beta \in \mathbb{Q}(i)$. So, $(b_1 + \alpha), (b_1 + \beta) \in \mathbb{Q}(i)$.

Then, we set $g(x) = b_n x^n + \cdots + (b_1 + \alpha)x + (b_0 + \beta) = c_n x^n + \cdots + c_1 x + c_0$.

$~\hspace{10mm}$ Which $c_n = b_n$, ..., $c_1 = (b_1 + \alpha)$, $c_0 = (b_0 + \beta)$. Where each $c_j \in \mathbb{Q}(i)$.

Then, $g(x)$ has degree of $n$ as well.

Also, $g(x) = c_n x^n + \cdots + c_1 x + c_0 = f(x) - \frac{(f(i) - f(-i))}{2i}x - \frac{(f(i) + f(-i))}{2}$.

$~\hspace{10mm}$ Where $g(x)$ is in the ring $\mathbb{Q}(i)[x]$

Next, I want to prove that $x^2 + 1$ divides g(x) in the ring $\mathbb{Q}(i)[x]$:
Which I need to find a $h(x) \in \mathbb{Q}(i)$ such that $g(x) = (x^2 + 1)h(x)$.

But I don't know how to find it. And I don't think it always work since if $g(x) = x + 1$. Then, I cannot find a $h(x)$.

答:
More conceptually, it's a special case of Lagrange interpolation [= a special case of CRT], viz.

$$\begin{align} &f(x) \,\equiv\ f(a)\, +\, (x-a)\ \dfrac{f(b)-f(a)}{b-a}\ \ \pmod{(x\!-\!a)(x\!-\!b)},\,\ a\neq b\\[.2em]
\overset{\textstyle a,b=\pm i}\Longrightarrow\ &f(x) \,\equiv\, f(-i) + (x+i)\ \dfrac{f(i)-f(-i)}{2i}\!\pmod{x^2+1}
\end{align}$$

Or, directly applying the CRT = Chinese Remainder Theorem formula

$$\begin{align}
   &\!\begin{array}{l}
      f(x)\,\equiv\,  f(i)\ \ \ \ \pmod{x-i}\\
      f(x)\,\equiv\, f(-i)\ \pmod{x+i}
   \end{array}\\[.4em]
 \iff\ & f(x) \,\equiv\, \dfrac{x+i}{2i}\, f(i) + \dfrac{x-i}{-2i}\, f(-i)\pmod{(x-i)(x+i)}\\[.6em]
 &\ \ \ \ \ \ \ \ \equiv\, \dfrac{f(i)+f(-i)}2 + \dfrac{f(i)-f(-i)}{2i}\,x\pmod{\!x^2+1}
\end{align}\qquad\quad$$

See here for a conceptual derivation of the common CRT formula applied above.


问:
finding a function with predetermined $f^{(n)}$s at $0$ and $1$.
Let $a_n,b_n$ be 2 arbitrary sequences of real numbers. Is there any $C^\infty$ function $f:\mathbb{R}\to\mathbb{R}$ such that for any $n$:
$$f^{(n)}(0)=a_n,\quad f^{(n)}(1)=b_n$$</p>

答:
Essentially a duplicate of

Order of growth of derivatives at given x

For a cutoff function $g(x)$ equal to $1$ near $0$ and vanishing outside $(-1/2,1/2)$, take $\sum (\frac{a_n}{n!}x^n g(nx) + \frac{b_n}{n!}(x-1)^n g(n(x-1))$.


问:
Find a probability such that all (not trivial) events are dependent.
Question :

Let's be $\Omega$ a finite set ($|\Omega |\geqslant 2$). Find a probability on $(\Omega, \mathcal{P}(\Omega))$ such that for all $(A,B)$ element of $(P(\Omega) \backslash \{ \emptyset, \Omega\})^2$, $A$ and $B$ are dependent.

My Attempt :

If we look at uniform probability, we easily get that it verifies the property if and only if the $|\Omega|$ is prime (if not we can take $A$ and $B$ such that $|A\cap B|=1$ and $|A|=p$, $|B|=q$ where $n=pq$)

Then if it is not the case, what I thought is I could modify this probability a bit. But the problem is I must keep the probability of all events (except $\emptyset$ of course) not null. I can't see how to make it.

Can someone help ?

答:
Let $|\Omega |=n$ and let $P$ be a  prime such that $p_1+p_2+...+p_n=P$ is a sum of positive integers.

Let $p_i/P$ be the probability of the sample space points.

Voila!


问:
Is $vw$ substitutable for $v$ in the following expression?
Is $vw$ substitutable for $v$ in the expression $∃y(x < vx ⇒ (∃w(w < v)))$ ? The new expression would be:
$$∃y(x < vwx ⇒ (∃w(w < vw)))$$
My answer would be yes, because, at least from a mathematical standpoint, the meaning of the expression remains unchanged. Is this a good way of justifying it?

答:
You have to consider the defintion of substitutable for (clause for quantifiers):

term $t$  is substitutable for $x$ in $\varphi$ iff [...] $\varphi := ∀yα$ [$∃yα$]  and either: a) $x$ is not free in $\varphi$ or: b) $y$ does not occur in $t$ and $t$ is substitutable for $x$ in $α$.

Consider the formula: $∃w(w < v)$. If we substitute $w$ in place of $v$ what we get will be $∃w(w < w)$ and thus it is not true in general that " the meaning of the expression remains unchanged".

The above substitution does not satisfies the clause of substitutable for because with $vw$ as $t$ the variable $w$ of the $∃w$ quantifier occurs in $t$.


问:
How do I find the intervals in which I have to calculate the integral?
I have to calculate the $\int_{E}1\,dx\,dy$ of the following set $C=\bigl\{(x,y)\in\mathbb{R}:1\leq x^2y \leq8, x\leq y\ \leq 27x\bigr\}$. I have to know in which interval I take  $x$ and in which I take $y$.
I have so many exercises like this, that I have to understand which the boundaries are from a given set, and I am struggling a lot. Sometimes at least one of them is clearer and I set the other as a function of the first one. Please If somebody could explain to me how I can generally do it, any sort of tip.

I have an end exam coming up and I have no idea what I'll do with this sort of exercise.

答: Here is an image of the domain of integration, if it may help:


问:
How can the $\tanh(x)$ activation function be written for two variables? $\tanh(x,y)$
The tanh activation function is:

$$\tanh \left( x \right) = 2 \cdot \sigma \left( 2 x \right) - 1$$

Is there a corresponding formula for variable pairs?

$$ \tanh(x,y) =?$$

My uneducated guess is that it would somehow include the bivariate covariance matrix $\Sigma$ instead of univariate $\sigma$.

答:
A function $a\in\Bbb R^{\Bbb R}$ is used as a neural network's activation function viz. the multivariate extension $z_i=a((Wx+b)_i)=a(\sum_jW_{ij}x_j+b_i)$, where the parameters $b_i,\,W_{ij}$ change by backpropagation. So $\tanh(x,\,y)$ would mean $\tanh(W_{ix}x+W_{iy}y+b_i)$.

In particular, $\tanh q=2\sigma(2q)-1$ remains true; it's just that the scalar $q$ is now obtained by the above dot product. Regarding $a$ as returning the vector $z$, a vector of $1$s, which I'll denote $e$ ($1$ is also common but might be confusing here), the vector version is $\tanh q=2\sigma(2q)-e$.


问:
Grothendieck point of view of algebraic geometry
Given a ring $R$ and $I\subseteq R[x_1,\dots ,x_n]$ an ideal, define the functor $V_R(I):\operatorname{Alg}_R\to \operatorname{Sets}$, that sends a $R$-algebra $A$ in the subset of points $\mathbf a \in A^n$ such that $f(\mathbf a)=0\ $for all $f\in I$. Recall that in the case $R=k$ an algebraically closed field, $V_k(I)(k)$ is in bijective correspondence with the maximal ideals of $k[x_1,\dots ,x_n]$.

Now, in the course, we talked about the geometric points of a ring $A$, defining them as the equivalence classes of ring homomorphisms $A\to K$, where $K$ is a field (not fixed). The motivation that we were given is that this is a generalization of the construction above, since in this way we can obtain a correspondence between homomorphisms and prime ideals, not only maximal ideals; moreover, one can consider any ring, not necessarily a $k$-algebra.

However, I don't understand in what sense this is a generalization: it seems to me that they are just different constructions. For example, even if I take $k[x_1,\dots ,x_n]$, $k$ an algebraically closed field, in order to capture all its spectrum I need to consider not only the homomorphisms in $k$, but also in some trascendental extension $k\subset K$; so we are not generalizing, we are just adding something in my opinion, because these two constructions (i.e. homomorphisms in $k$ fixed and homomorphisms in any field) don't reduce to the same when considering the case of finitely generated $k$-algebras. Am I right? Moreover, I have not clear the sense of arriving to the prime ideals, but not to any ideal (i.e. why geometric points of a ring $A$ are not defined as the classes of homomorphisms from $A$ to any ring). Thanks for any clarify, I know that these question may sound stupid but I'm quite new to this things.

答:
I want to talk about the meaning of spectrum in the classical sense. let $A$ be a $k$ algebra as you said for every Ideal $I$ you have a closed subset of $A(k)$ defined by $V(I)$. this subsets could be reducible meaning that it may be possible to write it non-trivially as the union of two closed subsets of $A(k)$.the irreducible closed subsets of $A(K)$ are in bijection with prime ideals of $A$.for example if $A$ is an integral domain $A(K)$ itself corresponds to the ideal $(0)$.

now that we know the meaning of prime ideals of $A$. why do we like to see them as point? in the classical language of algebraic geometry (and of course in the usual Geometry) there is a lot of situations you want to talk about things like that when a general point of a variety has some property. this is only make sense for irreducible varieties because reducible varieties consist of several parts and general points on different parts might behave differently.(just consider closed subset of a plane consisting of two different pairs.)

now the idea of Grothendieck was to add a generic point to each irreducible closed subset that controls the behavior of a general point on that set. so he has a point for every irreducible closed subset so equivalently for each prime ideal.

now this kills two birds with one stone: when you want to work with arbitrary rings $A$ it does not make sense to only considering $k$ points of $A$ for a fixed field $k$. you have to consider all maps $A\to K$ for every field $K$ and by taking kernels you recover prime ideals


答:
This extends @Empy2's answer.

Starting from $2x(x-1) = (x+y)(x+y - 1) := z(z-1)$, we complete the squares:

$$2(x^2-x+\frac14)-\frac12 = z^2-z + \frac14 - \frac14$$

$$2(x-\frac12)^2 = (z-\frac12)^2 + \frac14$$

$$2(2x-1)^2 = (2z-1)^2+1$$

Equations of the form $x^2 - Dy^2 = \pm1$, where $D$ is not a square, are the simplest Pell equations (as opposed to the generalized version). Their solutions are related to the convergents of $\sqrt D$ (here, $\sqrt 2$.)

http://oeis.org/A000129 gives some details of these convergents, whose numerators and denominators give the solutions:

$$1/1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, 8119/5741\dots$$

These fractions have a pattern:

$$1^2-2\times 1^2 = -1$$

$$3^2-2\times 2^2 = +1$$

$$7^2 - 2\times 5^2 = -1$$

$$17^2-2\times 12^2 = +1$$

In particular:
$$239^2 - 2\times169^2 = -1$$

which gives the solution $x=85, z = 120$, and is the only one in range ($199 < 2z - 1 < 999$).

Title: Is it possible to solve $\frac{x(x-1)}{(x+y)(x+y-1)} =\frac12$, with $x>y$ and $100 \leq x+y\leq500$, without writing some program?

Body:
I am reading a puzzle that ends up based on the given info to require the solution of the following equation:

$$\frac{x(x-1)}{(x + y)(x + y -1)} =\frac12$$

also knowing that:

$$x \gt y \qquad\text{and}\qquad 100 \leq x + y \leq 500$$  and $x >= 0 \land y >= 0$

The solution states that it requires a computer to figure out the values of $x$ and $y$ (which are $x=85$ and $y = 35$).

I was wondering is this only possible to be solved with a computer program or is there a way to progress the equation even further?

Trying out I could reach nowhere to be honest:

$$\begin{align}
2x(x-1) &= (x + y) (x + y -1) \\[4pt]
2x^2 - 2x &= x^2 + xy - x + yx + y^2 - y \\[4pt]
x^2 - x &= y^2 +2xy -y
\end{align}$$

but I don't see how this helps at all.

Is there any way to do some progression?


问:
Let G be a group and let H and N be subgroups of G such that N is a normal subgroup of G. Then, How come H∩N is a subgroup of H?
[ Theorem: Let G be a group and let H and N be subgroups of G such that N is a normal subgroup of G

Then, H∩N is normal subgroup of H ]

The proof of this theorem starts by saying that from the properties of intersection and union of subgroups, we can see that H∩N which is a subset of H is also a subgroup of H and then it builds upon this to go on to prove the stated theorem.

I'm having trouble understanding this line. Could somebody please elucidate it ?

答:
If $H_1,H_2\le G$, then $H_1\cap H_2\le G$.

Infact $1_G\in H1\le G$ and $1_G\in H_2\le G\implies1_G\in H_1\cap H_2$.

If $x\in H_1\cap H_2$, then it will exist the inverse of $x$ because both $H_1$ and $H_2$ are subgroups of $G$, so they are closed under inverse.

If $x,y\in H_1\cap H_2\implies xy\in H_1$ and $xy\in H_2$, so $xy\in H_1\cap H_2.$

Now we want to show that if $N\trianglelefteq G$ and $H\le G\implies H\cap N\trianglelefteq H$.

Let $x$ be an element of $H\cap N$ and let $h$ be a generic element of $H$. Since $N$ is normal in all $G$, it is also normal in any subgroup of $G$. In other words $g^{-1}ng\in N$ $\forall g\in G,\forall n\in N\implies h^{-1}xh\in N$ $\forall x\in H\cap N$ and $\forall h\in H$. But the product $h^{-1}xh$ also belongs to $H$, then $h^{-1}xh\in H\cap N\implies H\cap N\trianglelefteq H$.


问:
$\mathbb Z_2$-equivariant cohomology of tori
I consider the tori $X=(\mathbb R/ 2 \pi \mathbb Z)^d$ on which the group $\mathbb Z_2$ acts by $x \mapsto -x$ for the nontrivial element of $\mathbb Z_2$. This action has $2^d$ fixed points, which are of the form $x=(x_1,...,x_d)$ with each $x_i$ equal to $0$ or $\pi$. I would like to know what are the equivariant cohomology groups $H^{\bullet}_{\mathbb Z_2}(X, \mathbb Z)$ (although I suspect $\mathbb Z_2$ coefficients might be easier and would be happy to learn about these cohomology groups as well). I know that there is a spectral sequence with $E_2^{p,q}=(\mathbb{RP}^{\infty},H^q(X,\mathbb Z))$, but it is unclear to me how to extract useful information from that.

答:
Since in the fibration $X \hookrightarrow X \times_\mathbb{Z_2}E{\mathbb{Z}_2} \to B\mathbb{Z}_2 \simeq \mathbb{RP}^{\infty}$ the fiber's cohomology groups are freely generated, your situation calls for the Leray–Hirsch theorem (in Hatcher it is p. 432). You want then the inclusion $X \hookrightarrow X \times_\mathbb{Z_2}E{\mathbb{Z}_2}$ to induce a surjection on cohomologies. For that, it is enough to show the surjectivity of $H_\mathbb{Z_2}^1(X) \to H^1(X)$ (use multiplicativity). I can't now think of a clever way to show that but to investigate 1- and 2-dimensional cells seems to be enough. (Another way which might work is to replace absolute circles with pairs of intervals and their boundaries: in the relative case you won't have the first row in the spectral sequence, so the differential $d_2^{0,1}$ would vanish, which is equivalent to surjectivity.)

UPD: The spectral sequence reference was unnecessary, deleted.


答:
If this of some help for this arithmetical problem, here is a vizualisation of the domain (right hand side quarter) defined by inequations

$$x>y \ \ \text{and} \ \ x+y>100$$

(constraint $x+y \le 500$ is not taken into account)

with a red dot for the proposed solution and (in black) the curve of the hyperbola defined by the equation. What you have to show is that it the unique point of the hyperbola with integer coordinates in the domain.


问:
Generate all possible combinations of 3 digits without repetition
It's possible to generate all possible combinations of 3 digits by counting up from 000 to 999, but this produces some combinations of digits that contain duplicates of the same digit (for example, 099). I could obtain all combinations of 3 digits without repetition by counting up from 000 to 999, and then removing all numbers with duplicate digits, but this seems inefficient. Is there a more efficient way to accomplish the same task?</p>

答:
I think one solution to your question is given by Algorithm L in Volume 4 of "The Art of Computer Programming" by Knuth, section 7.2.1.3.  Here is a link to a pre-publication fascicle:

http://www.kcats.org/csci/464/doc/knuth/fascicles/fasc3a.pdf


问:
Check whether the following are eigenvalues and eigenvectors of a matri
"Check whether the following are eigenvalues and eigenvectors of a matrix. Justify your answer. If the answer is yes, give the matrix."

https://i.stack.imgur.com/ehdmC.png

I know how to calculate the matrix with the given eigenvalues and eigenvectors but I don't know how to check if they are eigenvalues and eigenvectors of a matrix without having/knowing the matrix.

答:
Hints:

The eigenvectors of a matrix corresponding to distinct eigenvalues must be linearly independent.

If $\ v_1, v_2, \dots, v_n\ $ are linearly independent eigenvectors of an $\ n\times n\ $ matrix, $\ M\ $, corresponding to eigenvalues $\ \lambda_1,\lambda_1,\dots,\lambda_n\ $ (not necessarily distinct), and $\ V\ $ is the matrix with columns $\ v_1,$$ v_2,$$ \dots,$$ v_n\ $, then
$$
   MV=V\text{diag}\big(\lambda_1,\lambda_1,\dots,\lambda_n\big)\ ,
 $$
and so
$$
M=V\text{diag}\big(\lambda_1,\lambda_1,\dots,\lambda_n\big)V^{-1}\ .
$$


问:
Is $ \bf a^tbc = cb^ta $ where $\bf a, b, c \in R^d$?

Is $ \bf a^tbc = cb^ta $ where  $\bf a, b, c \in R^d$?

$\bf a^tbc =(a^tb)c = (b^ta)c = c(b^ta) = (cb^t)a$

$\bf (cb^t)$ becomes $d*d$ matrix.

First form is some scalar multiplied by $\bf c$ vector while last form is matrix transformation on $\bf a$ vector. Is this consistent? how to think of this intuitively?

答:
As I've explained here, expressions of the form $a^tbc$ are fine if read as $(a\cdot b)c$. It doesn't make sense on a literal reading because $a^tb$ is a $1\times1$ matrix, but we can safely ignore that. There is therefore no issue with $(a\cdot b)c=(b\cdot a)c=c(b\cdot a)$. These are all ways of writing a vector whose $i$th component is $\sum_ja_jb_jc_i$.

Meanwhile, $cb^t$, with the transpose on the right, is a $d\times d$ matrix as you noted, since $a,\,b,\,c$ are column vectors. We can therefore also safely write $(a\cdot b)c=(cb^t)a$, equating a vector parallel to $c$ to $a$'s image under the square matrix $cb^t$. (This image, of course, is not in general parallel to $a$; note terms in the above sum contain factors of $a_j$ rather than $a_i$, and $c^tb$'s $ij$ entry is $c_ib_j$.)


问:
Evaluate $\int x \arcsin(x)dx$.

Evaluate:
$$\int x \arcsin(x)dx$$

My first guess was $u$ substitution but that didn't get me very far. I think using integration by parts is the correct way. Here's my attempt:

$$u = \arcsin(x), v' = x \Longrightarrow \int x \arcsin(x)dx = \arcsin(x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2\sqrt{1-x^2}}dx$$

But I am stuck on how to continue from here.

答:
Integrate by parts twice

\begin{align}
\int x \arcsin x\> dx &=\frac12 x^2 \arcsin x -\frac12
\int \frac{x^2}{\sqrt{1-x^2}}\,dx\\
&=\frac12 x^2 \arcsin x+\frac14
\int \frac x{\sqrt{1-x^2}} \>d\left( 1-x^2\right)\\
 &=\frac12 x^2 \arcsin x +\frac14 x \sqrt{1-x^2} -
\frac14\arcsin x+ C\\
\end{align}

Title: Evaluate $\int x \arcsin(x)dx$.


问:
Affine variety of a set
I am self-studying some commutative algebra out of Gregor Kemper's text, "A Course in Commutative Algebra''. He has an interesting exercise in chapter 1 and I was hoping I can get some help in understanding it. The problem goes as follows:

Consider the ideal $$I = (x_1^4 + x_2^4 + 2x_1^2x_2^2-x_1^2 - x_2^2) \subseteq \mathbb{R}[x_1,x_2]$$

(a) Determine $X := \mathcal{V}(I) \subseteq \mathbb{R}^2$ and draw a picture.

(b) Is $I$ a prime ideal. Is $I$ a radical ideal?

(c) Does Hilbert's Nullstellensatz hold for $I$?

Here are some definitions that have been presented in the text:

Define: Let $K[x_1 , \cdots , x_n ]$ be a polynomial ring over a field $K$. For a set $S \subseteq K [x_1 , \cdots , x_n ]$ the $\textbf{affine variety} $ given by $S$ is defined as $$\mathcal{V} := \{ (\xi_1, \cdots , \xi_n) \in K^n \: | \:  f(\xi_1, \cdots , \xi_n) = 0 \text{ for all } f \in S \}$$

Define: For an ideal $I \subseteq R$ of a ring $R$, the $\textbf{Radical Ideal}$ of $I$ is defined as $$\sqrt{I} := \{ f \in R \: | \: \text{ there exists a positive integer } k \text{ with } f^k \in I \}$$
$I$ is called a  $\textbf{Radical Ideal}$ if $I = \sqrt{I}$.

Hilbert's Nullstellensatz : Let $K$ be an algebraically closed field and let $I \subset  K[x_1, \cdots ,x_n]$ be a proper ideal in a polynomial ring. Then
$$\mathcal{V}(I) \neq \emptyset$$.

This exercise is not homework or part of an exam, but rather it is one I came across in my reading and would like to try to understand. I am very new to commutative algebra and was hoping for some help.

答:
Question: #Consider the ideal
$$I=(f:=x^4_1+x^4_2+2x^2_1x^2_2−x^2_1−x^2_2) \subseteq k[x_1,x_2]$$

with $k$ the real numbers.

(a) Determine $X:=V(I) \subseteq k^2$
and draw a picture.

It seems you may write $f$ as follows:

$$F1. \text{    }f(x_1,x_2)=(x_1^2+x_2^2)(x_1^2+x_2^2-1)$$

and hence $Z:=V(I)$ is the union of two varieties. A copy of the unit circle and the variety $V(x_1^2+x_2^2)$ which is the origin $(0,0)$. Hence the ideal is not prime as you can see from the factorization $F1$.

In the ring $A:=k[x_1,x_2]/(f)$ there are two non-zero elements

$f_1:=x_1^2+x_2^2\text{ and }f_2:=x_1^2+x_2^2-1$

and $f_1f_2=0$ in $A$. Hence the ring $A$ is not an integral domain and hence the ideal $I:=(f)$ is not a prime ideal.

Im unsure of what version of the nullstellensatz you are using, but let us use the following version (you may find this version in Atiyah-Macdonalds book):

HNS: Let $K$ be a field and let $A$ be a finitely generated $K$-algebra. Let $\mathfrak{n}\subseteq A$ be a maximal ideal. It follows $A/\mathfrak{n}$ is a finite field extension of $K$.

Corollary: Let $K$ be algebraically closed and let $K[x_1,..,x_n]$ be the polynomial ring over $K$ in $n$ variables. The maximal ideals in $K[x_i]$
are on the form

$$\mathfrak{n}:=(x_1-a_1,..,x_n-a_n)$$

for $a_i \in K$.

Proof: If $\mathfrak{n}$ is a maximal ideal with quotient map $\phi: K[x_i]/\mathfrak{n}\cong K$, it follows $\phi(x_i):=a_i\in K$ and hence
it follows $\mathfrak{n}=(x_1-a_1,..,x_n-a_n)$. QED.

Example. In $B:=\mathbb{R}[x]$ the maximal ideals are on the following form:

$$\mathfrak{m}:=(x-a)$$

for $a$ a real number or $p_z(x):=(x-z)(x-\overline{z})=x^2-2ax+a^2+b^2$
for $z:=a+ib\in \mathbb{C}-\mathbb{R}$ a complex number with non-trivial imaginary part.

Why? If a polynomial $f(x)$ in $B$ has the property that the
quotient $B/(f(x))$ is a field it follows this field is the real or complex number field. If the quotient is the field of real numbers, it follows $f(x)$ has a real zero $a$, hence $f(x)=(x-a)$. If the quotient is the field of complex numbers it follows $f(x)$ has a complex zero $z$. Since $f(x)$ has real coefficients it follows the conjugate $\overline{z}$ is another zero of $f(x)$.
Hence it follows $f(x)=(x-z)(x-\overline{z}):=p_z(x)$.

The ideal $\mathfrak{n}:=(p_z(x))$ is maximal but it is not on the form $(x-b)$ for a real number $b$. Hence Hilberts Nullstellensatz (HNS) does not hold for $B$ - but this is clear since the real number field is not algebraically closed. The residue field $B/\mathfrak{m}\cong \mathbb{R}$ is the field of real numbers. The residue field
$B/\mathfrak{n}\cong \mathbb{C}$ is the field of complex numbers. An explicit isomorphism is given by the map

$$\phi: B/\mathfrak{n}\rightarrow \mathbb{C}$$

where $\phi(\overline{x})=z$ or $\overline{z}$.

Note the following: It seems you have the HNS wrong. As proved in the Corollary above it follows the HNS implies that for $k$ an algebraically closed field and $C:=k[x_1,..,x_n]$ it follows the maximal ideals $\mathfrak{m}\subseteq C$ are on the following form:

$$m:=(x_1-a_1,..,x_n-a_n)$$

for $a_1,..,a_n\in k$. It also implies that for a proper ideal $I \subseteq C$ it follows $V(I)\neq \emptyset$. Why? If $I\neq (1)$ it follows $C/I\neq 0$ hence there is a maximal ideal $\mathfrak{m}\subseteq C/I$. By the HNS it follows $\mathfrak{m}=(x_1-a_1,..,x_n-a_n)$ with $a_i \in K$. This implies
that $a:=(a_1,..,a_n)\in K^n$ has the property that $a\in V(I)$. Note moreover that you may use the Hilbert Basis Theorem to conclude that $C/I$ has a maximal ideal - you do not need the "Axiom of choice".

Example. Let $k$ be the field of real numbers, $f:=x^2+y^2+1\in k[x,y]$ and let $D:=k[x,y]/(f)$. Let $\mathfrak{m}\subseteq D$ be a maximal ideal. The HNS implies that $\kappa(\mathfrak{m}):=D/\mathfrak{m}$ is a finite extension of $k$, hence $\kappa(\mathfrak{m})$ equals the field $k$ or the field of complex numbers. If $\kappa(\mathfrak{m})=k$ is the field of real numbers and if

$$\phi: \kappa(\mathfrak{m})\cong k$$
is an isomorphism, we must have $\phi(\overline{x})=a, \phi(\overline{y})=b$ with $a,b\in k$. This implies that $a^2+b^2=-1$ a contradiction. Hence the maximal ideals $\mathfrak{m}\subseteq D$ all have residue field the field of complex numbers. Let $x:=isin(\theta), y:=icos(\theta)$ for $\theta\in k$.
it follows

$$x^2+y^2=i^2cos^2(\theta)+i^2sin^2(\theta)=i^2(1)=-1.$$

Hence for any $\theta$ it follows $(isin(\theta), icos(\theta))$ is a point on your variety with residue field the complex number field. Using elementary properties of the trigonometric functions we get

$$xy=-sin(\theta)cos(\theta)=-\frac{1}{2}sin(2\theta).$$

Define the ideal $I:=(xy+\frac{1}{2}sin(2\theta), x^2+sin^2(\theta), y^2+cos^2(\theta))\in D$. If we define the map

$$\phi: D \rightarrow \mathbb{C}$$

by $\phi(x):=isin(\theta), \phi(y):=icos(\theta)$ we get a well defined surjective map with $I \subseteq \mathfrak{m}(\theta):=ker(\phi)$. Since $I\neq (0)$, I suspect $I=\mathfrak{m}(\theta)$. Hence $I$ seems to be the maximal ideal in $D$ corresponding to the (complex) point $(isin(\theta), icos(\theta))$. You get for any $\theta \in \mathbb{R}$ a maximal ideal $\mathfrak{m}(\theta)$ with residue field $\mathbb{C}$. The polynomial $f$ is another example where the HNS fails: By the above argument it follows $V(f)=\emptyset$ is empty.

Question: "b) Is $I$ a prime ideal. Is $I$ a radical ideal?"

On reducedness of the ideal $I$: The ideal $J:=(f_1,f_2)$ contains $1$ since $f_1-f_2=1$, Hence the ideals $I_1:=(f_1)$ and $I_2:=(f_2)$ are coprime and it follows from the Chinese remainder theorem (Matsumuras book "Commutative ring Theory" Theorem 1.4) that there is an isomorphism of rings

$$A:=k[x,y]/(f)\cong A_1\oplus A_2 :=k[x,y]/I_1 \oplus k[x,y]/I_2.$$

This is a ring theoretic way of expressing that your original variety is the disjoint union of two algebraic varieties. The ideal $I_2$ seems to be irreducible hence $A_2$ is an integral domain.
The ideal $I_1$ is contained in $\mathfrak{m}^2$ where $\mathfrak{m}:=(x,y)$,
and $\mathfrak{m}^2$ is non-reduced, but it seems $I_1$ reduced and hence $A_1$ is a reduced ring. The direct sum of reduced rings is reduced, hence it seems the original ideal $I:=(f)$ is reduced.

If the "notation" I'm using is confusing - indicate what it is that you do not understand and I'll change notation. Do not leave this to other people on the site - you asked the question and I wrote this answer as a consequence. I do not use $\overline{x}$ to denote the equivalence class of $x$ when this is clear from the context.

Title: Affine variety of a set


问:
Infinite dimensional vector space, and infinite dimensional subspaces.
I'm having trouble with this question:

Let  $V$ be an infinite dimensional vector space.

Prove that there exist subspaces $U_1,U_2,\dots$ of $V$ with $U_{n+1}\neq U_n$ for all $n\in\mathbb{N}$, such that $U_1\supset U_2\supset\cdots$ and $U_n$ is infinite dimensional for all $n\in\mathbb{N}$.

Thanks for your help.

答:
Let $v_1,v_2,\dots$ be infinitely many linearly independent vectors in $V$.

Define
$U_n:={\rm span}(v_n,v_{n+1},v_{n+2},\dots)$.


问:
Proof of cuts measurable
Prove that the set $ D=\{(x,x): x \in [0,\frac{1}{2}]\} $ has all cuts measurable but not measurable against $\sigma $-product body $\mathcal{A} \otimes \mathcal{A} $.

I started with that: $ D_{x_0}= \{x_0\}$ for $x_0\le \frac{1}{2} $ and $\emptyset $ for $x_{0} > \frac{1}{2}. $

Hence, the set D has all the cuts that are measurable because $\emptyset \in \mathcal{A} $ and $\{x_{0}\} \in \mathcal{A} $. However, I have a problem with proving that $D$ is not measurable in regard of $\sigma $-product body $\mathcal{A} \otimes \mathcal{A} $. Does anyone have any idea how to prove this?

答:
Ok, The measurable space is $([0,1],\mathcal{A})$ where $\mathcal{A}$ is the $\sigma$-field generated by the countable subsets of $[0,1]$.

Consider $f: [0,1] \rightarrow [0,1] \times [0,1]$  defined by $f(x)=(x,x)$. It is easy to see that $f$ is a measurable function (it is enough to consider the pre-image of rectangles $A \times B$ where $A$ and $B$ are countable or have countable complement).

Now, if $D \in \mathcal{A} \otimes \mathcal{A} $ then we should have
$ \left [0,\frac{1}{2} \right ] = f^{-1}(D) \in \mathcal{A}$.
So $D \notin \mathcal{A} \otimes \mathcal{A} $.

Remark: Since $\mathcal{A}$ is the $\sigma$-field generated by the countable subsets of $[0,1]$, we have that $\mathcal {A} = \{ E \subset [0,1] : E \textrm{ or } E^c \textrm { is countable }\}$

In fact, it is easy to check that  $\{ E \subset [0,1] : E \textrm{ or } E^c \textrm { is countable }\}$ is a $\sigma$-field and it is clealy the smallest $\sigma$-field containing the countable subsets of $[0,1]$. So $\{ E \subset [0,1] : E \textrm{ or } E^c \textrm { is countable }\}$ is the  $\sigma$-field generated by the countable subsets of $[0,1]$.

Title: Proof of cuts measurable


问:
Is it possible to solve $\frac{x(x-1)}{(x+y)(x+y-1)} =\frac12$, with $x>y$ and $100 \leq x+y\leq500$, without writing some program?
I am reading a puzzle that ends up based on the given info to require the solution of the following equation:

$$\frac{x(x-1)}{(x + y)(x + y -1)} =\frac12$$

also knowing that:

$$x \gt y \qquad\text{and}\qquad 100 \leq x + y \leq 500$$  and $x >= 0 \land y >= 0$

The solution states that it requires a computer to figure out the values of $x$ and $y$ (which are $x=85$ and $y = 35$).

I was wondering is this only possible to be solved with a computer program or is there a way to progress the equation even further?

Trying out I could reach nowhere to be honest:

$$\begin{align}
2x(x-1) &= (x + y) (x + y -1) \\[4pt]
2x^2 - 2x &= x^2 + xy - x + yx + y^2 - y \\[4pt]
x^2 - x &= y^2 +2xy -y
\end{align}$$

but I don't see how this helps at all.

Is there any way to do some progression?

答:
I am guesing $x,y$ are intergers and you have problems programming so here is my solution.

You have $x^2-x=y^2+2xy-y$ then $x^2-y^2-2xy=x-y \rightarrow  (x-y)^2-2y^2=x-y$.

let $z=x-y$, an positive interger. You will have:

$z^2-z-2y^2=0$, so $z$ (since it is positive) must be: $z=\frac{1+\sqrt{1+8y^2}}{2}$.

Since $2y-z=x+y$ you have that $100 \leq 2y+ \frac{1+\sqrt{1+8y^2}}{2} \leq 500$.

Now find a function plotter online, or Geogebra and plot (lets reverse the roles of $x$ and $y$, so now $y$ is named $x$ and vice versa) $f(x)=2x+ \frac{1+\sqrt{1+8x^2}}{2} $

You can plot the lines $y=100$ and $y=500$ and find wich intergers get plotted between those lines. This shortens your search.

Also you can plot $y > x$ (remember we revrsed the roles) and check which intergers image falls in the intersection of $y\geq 100 \cap y \leq 500 \cap y\geq x$.

This will give you values for $x$ (or likely value). This will shorten or end your search.

Hope it helps.


问:
If $P$ is a projection matrix such that $PX = X$ does it mean $X$ has a singular distribution?
Suppose ${\bf X}$ is a $n \times 1$ vector of random variables and ${\bf P}$ is $n \times n$ a projection matrix such that ${\bf PX} = {\bf X}$. Rank(${\bf P}) < n$. So does it mean that ${\bf X}$ has a singular distribution, for example singular multivariate normal distribution?

答:
Since $P$ is not full-rank, its range is a vector subspace of dimension less than n. Every vector subspace of dimension less than n has measure Lebesgue measure 0.

You say that every point in the support of the random variable $X$ is mapped by $P$ to itself. Thus, the support of the random variable $X$ is a subset of $\operatorname{ran}(P)$, which is a vector subspace of dimension less than n. Therefore, the support of $X$ has measure 0. Therefore, $X$ is a singular distribution.


问:
Show that if $g \circ f = \tilde{g} \circ f$ and $f$ is surjective, then $g=\tilde{g}$. Is the same statement true if $f$ is not surjective
Proof of the first part of the question :
Given $g \circ f = \tilde{g} \circ f$ and $f$ is surjective.
To prove $g=\tilde{g}$.

=> $f$ is surjective, which implies for every $y∈Y, ∃x∈X$, such that $f(x)=y$

=> $g∘f =\tilde{g}∘f⟹g∘f(x) =\tilde{g}∘f(x)⟹g(f(x))=\tilde{g}(f(x))⟹g(y)=\tilde{g}(y)$

=> from definition of "Equality of Set" since $g(y)=\tilde{g}(y)$, hence $g =\tilde{g}$.

答: Your proof for the first part is good. The statement is no longer true if $f$ is not surjective. For example, put $f, ~g, ~\tilde{g}:\mathbb R \to \mathbb R$ by $f(x)=0$, $g(x)=x$ and $\tilde{g}(x)=x^2$. Then $g \neq \tilde{g}$ but $g \circ f$ and $ \tilde{g}\circ f$ are same because $(g \circ f)(x) = 0 = (\tilde{g}\circ f)(x)$ for all $x \in \mathbb R$.


答:
It is not necessarily true. Let $I = \{1,2,3\}$ and take for example $
f,g,\tilde{g} \colon I \to I$ such that
\begin{alignat*}{3}
f(1)=1 & \quad  f(2)&&=2 \quad  f(3)&&=2 \\
g(1)=2 &\quad  g(2)&&=2 \quad  g(3)&&=3 \\
\tilde{g}(1) =2 &\quad  \tilde{g}(2)&&=2 \quad  \tilde{g}(3)&&=1.
\end{alignat*}
Then $f$ is not surjective,
\begin{align*}
g(f(1)) & = g(1) = 2,\\
g(f(2)) & = g(2) = 2, \\
g(f(3)) & = g(2) = 2,
\end{align*}
and
\begin{align*}
\tilde{g}(f(1)) & = \tilde{g}(1) = 2,\\
\tilde{g}(f(2)) & = \tilde{g}(2) = 2, \\
\tilde{g}(f(3)) & = \tilde{g}(2) = 2,
\end{align*}
so $g\circ f = \tilde{g}\circ f$, but $g \neq \tilde{g}$.

Title: Show that if $g \circ f = \tilde{g} \circ f$ and $f$ is surjective, then $g=\tilde{g}$. Is the same statement true if $f$ is not surjective

Body:
Proof of the first part of the question :
Given $g \circ f = \tilde{g} \circ f$ and $f$ is surjective.
To prove $g=\tilde{g}$.

=> $f$ is surjective, which implies for every $y∈Y, ∃x∈X$, such that $f(x)=y$

=> $g∘f =\tilde{g}∘f⟹g∘f(x) =\tilde{g}∘f(x)⟹g(f(x))=\tilde{g}(f(x))⟹g(y)=\tilde{g}(y)$

=> from definition of "Equality of Set" since $g(y)=\tilde{g}(y)$, hence $g =\tilde{g}$.


问:
Bounds for intervals of log-concave distributions.
I am working on a proof regarding log-cocave Distributions right now. The following statement looks right to me and it feels like the proof should be easy but I just don't see it.
Can you tell me if it's true and provide proof if possible? Thank you!

Let $\alpha$ be a log-concave distribution, which means that $\alpha(x) = \exp(-f(x))$ with convex $f$,  $t > 0$ and $0 <\lambda<1 $. If $\alpha(x_0+t) < \lambda \alpha(x_0)$, is it true that $\alpha(x_0 + rt) < \lambda^r\alpha(x_0)$ for all real $r > 1$ ?

答:
You have $\alpha(x) = \exp(-f(x))$ with a convex function $f$. So it is given that
$$ \tag 1
 f(x_0 + t) > f(x_0) + \ln \lambda
$$
and the goal is to show that
$$ \tag 2
 f(x_0 + rt) > f(x_0) + r \ln \lambda \, .
$$

This follows directly from the convexity condition  for $x_0 < x_0 + t < x_0 + r t$:
$$
 f(x_0+t) \le \frac{r - 1}{r} f(x_0)+ \frac{1}{r} f(x_0+rt) \, .
$$

Using $(1)$ we can estimate the left-hand side:
$$
 f(x_0) + \ln \lambda \le \frac{r - 1}{r} f(x_0)+ \frac{1}{r} f(x_0+rt) \, .
$$
and that simplifies to $(2)$.


问:
Simplifying a power series
$$\frac{n}{4}\sum_{k=1}^{\infty}k2^{1 - k} = n$$

More specifically I am curious how the below derivation is obtained. As in, why is the final result n.

Can someone please explain to me how the right-hand-side result is obtained?

Thanks.

答:
You are considering the series
$$
\sum_{k = 1}^\infty kx^{k-1}
$$
which, is the (term by term) derivative of
$$
\sum_{k = 1}^\infty x^{k}
$$
since $\frac{d}{dx} x^k = kx^{k-1}$. It should be noted that this series converges only if $|x| < 1$.

Title: Simplifying a power series


问:
Is second derivative of a convex function convex?
If $f$ is twice differentiable and convex, is it true that $f''$ is a convex function ?</p>

答:
I thought I'd take a personal challenge to find a counterexample on the entire real line. Here's a method for constructing such a function: let $g_1(x)$ be any nonconvex but positive real function such that the integrals $g_2(x)=\int_{0}^x g_1(z) dz$, and $g_3(x)=\int_{0}^x g_2(z) dz$ both exist for all $x\in\mathbb{R}$. Then $g_3$ is convex, but its second derivative $g_1$ is not.

If you'd like a specific example, consider
$$f(x) = x \mathop{\textrm{erf}}(x) + \tfrac{1}{\sqrt{\pi}} e^{-x^2}$$
I chose this deliberately, because
$$f'(x) = \mathop{\textrm{erf}}(x), \quad
f''(x) = \tfrac{2}{\sqrt{\pi}} e^{-x^2}.$$

The function:

The second derivative:


问:
Compute $ \mathbb{P}\left(X_{1}+\cdots+X_{n}=0\right) $
Assume $ \left(X_{n}\right)_{n} $ are independent random variables such that $ \mathbb{P}\left(X_{n}=1\right)=\frac{1}{2},\ldots,\mathbb{P}\left(X_{n}=-1\right)=\frac{1}{2}$. Define $S_{n}=\sum_{k=1}^{n}X_{n} $.

I have to compute the probability $ \mathbb{P}\left(S_{n}=0\right) $ .

I thought that it would just be $0 $ for an odd $ n $ and  $ 1/2^n $ for even $ n $ because there's exactly $2^n$ ways to choose the random variables that would equal $ -1 $, but I'm not sure this is correct.

Thanks in advance

答:
Hint:

Let $X_i = 2Y_i-1\:\forall\:1\le i\le n$. Then $Y_i$ are iid Bern$(\frac12)$ random variables. Now write $S_n$ in terms of $Y_i's$ and note that some of $n$ iid $Bern(p)$ random variables is $Bin(n,p)$.

Full soln:

Required probability reduces to:
\begin{align}
&P\left(\sum_{i=1}^nX_i=0\right)\\
=&P\left(2\sum_{i=1}^nY_i-n=0\right)\\
=&P\left(\sum_{i=1}^nY_i=\dfrac n2\right)\\
&=\begin{cases}
0&\text{if $n$ is odd}\\
\dbinom{n}{n/2}\dfrac{1}{2^n}&\text{if $n$ is even}
\end{cases}
\end{align}


问:
Approximate the function $\int_0^{1/2}x^{2}\ln(1+x^2)dx$ by 4 terms
The Function: $\int_0^{1/2}x^{2}\ln(1+x^2)dx$

So I tried approximating it doing the following.
First, I figured out that  the series of $\ln(1+x^2)$ is just  $\sum_{n=0}^\infty (-1)^n\frac{x^{2n+2}}{n+1}$. I've gotten it by taking the derivative of $\ln(1+x^2)$ and translated that into the geometric form of a series, and integrated that.

Now, I don't quite understand what allows me to simply take the first 4 terms and multiply them by $x^2$ to get the approximation I wanted (and obviously evaluating the bounds later). What allows us to completely ignore the fact that we haven't done anything with $x^2$ that we're simply allowed to use it on our end result now as is?

Would really appreciate some explanation, thanks in advance!

答:
What the exercise asks you is to approximate the integrand function and then compute the integral of the approximation. Since an infinite series will be involved, you just take the first four terms of that series.

$$\int_0^{0.5} x^2\ln(1+x^2)\ \text{d}x = \sum_{k = 0}^{+\infty} \frac{(-1)^k}{k+1}\int_0^{0.5} x^{2n+4}\ \text{d}x = \sum_{k = 0}^{+\infty} \frac{(-1)^k}{k+1}\frac{2^{-2 k-5}}{2 k+5}$$

The first four terms of the series are thence

$$\frac{1}{160} -\frac{1}{1792} +\frac{1}{13824} -\frac{1}{90112} = \frac{3185513}{553512960} \approx 0.00575508(...)$$

Notice that

$$\int_0^{0.5} x^2\ln(1+x^2)\ \text{d}x = \frac{1}{72} \left(22+\ln \left(\frac{125}{64}\right)-48 \arctan\left(\frac{1}{2}\right)\right) =  0.0057548(...)$$

Final Remark

$x^2$ is a polynomial function, and for that reason it cannot be approximated with a power series, like Taylor series, in the same way you do for non polynomial function.


问:
Comparing coefficients within fractions

Question: Find values for $a$, $b$, $n$ such that:
$\frac{-ab(n+1)x-(an+b)y}{(a+bn)x+(n+1)y} = \frac{10x-4y}{3x-y}$

Attempt: Tried using the following relationships by comparing coefficients and solving for $a$, $b$, $n$:

$-ab(n+1)=10, an+b=4, a+bn=3, n+1=-1$

This gives no values of $a$, $b$, $n$ that satisfy all four equations.

Dividing the numerator and denominator by $-(n+1)$ instead which gives
$\frac{abx+\frac{an+b}{n+1}y}{-\frac{a+bn}{n+1}x-y}$
and then comparing coefficients gives valid $a$, $b$, $n$. Why is this? I understand that this ensures the correct coefficient for $y$ in the denominator for all values, but I don't see how that is significant and how the first method does not also lead to the correct solution.

答:
HINT

The original equations are solved if there is a number $k$ such that  $$-ab(n+1)=10k,an+b=4k, a+bn=3k, n+1=-k.$$

You can use the fourth of these equations to eliminate $k$ and then you'll get a solution.

Title: Comparing coefficients within fractions


问:
A lemma for a new proof for the existence of a rational between two arbitrary real numbers
I had in mind to prove the theorem "There exists a rational number between each two arbitrary Real numbers" and towards the end, I happened to need to prove this: $$$$
If we have the two numbers $b$ and $c$ ($b \neq c$), we can always create a number arbitrary close to say, $b$, using just $b$ and $c$. $$$$
Let us name $A_0 = \frac{b+c}{2}$.
We take the average of $A_0$ and b, call it something like $A_1$ and continue this process. It is clear that each $A_{n+1}$ is closer to b than $A_{n}$. $$$$
To formalize the concept I defined the following, inductively: $$$$
Let $f(x)$ be an arbitrary function:
We define $f_1(x)=f(x)$ and $f_n(x)=f_{n-1}(f(x))$ which is just a more detailewd writing of $f_n(x) = fofofo...of(x) $ n times $$$$
So, what we wish to prove now is this: If $b$ and $c$ are real numbers and $f(x)= \frac{b+x}{2}$, prove $$$$
$$\forall \epsilon>0 \hspace{0.5cm} \exists n (|f_n(c)-b|<\epsilon)$$
I'd be very thankful if someone provides a proof or a hint to the proof since this is of a very different form than the theorems I've seen before.

答: You have that $|f(x)-b|=\frac{|x-b|}{2}$ by the definition of $f(x)$. So by induction, $|f_{n}(x)-b|=\frac{|x-b|}{2^n}$. So in your last formula you just have to pick $n$ such that $\frac{|c-b|}{2^n}<\varepsilon$.


答:
The first time around, you had four unknowns and four equations. The second time around, you only had three equations. So the second method was likely on general principles to be better able to give you an answer, because it involved solving a less strict set of simultaneous equations.

The first method forced $n$ to be something specific; the second method eliminated it from the set of equations, and hence allowed it to vary.

Of course, you might still not have found all the solutions! You'll need to prove that you've found all the solutions (if it's true and if that's something you want to prove). The solution set you got from the first method is a valid solution set; it's just smaller than the solution set you got from the second method. But maybe there's another transformation you could do to expand even the second method's solution set. (I don't know; I haven't tried.)

Title: Comparing coefficients within fractions

Body:
Question: Find values for $a$, $b$, $n$ such that:
$\frac{-ab(n+1)x-(an+b)y}{(a+bn)x+(n+1)y} = \frac{10x-4y}{3x-y}$

Attempt: Tried using the following relationships by comparing coefficients and solving for $a$, $b$, $n$:

$-ab(n+1)=10, an+b=4, a+bn=3, n+1=-1$

This gives no values of $a$, $b$, $n$ that satisfy all four equations.

Dividing the numerator and denominator by $-(n+1)$ instead which gives
$\frac{abx+\frac{an+b}{n+1}y}{-\frac{a+bn}{n+1}x-y}$
and then comparing coefficients gives valid $a$, $b$, $n$. Why is this? I understand that this ensures the correct coefficient for $y$ in the denominator for all values, but I don't see how that is significant and how the first method does not also lead to the correct solution.


问:
Social Golfer Problem - Quintets
I wrote an article on the Social Golfer Problem, which has questions like:

Each day, 16 people play Munchkin in foursomes simultaneously. How many days can they play with no two people playing with each other twice? tournament

I compiled best-known solutions for pairs, triplets, and quadruplets at the Social Golfer Problem Demonstration. To compile this data, I had to read through a few hundred books and papers. Very often, an answer would be given in new-to-me difficult notation, or it would reference a previous paper, or it would just be "obvious".

Now that I am considered an "expert", I've been asked about quintets and sextets, and for larger number of groups. I don't recall seeing solutions for any of the below problems in the papers I perused, but it's possible I missed them.

30 play in groups of 3 each day. No two play together more than once. How many days?

33 play in groups of 3 each day. No two play together more than once. How many days?

36 play in groups of 3 each day. No two play together more than once. How many days?

40 play in groups of 4 each day. No two play together more than once. How many days?

44 play in groups of 4 each day. No two play together more than once. How many days?

48 play in groups of 4 each day. No two play together more than once. How many days?

52 play in groups of 4 each day. No two play together more than once. How many days?

25 play in groups of 5 each day. No two play together more than once. How many days?

30 play in groups of 5 each day. No two play together more than once. How many days?

35 play in groups of 5 each day. No two play together more than once. How many days?

40 play in groups of 5 each day. No two play together more than once. How many days?

45 play in groups of 5 each day. No two play together more than once. How many days?

50 play in groups of 5 each day. No two play together more than once. How many days?

55 play in groups of 5 each day. No two play together more than once. How many days?

60 play in groups of 5 each day. No two play together more than once. How many days?

36 play in groups of 6 each day. No two play together more than once. How many days?

42 play in groups of 6 each day. No two play together more than once. How many days?

48 play in groups of 6 each day. No two play together more than once. How many days?

54 play in groups of 6 each day. No two play together more than once. How many days?

60 play in groups of 6 each day. No two play together more than once. How many days?

66 play in groups of 6 each day. No two play together more than once. How many days?

72 play in groups of 6 each day. No two play together more than once. How many days?

Does anyone have any solutions to any of these?

答: A new (Dec 2020) paper on this topic uses symmetry to approach the problem, and publishes solutions for groups up to 50 participants: Breakout Group Allocation Schedules and the Social Golfer Problem with Adjacent Group Sizes</p>

Title: Social Golfer Problem - Quintets


问:
Solving definite integral from -1 to 1
Please help me with this. I can't prove the result. Tried integral by parts or notations, nothing working

$$\int_{-1}^{1}{\frac{x^2}{e^x+1}}dx$$

答:
Note $\frac{1}{e^x+1}
= \frac12+\frac12\tanh^{-1}\frac x2 $ and the odd function $\tanh^{-1}\frac x2$ does not contribute to the integral over $(-1,1)$. Thus

$$\int_{-1}^{1}{\frac{x^2}{e^x+1}}dx
= \int_{-1}^{1}\frac{x^2}2=\frac13
$$

Title: Solving definite integral from -1 to 1


答:
hint

You need three equations
$$\lim_{x\to +\infty}f(x,y)=\frac{-ab(n+1)}{a+bn}=\frac{10}{3}$$
$$\lim_{y\to+\infty}f(x,y)=\frac{-(an+b)}{n+1}=4$$

and, if $ y=3x$, the denominator should be zero:

$$a+bn+3(n+1)=0$$

for example, you will find that $ab=10$.

Title: Comparing coefficients within fractions


问:
Degree of irreducible polynomial involving Primitive root of unity
The following question was asked in quiz of Field Theory ( now over ) and I am not sure about the answer of this question.

If z is a primitive 7th root of unity what is the degree of irreducible
polynomial over Q of $ z+ z^{-1}$?

I got the answer as 49 : let $x =z+ 1/z$ , $xz=z^2 +1$ Now, taking power 7 and LHS is devoid of z but RHS is $2+ 7( z^5 + 3z^3 + 5z + 5z^6 +3z^4+z^2)$ which cannot be removed of z if power is taken less than 7 as root is primitive.

So, I got the answer 49.

Am I right? If not kindly give the right solution.

答:
Hint: use that $\mathbf{Gal}(\Bbb Q(\xi_7)/\Bbb Q) \simeq \Bbb Z_6$.

 Since the extension is cyclic, there is exactly one subextension for each divisor of $6$, and we also know that $\omega := \xi_7 + \xi_7^{-1} \not \in \Bbb Q$. So, we are left to decide whether $\Bbb Q(\xi_7+\xi_7^{-1}) / \Bbb Q$ has order $2,3$ or $6$.

Hint: consider the map $\sigma  \colon \Bbb Q(\xi_7)/\Bbb Q \to \Bbb Q(\xi_7)/\Bbb Q$ defined by $\xi_7 \mapsto \xi_7^{-1}$.

 This map fixes $\omega$ an has order $2$ (because $\xi_7 \neq \xi_7^{-1}$). Hence $\Bbb Q(\omega)/\Bbb Q$ is a subextension of the fixed field of $\langle \sigma\rangle$, which has order $3$ and thus $\Bbb Q(\omega)/ \Bbb Q$ is actually cubic.


问:
Show that $P\left(\bigcup_{k=1}^\infty E_k\right)\le \sum_{k=1}^\infty P(E_k)$
For an infinite sequence of events $E_1,E_2,E_3,\ldots$ show that $$P\left(\bigcup_{k=1}^\infty E_k\right)\le  \sum_{k=1}^\infty P(E_k)$$

This is the question given to us. Now I know we can solve this by induction but is there any other method to solve this?

答:
I suppose you are talking about $\mathbb P\left(\bigcup\limits_{k=1}^{\infty} E_k   \right) \le \sum \limits_{k=1}^{\infty} \mathbb P\left( E_k   \right)$

This is true when the $E_k$ are mutually disjoint (countable addition of probability measure) and you get equality.

Define $F_k$ by $F_1=E_1$ and $F_k = E_k \backslash \bigcup\limits_{j=1}^{k-1} E_j$ so the $F_k$ are mutually exclusive. Then

$\mathbb P\left( F_k   \right)  \le \mathbb P\left( E_k   \right)$ since $F_k \subset E_k$

$\bigcup\limits_{k=1}^{\infty} F_k = \bigcup\limits_{k=1}^{\infty} E_k $ since any element of one is an element of the other

so $\mathbb P\left(\bigcup\limits_{k=1}^{\infty} E_k   \right) =\mathbb P\left(\bigcup\limits_{k=1}^{\infty} F_k   \right) = \sum \limits_{k=1}^{\infty} \mathbb P\left( F_k   \right) \le \sum \limits_{k=1}^{\infty} \mathbb P\left( E_k   \right)$

Which of these steps implicitly use induction, I leave to you


问:
show that $f$ is log concave $\frac{1}{\sum_{i=1}^{n}\frac{1}{x_i}}$
given $f(x)=\frac{1}{\sum_{i=1}^{n}\frac{1}{x_i}}$ over $\mathbb{R}^n_{++}$ show that that $f$ is log-concave.

I've got to $-\ln(\sum_{i=1}^{n}\frac{1}{x_i})$ and tried using the definition but didn't go well so I thought using maybe the hessian criterion but got complex aswell

any help?

答:
Let's do the 2d case as a hint:

So we want to show:

$$ln ( f ( θ x + ( 1 − θ ) y )) ≥ θ ln  f ( x ) + ( 1 − θ ) ln  f ( y ) $$

for $0<\theta <1$.

The left hand side is:

$$-ln( \frac{1}{θ x_{1} + ( 1 − θ ) y_{1}} + \frac{1}{θ x_{2} + ( 1 − θ ) y_{2}})$$

The right hand side is:

$$-\theta ln(\frac{1}{x_{1}}+\frac{1}{x_{2}}) - (1-\theta) ln(\frac{1}{y_{1}}+\frac{1}{y_{2}})$$

Getting common denominators and simplifying the left hand side is:

$$ln(θ x_{1} + ( 1 − θ ) y_{1}) + ln(θ x_{2} + ( 1 − θ ) y_{2}) - ln(θ x_{1} + ( 1 − θ ) y_{1} + θ x_{2} + ( 1 − θ ) y_{2} )$$

And the right hand side is:

$$\theta ln(x_{1}) + \theta ln(x_{2}) - \theta ln (x_{1}+x_{2}) + \gamma ln(y_{1}) + \gamma ln(y_{2}) - \gamma ln (y_{1}+y_{2})$$

where $\gamma = 1 - \theta$.

Now simplifying we get:

$$(x_{1}+x_{2})^{\theta}(y_{1}+y_{2})^{\gamma}(θ x_{1} + ( 1 − θ ) y_{1})(θ x_{2} + ( 1 − θ ) y_{2}) ≥ (x_{1}x_{2})^{\theta}(y_{1}y_{2})^{1-\theta}( θ x_{1} + ( 1 − θ ) y_{1} + θ x_{2} + ( 1 − θ ) y_{2} )$$


问:
On a core of the Neumann Laplacian
Let $\Omega$ be a bounded domain of $\mathbb{R}^d$. We define the Sobolev space $H^1(\Omega)$ by
\begin{align*}
H^1(\Omega)=\{f \in L^2(\Omega,m) \mid |\nabla f| \in L^2(\Omega)\}.
\end{align*}
Here, $m$ denotes the Lebesgue measure on $\Omega$, and $\nabla f$ is the distributional gradient of $f$. The Neumann Laplacian $(L,\text{Dom}(L))$ on $\Omega$ is defined as
\begin{align*}
\text{Dom}(L)&=\left\{f \in L^2(\Omega,m) :  H^1(\Omega) \ni g \mapsto \int_{\Omega}\nabla f\cdot \nabla g\,dm \text{ is continuous on $L^2(\Omega,m)$}\right\} \\
-\int_{\Omega}g Lf\,dm&=\int_{\Omega} \nabla f\cdot \nabla g\,dm,\quad f \in \text{Dom}(L),\,g \in H^1(\Omega).
\end{align*}

Question. When $\partial \Omega$ is Lipschitz, we define
\begin{align*}
C=\left\{g \in C^2(\overline{\Omega}) :\frac{\partial g}{\partial n}(x)=0,\quad \sigma\text{-a.e.} \right\}
\end{align*}
Here, we denote by $n$ the inward unit normal vector on $\partial \Omega$, and $\sigma$ is the surface measure on $\partial D$. Can we show that
$\{(g,Lg) \mid g \in C\}$ is a dense subspace of $\{(f,Lf) \mid f \in \text{Dom}(L)\}$ with respect to the $L^2$-norm? This means that $C$ is a core of $(L,\text{Dom}(L))$.

Although this fact should be valid, I do not know the proof.

Please let me know if you have any references.

答:
What helped me understand how things work with cores is the following consideration:

The set of finite linear combinations of eigenfunctions of a selfadjoint operator (say with compact resolvents, for simplicity) forms a core for the operator.

Now take your question for example. Let $$E=\{e\ : \ e \text{ is a finite linear combination of eigenfunctions of L}\}\subseteq \mathrm{D}(L).$$ Now, $$\{(e,L e) \ : \ e \in E\}$$
is a dense subset of $$\{(f,Lf) \ : \ f \in \mathrm{D}(L)\},$$
because of the density of the eigenfunctions in the domain. In particular, your question is now whether $$E \subseteq C =C^2_\mathrm{neu}(\Omega).$$
Now we have reduced this fairly abstract problem to a very concrete one. Pick any $\lambda\in(-\infty,0]$ (the Laplacian is a negative operator) and suppose that $e_\lambda \in L^2(\Omega), \|e_\lambda\|=1$ solves
$$\Delta e_\lambda = \lambda e_\lambda.$$
What is the regularity of $e_\lambda$?

This kind of question is what Laplacians are build for, and maybe you already know that eigenfunctions are smooth. If not, to see this we can restate the problem as
$$e_\lambda=(-\Delta +1)^{-1}\big( (1-\lambda) e_\lambda\big).$$
Now one can estimate
$$\|e_\lambda\|_{H^2_{\mathrm{neu}}} = \bigg\|(-\Delta +1)^{-1} \big( (1-\lambda) e_\lambda\big)\bigg\|_{H^2_{\mathrm{neu}}} \leq C(1-\lambda) \|e_\lambda\|_{L^2} =C(1-\lambda) ,$$
where the constant $C>0$ depends on how you define the $H^2_{\mathrm{neu}}$ norm (choose your favourite version of Sobolev space). Iterating this inequality you find that $e_\lambda \in H^{2m}_{\mathrm{neu}}$ for any $m\in\mathbb{N}$. Then by Sobolev embedding you find that $e_\lambda \in C^{\infty}_{\mathrm{neu}}(\Omega)\subseteq C$.

A final note: this proof uses a lot of nontrivial tools, such as Sobolev embeddings, spectral decompositions etc.. With it I am not aiming for a simple direct proof (no doubt possible with some approximation via $C^2$ functions), but I hope it gives the idea of what's going on.

Title: On a core of the Neumann Laplacian


问:
Solution to cauchy's problem

Find the solution for the following Cauchy's problem

$y(1-y^2)\frac{\partial z}{\partial x}-(1+x-y^2x-y^2)\frac{\partial
> z}{\partial y}=0, x\geq 0, y\in\mathbb{R}$

$z(x,0)=x, x\geq0$

I found the first integral: $\phi _1=x+\frac{x^2}{2}-\frac{y^2}{2}$

Could you please tell me how to continue? I'm stuck

答:
Your characteristic equation is correct.
The general solution is :
$$z(x,y)=F\big(x+\frac{x^2}{2}-\frac{y^2}{2}\big)$$
$F(X)$ is an arbitrary function (to be determined according to the condition).

Condition : $z(x,0)=x=F\big(x+\frac{x^2}{2}\big)$

With $\quad X=x+\frac{x^2}{2} \quad\implies\quad  x=-1\pm\sqrt{2X+1}$

$$x=F(X)=-1\pm\sqrt{2X+1}$$
Now the function $F(X)$ is known. We put it into the above general solution where $X=x+\frac{x^2}{2}-\frac{y^2}{2}$
$$z(x,y)=-1\pm\sqrt{2\big(x+\frac{x^2}{2}-\frac{y^2}{2}\big)+1}$$
$$z(x,y)=-1\pm\sqrt{(x+1)^2-y^2}$$
The sign is determined according to $z(x,0)=x$
$$\boxed{z(x,y)=-1+\sqrt{(x+1)^2-y^2}}$$

Title: Solution to cauchy's problem


问:
How to find $\;\lim_{x\to 0} \frac{x}{2-\sqrt{4}-2x}\;$ without using L'Hôpital's rule?
Well, I'm studying calculus and doing some exercises. First of all, from the answers that were given by my teacher the result of this limit should be $4$. I'm beginning my calculus class now but I've already studied calculus by myself before so I know how to apply l'Hôpital's Rule to this limit, following l'Hôpital's I get $-1$, the step-by-step resolution answer from WolframAlpha gives me $-1$ as well.

I thought of all I know about factoring and simplifying equations but I'm not able to actually solve it without l'Hôpital's. Any hint or way about how to do it?

$$\lim\limits_{x\to 0}  \frac{x}{2-\sqrt{4}-2x}$$

答:
Note: The function in the OP has gone through several edits, and is now not the function discussed below.

Hint: Multiply top and bottom by $2+\sqrt{4-2x}$. Very nice things happen.

Remark: An interesting alternative is to consider the reciprocal. We will also change notation a bit, and look for
$$\lim_{h\to 0}-\frac{\sqrt{4-2h}-2}{h}.\tag{$1$}$$
We recognize
$$\lim_{h\to 0}\frac{\sqrt{4-2h}-2}{h}.$$
as the derivative of $f(x)=\sqrt{4-2x}$ at $x=0$. If we are already familiar with differentiation "rules," we can compute the derivative of $\sqrt{4-2x}$, set $x=0$, and insert the minus sign from $(1)$. Now take the reciprocal to get the answer to the original problem.

Note that if the function is indeed the one we have calculated with, the limit is neither $4$ nor $-1$.


问:
A counter example on uniform continuity

Let $f:(0,1]\rightarrow \mathbb{R} $ be a continuous function and bounded is it uniformly continuous?

I know this isn't true, but I can't find a good counter example I was thinking something like this

$f(x)=\min\{3,1/x\}$ can someone give a better example.

答: You can simply take $f(x)=\sin\left(\frac1x\right)$. It is not uniformly continuous because, for any $\delta>0$, there are $x,y\in(0,1]$ such that $|x-y|<\delta$ and that $\bigl|f(x)-f(y)\bigr|=2$.

Title: A counter example on uniform continuity


问:
what is the surface integral of graphs
The surface integral of a vector field  $F$ on $\mathbb R^3$  over a parametrized surface $\Phi:D\to \mathbb R^3$ is the number
$$ \int \int _{S} F\cdot dS = \int \int _{D} F(\Phi (u,v)) \cdot (\Phi_u \times \Phi_v) du dv $$

(Here, $S:=\Phi(D)$)

My Question is: How to prove the following assertion:
"If $F= Pi + Qj + Rk$
is a vector field in space and $S$ is the surface $z= f(x,y)$ where $f$ is a function on $D,$ the surface integral of $F$ over $S$ is:
$$ \int \int_{S} F\cdot S = \int \int_{D} (-Pf_x-Qf_y+R) dx dy. $$"

答:
You can always parameterize a graph in the obvious way:
$$ \Phi(x,y) = (x,y,f(x,y)) $$

Then you have $\Phi_x = (1,0,f_x)$ and $\Phi_y=(0,1,f_y)$, with cross product
$$ \Phi_x \times \Phi_y = (-f_x,-f_y,1) $$

Taking the dot product with $F=(P,Q,R)$, you get the integrand: $(-Pf_x - Qf_y + R)$.

Title: what is the surface integral of graphs


问:
An inequality consequence of Berry-Esseen's inequality
Let $\Phi$ be the cumulative distribution function of the standard gaussian random variable and let $(X_i)_{i\in\mathbb{N}}$ be a sequence of i.i.d random variables $E[X_i]=\mu, Var[X_i]=\sigma^2$. Then call $Z_n = \frac{(\sum\limits_{i=1}^n X_i)/n-\mu}{\sigma/\sqrt{n}}.$

Now if $Z_n$ has bounded third moment, Berry-Esseen's inequality tells us that $$\sup\limits_{x\in\mathbb{R}} |P(Z_n\le x)-\Phi(x)|\le \frac{k}{\sqrt{n}}$$ where $k$ is some real constant. Then apparently we have for a real number $c$ that $$P(|Z_n|\ge c) \le 2(1-\Phi(c))+2\frac{k}{\sqrt{n}}.$$ I think the term $2(1-\Phi(c))$ would be the result if $Z_n$ was $\mathcal{N}(0,1)$ and the second term is the correction term because it is not (because $n<\infty$). But other than that I am not sure how to derive the above (second) inequality.

答:
\begin{align}
\mathsf{P}(|Z_n|\ge c)&=\mathsf{P}(Z_n\le -c)+\mathsf{P}(Z_n\ge c) \\
&\le \Phi(-c)+(1-\Phi(c))+2k/\sqrt{n} \\
&=2(1-\Phi(c))+2k/\sqrt{n}.
\end{align}


问:
Intersections of $y=\cos(2x)$ and $y=e^{3x}$ such that $x \in [-\pi, \frac{\pi}{2}]$
I am having a hard time finding the intersections of $y = \cos(2x)$ and $y = e^{3x}$ such that $x \in [-\pi, \frac{\pi}{2}]$ through solving manually. But through graphing, I know that the intersections are (0,1) and (-0.7292,0.1122). I need those to find the area bounded by the two curves. Can someone help me find it through solving?

答:
$$\cos 2x = e^{3x}$$
it can be solved finding the zeros of the function
$$f(x)=\cos 2x -e^{3x}$$
which is not an algebraic equation.
We can find approximate roots with the Newton's method

looking at the figure below, one root is about $x_0=-0.8$

So we define for $n\ge 0$
$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$
and reiterate the process until the error is less than $10^{-6}$ if we want $6$ exact decimals.

$$
\begin{array}{c|r|r}
 n & x_n & error\\
\hline
 0 & -0.8 &  \\
 1 & -0.730563 & 0.0694371 \\
 2 & -0.729185 & 0.00137768 \\
 3 & -0.729184 & 8.3\times 10^{-7}\\
\end{array}
$$

In a similar way we can find the second root, which is about $z_0=-2.4$

$$
\begin{array}{c|r|r}
 n & x_n & error\\
\hline
 0 & -2.4 &  \\
 1 & -2.35651 & 0.0434943 \\
 2 & -2.35662 & 0.000113975 \\
 3 & -2.3566196667 & 3.4\times 10^{-11}\\
\end{array}
$$

The two solutions in the given interval are:
$$\alpha_1=-2.3566197,\alpha_2=-0.729184$$


问:
Infinite dimensional vector space, and infinite dimensional subspaces.
I'm having trouble with this question:

Let  $V$ be an infinite dimensional vector space.

Prove that there exist subspaces $U_1,U_2,\dots$ of $V$ with $U_{n+1}\neq U_n$ for all $n\in\mathbb{N}$, such that $U_1\supset U_2\supset\cdots$ and $U_n$ is infinite dimensional for all $n\in\mathbb{N}$.

Thanks for your help.

答:
We assume the axiom of choice, of course.

Since $V$ is infinitely dimensional, it has an infinite basis, and therefore there is a countably infinite set of linearly independent vectors, write then as $\{v_n\mid n\in\Bbb N\}$.

Now you can easily find a strictly decreasing sequence of infinite sets, $A_n\subseteq\Bbb N$. Let $U_n=\operatorname{span}(\{v_k\mid k\in A_n\})$. Show that from the reverse inclusion of the $A_n$'s, the sequence of $U_n$'s is decreasing, and it is strictly decreasing because we took the $v_n$'s to be independent.

Note that by choosing your $A_n$'s carefully you can control whether or not $\bigcap U_n$ is the trivial subspace, finite dimensional subspace, or an infinite dimensional subspace of $V$.

Without the axiom of choice it is consistent that there is a vector space which is not spanned by a finite set, but every proper subset is of finite dimension, so you really have to use the axiom of choice for this.


问:
How do you define a non-integer as even or odd?
Note that in this question, I will use the word non-integers to mean all rational non-integers.

I've been working on problems on non-integers lately, and a few thoughts popped into my head.

How do you define non-integers as even or odd?

How do you define non-integers as prime or composite?

I thought of some of the possibilities. Do you round to the nearest integer and define parity on that? Do you take away the decimal point to find primes and composites? Or are non-integers described as neither even or odd, neither prime or composite?

Any help on this would be appreciated greatly.

答:
Disclaimer: In this answer I mostly focus on prime numbers, i.e. the second question of the orignal post. Only in the end I will return to the question about odd or even. This bit is not connected to the concepts introduced before. Please do not confuse the concept or prime numbers and unique factorization with the concept of parity.

Many people have wondered about these questions in abstract algebra already. What makes a number even, what makes it odd? What is the property that will define prime numbers?

But even more importantly: What would the concepts help us with?

Let's begin with primes numbers and the concept itself. Prime numbers are the multiplicative building blocks of any integer. For example $28=2\cdot 2\cdot 7$, where $2$ and $7$ are both prime numbers. This decomposition is, as you will probably know, mostly unique - in fact, since the multiplication on the integers is commutative, the order of the factors does not matter at all. We could do the following though: $28=(-2)\cdot 2\cdot (-7)$. Here both $-2$ and $-7$ are also prime - if we didn't allow negative numbers to be prime, then we couldn't build negative numbers as products of primes.

So the question becomes: What is the difference between the two representations of $28$? Are $2 $ and $-2$ fundamentally different? Turns out, they're not. What makes them similar? Well, for a start: $-2=(-1)\cdot 2$. So, what is special about $-1$ as an integer? We have to look at a different number first: $1$. We know that the number $1$ is actually really special, since $1\cdot n=n$ for every integer $n$. So multiplication with $1$ doesn't change anything. That's cool.

Now back to $-1$. Why is $-1$ special? Well - Except for $1$, the integer $-1$ is the only integer, which can be multiplied by a different integer to give us $1$ - in a formula:
$$(-1)\cdot x=1$$
can be solved for $x$. In particular, $x=-1$. Now you might try to argue that $2x=1$ can also be solves, solution would be $x=1/2$, but here $x$ is not an integer anymore. So, $-1$ is indeed special. These numbers, that can be multiplied by something to give $1$ are called units. In the prime decomposition, mutiplying by units doesn't give us fundamentally different primes - in fact we call $2$ and $-2$ associated primes, because we can reach one of the two by multiplying the other with a unit.

Our uniqueness result is now more precise: If we don't care about units ($-1$ or $1$), then the decomposition is unique up to associated primes.

Great! We found a really useful result for primes. We have one problem remaining though. What about the units? Does $1$ have a unique representation as a product of primes? Does $-1$? Well, here mathematicians did something weird at first glance: We define an empty product. A product over no factors. This is simply defined to be $1$. So, the uniqueness even works for units, but unfortunately not as pretty. We say
$$28=2^2\cdot 7\\
12=2^2\cdot 3\\
-45=3^2\cdot (-5)\\
1=2^0\cdot 3^0\cdot 5^0 \cdots.$$

This gave us a great idea what prime numbers do. And we can now try to define prime numbers on the rationals. First we remember that units are weird and cannot be primes, they are different. So let's figure out the units of the rational numbers.

Well, $1$ is still the same. Now we wonder: For which rational number $q$ is the equation $q\cdot x=1$ solvable over the rational numbers? First of all, $0$ doesn't work, we don't consider $0$ for multiplicative purposes (because $0\cdot x=0$ for all $x$). But then I can simply define
$$x=\frac{1}{q}\implies q\cdot x=1.$$
But then $q$ is a unit. But $q$ was any non zero rational number. So in fact, non zero rationals are units? Yes. We call this a field. Over a field the concept of primes doesn't make any sense, since we can't have multiplicative building blocks.

The same problem arises with odd and even: What makes an integer even? The fact the we can divide by $2$ and still be an integer. $3$ is not even, because $2x=3$ is not solvable over the integers, but $4$ is even, because $2\cdot 2=4$, because $2\cdot x=4$ is solvable over the integers.

What happens over the rationals though? There $2x=q$ can always be solved for any given $q$. Same thing for numbers divisible by $3$ - $3x=q$ can be solved by taking $x=q/3$. So the concepts of odd and even don't really make any sense over the rationals.

You did have a few ideas though. First of all - ideas are great, don't be discouraged by this. The ideas might actually be valid definitions for those concepts. But valid doesn't always mean useful. What do we want to do with the concept of primes closest to the rational number? What things could be done here? Is there anything we could get from that idea? If you find something, great - your definition is actually very fruitful. If not though, you might want to try to find a different concept to generalise.

I hope this was at all helpful.


问:
Continuity in the Mean Value Theorem and the Implicit Function Theorem
Let $f:\mathbb{R}^n\to\mathbb{R}^n$ be a $\mathcal{C}^2$ function of $x\in\mathbb{R}^n$. Then we can use the mean value theorem (MVT) to formulate the following equality:
$$f(x_1)-f(x_2)=\frac{\partial f}{\partial x}(x_2+\lambda(x_1-x_2))(x_1-x_2),$$
with $\lambda\in(0,1)$. This equality is point-wise true by the MVT for every pair $(x_1,x_2)$. Furthermore, it is easy to show with a simple example that for a pair $(x_1,x_2)$, there might be multiple $\lambda$ that yield the equality.

I'm currently looking in the continuation of $\lambda$ when $x_1,x_2$ vary continuously. There are also quite simple examples to formulate that show that continuous variation of $x_1,x_2$ yield discontinuous continuations of $\lambda$. (Check e.g. this example, and move point B to 4)

Now I want to show that for every pair $(x_1,x_2)$ you can find a neighborhood $\mathfrak{B}$ around $(x_1,x_2)$ for which there exists a continuous function $\lambda(x_1,x_2)$ such that for all $(x_1,x_2)\in\mathfrak{B}$, we have that $f(x_1)-f(x_2)=\frac{\partial f}{\partial x}(x_2+\lambda(x_1,x_2)(x_1-x_2))(x_1-x_2)$ and for any continuous trajectory that takes values in $\mathfrak{B}$, the function $\lambda(x_1,x_2)$ is continuous. Trying to counterexample this with multiple examples in geoabgebra (and others) did not help. Up until now I found that this must be true in my setting, I just don't know how to prove this...

What I tried:
I tried to show this with the implicit value theorem by analyzing the implicit function
$$\mathscr{F}(x_1, x_2, \lambda)= f(x_1)-f(x_2)-\frac{\partial f}{\partial x}(x_2+\lambda(x_1-x_2))(x_1-x_2),$$
However, the classic version requires a differentiability condition: $\frac{\partial^2 f}{\partial x^2}\neq 0$ at the point where $\mathscr{F}=0$, which I cannot guarantee perse - plus this seems like a very restrictive assumption for my case. I found a variation of the theorem in [1]. But this theorem requires a one-to-one relationship, which is for me not the case, as there might be multiple lambda's. This was the closed I could get, but no concrete results yet...

Can anybody help me with this problem?

[1] Jittorntrum (1978) - TECHNICAL NOTE: An Implicit Function Theorem

答:
I know you are not too enthusiastic about counterexamples, but there is a problem with your MVT. Explicitly, there may be no such $\lambda(x_1,x_2)$.

A simple example: Let $n=2$ and set $f(x,y)=(x^2,y^3)$. Obviously, $f$ satisfies your assumptions and has Jacobi matrix $$ \frac{\partial f}{\partial(x,y)} = \begin{pmatrix}2x & 0 \\0 & 3y^2\end{pmatrix}. $$ Now for $(x_1,y_1)=(0,0)$ and $(x_2,y_2)=(1,1)$ we are looking for a solution of
$$f(1,1)-f(0,0)=\begin{pmatrix}1\\1\end{pmatrix} \overset{?}{=} \begin{pmatrix}2\lambda & 0\\3\lambda^2 & 0\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix} =\begin{pmatrix}2\lambda\\3\lambda^2\end{pmatrix}, $$
which can easily be seen does not exist.


问:
Prove that $\overline{e^{ix}}=e^{-ix}$
I am trying to construct properly the imaginary exponential as an extension of the real exponential. For this, I need to show that  $\overline{e^{ix}}=e^{-ix}$.

I know that there is a very simple proof using an infinite sum of powers. But I am really looking for something purely algebraic using mainly if not only the properties of the real exponential extended to complex numbers.

One can easily show that $1=e^{ix}\cdot e^{-ix}$ and therefore that if $z=e^{ix}$, then $e^{-ix}=\lambda\cdot \overline z$ with $\lambda\in\mathbb R$, but how can we show that $\lambda=1$ ?

Once again, I am really looking for a proof that doesn't use any infinite sum.

答:
Let $P(f) = f(x + iy) - f(x-iy)$ and $H(f) = f(x+iy) + f(x-iy)$. These transformations are linear, i.e. $P(\alpha f + \beta g) = \alpha P(f) + \beta P(g)$, and $H(\alpha f + \beta g) = \alpha H(f) + \beta H(g)$.

Now, for $x$ real, we have:
$$
\overline{e^x} = e^{\overline{x}}.
$$
Applying $P$ on the LHS yields
$$
P[\overline{e^x}] = \overline{P[e^x]} = \overline{e^{x+iy}} - \overline{e^{x-iy}},
$$
and similarly on the RHS
$$
P[e^{\overline{x}}] = e^{\overline{x+iy}} - e^{\overline{x-iy}} = e^{x-iy} - e^{x+iy}.
$$

Thus:
$$
\overline{e^{x+iy}} - \overline{e^{x-iy}} = e^{x-iy} - e^{x+iy} \implies \overline{e^{x+iy}} + e^{x+iy} = \overline{e^{x-iy}} + e^{x-iy} \implies \Re{e^{x+iy}} = \Re{e^{x-iy}}
$$

Applying $H$ on the LHS we get
$$
H[\overline{e^x}] = \overline{H[e^x]} = \overline{e^{x+iy}} + \overline{e^{x-iy}},
$$
and similarly on the RHS
$$
H[e^{\overline{x}}] = e^{\overline{x+iy}} + e^{\overline{x-iy}} = e^{x-iy} + e^{x+iy},
$$
so
$$
\overline{e^{x+iy}} + \overline{e^{x-iy}} = e^{x-iy} + e^{x+iy}\implies e^{x+iy} - \overline{e^{x+iy}} = \overline{e^{x-iy}} - e^{x-iy}  \implies \Im{e^{x-iy}} = -\Im{e^{x+iy}}.
$$

Therefore for $z = x+iy$, $e^z$ is a conjugate of $e^{\overline{z}}$, i.e. $\overline{e^z} = e^{\overline{z}}$. The result $\overline{e^{ix}}=e^{-ix}$ follows trivially.

Title: Prove that $\overline{e^{ix}}=e^{-ix}$


问:
Differentiability of $f(f(x))$ for $f(x) = |\log2 – \sin x|$

For $x\in\Bbb{R}$, $f(x) = |\log2 – \sin x|$ and $g(x) = f(f(x))$, then:

$g$ is differentiable at $x = 0$, and $g'(0) = –\sin(\log 2)$

$g$ is not differentiable at $x = 0$

$g'(0) = \cos(\log2)$

$g'(0) = –\cos(\log2)$

Please someone help me with this question.

Sorry to say but I tried hard but was not able to find any thing. So I tried different things which were useless and I am not sharing those very very useless solution with you. Pls help me

答:
Since the absolute value function isn't everywhere-differentiable, we will get rid of them.  Luckily everything is continuous, so we can plug in $x=0$ to find out the signs we need.

In particular, note that $\log(2)-\sin(0)$ is positive, and so $f(x)=\log(2)-\sin(x)$ (without the absolute value) in a neighborhood of zero.  Hence, $g(x)=|\log(2)-\sin(\log(2)-\sin(x))|$ in the same neighborhood.

But we still need to get rid of the other absolute value.  As before, we note that $\log(2)-\sin(\log(2)-\sin(0))$ is also positive, and so $$g(x)=\log(2)-\sin(\log(2)-\sin(x))$$ in a (possibly smaller) neighborhood of zero.  Now you can compute the derivative in the usual way, using the chain rule.


问:
About the existence of the diagonal set of Cantor
The classic proof of the Cantor set start with the assumption that the set

$$B=\{x\in A:x\notin f(x)\}$$

exists, where $f: A\to\mathcal P(A)$ is a bijective function. I understand the proof but I dont understand the assumptions where you start to make this proof.

To be clear, why a set $B$ can be constructed? How you can justify this assumption? To me the proof of the Cantor theorem is far to be clear or complete if there are not an explanation about why $B$ must be possible.

Then, can someone explain to me or justify, via other theorems if possible, why $B$ must exist? Thank you in advance.

P.S.: can someone explain to me the downvotes in the question?

答:
Here is an example on a finite set:
$S = \{1,2,3\}$,
$P\left(S\right)=\{\phi,S,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\}\}$
take for example:
\begin{eqnarray}
f\left(1\right)=\{1,2\}\\
f\left(2\right)=\{1,3\}\\
f\left(3\right)=\{1\}
\end{eqnarray}
$B=\{x\in S; x\notin f\left(x\right)\}$
\begin{eqnarray}
1\notin B\\
2\in B\\
3\in B
\end{eqnarray}
$B=\{2,3\}\in P\left(S\right)$


问:
Confusion about the definition of maximal ideal
In my book, the definition of a maximal ideal is as follows:

Let $R$ be a commutative ring. A maximal ideal of $R$ is an ideal $I$ such that:

$I \neq R$.

There exists no ideal $J$ of $R$ such that $I \subsetneq J \subsetneq R$.

When trying to prove the following statement:

$I$ is a maximal ideal of $\mathbb{Z} \implies I = (p)$ with $p$ a prime number.

I came up with an error in my reasoning, but do not understand at all why it is an error. Because all ideals of $\mathbb{Z}$ are of the form $(n)$ with $n \in \mathbb{Z}$, I know that $I = (n)$. So I tried formulating the definition of maximal ideal as follows:

$\forall n' \in \mathbb{Z}: (n') = \mathbb{Z} \ $ or $ \ (n') \subset (n)$.

But this must obviously be false because then $n$ would be a divisor of all $n' \in \mathbb{Z}$, so that $n = 1$ and thus $(n) = \mathbb{Z}$. This cannot be the case since $I$ is a maximal ideal. Where did my reasoning go wrong ?

答: You seem to be assuming that all the ideals of $\mathbb Z$ are linearly ordered, but they're not.


问:
Finding displacement as a function of time given a Force F(x)
If I have a force $F = 4 - 2x$ and want to find the displacement as a function of time i.e. $x(t)$ I'm confused why the answer wouldn't just be $4 - 2x$ integrated twice. Also for this question $m = 1$ so I don't need to worry about it for now.

If anyone could help me I would greatly appreciate it. I've tried searching for an answer but I don't think I'm searching correctly and I'm really disappointed I can't answer this

答: The displacement is just the acceleration integrated twice, but it is integrated with respect to time.  To get the velocity you would integrate once, so you have $v=v_0+\int_o^\tau x(\tau) d\tau$.  We don't (yet) know what $x(\tau)$ is so we cannot do the integral.  The differential equation approach solves this problem.  If we write $F=x''$ because you have $m=1$ you get $x''=4-2x$.  You still need starting conditions before you get a complete solution.


问:
Three line equations form a triangle. How do I find the equations for the heights?
The line equations:
$$x+2y-1=0$$
$$5x+4y-17=0$$
$$x-4y+11=0$$
form a triangle,

How can I find the line equations for the heights of that triangle, without searching for the coordinates of the vertices?
I know it has to do with the intersection of the lines, but I don't know how to find the symmetrical line of that angle.

Thanks in advance

答: You can find one coordinate for one vertice (lets say where the first two lines intersect) and then find the equation of the line perpendicular to $x-4y+11=0$ that passes through the point you just calculated.


问:
Refinements of the inequality $f(x)=x^{2(1-x)}+(1-x)^{2x}\leq 1$ for $0<x<0.5$
Hi it's related to Showing the inequality $f(x)=x^{2(1-x)}+(1-x)^{2x}\leq 1$ for $0<x<1$

We want to show 1:

Let $0<x<0.5$ such that  then we have :
$$f(x)=x^{2(1-x)}+(1-x)^{2x}\leq q(x)=(1-x)^{2x}+2^{2x+1}(1-x)x^2\leq 1$$

The Lhs is equivalent to :

$$x^{2(1-x)}\leq h(x)=2^{2x+1}(1-x)x^2$$

Or :
$$\ln\Big(x^{2(1-x)}\Big)\leq \ln\Big(2^{2x+1}(1-x)x^2\Big)$$

Making the difference of these logarithm and introducing the function :

$$g(x)=\ln\Big(x^{2(1-x)}\Big)-\ln\Big(2^{2x+1}(1-x)x^2\Big)$$

The derivative is not hard to manipulate and we see that it's positive and $x=0.5$ is an extrema .The conclusion is :

$$g(x)\leq g(0.5)=0$$

And we are done with the LHS.

For the Rhs I use one of the lemma (7.1) due to Vasile Cirtoaje we have :

$$(1-x)^{2x}\leq p(x)=1-4(1-x)x^{2}-2(1-x)x(1-2 x)\ln(1-x)$$

So we have :

$$q(x)\leq p(x)+h(x)$$

We want to show that :

$$p(x)+h(x)\leq 1$$

Wich is equivalent to :

$$-2(x-1)x((4^x-2)x+(2x-1)\ln(1-x))\leq 0$$

It's not hard so I omitt here the proof of this fact .

We are done .

Question :

Is it right ?

Thanks in advance .

Regards Max.

1 Vasile Cirtoaje, "Proofs of three open inequalities with power-exponential functions",
The Journal of Nonlinear Sciences and its Applications (2011), Volume: 4, Issue: 2, page 130-137.
https://eudml.org/doc/223938

答:
What you've done looks correct to me.

$$g(x)=\ln\Big(x^{2(1-x)}\Big)-\ln\Big(2^{2x+1}(1-x)x^2\Big)$$The derivative is not hard to manipulate and we see that it's positive and $x=0.5$ is an extrema. The conclusion is :$$g(x)\leq g(0.5)=0$$And we are done with the LHS.

I think that this is correct. (A proof that $g'(x)\gt 0$ : We have $g'(x)=\dfrac{ F(x)}{1-x}$ where
$F(x)=x (2 + \ln(4)) + 2 (x - 1) \ln (x) - 1 - \ln(4)$, $F'(x)=\underbrace{\dfrac{2(2x-1)}{x}}_{\lt 0}+\underbrace{\ln(4x^2)}_{\lt 0}\lt 0$ and $F(0.5)=0$. These imply $F(x)\gt 0$ from which $g'(x)\gt 0$ follows.)

For the Rhs I use one of the lemma (7.1) due to Vasile Cirtoaje we have :$$(1-x)^{2x}\leq p(x)=1-4(1-x)x^{2}-2(1-x)x(1-2 x)\ln(1-x)$$

I think that you have correctly got the inequality by setting $a=1-x$ and $c=2x$ in the lemma.

So we have :$$q(x)\leq p(x)+h(x)$$We want to show that :$$p(x)+h(x)\leq 1$$Wich is equivalent to :$$-2(x-1)x((4^x-2)x+(2x-1)\ln(1-x))\leq 0$$

I've found no errors here. (I don't check the last inequality since you've written "It's not hard so I omitt here the proof of this fact".)


Title: How to find $\;\lim_{x\to 0} \frac{x}{2-\sqrt{4}-2x}\;$ without using L'Hôpital's rule?

Body:
Well, I'm studying calculus and doing some exercises. First of all, from the answers that were given by my teacher the result of this limit should be $4$. I'm beginning my calculus class now but I've already studied calculus by myself before so I know how to apply l'Hôpital's Rule to this limit, following l'Hôpital's I get $-1$, the step-by-step resolution answer from WolframAlpha gives me $-1$ as well.

I thought of all I know about factoring and simplifying equations but I'm not able to actually solve it without l'Hôpital's. Any hint or way about how to do it?

$$\lim\limits_{x\to 0}  \frac{x}{2-\sqrt{4}-2x}$$

Body: $$\lim\limits_{x\to 0}  \frac{x}{\underbrace{2-\sqrt{4}}_{2 - 2 = 0}-2x} = \lim_{x \to 0} \frac x{-2x} = -\frac 12$$


问:
Is $ \bf a^tbc = cb^ta $ where $\bf a, b, c \in R^d$?

Is $ \bf a^tbc = cb^ta $ where  $\bf a, b, c \in R^d$?

$\bf a^tbc =(a^tb)c = (b^ta)c = c(b^ta) = (cb^t)a$

$\bf (cb^t)$ becomes $d*d$ matrix.

First form is some scalar multiplied by $\bf c$ vector while last form is matrix transformation on $\bf a$ vector. Is this consistent? how to think of this intuitively?

答:
You are mixing matrix multiplication and scalar multiplication.

Let us use different symbols to see what is going on. For a scalar $\lambda\in K$ and a vector $v\in K^n$ there is the scalar multiple $\lambda\odot \mathbf v$ (typically written as $\lambda\mathbf v$). Since matrices form a vector space as well, any $n\times m$ matrix $\mathbf X$ can also be scaled by $\lambda\in K$ as $\lambda\odot \mathbf X$ (typically written as $\lambda\mathbf X$ as well).

Now for an $n\times m$ matrix $\mathbf A$ and a $m\times k$ matrix $\mathbf B$ there is the matrix product $\mathbf A\cdot \mathbf B$, a $n\times k$ matrix (typically written as $\mathbf A\mathbf B$). When this is a $1\times 1$ matrix, we may treat it as a scalar.

We may also treat vectors in $K^n$ as column vectors, that is $n\times 1$ matrices.

Using this notation, your question concerns the equation
$$
(\mathbf a^t \cdot \mathbf b)\odot \mathbf c = \mathbf c \cdot \mathbf b^t \cdot \mathbf a.
$$

Note that treating $\mathbf c$ as a $n\times 1$ matrix and scalars $\lambda$ as $1\times 1$ matrices, we do indeed have
$$
\lambda\odot \mathbf c = \begin{pmatrix} \lambda c_1 \\ \vdots \\ \lambda c_n\end{pmatrix} = \mathbf c \cdot (\lambda) = \mathbf c \cdot \lambda.
$$
Hence,
$$
(\mathbf a^t \cdot \mathbf b)\odot \mathbf c = (\mathbf b^t \cdot \mathbf a)\odot \mathbf c = \mathbf c\cdot \mathbf b^t\cdot \mathbf a.
$$

What you see here is that the linear map $\mathbf x \mapsto (\mathbf x^t\cdot \mathbf b)\odot \mathbf c$ is described by the matrix $\mathbf A=\mathbf c\cdot \mathbf b^t$ as $\mathbf x\mapsto \mathbf A\cdot \mathbf x$.

In general, the matrix product $\mathbf A\cdot\mathbf x$ of a matrix and a column vector is obtained as the linear combination of the columns of $\mathbf A$ with the coefficients from $\mathbf x$. Now note that the columns of $\mathbf c \mathbf b^t$ are all scalar multiples of $\mathbf c$, so any linear combination of those will be a scalar multiple of $\mathbf c$ as well. That explains why $(\mathbf c\cdot\mathbf b^t)\cdot \mathbf a$ is always a scalar multiple of $\mathbf c$.


问:
Solve equation $y' - 8x\sqrt{y} = \frac{4xy}{x^2 - 1}$
$y' - 8x\sqrt{y} = \frac{4xy}{x^2 - 1}$

I think I should reduce it to the Bernoulli equation with n = $\frac{1}{2}$, but have no idea how.

答:
$$y' - 8x\sqrt{y} = \frac{4xy}{x^2 - 1}$$
This is Bernoulli differential equation. Divide by $\sqrt y$:
$$\dfrac {y'}{\sqrt y}  -\frac{4x\sqrt y}{x^2 - 1}=8x$$
Note that $(\sqrt y)'=\dfrac 12 \dfrac {y'}{\sqrt y}$

So that you can substitute $v=\sqrt y$:
$$2v'  -\frac{4xv}{x^2 - 1}=8x$$
$$v'  -\frac{2xv}{x^2 - 1}=4x$$
It's a linear first ode that you should be able to solve.


问:
Show that if $g \circ f = \tilde{g} \circ f$ and $f$ is surjective, then $g=\tilde{g}$. Is the same statement true if $f$ is not surjective
Proof of the first part of the question :
Given $g \circ f = \tilde{g} \circ f$ and $f$ is surjective.
To prove $g=\tilde{g}$.

=> $f$ is surjective, which implies for every $y∈Y, ∃x∈X$, such that $f(x)=y$

=> $g∘f =\tilde{g}∘f⟹g∘f(x) =\tilde{g}∘f(x)⟹g(f(x))=\tilde{g}(f(x))⟹g(y)=\tilde{g}(y)$

=> from definition of "Equality of Set" since $g(y)=\tilde{g}(y)$, hence $g =\tilde{g}$.

答:
The other two answers haven't tackled the general case. So here it is.

In general, if $Y$ is the codomain of $f$ and there is a $y\in Y$ that is not in the image of $f$, then $g\circ f=\tilde g\circ f$ says nothing about whether $g(y)=\tilde g(y)$. Thus if $f$ is not surjective, we can never conclude $g=\tilde g$ without more information of some kind.

Your proof in the case where $f$ is surjective is fine.

Also note that there is a completely dual property for injectivity. If we have $f\circ g=f\circ \tilde g$, then in an entirely analogous way, $f$ being injective lets us conclude that $g=\tilde g$, and if $f$ is not injective we can find a similar "lack of information" as in the not-surjective case that stops us from concluding.