问:
$|\mu^*(A)-\mu^*(B)| \leq \mu^*(A \Delta B).$
Let $A,B \subset \mathbb{R}^d$ and at least one of $\mu^*(A)$ and $\mu^*(B)$ is finite.
Show that $|\mu^*(A)-\mu^*(B)| \leq \mu^*(A \Delta B).$

答:
See if this works.

If $\mu^*(A)=+\infty$ and $\mu^*(B)<+\infty$, then $|\mu^*(A)-\mu^*(B)|=+\infty$ and $\mu^*(A\triangle B) \geq \mu^*(A\backslash B)$. So, since $\mu^*(A\backslash B)=+\infty$ (you can argue by contraddiction), $|\mu^*(A)-\mu^*(B)|=\mu^*(A\triangle B)$ follows.

If $\mu^*(A),\mu^*(B)<+\infty$, you have
$$\mu^*(A\triangle B) \geq \mu^*(A \cup B) - \mu^*(A \cap B)$$
and since
$$ \mu^*(A \cap B) \leq \mu^*(A),\mu^*(B)\leq \mu^*(A \cup B)$$
the required inequality follows.


问:
Equivalent definition about bounded set in metric space
I have read some book they said that "A nonempty subset $A$ of metric space $X$ is bounded if $\sup \{ d(x,y) : x, y \in A \} < \infty$" and another book they said that "A nonempty subset $A$ of metrix space $X$ is bounded if there is a ball $B(x,r)$ such that $A \subseteq B(x,r)$"

I want to show that these two definitions are equivalent. First, I let $A$ be a nonempty subset of metric space $X$ and $d=\sup \{ d(x,y) : x,y \in A \}$. Since $A$ is nonempty set, there is $a \in A \subseteq X$. We let $r=d+1$ and $y \in A$. Then
$$d(y,a) \leq d < d+1=r$$
Thus $A \subseteq B(a,r)$.

I cannot complete another hand. Please help me.

答:
If $A \subseteq B(x,r)$ then $D(A)$ = $\sup \{d(x,y): x,y \in A \}$ $\le2r<\infty$.

I think this does the job :)

the simple rationale being that the diameter of a set cannot be greater than the diameter of the sphere that contains it.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
For $A \in \mathcal {E}$ exist disjoint intervals $I_1,...,I_n$ with $A =I_1 \cup ... \cup I_n$
$\mathcal {E}$ is the union of finite intervals in $\mathbb{R}^d$. For $A \in \mathcal {E}$ exist disjoint intervals $I_1,...,I_n$ with $A =I_1 \cup ... \cup I_n$.

How do I show this?

答:
Lemma: For any two finite intervals $I,J$ in $\mathbb{R}^d$, $I \setminus J$ is a finite disjoint union of finite intervals: $I \setminus J = \bigcup_{i=1}^{m} K_i$.  You can see this vividly in $\mathbb{R}^2$ by drawing some pictures. And the pictures will indicate how to prove it.

For the main result do an induction on $n$. The key step is $n=2$: Given $A = I_1 \cup I_2$, write $A$ as $A =  I_1 \cup (I_2 \setminus I_1)$, then by the lemma $A = I_1 \cup \bigcup_{i=1}^{m} K_i$ where $K_i$ are disjoint intervals. Each $K_i$ is also disjoint from $I_1$ because they are subsets of $I_2 \setminus I_1$.


问:
What is a "symmetric Boolean function"?
Below are two different definitions of a "symmetric boolean function".

Question: Are these equivalent and how?

Definition 1:

The $k$th elementary symmetric function of $n$ variables is the sum of all possible products of $k$ distinct variables = $\Sigma x_{i_1} \cdot \cdot \cdot x_{i_k}$ where the subscripts are taken from the first $n$ integers. (Cunkle, p. 834)

Definition 2:

A symmetric Boolean function is a Boolean function whose value does not depend on the permutation of its input bits, i.e., it depends only on the number of ones in the input. (wikipedia)

References:

Cunkle, C. (1963). Symmetric Boolean Functions. The American Mathematical Monthly, 70(8), 833-836. doi:10.2307/2312663

答:
The first definition does not specify the domain of the $x_i$ as booleans. It uses sums and products so it makes sense for the domain of each of the $x_i$ to be anything (call it $R$) where you can add and multiply and the value outputted is in that same domain. That is like the real numbers or the complex numbers. So the first definition takes $n$ inputs from $R$ and outputs something in $R$. If you permute the inputs you can see the summands just get shuffled around but the result is the same. That is why it is called symmetric.

But the second definition says instead of $R$ the inputs and outputs are all boolean.

Now consider the case when $R$ is booleans where sum means the xor operation and multiplication is the and operation. Then you can use the first definition to get a symmetric Boolean function of the second definition.

Edit :
Clarification on last paragraph for the comment
Restricting 1st definition to booleans gives examples of the second definition. Did not mean to imply that it gave the full second definition. Get a symmetric Boolean function of the second definition, not all of them.

Title: What is a "symmetric Boolean function"?


问:
$\int^1_0 \frac{xdx}{x^2+2x+1}$
I need some suggestion how to solve this integral.

$$\int^1_0 \frac{xdx}{x^2+2x+1}$$
I think about to do the following step :

$$\frac{1}{2}\int^1_0\frac{2x+2-2dx}{x^2+2x+1}$$$$ t=x^2+2x+1 \rightarrow 2x+2dx=dt$$
then the integral will be : $$\frac{1}{2}\int^1_0 \frac{-2dt}{t}$$
its a correct way to solve it?

Thanks!

答:
Hint:

Notice that $x^2+2x+1=(x+1)^2$

$\int^1_0 \dfrac{xdx}{x^2+2x+1}=\int_0^1\dfrac{A}{x+1}+\int_0^1\dfrac{B}{(x+1)^2}$

What is wrong with your approach?

After you have $dt=2(x+1)dx$, your expression will be $\dfrac{1}{2} \int_0^1 \dfrac{2(\sqrt{t}-1)}{2\sqrt{t} \cdot t} dt$

Title: $\int^1_0 \frac{xdx}{x^2+2x+1}$


问:
What is a good rule to understand how inequity shading works?
I am learning mathematics for my upcoming IGCSE exams and we have a topic of Inequality where we are sometimes asked to 1) solve some inequality equation 2) shade a region that all the already given inequalities satisfy 3) derive 2-3 inequalities that define an already-shaded region. The last part is where I have some troubles. I want to show two problems of such type that I am having trouble with.

For this question, the answers are:

x<=3

y>=6-2x

y<=5+x

By the looks of it, it is my understanding that if the region shaded is below or behind the line then we use the "<=" or "<" sign. Because for the question, x<=3 was an answer since the shaded parts were behind x=3. And that if the region shaded is above or in front of the line, then we use the ">=" or ">" sign. For example, y>=6-2x was an answer since the shaded parts were above y=6-2x. This is my understand and I can be wrong on this one, it is my request to correct me if I am.

But I also encountered another question which seemed to contradict this point.

For this question, the answers were:

x>=0

x+y<=5

x-y<=5

So here, according to what I understand, x>=0 is correct since the shaded region is in front of it. And, x+y<=5 is also correct since the shaded region was below it. But why is x-y<=5 the third answer instead of x-y>=5? From the question, is it clear that the shaded region is above x-y=5 line so shouldn't the third answer be x-y>=5? Why is it x-y<=5?

This is my question and I hope I was able to clearly state the trouble that I am having. It will be much appreciated if anyone can help me with when to use ">=" and when to use "<=".

答:
Rules that always work are:

If the equation of the line is in the form $x = ay + b,$
then the region $x \leq ay + b$ will be on the left of the line and the
region $x \geq ay + b$ will be on the right of the line.

If the equation of the line is in the form $y = ax + b,$
then the region $y \leq ax + b$ will be below the line and the
region $y \geq ax + b$ will be above the line.

These rules work because the lesser value of $x$ is always on the left of the greater value of $x$, and the lesser value of $y$ is always below the greater value of $y.$

But these rules do not always work when you mix $x$ and $y$ together on the same side of an inequality.

In the case of $x - y \leq 5,$ the left hand side can become less if you decrease $x$
(moving to the left), but it can also become less if you increase $y$
(moving upward).
To disambiguate this, you can rewrite the inequality with just one variable on each side, for example,
$x \leq y + 5$ is completely equivalent to $x - y \leq 5,$
but because it is in the form $x \leq ay+b$ we see that the shaded region must be on the left side of the line, which it is.

Another way to figure out which side to shade is:

Pick a point that is not on the line.
Take the $x$ and $y$ coordinates of that point and plug them into the inequality.

Is the inequality true with these values plugged in?
If so, the point you chose is on the side where the shading should be.
If not, the shading should be on the opposite side.


问:
SLLN for iid $X_n$
For i.i.d $X_n$'s with $E[(X_1)_-]<+\infty$ and $E[(X_1)_+]=+\infty$, I want to prove that $$\frac{1}{n}\Sigma_{i=1}^nX_i\xrightarrow{\text{a.e}} +\infty$$ as $n\xrightarrow{}+\infty$.

I know that I have to use the Strong Law of Large numbers but I am confused on how to implement it. I think I have to find out that the mean of $\Sigma X_i$ is infinity.

I also found in my notes that $n^{-1}S_n\xrightarrow{\text{a.e}}\mu$ as $n\xrightarrow{}\infty$ whenever $E[X_1]<\infty$ and $S_n=\Sigma_{i=1}^nX_i$.

I would appreciate any answers to this.

答:
Let $X_i^M(\omega) = \min\{X_i(\omega), M\}$, and $S_n^M=\sum_{i=1}^nX_i^M$. Then almost surely:
$$\liminf_{n\rightarrow\infty}\frac{S_n}n\ge\lim_{n\rightarrow\infty}\frac1nS_n^M = E(X_1^M)$$
By monotone convergence, $E(X_{1,+}^M)\rightarrow\infty$ as $M\rightarrow\infty$, and since $E(X_{1,-}^M)=E(X_{1,-})<\infty$, the claim follows.

Title: SLLN for iid $X_n$


问:
Derivative of indefinite integral
I'm trying to understand the following derivative, where $r\geq0$:

$$
\psi(r):=\int_r^\infty e^{-(y-r)}\Bigl(\int_0^y \psi(y-x)dx\Bigr)dy
$$
$$
\Longrightarrow\\
$$
$$
\psi'(r)=-\int_0^r \psi(r-x)dx +\int_r^\infty -e^{-(y-r)}\Bigl(\int_0^y \psi(y-x)dx\Bigr)dy
$$

More specifically, I was under the impression that, by the fundamental theorem of calculus, the derivative of $\psi$ was simply the function inside the integral,
$$
e^{-(y-r)}\Bigl(\int_0^y \psi(y-x)dx\Bigr)
$$
evaluated at $r$. (Or rather, the negative thereof, since the $r$ was the lower bound.)

However, that would just be the first summand in the above result. The second summand I can't explain. Where does it come from? And where does my application of the FtoC fail?

答:
This should follow directly from the Leibniz integral rule. With

$$\psi(r)=\int_r^\infty \underbrace{e^{-(y-r)} \left( \int_0^y \psi(y-x)\,\mathrm dx\right)}_{f(r,y)} \,\mathrm dy$$

the derivative would be

$$\begin{align}
\frac{\mathrm d\psi}{\mathrm dr}&=-f(r,r)\frac{\mathrm dr}{\mathrm dr}+\int_r^\infty \frac{\partial f(r,y)}{\partial r}\,\mathrm dy\\[1ex]
&=-e^{-(r-r)}\int_0^r \psi(r-x)\,\mathrm dx + \int_r^\infty e^{-(y-r)}\left(\int_0^y\psi(y-x)\,\mathrm dx\right)\,\mathrm dy
\end{align}$$

though there seems to be a sign error either here or in the result you've provided.

Title: Derivative of indefinite integral


问:
How to proof: Suppose $R$ and $S$ are equivalence relations on a non-empty set $A$, then $R \setminus S$ is not transitive
I've able to proof that $R \setminus S$ is symmetric and not reflexive, but don't know how to start the proof of $R \setminus S$ is not transitive.

答: This isn’t quite true: if $R\subseteq S$, then $R\setminus S=\varnothing$, which is vacuously transitive. However, as long as $R\setminus S\ne\varnothing$, you can let $\langle a,b\rangle\in R\setminus S$. Then $\langle b,a\rangle\in R\setminus S$. Can you construct a counterexample to transitivity from these ordered pairs?


答: You followed the general strategy. Alternatively, you can remark that $x^2+2x+1=(x+1)^2$,, put $u=x+1$ and reduce to $$\int \frac{u-1}{u^2}\, du,$$ which is really simple.</p>

Title: $\int^1_0 \frac{xdx}{x^2+2x+1}$

Body:
I need some suggestion how to solve this integral.

$$\int^1_0 \frac{xdx}{x^2+2x+1}$$
I think about to do the following step :

$$\frac{1}{2}\int^1_0\frac{2x+2-2dx}{x^2+2x+1}$$$$ t=x^2+2x+1 \rightarrow 2x+2dx=dt$$
then the integral will be : $$\frac{1}{2}\int^1_0 \frac{-2dt}{t}$$
its a correct way to solve it?

Thanks!


问:
what's the distribution of CDF(X) when X belong to normal distribution
Say I sample X from a standard normal distribution.
Then send those X to a standard normal CDF and get Y, then I plot the histogram of Y

it looks like they belong to a uniform distribution, but how do I express this in formula?

code:

```<code>import numpy as np
import scipy
import matplotlib.pyplot as plt
import seaborn as sns

x = np.random.normal(0,1,100000)
norm_cdf = scipy.stats.norm.cdf(x)

# plot the cdf
plt.subplot(211)
sns.distplot(x)
plt.subplot(212)
sns.distplot(norm_cdf)
plt.show()
</code>```

答:
You are defining $Y=F(X)$. As $F$ takes values in $[0,1]$, so does $Y$.

Consider
$$
\begin{align}
P(Y \le y ) &= P(F(X) \le y) \\
&= P(X \le F^{-1}(y))\\
&= F(F^{-1}(y)) \\
&= y
\end{align}
$$
which is the CDF of the uniform distribution. In summary you have $Y\sim U(0,1)$ as you speculated. It's called the Probability Integral Transform.


问:
What is an example of zero exponent in nature?
This concept is difficult for non-professionals to grasp, and I admit that I can't even conceive of how this exists in nature, as opposed to proving that 2+2=4 by a more traditional explanation such as grouping objects together.

Does some number taken to the zero power exist in nature?  If so, please describe it.

答:
Yes, it does. If you have an exponential function to, for instance, describe a population that doubles daily and starts at $a$
$$p(x) = a2^x$$

then taking $2^0$, should naturally equal $1$, because at time $0$, the population is exactly the starting population $a$.


问:
Systems of Congruences
\begin{cases} \overline{xyz138} \equiv 0 \mod7 \\ \overline{x1y3z8} \equiv 5 \mod11 \\ \overline{138xyz} \equiv 6 \mod13 \end{cases}

I worked my way up to this:

\begin{cases} 2000x+200y+20z \equiv 2 \mod7 \\ 100000x+1000y+10z \equiv 4 \mod11 \\ 2000x+200y+20z \equiv 7 \mod 13 \end{cases}

I tried to subtract one equation from the other,  but got nothing. No idea what to do.

答:
$$\overline{xyz138}=1000\overline{xyz}+138\equiv 6\overline{xyz}-2\pmod 7\\ \overline{138xyz}=138000+\overline{xyz}\equiv 5+\overline{xyz}\pmod{13}\\ \overline{x1y3z8}\equiv 10x+1+10y+3+10z+8\equiv 1-x-y-z\pmod{11}$$

Denote $a:=\overline{xyz}$, then you have the linear system

$$\begin{cases}6a-2\equiv 0\pmod 7\\5+a\equiv 6\pmod{13}\end{cases}$$

along with $x+y+z\equiv 7\pmod{11}$

Apply CRT to solve modulo $7\times 13$ and then check the values of $a$ that satisfy the congruence modulo $7\times 13$ against the final constraint. There are only $11$ values of $a$ to check.

Title: Systems of Congruences


答:
First, let $\mu^*(A)$ and $\mu^*(B)$ both be finite. Then, since $A \subseteq A \cup B = (A\Delta B) \cup B$, the following holds:
$$
\mu^*(A) \leq \mu^*(A \cup B) = \mu^*((A\Delta B) \cup B) \leq \mu^*(A \Delta B) + \mu^*(B) \iff \mu^*(A) - \mu^*(B) \leq \mu^*(A \Delta B)
$$
Doing the same estimation for $\mu^*(B)$ (use that $A \cup B = (A\Delta B) \cup A$) yields
$$
\mu^*(B) - \mu^*(A) \leq \mu^*(A \Delta B),
$$
so we have
$$
\lvert \mu^*(A) - \mu^*(B) \rvert \leq \mu^*(A \Delta B)
$$
in total.

If without loss of generality $\mu^*(A) = \infty$ but $\mu^*(B) < \infty$, then
$$
\infty = \mu^*(A) - \mu^*(B) \leq \mu^*(A \Delta B)
$$
still holds, which means $\mu^*(A \Delta B)= \infty$.

Body:
Let $A,B \subset \mathbb{R}^d$ and at least one of $\mu^*(A)$ and $\mu^*(B)$ is finite.
Show that $|\mu^*(A)-\mu^*(B)| \leq \mu^*(A \Delta B).$


问:
Integral of fractional expression $\int^3_0 \frac{dx}{1+\sqrt{x+1}}$
I want to solve this integral and think about call $\sqrt{x+1} = t \rightarrow t^2 = x+1$
$$\int^3_0 \frac{dx}{1+\sqrt{x+1}}$$
Now the integral is : $$\int^3_0 \frac{2tdt}{1+t}$$ now I need your suggestions.

Thanks.

Body:
$$\begin{align}
\int^2_0 \frac{2t\, dt}{1+t} &= 2\int^2_0 \frac{1+t-1}{1+t}\,dt \\
&= 2\int^2_0 \left( \frac{1+t}{1+t}-\frac{1}{1+t}\right)\,dt \\
&= 2\int^2_0\left( 1-\frac{1}{1+t}\right)\,dt \\
&= 2[ t-\log({1+t})]_0^2 \\
&= 2[ 2-\log 3] \\
\end{align}$$


答:
In the case of radioactive decay, the amount of non-decayed substance is

$$f(t)=A2^{-\frac{t}{T}}$$

where $A$ is the initial quantity, $t$ is time, and $T$ is the half-time of the decay. Plugging $t=0$ yields $f(0)=A$, the initial quantity, as expected.

Title: What is an example of zero exponent in nature?

Body:
This concept is difficult for non-professionals to grasp, and I admit that I can't even conceive of how this exists in nature, as opposed to proving that 2+2=4 by a more traditional explanation such as grouping objects together.

Does some number taken to the zero power exist in nature?  If so, please describe it.


问:
Systems theory: show necessary condition for "reachable system"
Do not worry if you are new to systems theory, I will define all relevant terminology below, the mathematical problem should be clear!

Suppose the following system is given

\begin{align}
\dot{x} = Ax,\quad x\in \mathbb{R}^{n}
\end{align}

where $A$ is a constant. You are asked to add one input $u\in \mathbb{R}$ to make the system reachable, i.e. you design $b\in \mathbb{R}^{n}$ such that

\begin{align}
\dot{x} = Ax + bu
\end{align}

is reachable.

a) Show that a necessary condition for such $b$ to exist is that $\text{Rank}(A)\geq n-1$.

So, what we have is a so called time invariant system since $A$ is constant ($n$-by-$n$) matrix. Moreover, if the system $\dot{x}=Ax+bu$ is reachable then systems theory tells us that the reachability matrix

\begin{align}
\Gamma = \begin{bmatrix}b,Ab,A^{2}b,\dots,A^{n-1}b\end{bmatrix}
\end{align}

is of full rank. Also $\Gamma$ is a $n$-by-$n$ matrix. Now, correct me if I am wrong, what we want to show is that if the system is reachable, i.e. $\text{Rank}(\Gamma)=n$, then this implies ($\implies$) that $\text{Rank}(A)\geq n-1$? Or is the implication the other way around ...?

Furthermore, how may I proceed from this to showing that $A$ has at least rank of $n-1$? My linear algebra is weak, and I do not know what to do!

答:
You have correctly interpreted the question. Indeed, we want to show that if the system is reachable (AKA controllable), then the rank of $A$ is at least $n-1$. Actually, I would suggest that it is (slightly) easier to do this problem by using the "Hautus test" instead of the reachability matrix. The Hautus test says that a system is reachable iff the matrix $[A - \lambda I  \quad b]$ has full rank for all $\lambda \in \Bbb C$. For our purposes, it suffices to consider what this means for $\lambda = 0$.

However, let's stick to the reachability matrix criterion. We can prove this by contradiction: suppose that the rank of $A$ is not at least $n-1$.

Recall that the rank of a matrix $M$ is equal to the dimension of its column space, which is equivalently the dimension of the span of its columns. The rank of $\Gamma$ is the dimension of the span of its columns, but note that each of the columns $Ab,A^2b,\dots,A^{n-1}b$ is of the form $Ax$ for some vector $x$. That is, all of these vectors are elements of the columns space of $A$. Let $\{x_1,\dots,x_{k}\}$ be a basis for the span of the columns $Ab,A^2b,\dots,A^{n-1}b$; this span is a subspace of the column space of $A$, which means that $k$ (the dimension of the span) is at most $n-2$.

We see that the set $\{b,x_1,\dots,x_k\}$ gives us a set that spans the column space of $\Gamma$. However, this set has at most $(n-2) + 1 = n-1$ columns, so the column space of $\Gamma$ has dimension at most $n-1$. So, the dimension of this space (rank of $\Gamma$) cannot be equal to $n$ after all.

This whole proof can be made much quicker if we can appeal to the fact that for any matrix $M$, we have
$$
\operatorname{rank}([M \mid b]) \leq \operatorname{rank}(M) + 1.
$$


答:
We have following divisibility rules :

$\overline{abcdef} \equiv \overline{def} - \overline{abc} \pmod 7$

$\overline{abcdef} \equiv a-b+c-d+e-f \pmod {11}$

$\overline{abcdef} \equiv \overline{def} - \overline{abc} \pmod {13}$

So one obtains
$$ \overline{xyz} \equiv 5 \pmod 7$$
$$ x+y+z \equiv 6 \pmod {11}$$
$$ \overline{xyz} \equiv 1 \pmod {13}$$

Can you take it from here?

Title: Systems of Congruences

Body:
\begin{cases} \overline{xyz138} \equiv 0 \mod7 \\ \overline{x1y3z8} \equiv 5 \mod11 \\ \overline{138xyz} \equiv 6 \mod13 \end{cases}

I worked my way up to this:

\begin{cases} 2000x+200y+20z \equiv 2 \mod7 \\ 100000x+1000y+10z \equiv 4 \mod11 \\ 2000x+200y+20z \equiv 7 \mod 13 \end{cases}

I tried to subtract one equation from the other,  but got nothing. No idea what to do.


问:
How to find the missing angle on this figure
I tutored a student today and we were both stumped on the following question:

My thought process was that since $\angle$J is an inscribed angle, we can say that arc ILK = $222^{\circ}$ I don't think that gets me anywhere as I am not sure how to work with the lines $y - 2$ and $6$. Any hints would be greatly appreciated.

P.S. I know taking snapshots makes it rather difficult to track the problem if others had questions about it, but it came from my student's school portal, so I do not have direct access to its location.

答:
Since $\;I\hat{J}K\;$ and $\;I\hat{L}K\;$ are opposite angles in a cyclic quadrilateral, their sum is $\;180^\text{o}$, hence

$I\hat{L}K=180^\text{o}-I\hat{J}K=180^\text{o}-111^\text{o}=69^\text{o}.$

Moreover

$y-2=6$

because tangent segments to a circle from the same external point $\;M\;$ are congruent.

Therefore

$y=8\;.$


答:
If your workings are correct, you can rewrite your equations as:

\begin{cases}
20(\overline{xyz}) &\equiv 2 \pmod 7\\
10(x+y+z) &\equiv 4 \pmod {11}\\
20(\overline{xyz}) &\equiv 7 \pmod {13}
\end{cases}

which reduces to

\begin{cases}
\overline{xyz} &\equiv 5 \pmod 7\\
x+y+z &\equiv 7 \pmod {11}\\
\overline{xyz} &\equiv 1 \pmod {13}
\end{cases}

By CRT, $\overline{xyz} \equiv 40 \pmod {91}$, so there are not many numbers to check.

Title: Systems of Congruences


问:
Total Area of the Region Bounded by the Curves - Need some assistance
*This is not a homework question. This is from a test from a previous year, in an attempt to learn the material.

Find the total area of the region bounded by the curves y = 3/x and y = 4-x

To start off, I figured I'd need to find the points of intersection, in order to find the actual area I'm trying to solve.

On the right side of the x axis, x = 1 and x = 3 are the points of intersection. On the negative side, they will never intersect. This leaves me confused as to the actual area I'm attempting to solve.

Pretty much, I don't even know where to start, and even if I find that, I'm doubtful I'll be able to finish.

My current assumption, I'm trying to find the area between the interval [1,3]?

Update : This is a guess, but seems semi correct? The integral from 1 to 3 of (3/x) - (4-x)

Take the anti derivatives, separate, end up with

the integral from 1 to 3 of ln abs(3x)   - the integral from 1 to 3 of (4x - x^2 / 2)

.. the integral from 1 to 3 of (4x - x^2 / 2) seems to equal 4, assuming my math was correct, meaning that the area would be equal to (ln9-ln3) - 4? That would give a negative number, so I'd guess it would actually be 4 - (ln9-ln3), which is roughly 2.9

Am I going about this the completely wrong way?

答:
First of all where do the two graphs intersect? If you set them equal it will lead you to the two solutions $1$ and $3$. First of all I would evaluate the area in the interval [1,3] under the function $f_1(x) := 4-x$. Then find the area under the function $f_2(x) := 3/x.$
I think these steps should be very intuitive, even without calculus/analysis knowledge.

Let's move on to the main dish - the integrals. Let's call our desired area $A$. Then
$$A = \int_{1}^{3}4-x\,\mathrm{d}x \, - \int_{1}^{3}3/x\,\mathrm{d}x.$$

I think the first integral shouldn't be a problem. You can spit it up into two futher integrals and evaluate these with basic calculus. The second integral shouldn't be hard too if you know that the antiderivative of $1/x$ is $\log\lvert x\rvert + C$. [Bonus: Don't forget that you will get a small problem if u take the antidervative of $1/x$ while integrating over $0$.]


问:
Question regarding the existence of a general solution to the functional equation $f \circ f = af$
I would like to know how to determine all functions $f: \mathbb{R} \to  \mathbb{R}$ satisfying $f\circ f =a f$ (where $a$ is any real number)

Is this problem completely solved (without or with additional assumptions on $f$ )

A friend sent me the problem with $a=3$ , if we assume $f$ bijective then $f $ is linear $3x$

Thanks for sharing any relevant information

答:
Well, clearly $f(x) = 3x$ for all $x$ in the image of $f$. In particular, the image of $f$ is a set $S$ with $3S \subseteq S$. There are many such sets: all of $\mathbb R$, the positive reals, the set $\{27, 81, 243, 729, \dots\}$, the rationals, and many more.

(In general, we can take $S$ to be any union of the sets $\{y, 3y, 9y, 27y, \dots\}$ for one or more real numbers $y$.)

If $S$ is such a set, then we can pick an arbitrary $f : \mathbb R \setminus S \to S$ and extend it to all of $\mathbb R$ by $f(x) = 3x$ for all $x \in S$.

The picture does not change much if $3$ is replaced by $a$.


问:
A special Holder-Lipschitz function
I am interested in extending classical inequalities in the domain of the real numbers to the appropriate setting within the complex numbers. Here is one example I couldn't solve yet.  Let  $\mathbf D$ be the closed unit disk, $0<\alpha<1$, and $L:= \displaystyle\sup_{z,w\in \mathbf D\atop z\not=w}\frac{|(1-z)^\alpha -(1-w)^\alpha|}{|z-w|^\alpha}$, where one takes the main branch of the $\alpha$-root on $\mathbb C\setminus\;]-\infty,0]$. Is $L<\infty$? Whereas this is easily seen to hold for real numbers $x,y\in [-1,1]$, that is
  one has   $|(1-x)^\alpha-(1-y)^{\alpha}|\leq |x-y|^{\alpha}$,  I see no elementary way to determine $L$. May be some one on this forum has an idea?</p>

Title: A special Holder-Lipschitz function

Body:
$L=\max\{1, 2^{1-\alpha}\sin(\alpha\pi/2)\}$.
Proofs are in Comput. Methods Funct. Theory 20, 667–676 (2020)
https://doi.org/10.1007/s40315-020-00344-7
and Elemente der Math. 75 (2020), 38--41.


问:
If an integer $n$ is such that $7n$ is the form $a^2 + 3b^2$, prove that $n$ is also of that form.
If an integer $n$ is such that $7n$ is the form $a^2 + 3b^2$, prove that $n$ is also of that form.

I thought that looking at quad residues mod $7$ might??? help. But that didn't take me anywhere so apart from that I'm at a loss. I would appreciate any help. Thanks in advance.

答:
$\text{Make use of the identity }$$$\color{red}{(x^2+3y^2)(u^2+3v^2) = \underbrace{(xu\pm3yv)^2 + 3(xv\mp yu)^2}_{a^2 + 3b^2}}$$

$\text{Now we have}$
$$7 = 2^2 + 3 \times 1^2$$
$\text{i.e., }x=2 \text{ and }y=1$.

$\text{Use this to obtain $u$ and $v$.}$

$\textbf{EDIT}$:

$\text{Here is the full solution. The main claim is that if $7 \vert (a^2+3b^2)$, then}$

Either $7 \vert (a+2b)$ and $7 \vert (2a-3b)$

Or $7 \vert (a-2b)$ and $7 \vert (2a+3b)$

This can be checked by listing out the different possibilities for $a$ and $b$.

If $7 \vert (a+2b)$ and $7 \vert (2a-3b)$, then set $u = \dfrac{2a-3b}7$ and $v = \dfrac{a+2b}7$.

Else, if $7 \vert (a-2b)$ and $7 \vert (2a+3b)$, then set $u = \dfrac{3b-2a}7$ and $v = \dfrac{a+2b}7$.

This guarantess that $u$ and $v$ are integers. Hence, $\text{$n$ can be written as $u^2+3v^2$.}$


问:
Basic notation in limits
What is the difference between $f'(a^+)$, $f'_+(a)$, $\lim_{x\to a^{+}}f'(x)$, and $\lim_{x\to a^{+}}\frac{f(x)-f(a)}{x-a}$? I assume they're all the same, but I'm not sure; don't they all essentially mean "the slope of the graph of f(x) just after the point x=a"?

Edit: I could used the left-hand limit with the negative sign, I know; I just used this for convenience. Other than that, is there any difference between any of these?

答:
The third one, $\lim_{x \to a^+} f'(x)$, is different! If that limit exists and $f$ is continuous at $a$, then it equals the right-hand derivative $f'_+(a)$, as can be shown using the mean value theorem for derivatives. But it may happen that it doesn't exist even if $f'_+(a)$ exists, as shown by the standard examples of differentiable but not continuously differentiable functions. At it may also exist if $f$ is discontinuous at $a$, say $f(x)=0$ for $x \le a$ and $f(x)=1$ for $x>a$, where $\lim_{x \to a^+} f'(x) = \lim_{x \to a^+} 0 = 0$ but $f'_+(a)$ doesn't exist.

The second and the fourth one denote and define, respectively, the right-hand derivate.

Whether the first one, $f'(a^+)$, is supposed to mean the third or the second/fourth alternative is unclear to me. Usually $g(a^+)$ means the right-hand limit of $g$, and with that interpretation, $f'(a^+)$ would mean the right-hand limit of $f'$, i.e., the third alternative. But some people might also use it to denote the right-hand derivative. (And in many cases that's the same thing, so it might not matter much, but as I wrote above, it's not quite equivalent.)

Title: Basic notation in limits


答: The trivial solution:  $f=0$ also holds (if $f$ not bijective).

Title: Question regarding the existence of a general solution to the functional equation $f \circ f = af$

Body:
I would like to know how to determine all functions $f: \mathbb{R} \to  \mathbb{R}$ satisfying $f\circ f =a f$ (where $a$ is any real number)

Is this problem completely solved (without or with additional assumptions on $f$ )

A friend sent me the problem with $a=3$ , if we assume $f$ bijective then $f $ is linear $3x$

Thanks for sharing any relevant information


问:
Minimise $|f''(x)|$ on an interval when you know the values of the function and the values of the derivative at the endpoints of the interval only.
Given $a, b,c \in \mathbb{R}$, let $A_{abc}$ be the set of twice-differentiable real functions $f:[0,1] \to \mathbb{R}$ such that $f(0) = 0$, $f(1) = a, f'(0) = b$ and $f'(1) = c.\ $ Find the infimum of {$\max_{x \in [0,1]}$$|f''(x)|: f \in A_{abc}$ }$\subset \mathbb{R}$ in terms of $a, b$ and $c$.

I'm not sure I've written that correctly: I'm sure I'll have to re-write that more clearly. Anyway, the title should be somewhat clear as to what I'm after.

Example 1:

$$$$
Example 2:

I think we should also add the constraint that $ f'(0) \geq 0$, because if we then reflect everything the functions in the x-axis, we get all the cases where $f'(0) < 0$, so we don't lose any generality by assuming $ f'(0) \geq 0$.

Anyway, I'd be very surprised if this wasn't some sort of duplicate as it's such a natural optimisation question: I just think I'm not using the right lingo.

答:
Not a complete answer, but some lower bounds. As Alex already pointed out, it suffices to consider the case $a=0$. I will also denote the sought infimum by $I$.

From Taylor's formula at $x=0$ we have
$$
 0 = f(1) = f(0) + f'(0) + \frac 12 f''(\xi) = b + \frac 12 f''(\xi)
$$
for some $\xi \in (0, 1)$ and therefore $I \ge 2|b|$. Similarly, Taylor's formula at $x=1$ gives $I \ge 2|c|$, so that
$$
 \boxed{I \ge 2 \max(|b|, |c|) \, .}
$$

If $b$ and $c$ have the same sign then a better bound is possible. Without loss of generality assume that $b, c > 0$. With $M = \max |f''(x)| $ we have, again using Taylor's formula
$$
  bx - \frac M2 x^2 \le
 f(x) \le c(x-1) + \frac M2 (x-1)^2
$$
which gives
$$
 M \ge 2 \frac{bx + c(1-x)}{x^2 + (1-x)^2}
$$
for all $x \in (0, 1)$. Setting $x = b/(b+c)$ gives $M \ge 2(b+c)$, so that
$$
 \boxed{I \ge 2|b+c| \text{ if } bc > 0 \, .}
$$


问:
Find last three nonzero digits of $1^1 \cdot 2^2 \cdot 3^3 \cdot ... \cdot 25^{25}$

Find the last three nonzero digits of $1^1 \cdot 2^2 \cdot  3^3 \cdot  ... \cdot  25^{25}$.

I had been sitting with this problem whole day now, what I had tried so far:

Dividing the product by $10^{100}$ (since there are $5^{100}$ (taking fives from factors as well)). And playing with Chinese remainder theorem.

$1^1 \cdot  2^2 \cdot  3^3   \dots   25^{25} \equiv 0 \pmod{8}$

But I don't see any way apart from brute forcing the modulus $125$.

I had also noticed that the product can be rewritten as: $$ \frac{25!}{0!}\cdot  \frac{25!}{1!} \cdot  \frac{25!}{2!} \cdot  \frac{25!}{3!} \cdot  \cdot  \cdot  \frac{25!}{24!}$$
But I can't seem to get any useful information from that.

Body:
We can use a $\sharp$ algorithm  that 'digests'

$\quad \displaystyle \prod_{k=1}^{25}k^k$

from right to left.

$\; 25^{25} \cdot 1 \, \sharp \, 24^{24} \cdot 25^{25} \cdot 1 \, \sharp \,  (24\cdot 25)^{24} \cdot (25 \cdot 1)  \, \sharp$

$\; 6^{24} \cdot 25 \, \sharp \, 23^{23} \cdot 6^{24} \cdot 25 \, \sharp \,  (23\cdot 6)^{23} \cdot (6 \cdot 25)  \, \sharp$

$\; 138^{23} \cdot 15 \, \sharp \, 22^{22} \cdot 138^{23} \cdot 15 \, \sharp \,  (22\cdot 138)^{22} \cdot (138 \cdot 15)  \, \sharp$

$\; 36^{22} \cdot 207 \, \sharp \, 21^{21} \cdot 36^{22} \cdot 207 \, \sharp \,  (21\cdot 36)^{21} \cdot (36 \cdot 207)  \, \sharp$

$\; 756^{21} \cdot 452 \, \sharp \, 20^{20} \cdot 756^{21} \cdot 452 \, \sharp \,  (20\cdot 756)^{20} \cdot (756 \cdot 452)  \, \sharp$

$\; 512^{20} \cdot 712 \, \sharp \, 19^{19} \cdot 512^{20} \cdot 712 \, \sharp \,  (19\cdot 512)^{19} \cdot (512 \cdot 712)  \, \sharp$

To speed it up we'll skip the middle step.

$\; 728^{19} \cdot 544 \, \sharp \,  (18\cdot 728)^{18} \cdot (728 \cdot 544)  \, \sharp$

$\; 104^{18} \cdot 32 \, \sharp \,  (17\cdot 104)^{17} \cdot (104 \cdot 32)  \, \sharp$

$\; 768^{17} \cdot 328 \, \sharp \,  (16\cdot 768)^{16} \cdot (768 \cdot 328)  \, \sharp$

$\; 288^{16} \cdot 904 \, \sharp \,  (15\cdot 288)^{15} \cdot (288 \cdot 904)  \, \sharp$

$\; 432^{15} \cdot 352 \, \sharp \,  (14\cdot 432)^{14} \cdot (432 \cdot 352)  \, \sharp$

$\; 48^{14} \cdot 64 \, \sharp \,  (13\cdot 48)^{13} \cdot (48 \cdot 64)  \, \sharp$

$\; 624^{13} \cdot 72 \, \sharp \,  (12\cdot 624)^{12} \cdot (624 \cdot 72)  \, \sharp$

$\; 488^{12} \cdot 928 \, \sharp \,  (11\cdot 488)^{11} \cdot (488 \cdot 928)  \, \sharp$

$\; 368^{11} \cdot 864 \, \sharp \,  (10\cdot 368)^{10} \cdot (368 \cdot 864)  \, \sharp$

$\; 368^{10} \cdot 952 \, \sharp \,  (9\cdot 368)^{9} \cdot (368 \cdot 952)  \, \sharp$

$\; 312^{9} \cdot 336 \, \sharp \,  (8\cdot 312)^{8} \cdot (312 \cdot 336)  \, \sharp$

$\; 496^{8} \cdot 832 \, \sharp \,  (7\cdot 496)^{7} \cdot (496 \cdot 832)  \, \sharp$

$\; 472^{7} \cdot 672 \, \sharp \,  (6\cdot 472)^{6} \cdot (472 \cdot 672)  \, \sharp$

$\; 832^{6} \cdot 184 \, \sharp \,  (5\cdot 832)^{5} \cdot (832 \cdot 184)  \, \sharp$

$\;416^{5} \cdot 88 \, \sharp \,  (4\cdot 416)^{4} \cdot (416 \cdot 88)  \, \sharp$

$\;664^{4} \cdot 608 \, \sharp \,  (3\cdot 664)^{3} \cdot (664 \cdot 608)  \, \sharp$

$\;992^{3} \cdot 712 \, \sharp \,  (2\cdot 992)^{2} \cdot (992 \cdot 712)  \, \sharp$

$\;984^{2} \cdot 304 \, \sharp \,  (1\cdot 984)^{1} \cdot (984 \cdot 304)  \, \sharp$

$\;984^{1} \cdot 136$ $\quad \sharp \text{ has finished executing.}$

Final calculation,

$\quad 984 \cdot 136 \equiv 824 \pmod{1000}$


问:
Exercise on Radon–Nikodym derivative
Consider the measure space $(\mathbb{N}, P(\mathbb{N}))$  and two measures $\mu$ and $\nu$, where $\mu$ is the counting measure and $\nu$ is defined as
$$ \nu(E) = \sum_{n \in E}(n+1)$$

Compute $d\mu/d\nu$ if its possible.

--

How would you proceed?

Thanks in advice!

Julian

答:
By definition the Radon-Nikodym derivative is a function $f$ such that
$$\nu (E)=\int _Efd\mu,$$ in the counting measure this means
$$\nu (E)=\int _Efd\mu =\sum _{n\in E}f(n),$$
so is not $f(n)=n+1.$

Edit: Taking the inverse problem, $d\mu /d\nu =\frac{1}{n+1}.$

Title: Exercise on Radon–Nikodym derivative


问:
$\int^1_0 \frac{xdx}{x^2+2x+1}$
I need some suggestion how to solve this integral.

$$\int^1_0 \frac{xdx}{x^2+2x+1}$$
I think about to do the following step :

$$\frac{1}{2}\int^1_0\frac{2x+2-2dx}{x^2+2x+1}$$$$ t=x^2+2x+1 \rightarrow 2x+2dx=dt$$
then the integral will be : $$\frac{1}{2}\int^1_0 \frac{-2dt}{t}$$
its a correct way to solve it?

Thanks!

答:
You started well, but ended it wrong. The numerator is $(2x+2)dx-2dx=dt-2dx$, in the second term I left $dx$ because the change of variables is not useful to it, so you are left with $\frac{1}{2}\int^4_0\frac{dt}{t}-\int^1_0\frac{dx}{x^2+2x+1}$.
Notice also that you also forgot to change the limits of the first integral, that is now over $t$, that is why it ranges from 0 to 4 instead than from 0 to 1.
The second integral is easy, but let me know if you cannot solve it (hint: is equal to $\int^2_1\frac{dt}{t^2}$)</p>

Title: $\int^1_0 \frac{xdx}{x^2+2x+1}$


问:
How many parameter-free definable subsets are there of the structure $(\mathbb{R}, *)$?
This is related to a question where I asked how many subsets of $\mathbb{R}$ are parameter-free definable in the structure $(\mathbb{R}, +)$. Now, I am asking how many subsets of $\mathbb{R}$ are parameter-free definable in $(\mathbb{R}, *)$. I believe there are only 32 of them. Is this correct?

答:
A good first step, which is almost always easier, is to count automorphism orbits. (Note that this is also how the answer to your previous question went.)

In the structure $(\mathbb{R};*)$, this is easy to do: the orbits are exactly $\{0\}$, $\{1\}$, $\{-1\}$, $\mathbb{R}_{>0}\setminus\{1\}$, and $\mathbb{R}_{<0}\setminus\{-1\}$. Checking that the latter two are in fact orbits is a bit annoying, but it's not hard. My suggestion is to first think about the orbits of the structure $(\mathbb{R};+)$, which is isomorphic to $(\mathbb{R}_{>0};*)$ and is easier to think about.

This means that there are exactly $32$ automorphism-closed subsets of $(\mathbb{R};*)$. Since every parameter-freely-definable set is automorphism-closed, this gives an upper bound for the OP; it's easy to check that in fact each of the five orbits above (and hence each automorphism-closed set) is parameter-freely definable, so we're done.


问:
Continuity and sequences , with only existence of one sequence
In a true/false exercise, I saw the following proposition: if there is ($x_n$) which converges to x and such that f($x_n$) converges to f(x) when n tends towards infinity then f is continuous in x. I think this proposition is false since we have the existence of a single sequence and not the respected property for any sequence but I can't find a counter-example. Could you help me, please?

答: Consider $f(x) = 0$ if $x \neq 0$ and $f(0)=1$. Of course $f$ is not continuous at $x=0$, but for the constant sequence $x_n = 0$, you have $f(x_n)=1$ which converges to $f(0)=1$.


问:
$1^n +2^n + \cdots +(p-1)^n \mod p =$?
Calculate for every positive integer $n$ and for every prime $p$ the expression
$$1^n +2^n + \cdots +(p-1)^n \mod p$$

I need your help for this. I don't know what to do, but I'll show you what I know.

Wilson's theorem

The identity $X^p-X = \prod_{a \in \mathbb{F}_p-1}(X-a)$

$\exists a \in \mathbb{Z}, \ a^2 \equiv -1 \mod p \qquad \iff \qquad p \equiv 1 \mod 4$

Now can you please provide me a hint?

答:
I don't see the simple answer plainly stated so I will state it here. For $p=2$, the only term in the sum is $1^n$ which is identified in a previous answer as $\equiv 1 \bmod 2$, but for consistency with the larger scheme of things, might better be stated as $\equiv -1 \bmod 2$

For odd $p$ and odd $n$, there are an even number (i.e., $p-1$) of addends, and the sum may be arranged as $1^n+(-1)^n+2^n+(-2)^n+\dots \equiv 0 \bmod p$

If $n$ is even, and if $(p-1)\mid n$, then every addend is a $(p-1)$ power of an integer, so every addend is $\equiv 1 \bmod p$ and the $(p-1)$ addends sum to $(p-1)\equiv -1 \bmod p$

If $n$ is even, and $p\equiv 1 \bmod 4$, and if $(p-1)\not \mid n$, then since quadratic residues in this case come in pairs $a, (p-a)$, the sum is $\equiv 0 \mod p$

If $n$ is even, and $p\equiv 3 \bmod 4$, and if $(p-1)\not \mid n$, I am without a proof, but empirically for several small primes, the sum is $\equiv 0 \mod p$

Overall, if $(p-1)\mid n$, the sum $\equiv -1 \bmod p$; otherwise, the sum $\equiv 0 \bmod p$

Title: $1^n +2^n + \cdots +(p-1)^n \mod p =$?


问:
Sum of all possible valuse of $\frac{a}{b}+\frac{c}{d}$?
If a,b,c,d are real numbers and $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=17$ and $\frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b}=20$, then find the sum of all possible valuse of $\frac{a}{b}$+$\frac{c}{d}$ ?

I tried this problem for a while but made no progress. I don't know how $\frac{a}{b}+\frac{c}{d}$ can take only certain values. The answer was given to be $17$. Can someone help me with this?

答:
Hint: Note that
$$
\frac ac + \frac ca + \frac bd + \frac db = 20 \implies\\
\frac ab \cdot \frac bc + \frac cd \cdot \frac da + \frac bc \cdot \frac cd + \frac da \cdot \frac ab = 20.
$$
With that in mind, compare the expanded sums
$$
\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\right)^2, \quad
\left(\frac{a}{b}-\frac{b}{c}+\frac{c}{d}-\frac{d}{a}\right)^2.
$$


答:
\begin{align}
  17 &= \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \\
  &= \frac{a}{b}+\frac{c}{d}+\frac{b}{c}+\frac{d}{a} \\
  &= \color{red}{\frac{ad+bc}{bd}}+\color{blue}{\frac{ab+cd}{ac}} \\
  &= \color{red}{x}+\color{blue}{y} \\
  20 &= \frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b} \\
  &= \frac{a}{c}+\frac{b}{d}+\frac{d}{b}+\frac{c}{a} \\
  &= (ad+bc) \left( \frac{1}{cd}+\frac{1}{ab} \right) \\
  &= \frac{(ad+bc)(ab+cd)}{abcd} \\
  &= \color{red}{\frac{ad+bc}{bd}} \times \color{blue}{\frac{ab+cd}{ac}} \\
  &= \color{red}{x} \color{blue}{y} \\
\end{align}

 Now, $$20=x(17-x) \implies x^2-17x+20=0$$
 Note that $x$ and $y$ are conjugate pair and also symmetric in roles.

Title: Sum of all possible valuse of $\frac{a}{b}+\frac{c}{d}$?

Body:
If a,b,c,d are real numbers and $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=17$ and $\frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b}=20$, then find the sum of all possible valuse of $\frac{a}{b}$+$\frac{c}{d}$ ?

I tried this problem for a while but made no progress. I don't know how $\frac{a}{b}+\frac{c}{d}$ can take only certain values. The answer was given to be $17$. Can someone help me with this?


问:
Reducibility of $x^{2n} + x^{2n-2} + \cdots + x^{2} + 1$
Just for fun I am experimenting with irreducibility of certain polynomials over the integers. Since $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$, I thought perhaps $x^6+x^4+x^2+1$ is also reducible. Indeed:
$$x^6+x^4+x^2+1=(x^2+1)(x^4+1)$$
Let $f_n(x)=x^{2n}+x^{2n-2}+\cdots + x^2+1$. Using Macaulay2 (powerful software package) I checked that:
$$f_4(x) = (x^4-x^3+x^2-x+1)(x^4+x^3+x^2+x+1)$$
 We get that
$$ f_5(x)=(x^2+1)(x^2-x+1)(x^2+x+1)(x^4-x^2+1)$$
The polynomial is reducible for $n=6, 7, 8, 9$ as far as I checked. I suspect that $f_n(x)$ is reducible over integers for all $n\ge 2$. Is this true?

Thanks!

答:
I am a bit surprised that the answers given so far that invoke cyclotomic polynomials don't go all the way to exhibit a factorisation of $1+x^2+\cdots+x^{2n}$. Using $x^m-1=\prod_{d\mid m}\Phi_d(x)$ one gets
$$
 1+x^2+\cdots+x^{2n}=\frac{x^{2n+2}-1}{x^2-1}=\prod_{\substack{d\mid 2(n+1) \\ d\notin\{1,2\}}}\Phi_d(x),
$$
and since cyclotomic polynomials are irreducible over the rational numbers, this is precisely the factorisation of your polynomial in $\Bbb Q[x]$. The product is empty for $n=0$, has a single factor $\Phi_4(x)=x^2+1$ for $n=1$; in all other cases it has at least two factors, namely the polynomials $\Phi_{n+1}(x)$ and $\Phi_{2(n+1)}(x)$. So in particular it is always reducible for $n\geq2$.</p>


问:
Determining the change in $P(t)$ over the infinitesimal time $dt$
A certain security has a price given by the following stochastic process: $$P(t) = S(t)e^{(r-q)r}, \hspace{20 pt} 0\le t \le T, \tau = T - t$$ where $S(t)$ is the price of a security following geometric Brownian motion. Can someone please help me determine the change in $P(t)$ over the infinitesimal time $dt.$ Also, is $P(t)$ a geometric Brownian motion?

答:
Geometric Brownian motion is given as a differential equation.
See here.

You can differentiate the function $P$ to get the desired $\frac{dP}{dt}$. You get $S'=\frac{dS}{dt}$ from the differential equation defining $S$.


问:
Using l'Hopital's Rule on two trigonometric functions
I have to use the l'Hopital's Rule to find the limit of:
$$\lim_{x \to \pi/2} \frac{\cos(x)}{1 - \sin(x)}$$

If I plug in $\pi/2$, I can see that this is a $0/0$ case, so I must use the rule.

The derivative of the numerator and the derivative of the denominator would come out to be this I believe:

$$\lim_{x \to \pi/2} \frac{-\sin(x)}{-\cos(x)}$$

which is just $\tan(x)$

if I plug in $\pi/2$, to the derivative, I get $1/0$, which is undefined. However the correct answer seems to be $-\infty$ and I have no idea why

答: In the context of limits $\frac{1}{0}=\infty$, however the limit should be $x \to (\pi/2)^{+}$ to be $-\infty$; written like this the limit doesn't exist because it is different if you approach to $\pi/2$ from left or from right.


问:
how to prove helly property of a tree
how can we prove by induction that a set of subtrees satisfies the Helly property of a tree T? i.e
if a set of subtrees of T has the property that two of these subtrees have a common non-empty intersection, then the intersection of all the subtrees is also not empty

I have drawn out some trees to prove that this is true but I'm not sure how to formally write out the proof

答:
First show that trivially it’s true if one of the subtrees has just one vertex. Then prove the following result by induction on the number of vertices in $T$:

Let $T$ be a tree, and let $\mathscr{T}$ be a finite family of subtrees of $T$ such that each $S\in\mathscr{T}$ has at least two vertices, and every pair of trees in $\mathscr{T}$ has non-empty intersection; then $\bigcap\mathscr{T}\ne\varnothing$.

Suppose that this is true for all trees with at most $n$ vertices, and let $T$ be a tree with $n+1$ vertices. Let $\mathscr{T}$ be a family of subtrees of $\mathscr{T}$ satisfying the conditions of the theorem. Let $v_0$ be a pendant vertex of $T$, let $v_1$ be its unique neighbor in $T$, and let $T'=T-v_0$. Let $\mathscr{T}'=\{S-v_0:S\in\mathscr{T}\}$.

Show that if $S_0',S_1'\in\mathscr{T}'$, then $S_0'\cap S_1'\ne\varnothing$.

Now apply the induction hypothesis to $T'$ and the family $\mathscr{T}'$ to conclude that $\bigcap\mathscr{T}'\ne\varnothing$ and therefore $\bigcap\mathscr{T}\ne\varnothing$.

Title: how to prove helly property of a tree


问:
Problem with $\log(\exp(z)) \neq z$ and solving equations
In the real numbers, we always have that $\log(\exp(x)) = x$ for all $x\in\mathbb{R}$. However, as How to prove $\log(\exp(z))\neq z$ if $z\in \mathbb{C}$ shows, we can find some $z\in\mathbb{C}$ such that $\log(\exp(z))\neq z$ depending on the branch of $\log$ that we use. So if I have an equation $f(z) = \exp(g(z))$, and I want to isolate $g(z)$, I would just take logarithms and be done if I were working in $\mathbb{R}$. But if I am working in $\mathbb{C}$, I can't be assured that $\log(\exp(g(z)) = g(z)$, so how could I accomplish isolating $g(z)$? I assume that I would have to choose a specific branch cut to accomplish this but am not certain which one to choose.

NOTE: Assume that I have no prior knowledge of what values $f(z)$ takes on.

答:
$$\log f(z) = \log |f(z)| + i \arg f(z)$$

$$\log |f(z)| = \log \exp( \Re(g(z))) = \Re(g(z)).$$

$$\arg f(z) = \arg \exp(g(z)) = \arg \exp(i \Im g(z)).$$

Yes, you'll need to define your branch to compute the argument, and then

$$ \log f(z) = \Re(g(z)) + i \arg \exp(i \Im g(z)).$$

$$ \log f(z) = \Re(g(z)) + i  (\Im g(z)+ 2k\pi )\quad (k \text{ depends on the branch}).$$

For the example you allude to:

$$g(z) = z = x+ i y,$$
$$\log f(z)  = x + i \arg \exp( i y) = x + i (y + 2k\pi), \quad (k \text{ depends on the branch}).$$


问:
Computing a Fourier transform of $t \mapsto e^{t^2/2}\frac{d^k}{dt^k}e^{-t^2}$
Consider a real-valued function $h_k(t)$ defined by
$$
t \mapsto \exp\left(\,{t^{2} \over 2}\right)\,
\frac{\mathrm{d}^{k}}{\mathrm{d}t^{k}}e^{-t^2}\quad
\mbox{where}\quad k \in \mathbb{N}_0.
$$
I need to show that
$$\mathcal{F}h_k = (-i)^k\sqrt{2\pi}h_k$$
where $\mathcal{F}$ is a Fourier transform, i.e $\mathcal{F}: L^1(\mathbb{R}) \to L^{\infty}(\mathbb{R})$ is defined by $f(x) \mapsto \hat{f}(\xi) = \int_{\mathbb{R}}f(x)e^{-ix\xi}\ dx$. First, as a hint, I need to show that $h_{k+1}(t) = h^{\prime}_k(t) -th_k(t)$ for all $t \in \mathbb{R}$. I manage to show this hint, but the problem is, how is this helpful. There is $h_{k+1}$ in the equation, but I only want $\mathcal{F}h_k$. The operator $\mathcal{F}$ is linear, so
$$\mathcal{F}h_{k+1} = \mathcal{F}h^{\prime}_k - \mathcal{F}(th_k).$$
I know that $\mathcal{F}h^{\prime}_k(x) = \widehat{h^{\prime}_k}(\xi) = i\xi\widehat{h_k}(\xi)$ and $-\mathcal{F}(th_k) = \widehat{-th_k}(\xi) = -i(\widehat{h_k})^{\prime}(\xi)$. So,
$$\widehat{h_{k+1}}(\xi) = i\xi\widehat{h_k}(\xi)-i(\widehat{h_k})^{\prime}(\xi).$$
I have no idea what to do next. If anyone has an idea, could you please share your idea on how to solve this problem?

答:
First of all see how it works with the generating function of the Hermite functions

Your exercice is saying that if $\hat{h} = ch$ then $\widehat{ h'-th} = it \hat{h}-i \hat{h}' = it c h- ich'$


问:
Show that a linear transformation is surjective if and only if it sends generating sets into generating sets.
I'm struglying to answer this question, I think the questions is kind ambiguos since it doesn't give from where to where is $T$

Can someone explain me?

答:
Let V and W be vector spaces and $T:V\rightarrow W$ be a linear map

Assume $T$ sends generator set $\{v_1,...,v_n\}$  to another generator set $\{w_1,...,w_m\}$ and $n \geq m$ so we can say $T(v_i)=w_i$

Let $w\in W$ then there exists real numbers $c_i$ such that

$w=c_1w_1+...+c_mw_m=c_1T(v_1)+...+c_mT(v_m)=T(c_1v_1)+...+T(c_mv_m)=T(c_1v_1+...+c_mv_m)$

that means for all $w\in W$ there exists and $v \in V$ such that $T(v)=w$ so T is
surjective

Assume T is surjective then for all $w \in W$ there can be found a $v\in V$ such that

$T(v)=w$ $\Rightarrow$ $T(c_1v_1+...+c_nv_n)=c_1T(v_1)+...+c_nT(v_n)=w$

which means $\{T(v_1),...,T(v_n)\}$ generates all $w \in W$ so T sends a generator set to another generator set


问:
Show that the ‖‖$u$+‖‖$v$ vector is perpendicular to the ‖‖$u$−‖‖$v$ vector.
Show that the ‖‖$u$+‖‖$v$ vector is perpendicular to the ‖‖$u$−‖‖$v$ vector.

What should I do here? If it's perpendicular, the dot product of the two vectors should be 0. But I could not use this.

答:
There is another way, if you insist. Recall that $x \perp y$ if and only if $\|x+y\|^2 = \|x\|^2 + \|y\|^2$ (the Pythagorean theorem and its converse).

Applying this to $x = \|v\|u+\|u\|v$ and $y = \|v\|u-\|u\|v$, the parallelogram law gives
\begin{align*}
\big\|\|v\|u+\|u\|v\big\|^2 + \big\|\|v\|u-\|u\|v\big\|^2 &= 2(\|\|v\|u\|^2+\|\|u\|v\|^2) \\
&= 4\|u\|^2\|v\|^2 \\
&= \big\|2\|v\|u\big\|^2\\
&=\big\|(\|v\|u+\|u\|v)+(\|v\|u-\|u\|v)\big\|^2
\end{align*}
so $(\|v\|u+\|u\|v) \perp (\|v\|u-\|u\|v)$.


问:
Why is a geometric theory called “geometric”?
In topos theory, the notion of “being geometric” often comes up. Some examples are: geometric morphisms, geometric logic, and geometric theories. For instance, here's a quote from Steve Vickers' Locales and Toposes as Spaces:

... the logic is not at all ordinary classical logic. It is an infinitary
positive logic known as geometric logic.

I have a vague intuition on this notion of geometricity: in locale theory one sees open sets as observable properties, and my understanding of geometricity is that when we categorify from opens to sheaves, the notion of observability generalises to geometricity. In this context, “geometric morphisms” make sense as they respect the geometric structure as a continuous function respects the observability structure. However, I don't quite understand the use of the adjective “geometric” for this.

How do I connect the intuition of observability with the notion of geometricity? If this approach doesn't make much sense, what is a good way to think about geometricity?

PS: I am aware that the paper of Vickers I linked to essentially aims to explain the spatial intuition in detail. However, it seems to take for granted the notion of being “geometric” without a familiarity with which I find hard to follow the paper.

答: To summarize the comments: geometric logic constitutes the logic, models of whose theories are preserved by geometric morphisms between topoi. Geometric morphisms are those appropriate to toposes viewed as generalized spaces, for instance, identifying the topos of sheaves on a topological space, or on a locale, with the space itself. Historically, toposes were first introduced by Grothendieck's school to model generalized algebro-geometric spaces, which explains why these morphisms are called geometric, rather than topological.


问:
Second degree linear derivate with exponential function, which general solution?
Suppose $y''-4y=xe^{2x}$.

Solution of the homogenous equation is $y_H(x) = C_{1} e^{2x} + C_{2} e^{-2}$, after solving the characteristic equation with the guess $y=e^{rx}$.

Now my instructor insists that the one solution is either

$y(x) = Ax^{2} e^{2x} + Bxe^{2x}$

or

$y(x) = Ax^{2}e^{2x} + Bxe^{2x}+Ce^{2x}$

...and then I get the solution $y(x) = y_{H}(x) +\frac{1}{8}x^{2}e^{2x}-\frac{1}{16}xe^{2x}$

but I cannot understand the choices. Why do just the forms work? I have tried many different forms there and get lost/wasted a lot time -- and my instructor just says that it is a "guess" and smiles. How can I find the one solution without guessing? For one solution, things such as $y(x)=e^{2x}$ or $y(x)=xe^{2x}$ work so I am a bit lost why the more complex form is "the solution" and why it provides all solutions.

For example, the case $y=e^{2x}$. We get $y'=2e^{2x}$ and $y''=4e^{2x}$ so

$y''-4y = 4e^{2x} - 4 e^{2x} = 0$

so $RHS = xe^{2x} = 0$ so $x=0$, a solution?! If not why? Ok it is naive, it does not contain all cases but how can I know that some form contains all?

答:
Perhaps it would help to see the general method.

You want to solve an equation $Ly=x^n e^{\lambda x}$, where $L$ is a constant linear differential operator - in your case $Ly=y''-4y$. Let us solve $Ly=e^{(\lambda+\epsilon) x}$ instead: if we substitute $y=c(\epsilon)e^{(\lambda+\epsilon) x}$, we get $c(\epsilon)p(\lambda+\epsilon)=1$ (where $p$ is the characteristic polynomial of $L$, in your case $p(t)=t^2-4$), i.e. $c(\epsilon)=1/p(\lambda+\epsilon)$.
We now expand
$$L\frac{e^{(\lambda+\epsilon) x}}{p(\lambda+\epsilon)}=e^{(\lambda+\epsilon) x}$$
to a power series in $\epsilon$, look at the term at $\epsilon^n$, and get a solution of $Ly=x^n e^{\lambda x}/n!$.

In your case $\lambda=2$, $p(t)=t^2-4$, and $c(\epsilon)=1/(\epsilon(4+\epsilon))$. We thus get

$$L\bigl((\epsilon^{-1}4^{-1}+(4^{-1}x-4^{-2})+\epsilon(4^{-1}x^2/2-4^{-2}x+4^{-3}) +\dots )e^{2 x}\bigr)=$$
$$=(1+\epsilon x+\epsilon^2 x^2/2 +\dots)e^{2 x}.$$
Looking at the term at $\epsilon$, we get
$$L((4^{-1}x^2/2-4^{-2}x+4^{-3})e^{2 x})=xe^{2 x},$$
i.e. $y=(4^{-1}x^2/2-4^{-2}x+4^{-3})e^{2 x}$ is a solution of your equation (we can drop $4^{-3}e^{2 x}$ from the solution, as it solves the homogeneous equation).


问:
How do you use the gradient to find $n$?
$$z = \sqrt{1 - x^2 - y^2}$$

$$W = x^2 + y^2 + z^2 - 1$$$$\overrightarrow{N} = \nabla W   = \langle 2x,2y,2z \rangle$$

$$\overrightarrow{n} = \frac{\overrightarrow{N}}  {\|\overrightarrow{N}\|} = \frac{\langle x,y,z \rangle}{\sqrt{x^2 + y^2 + z^2}} = \langle x,y,z \rangle$$

My question is how did it come out to be ? And shouldn't you be plugging in $\langle 2x,2y,2z \rangle$ in the denominator for $x,y,z$ inside the sqrt, so shouldn't it be $\sqrt{(2x)^2 + (2y)^2 + (2z)^2}$? And then shouldn't it equal $$\left \langle \frac{x}{\sqrt{(2x)^2 + (2y)^2 + (2z)^2}}, \frac{y}{\sqrt{(2x)^2 + (2y)^2 + (2z)^2}}, \frac{z}{\sqrt{(2x)^2 + (2y)^2 + (2z)^2}} \right \rangle $$ and not just $\langle x,y,z \rangle$
?

The original question was evaluate integral of C of
$$$$$$\overrightarrow{F} *d \overrightarrow{r} $$ where $$$$$$\overrightarrow{F}(x,y,z) = <e^(-x^2)+sinx, e^(-y^2)-2yz, x^2+ e^(-z^2)=2z> $$ and C is the boundary of the surface $$z = \sqrt{1 - x^2 - y^2}$$ oriented Counterclockwise.

答:
It would be a good idea to note what you mean by$\overrightarrow{n}$.
Assuming that you meant the direction orthogonal to $W$, then your claim sounds valid.
The norm of $\overrightarrow{N}$ is not always 1, so it shouldn't simplify like that all the time.
The only case where $\overrightarrow{N} \over ||\overrightarrow{N}||$ equals $\overrightarrow{N}$ itself is only when $\overrightarrow{N}$ itself is a unit vector, as in $\sqrt{(2x)^2+(2y)^2+(2z)^2}=1$

My suggestion is, write down the exact problem that you are working on so that everyone can see better what you are really trying to do.


问:
Proof of formula for permutation without repeating elements?
My book wants me to find every unique word of "ABRAKADABRA" and since I know the formula:

Factorial of the set length divided by the product of the integer values of how many time an element has repeated itself, in this case = 11!/(5!*2!*2!)

Though I realise I can't rationalise this in my head. I tried painting a choice tree of something like the word "ABBA" but did not get a better understanding.

答:
For the example AABBB:

If the letters were distinct (say, $A_1, A_2, B_1, B_2, B_3$), there would be $5!$ permutations. However, if you want the two $A$s to be indistinguishable, then you've counted words like $A B_2 B_1 A B_3$ twice: once as $A_1 B_2 B_1 A_2 B_3$ and once like $A_2 B_2 B_1 A_1 B_3$. To account for this, you divide by $2!$.

Similarly, if the $B$s are indistinguishable, you have counted words like $ABBAB$ six times: $AB_1 B_2 A B_3$, $A B_1 B_3 A B_2$, $A B_2 B_1 A B_3$, $A B_2 B_3 A B_1$, $A B_3 B_1 A B_2$, $A B_3 B_2 A B_1$. To account for this, you divide by $3!$.


答:
Welcome to MSE. For a second, lets say that you paint your letters of different color. Say $\color{red}{A}B\color{blue}{B}\color{green}{A}$ then the letters with the same symbol now look different because of their color. No? In how many ways can you arrange them? $4!$ seems the right choice because you can distinguish between every letter and the number of orderings in $n$ distinct objects is $n!$

 Now, erase the colors, notice that for a particular arrangement, you flip the colors in every possible way! So for an arrangement like $BAAB$ you actually had $B\color{red}{A}\color{green}{A}\color{blue}{B},B\color{green}{A}\color{red}{A}\color{blue}{B},\color{blue}{B}\color{red}{A}\color{green}{A}B$ and $\color{blue}{B}\color{green}{A}\color{red}{A}B.$ Notice then that if you fix a symbol you did the number of distinct letter different arrangements. By the multiplication principle, you did the product of the factorials of each individual symbol. So you have to divide by the total arrangements to get $1$ per group of those. That is, then how you get $\frac{4!}{\underbrace{2!}_{B}\cdot \underbrace{2!}_{A}}.$

Title: Proof of formula for permutation without repeating elements?

Body:
My book wants me to find every unique word of "ABRAKADABRA" and since I know the formula:

Factorial of the set length divided by the product of the integer values of how many time an element has repeated itself, in this case = 11!/(5!*2!*2!)

Though I realise I can't rationalise this in my head. I tried painting a choice tree of something like the word "ABBA" but did not get a better understanding.


问:
Show that if a linear transformation sends bases to bases, then it is bijective.
I want to show that if a linear transformation sends bases to bases, then it is bijective.

I have the idea, but can't find a way to prove this.

Can someone help me or give a me a hint?

答:
Let $\{x_1, \ldots, x_n\}$ be a basis in the domain such that $\{Ax_1, \ldots, Ax_n\}$ is a basis in the codomain.

For injectivity, we wish to prove that $A$ has trivial kernel. Assume $Ax = 0$ and write $x = \sum_{i=1}^n \alpha_ix_i$. We have
$$0 = Ax = A\left(\sum_{i=1}^n \alpha_ix_i\right) = \sum_{i=1}^n \alpha_i Ax_i$$
which implies $\alpha_1 = \cdots = \alpha_n = 0$ since $\{Ax_1, \ldots, Ax_n\}$ is linearly independent. Hence $x = 0$.

For surjectivity, any $y$ in the codomain can be written as a linear combination $y = \sum_{i=1}^n \alpha_i Ax_i$ since $\{Ax_1, \ldots, Ax_n\}$ spans the codomain. Then
$$A\left(\sum_{i=1}^n \alpha_ix_i\right) = \sum_{i=1}^n \alpha_i Ax_i = y$$
so $y$ is in the image of $A$.

Hence $A$ is bijective.


问:
Haken's Algorithm for Unknot Recognition
I'm an undergrad studying knot theory and I'm exploring the problem of unknot recognition. I believe I understand Haken's algorithm at a high level but I'm having trouble understanding why it is correct.

From my understanding, the algorithm begins: Construct a finite combinatorial triangulation $T$ of $S^3$ which contains a polygonal knot $K$ in its 1-skeleton which has diagram $K$, and also contains a good compact triangulated $3$-manifold $M_K \cong S^3 - R_K$, where $R_K$ is a regular neighborhood of $K$. Then, if there are $t$ tetrahedra in $M_K$, some vector in $\mathbb{Z}^{7t}$ corresponds to a normal surface which is a spanning disk of $K$ iff $K$ is unknotted.

Why can we guarantee that a spanning disk must be one of the vectors in $\mathbb{Z}^{7t}$? And why can we guarantee that a connected, essential disk corresponding to any vector in $\mathbb{Z}^{7t}$ is necessarily spanning?

答:
Joel Hass gave some nice lectures about these things here.  The main ideas are (1) taking a disk and manipulating it to get a new disk that is a normal surface with respect to the triangulation, therefore there is a spanning disk among the vectors, (2) finding a collection of fundamental vectors whose positive linear combinations span all spanning surfaces, and (3) one of these fundamental vectors will be the disk if there is one.

For your second question, it's because of the gluing conditions and boundary conditions, if I remember correctly.  You need the normal coordinates for each tetrahedron to match up along faces, so then surfaces always continue from one tetrahedron to the next, and you need the normal coordinates to match a specific pattern in the boundary of the knot exterior (and you need to run the algorithm for each possible pattern -- though please correct me if I've misremembered).


问:
Oscillating solutions of linear o.d.e $y'''+y'=2+\sin x$
Is solve the differential equation $$y'''(x)+y'(x)=2+\sin x, \ x\geq 0$$ and I get that the general soloution has the form
$y(x)=c_1+c_2\cos x+c_3\sin x+ 2x-\frac{1}{2}x\sin x,\ x\geq 0$.
I see that the part soloution $f(x)=2x-\frac{1}{2}x\sin x$ is not oscillating, because $f(x)=0 \iff x=0$. However how can I proof that this differential equation has or hasn't oscillating solutions?

答:
I don't know if this is what you wanted.
Since
$$ c_2\cos x+c_3\sin x\ge-\sqrt{c_2^2+c_3^2},2x-\frac12 x\sin x\ge 2x-\frac1 2x=\frac32x $$
for $x\ge0$, so
$$ y(x)=c_1\cos x+c_2\sin x+2x-\frac12 x\sin x\ge\frac3 2x+c_1-\sqrt{c_2^2+c_3^2}. $$
Thus for $x>|c_1-\sqrt{c_2^2+c_3^2}|$, $y(x)>\frac12x>0$; namely $y(x)$ is not oscillating.


问:
Proving that two measures are equal
Let $\lambda$ be the lebesgue measure on  $\mathbb{R}$, Let $\mathcal{M}(\mathbb{R}^n)$ be the system of lebesgue measurable sets of $\mathbb{R}^n$ and $t>0$.

Show that $\mu: \mathcal{M}(\mathbb{R}^n)\to [0,\infty], \mu(A):=t^{-n}\lambda(tA)$ defines a measure and that $\lambda(A)=\mu(A)$ for all $A\in \mathcal{M}(\mathbb{R}^n)$

My work:

We have:$$\mu(\emptyset)=t^{-n}\cdot \lambda(t\cdot \emptyset)=t^{-n}\cdot \lambda(\emptyset)=t^{-n}\cdot 0=0 \quad \checkmark$$ and $$\mu\left(\bigcup\limits_{n=1}^{\infty}A_n\right)=t^{-n}\lambda \left(\bigcup\limits_{n=1}^{\infty}tA_n\right)\\ =t^{-n}\sum \limits_{n=1}^{\infty}\lambda(tA_n)=t^{-n}\sum \limits_{n=1}^{\infty}t^n\cdot t^{-n}\lambda(tA_n) =t^{-n}\sum \limits_{n=1}^{\infty}t^n\mu(A_n)$$  Hence:$$\mu\left(\bigcup\limits_{n=1}^{\infty}A_n\right)=\sum \limits_{n=1}^{\infty}\mu(A_n) \quad \checkmark  $$

All in all, $\mu$ is a measure. But how can I show the second part, namely that $\lambda(A)=\mu(A)$ for all $A\in \mathcal{M}(\mathbb{R}^n)$?

答:
The Lebesgue measure is defined by the restriction of the lebesgue outer measure to lebesgue measurable sets. The outer measure is defined by

$$ \lambda^*(A) = \inf \{ \sum_{j} \text{Vol} (I_j)  \; \vert  \; A \subset \bigcup_{j \in \mathbb N} I_j \}$$

where the $I_j$'s are compact rectangles which cover $A$.

Let $A$ be a lebesgue measurable set and $t > 0$. We start by showing that $tA$ is lebesgue measurable. To do this we establish the fact that
$$ \lambda^*(tA) = t^n \lambda^*(A) $$

This follows by definition of the outer measure

\begin{align*}
\lambda^*(tA) &= \inf \{ \sum_{j} \text{Vol} (I_j)  \; \vert  \; tA \subset \bigcup_{j \in \mathbb N} I_j \} \\
&= \inf \{ \sum_{j} \text{Vol} (I_j)  \; \vert  \; A \subset \bigcup_{j \in \mathbb N} \frac{1}{t}I_j \} \\
&= \inf \{ t^n \sum_{j} t^{-n}\text{Vol} (I_j)  \; \vert  \; A \subset \bigcup_{j \in \mathbb N} \frac{1}{t}I_j \} \\
&= \inf \{ t^n \sum_{j}\text{Vol} (\frac 1 tI_j)  \; \vert  \; A \subset \bigcup_{j \in \mathbb N} \frac{1}{t}I_j \} \\
&= t^n \inf \{\sum_{j}\text{Vol} (\frac 1 tI_j)  \; \vert  \; A \subset \bigcup_{j \in \mathbb N} \frac{1}{t}I_j \} \\
&= t^n \inf \{\sum_{j}\text{Vol} (\tilde{I}_j)  \; \vert  \; A \subset \bigcup_{j \in \mathbb N} \tilde{I}_j\} \\
&= t^n \lambda^*(A).
\end{align*}

We now show that $tA$ is lebesgue measurable if $A$ is measurable. By definition we must show that for any $E \subset \mathbb R^n$

$$ \lambda^*(E) = \lambda^*(E \cap tA) + \lambda^*(E \cap (tA)^c)$$

Since $(tA)^c = tA^c$ we have

\begin{align*}
\lambda^*(E \cap tA) + \lambda^*(E \cap (tA)^c) &= \lambda^*(E \cap tA) + \lambda^*(E \cap tA^c) \\
&= \lambda^*(t \frac 1 t E \cap tA) + \lambda^*(t \frac 1 t E \cap tA^c) \\
&= \lambda^*\big( t  (\frac 1 t E \cap A) \big) + \lambda^*\big( t(\frac 1 t E \cap A^c)\big) \\
&= t^n \lambda^*( \frac 1 t E \cap A) + t^n \lambda^*(\frac 1 t E \cap A^c)\\
&= t^n \big( \lambda^*( \frac 1 t E \cap A) + \lambda^*(\frac 1 t E \cap A^c)\big) \\
&= t^n \lambda^*(\frac 1 t E) = \lambda^*(E).
 \end{align*}

Therefore $tA$ is Lebesgue measurable. Since $tA$ and $A$ are both Lebesgue measurable we have by definition that

$$ \lambda^*(tA) = \lambda(tA) \quad \text{and} \quad \lambda^*(A) = \lambda(A)$$

Hence $ \lambda(tA) = t^n \lambda(A)$ which is equivalent to  $\lambda(A) = \mu(A)$.

Title: Proving that two measures are equal


问:
Help with arguing against symmetry proofs.
I am asking for the help of mathematicians here to show my colleague that her arguments are wrong. I tried my best to explain that inequalities and equations can be solved like that, but maybe my arguments were not good enough.

Some of the examples of her solutions:

Example 1. Over the real numbers, if $x+y=2$, what is the maximum value of $xy$?

Solution: By symmetry, there must be a maximum or minimum when $x=y$ so $x=y=1$ and $xy=1$. This is clearly a maximum, since if say $x=0$ and $y=2$, you get a lower value.

Example 2. Let $x, y$ and $z$ be positive real numbers such that $x + y + z = 1$. Prove that $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} \geq 27$.

Solution: By symmetry, there is a minimum at x=y=z hance all = 1/3 and the expression equals 27.

Example 3. For positive numbers $a,b,c$ prove that $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}} \leq \frac{3\sqrt{2}}{2}$?

Solution: Another of those symmetry arguments.  The LHS is a maximum for $a=b=c$ and then equals the RHS.

It is very clear, that these are not even close to be called solutions. When I (or my other collegaues) try to explain that it not that simple to show that extreme points will be at $x=y=z$ we get answers like

It's a symmetry thing.  If you imagine that say $a$ tends to 0 or infinity you see that the LHS is going to decrease.

have repeatedly said that symmetry arguments do not always work.  However, it must be obvious to any bright 18 year old that the argument does work in this particular case.

Much to my amazement, many people in this group do not understand how to solve problems using symmetry. In England, this is taught in schools.

I am not sure what will help us to explain that it does not work like that. If we will get many reponces here, I believe it will help to convince my colleagues. Also, if you are from UK or have degree from prestigous college it will be helpfull to mention that.

答:
The issue is that your colleague seems to be confusing a useful heuristic with a proof.   If the principle of why something 'must be true' by symmetry cannot be articulated and only applied ex post, then that's a red flag that it's not a proof.  One approach is to suppose your colleague is right in 2 simple separate cases, then combine them (via multiplication in this case) and show that the 'principle' breaks.

A better way to view 'symmetry arguments' when looking at inequalities involving symmetric functions like example 1, is through majorization.  The deeper reason example 1 holds is that $x_1\cdot x_2$ is symmetric and in fact a Schur concave function and the constraint $x_1+x_2=2$  ensures that $\mathbf x =\mathbf 1$ is majorized by any other candidate.  By Schur concavity this is a maximum.

Alternatively consider $\mathbf x \in \mathbb R^2_{\geq 0}$ and $\mathbf x\mapsto x_1^4 + x_2^4$ with the constraint $x_1+x_2=2$.  This is symmetric and in fact a Schur convex function.  The candidate $\mathbf x =\mathbf 1$ is a minimum since it is majorized by all other allowed choices of $\mathbf x$ and Schur convexity ensures that is a minimum.

Finally consider consider $\mathbf x \in \mathbb R^2_{\geq 0}$ and $\mathbf x\mapsto  (x_1\cdot x_2)\cdot(x_1^4 + x_2^4)$ with the constraint $x_1+x_2=2$.   This is a symmetric function but neither (Schur) concave nor convex. Being majorized by all choices of $\mathbf x$ (what your colleague calls the symmetric solution) doesn't tell you anything definitive.  Hence applying a heuristic like "It's a symmetry thing. If you imagine that say  tends to 0 or infinity you see that the LHS is going to decrease" misleads you despite the fact that this is a symmetric function.  And $\mathbf x =\mathbf 1$ is neither a maximum or a minimum, as $0$ is a clear minimum e.g. as $x_1\to 0$ but e.g. $x_1=1.5, x_2=0.5$ is much larger than the constant solution.

conclusion: your colleague's heuristic is true when dealing with symmetric functions that are Schur convex (concave). But not all functions are Shur convex (concave); understanding this is vital.  The theory of majorization is a well developed mathematical theory quite distinct from declaring things are obvious.


答:
You're correct that through L'Hopital's rule, you get

$$\lim_{x \to \frac{\pi}{2}} \tan x$$

As you say, $\tan \left(\frac{\pi}{2}\right)$ is undefined, but that doesn't mean you should stop. What you should do is look at the graph of $f(x)=\tan x$. You'll see that as $x$ approaches $\frac{\pi}{2}$ from the left, the graph goes to inifinity; as $x$ approaches $\frac{\pi}{2}$ from the right, the graph goes to negative infinity. As a result, the limit does not exist.

Title: Using l'Hopital's Rule on two trigonometric functions

Body:
I have to use the l'Hopital's Rule to find the limit of:
$$\lim_{x \to \pi/2} \frac{\cos(x)}{1 - \sin(x)}$$

If I plug in $\pi/2$, I can see that this is a $0/0$ case, so I must use the rule.

The derivative of the numerator and the derivative of the denominator would come out to be this I believe:

$$\lim_{x \to \pi/2} \frac{-\sin(x)}{-\cos(x)}$$

which is just $\tan(x)$

if I plug in $\pi/2$, to the derivative, I get $1/0$, which is undefined. However the correct answer seems to be $-\infty$ and I have no idea why


答:
I assume that we consider finite dimensional spaces to keep the notation approachable. That said, the proof can be extended to the case of arbitrary dimension without too much effort.

Suppose that $T:V \to W$. Let $\{v_1,\dots,v_n\}$ be a basis of $V$. By the information in the question, $\{T(v_1),\dots,T(v_n)\}$ is a basis of $W$. I claim that there is a unique linear transformation $S$ for which
$$
S(Tv_j) = v_j, \quad j = 1,\dots, n.
$$
In particular: because $\{T(v_1),\dots,T(v_n)\}$ is a basis, there is indeed such a linear transformation, and $S$ is uniquely specified by this condition.

With that, it suffices to note that $S \circ T$ and $T \circ S$ are identity maps. Since $T$ has an inverse, it must be bijective.

Title: Show that if a linear transformation sends bases to bases, then it is bijective.

Body:
I want to show that if a linear transformation sends bases to bases, then it is bijective.

I have the idea, but can't find a way to prove this.

Can someone help me or give a me a hint?


答:
There are $11$ positions. Choose the positions that will be $A$. There are $\dbinom{11}{5}$ ways to choose. Regardless of the positions chosen for the $A$'s, there are $6$ positions remaining. From among them, choose positions for $B$. There are $\dbinom{6}{2}$ ways to place them. Regardless of how the A's and B's are placed, there are 4 positions remaining. Choose how to place the single $D$: There are $\dbinom{4}{1}$ way to place the $D$. Then, there are $\dbinom{3}{1}$ way to place the $K$, and finally $1$ way to place the $R$'s in the remaining two positions. That gives a total number of permutations of:

$$\require{cancel} \dbinom{11}{5}\dbinom{6}{2}\dbinom{4}{1}\dbinom{3}{1} = \dfrac{11!}{5!\cancel{6!}}\cdot \dfrac{\cancel{6!}}{2!\cancel{4!}} \cdot \dfrac{\cancel{4!}}{1!\cancel{3!}}\cdot \dfrac{\cancel{3!}}{1!2!} = \dfrac{11!}{5!2!1!1!2!}$$

This method can be generalized for multiset permutations. Would you want to see a generalization, as well?


问:
$f(x) + f(-x)$ is aways even
I was having some troubles to help some friends with this question: Let $f:\mathbb{R} \to \mathbb{R}$ a function. Is true that $g(x) = f(x) + f(-x)$ is aways symmetric to the $y$ axis? If $f$ is even or odd, this is trivial, but if the function is neither even or odd? When it is a polynomial neither ever or odd, it's pretty easy to see that this is true, but there is some function continuous in $\mathbb{R}$ that this is not true? If doesn't, how to prove that it is true for every $f$? Thank you!

Title: $f(x) + f(-x)$ is aways even

Body:
$$g(x)=f(x)+f(-x)$$

$$g(-x)=f(-x)+f(x)$$


问:
Prove that a cubic function is either convex either concave along ]a,b[ ; a and b 2 of its roots
I am working on a real analysis problem that involves Rolle and the mean value theorems ...

I got to the final step that consists to prove that f, a cubic function , is either convex either concave along ]a,b[, where a and b are two consecutive roots.

I tried to prove it by calculating the expression of f''(x), but this seems unfruitful.

Could someone please provide a hint?

答: It's false. Consider, for example, $f_\alpha(x)=x^3-x+\alpha$. The second derivative is $6x$, which changes sign at $x=0$. So if $f$ has three distinct roots, which it certainly does for some values of $\alpha$, the only way for your conjecture to be true is if $x=0$ is a root of $f$. But the roots of $f$ depend on $\alpha$, so this can't possibly work.


问:
Find a $3\times 3$ matrix $P$ such that $g_{j} = \sum_{i=1}^{3}P_{ij}f_{i}$
Let $V$ be the vector space over the complex numbers of all functions from $\textbf{R}$ into $\textbf{C}$, i.e., the space of all complex-valued functions on the real line. Let $f_{1}(x) = 1$, $f_{2}(x) = e^{ix}$ and $f_{3}(x) = e^{-ix}$.

(a) Prove that $f_{1}$, $f_{2}$ and $f_{3}$ are linear independent.

(b) Let $g_{1}(x) = 1$, $g_{2}(x) = \cos(x)$ and $g_{3}(x) = \sin(x)$. Find a $3\times 3$ matrix $P$ such that
\begin{align*}
g_{j} = \sum_{i=1}^{3}P_{ij}f_{i}
\end{align*}

MY ATTEMPT

(a) The given functions are indeed linear independent. In fact, we have
\begin{align*}
& af_{1} + bf_{2} + cf_{3} = a + b\cos(x) + c\cos(x) + i(b\sin(x) - c\sin(x)) = 0
\end{align*}

If $x = 0$, then $a + b + c = 0$. If $x = \pi/2$, then $b = c$ and $a = 0$. Consequently, $a = b = c = 0$, from whence we conclude that $f_{1}$, $f_{2}$ and $f_{3}$ are linear independent.

(b) Such sought matrix $P$ is the change of basis matrix, that is to say
\begin{align*}
P = \begin{bmatrix}
1 & 0 & 0 \\
0 & \frac{1}{2} & \frac{1}{2i} \\
0 & \frac{1}{2} & -\frac{1}{2i}
\end{bmatrix}
\end{align*}

Could someone double-check if the matrix $P$ is expressed correctly?

答:
(a) Your reasoning is not sound. Scalars $a,b,c$ are complex, so $a = -1 - i$, $b = 1$, $c = i$ satisfy your two equations also. A hint is that you are on the right track—just go one step further.

(b) I got the same $P$.


问:
$\int^1_0 \frac{xdx}{x^2+2x+1}$
I need some suggestion how to solve this integral.

$$\int^1_0 \frac{xdx}{x^2+2x+1}$$
I think about to do the following step :

$$\frac{1}{2}\int^1_0\frac{2x+2-2dx}{x^2+2x+1}$$$$ t=x^2+2x+1 \rightarrow 2x+2dx=dt$$
then the integral will be : $$\frac{1}{2}\int^1_0 \frac{-2dt}{t}$$
its a correct way to solve it?

Thanks!

Title: $\int^1_0 \frac{xdx}{x^2+2x+1}$

Body:
$$\int^1_0 \frac{xdx}{x^2+2x+1}$$
$$\int^1_0 \frac{x}{(x+1)^2}\, dx$$
$$\int^1_0 \frac{x+1-1}{(x+1)^2}\, dx$$
$$\int^1_0 \frac{1}{(x+1)}-\frac{1}{(x+1)^2}\, dx$$
$$[\log(x+1)+\frac{1}{x+1}]_0^1$$
$$\log 2-\frac{1}{2}$$


问:
Multiplication of three partial derivatives giving unexpected results.
Given the equation
$PV=nRT$
if we evaluate $\frac{\partial{P}}{\partial{V}} * \frac{\partial{V}}{\partial{T}} * \frac{\partial{T}}{\partial{P}}$ through taking all three partials and multiplying, we get $-1$. Why can we not cancel the fractions and get $1$ instead?

答:
When doing partial differentials, the terms are not unique.  That is, $\partial P$ is not a unique value, and means different things depending on what it is in ratio with.  In other words, the $\partial P$ in $\frac{\partial P}{\partial V}$ is a different term than the $\partial P$ in $\frac{\partial T}{\partial P}$.

To flesh this out a little more, the total differential of the ideal gas law is:

$$P\,dV + V\,dP = nR\,dT$$

A partial differential of $P$ with respect to $V$ happens when we only allow $P$ and $V$ to change.  That means the other differential, $dT$, is equal to zero (another way for saying that $T$ is not changing).  To keep track of these, we can subscript these partials with the variables that they are changing with:

$$P\,\partial_PV + V\,\partial_V\!P = 0 \\
\partial_V\!P = -\frac{P}{V}\partial_PV$$

Now, if you have the partial differential of $T$ with respect to $P$, you are now holding $V$ constant, so $dV = 0$.  That will give you the partial:

$$ V\,\partial_T\!P = nR\,\partial_PT  \\
\partial_T\!P = \frac{nR}{V}\,\partial_PT
$$

As you can see, $\partial_T\!P$ and $\partial_V\!P$ are two very different quantities with very different formulas!


答: This is not true. For $f(x)=x(x-2)(x-3)$, then $f''(x)= 6x-10$ vanishes at $x=5/3 \in (0,2)$.

Title: Prove that a cubic function is either convex either concave along ]a,b[ ; a and b 2 of its roots

Body:
I am working on a real analysis problem that involves Rolle and the mean value theorems ...

I got to the final step that consists to prove that f, a cubic function , is either convex either concave along ]a,b[, where a and b are two consecutive roots.

I tried to prove it by calculating the expression of f''(x), but this seems unfruitful.

Could someone please provide a hint?


答: Your result is right. Your intuition is for multiplying derivatives defined with the same quantities held fixed. But in your calculation, $\partial P/\partial V$ is on a surface of constant $T$ etc.

Title: Multiplication of three partial derivatives giving unexpected results.

Body:
Given the equation
$PV=nRT$
if we evaluate $\frac{\partial{P}}{\partial{V}} * \frac{\partial{V}}{\partial{T}} * \frac{\partial{T}}{\partial{P}}$ through taking all three partials and multiplying, we get $-1$. Why can we not cancel the fractions and get $1$ instead?


问:
Proof of more general strong law of large numbers for dependent random variables
I want to show the following version of the strong law of large numbers. This is the longest proof I've ever attempted, and I feel a bit overwhelmed.

Let $X_i$ be a sequence of real-valued
random variables with $E(X_i^2)<\infty$ for all $i$. If there is a sequence $c_k\subseteq[0,\infty)$ such that $C=\sum_{k=0}^\infty c_k <\infty$ and $\operatorname{Cov}(X_i,X_j)\le c_{|i-j|}$, then
$$\lim_{n\rightarrow\infty}\frac1n\sum_{i=1}^n(X_i-E(X_i))=0$$

I have the following proof outline provided to me:

Show $V(\sum_{a+1}^bX_i)\le 2(b-a)C$ (Done)

Subsequence: Show
$$\lim_{k\rightarrow\infty}\frac1{k^2}\sum_{i=1}^{k^2}(X_i-E(X_i))=0\text { a.s.}$$

Fluctuation: Show
$$\lim_{k\rightarrow\infty}\frac1{k^2}\max\left\{\left\lvert\sum_{i=k^2+1}^m(X_i-E(X_i))\right\rvert:m\in\{k^2,...,k^2+2k\}\right\} = 0\text{ a.s.}$$

Deduce the claim

I have successfully shown (1). I'm not sure it's correct, but I think I've shown (2) like so: For sufficiently small $\delta$ we have
$$\begin{aligned}P\left(\left\lvert\lim_{k\rightarrow\infty}\frac1{k^2}\sum_{i=1}^{k^2}(X_i-E(X_i))\right\rvert > 0\right)
&= P\left(\left\lvert\lim_{k\rightarrow\infty}\frac1{k^2}\sum_{i=1}^{k^2}(X_i-E(X_i))\right\rvert \ge \delta\right)\\
&\le \lim_{k\rightarrow\infty}\frac1{k^4}\frac{V(\sum_{i=1}^{k^2}(X_i-E(X_i))}{\delta^2} = 0
\end{aligned}$$
Is this valid? I'm not sure about the way I rewrote the inequality and pulled the limit out of the variance.

I have no ideas for how to show (3), I don't know how to "break" that maximum. Also, the final step is not obvious to me, again because the max-statement confuses me.

How can I close out this proof? Does someone know a reference for a proof of this law?

答:
First of all, let us make a simplification. Set $Y_i=X_i-EX_i$ and note that $EY_i=0$ and $Cov(Y_i, Y_j)=Cov(X_i, X_j).$ The conclusion of your theorem can be recasted as $\frac{1}{n}\sum_i^n Y_i\to 0$ almost surely.

The first step in such proofs is to show a weaker result, namely, we establish that $S_n:=\frac{1}{n}\sum_i^n Y_i\to 0$ in probability. The usual tool for showing the convergence in probability is Chebyshev's inequality. In fact, once you have a quantitative estimate coming from Chebyshev's inequality, one can write an explicit sequence $k_n$ such that $S_{k_n}\to 0$ almost surely (this argument usually requires Borel-Cantelli). In your case (we will see), suffices to take $k_n=n^2.$

Once you have got the a.s. convergence along subsequence $k_n,$ you want to strengthen it along the full sequence. The argument for this part usually goes as follows. For any $m,$ let $k_{n_m}$ be such that $k_{n_m}\le m\le K_{n_{m+1}}.$ Now, you know that $|S_m|\le |S_m-S_{k_m}|+|S_{k_m}|.$ We already know that $S_{k_m}\to 0$ a.s., therefore, suffices to show that $|S_m-S_{k_m}|\to 0$ almost surely. This is the fluctuation. In simple cases, like the present, it is often handled by using Chebyshev's inequality and applying Borel-Cantelli.

Let us now see the above principles in action. You already have $V(\sum_{i=m+1}^{n}Y_i)\le 2(n-m)C.$ Use the Chebyshev's inequality to obtain $$P\left(\frac{1}{n}|\sum_{i=1}^{n}Y_i|\ge \delta\right)=\frac{2nC}{n^2\epsilon^2}=\frac{2C}{n\epsilon^2}.$$

This already tells us that $S_n\to 0$ in probability. It also tells us that if we take $k_n=n^2,$ then $P(S_{k_n}\ge \epsilon)\le \frac{2C}{n^2\epsilon^2}.$ Applying Borel-Cantelli, we obtain therefore that $S_{k_n}\to 0$ almost surely. (Note that any sequence $k_n$ such that $\sum k_n^{-1}<\infty$ would work.) This corresponds to Step 2 in your outline.

Before going to the third step, first note that $$|S_m|\le |S_m-S_{k^2}|+|S_{k^2}|$$ where $k$ is such that $k^2\le m<(k+1)^2.$

The second term namely $S_{k^2}\to 0$ almost surely. So we are concerned about the first term. Let us look at it more closely. Observe that
$$|S_m-S_{k^2}|\le |\sum_{i=1}^{k^2}Y_i(\frac{1}{k^2}-\frac{1}{m})|+\frac{1}{m}|\sum_{i=k^2+1}^{m}Y_i|\le |S_{k^2}|+\frac{1}{m}|\sum_{i=k^2+1}^{m}Y_i|.$$

Once again $S_{k^2}\to 0$ almost surely. Therefore, we now focus on the 'fluctuation' $\frac{1}{m}|\sum_{i=k^2+1}^{m}Y_i.$ Apply Chebushev's inequality to obtain $$P(\frac{1}{m}|\sum_{i=k^2+1}^{m}Y_i|>\epsilon)\le \frac{Var(\sum_{i=k^2+1}^{m}Y_i)}{m^2\epsilon^2}\le \frac{2C(m-k^2)}{m^2\epsilon^2}\le \frac{4Ck}{k^4\epsilon^2}\le \frac{C_{\epsilon}}{k^3}.$$

Now Borel-Cantelli lemma gives you that $\frac{1}{m}|\sum_{i=k^2+1}^{m}Y_i|\to 0$ almost surely. Thus completing the proof.


问:
Systems of Congruences
\begin{cases} \overline{xyz138} \equiv 0 \mod7 \\ \overline{x1y3z8} \equiv 5 \mod11 \\ \overline{138xyz} \equiv 6 \mod13 \end{cases}

I worked my way up to this:

\begin{cases} 2000x+200y+20z \equiv 2 \mod7 \\ 100000x+1000y+10z \equiv 4 \mod11 \\ 2000x+200y+20z \equiv 7 \mod 13 \end{cases}

I tried to subtract one equation from the other,  but got nothing. No idea what to do.

答:
There are some basic division tricks.  $7*13*11=1001$ so to have $xyz138\equiv 0 \pmod 7$ implies that $xyz138 \equiv xyz138 - 1001*xyz = 138-xyz \equiv 0$ so $xyz \equiv 138\pmod 5\pmod 7$.

Likewise $138xyz \equiv 138xyz - 1001*138\equiv xyz -138\equiv 6 \pmod {13}$ so $xyz \equiv 144\equiv 1 \pmod {13}$.

Both together means $xyz \equiv 5+7k \equiv 1+13j\pmod{7\times 13}$ for some $k,j$. So $xyz \equiv 40\pmod{91}$

We can do the same trick for $11$ to get $x1y-3z8 \equiv 5\pmod {11}$.  Or we can do the other trick that $x1y3z8\equiv 1+3+8 - (x+y+z)\equiv 5\pmod {11}$.

So we have $x+y+x \equiv 7\pmod 11$.

So $xyz = 40 + 91k$ for some $k$ where $x+y+z\equiv 7\pmod {11}$.

$40$ yield $4+0\equiv 4\pmod {11}$.

$131\implies 1+3+1\equiv 5\pmod {11}$.

$222\implies 2+2+2\equiv 6\pmod{11}$.

$313\implies 3+1+3\equiv 7\pmod {11}$.

So that'll do

$313138\equiv 0\pmod 7$

$138313 \equiv 6\pmod 13$

And $311338 \equiv 5\pmod{11}$

Title: Systems of Congruences


问:
Proof that an infinite set can be written as the countably infinite union of pairwise disjoint infinite subsets without using the axiom of choice.
I have the following theorem

Any infinite set can be written as the countably infinite union of pairwise disjoint infinite subsets

I have found a couple of proofs for this but I was wondering if there is some proof that doesn't use the axiom of choice (and it's equivalent statements).
Does there exist such a proof?

答: No. An amorphous set is infinite but cannot be written even as the union of two disjoint infinite sets, and it is consistent with $\mathsf{ZF}$ without choice that amorphous sets exist.


问:
What is so strong about the inner product?
Hilbert spaces possess some special properties, such as being uniformly convex and isometrically isomorphic to its dual space. Why does the assumption of an inner product lead to such strong statements, going so far as to identify all (separable) Hilbert spaces with $l_2(\mathbb{N})$? As a student, and at first glance, one would not perceive that they would differ so dramatically from the Banach spaces considered in the previous chapters with the weaker assumption of a norm.

答:
One possible answer is that out of the $L_p$ spaces, the p-norm $\|\cdot\|_p$ satisfies the Parallelogram Law:

$$2\|u\|^2+2\|v\|^2=\|u+v\|^2+\|u-v\|^2$$

only for $p=2$. It is a well-known result that by satisfying the above, one can define an inner product via:

$$(u,v) := \frac{1}{2}\left(\|u\|^2+\|v\|^2-\|u-v\|^2\right).$$

Conversely, you can show that the induced norm $\|u,u\|:=\sqrt{(u,u)}$ must satisfy the Parallelogram Law.


问:
Proof that $\frac{f(x)}{x}$ is increasing in $x$ when $f$ has increasing returns to scale.

Let $f(x):(0,\infty)\to\mathbb{R}_+$, increasing and differentiable, and
consider the ratio $g(x)=\frac{f(x)}{x}$. I want to show that if $f$
has increasing returns to scale ($f(\lambda x)>\lambda f(x)$ for any
$\lambda>1$) then $g'(x)>0$, and if $f$ has decreasing returns to
scale ($f(\lambda x)<\lambda f(x)$ for any $\lambda>1$), then
$g'(x)<0$.

This seems intuitive to me, as under increasing returns to
scale, the numerator of $g$ grows more than linearly, while under decreasing
returns to scale, it grows less than linearly. However, I'm having
trouble with the formal proof. Here's my attempt:

Let $f$ have increasing returns to scale. Then, $g(\lambda x)=\frac{f(\lambda x)}{\lambda x}>\frac{f(x)}{x}=g(x)$ for any $\lambda>1$. Consider the continuous invertible transformation defined by $\lambda x=x+h$, so that $h=(\lambda-1)x$. Then, we can write

$$g'(x)=\lim_{h\to 0^+}\frac{g(x+h)-g(x)}{h}=\lim_{\lambda\to 1^+}\frac{g(\lambda x)-g(x)}{(\lambda-1)x}>\lim_{\lambda\to 1^+}\frac{g(x)-g(x)}{(\lambda-1)x}=0,$$

so that $g'(x)>0$. I take the limit from the right because $g$ is differentiable and the limit from the right and left must coincide. What tells me there's something wrong with my proof is taking the limit from the left:

$$g'(x)= \lim_{h\to 0^-}\frac{g(x+h)-g(x)}{h}=\lim_{\lambda\to 1^-}\frac{g(\lambda x)-g(x)}{(\lambda-1)x}<\lim_{\lambda\to 1^-}\frac{g(x)-g(x)}{(\lambda-1)x}=0,$$

where the denominator is negative if $\lambda\to 1^-$, which flips the inequality. Any help spotting the mistake or coming up with a proof of this statement is greatly appreciated.

Update/Answer: As the answer below by @EtienneBfx indirectly points out, increasing returns to scale apply for $\lambda>1$, so that the inequality in the limit from the right applies ($\lambda>1$), but not the inequality in the limit from the left $(\lambda<1)$. That eliminates the contradiction.

答:
If you have $\forall \lambda > 0$, $g(\lambda x) \geq g(x)$ then you have $g(x)=g(x*1) \geq g(1)$ and $g(1)=g(x*\frac{1}{x}) \geq g(x)$.

So I guess that $g$ is a constant fonction. If the inequality are stricts there is no solution!

Maybe the exercice is with $\lambda > 1$...


问:
Help needed with the solution about Differential Equations.
$$(x^2 + 4y^2)\dfrac {dx}{dy} = xy$$

In my opinion, I should separate dx and dy into opposite sites then look for mx = my or not.

Is this true? Or is it that the constant C only influences F(x) and so that makes it an important only to that part of the solution?

Thanks for help, I appreciated.

Body:
$$(x^2 + 4y^2)\dfrac {dx}{dy} = xy$$
$$(x^2 + 4y^2) = xyy'$$
$$(x^2 + 4y^2)=\dfrac x2 (y^2)'$$
$$x^2 + 4v=\dfrac x2 v'$$
$$uv' -4v=u$$
$$\left ( \dfrac v {u^4}\right)'=\dfrac 1 {u^4}$$
Where $v=y^2,u=x^2$. Integrate both sides.


问:
Is there a clever solution to Arnold's "merchant problem"?
There is a problem that appears in an interview$^\color{red}{\star}$ with Vladimir Arnol'd.

You take a spoon of wine from a barrel of wine, and you put it into your cup of tea. Then you return a spoon of the (nonuniform!) mixture of tea from your cup to the barrel. Now you have some foreign substance (wine) in the cup and some foreign substance (tea) in the barrel. Which is larger: the quantity of wine in the cup or the quantity of tea in the barrel at the end of your manipulations?

This problem is also quoted here.

Here's my solution:

The key is to consider the proportions of wine and tea in the second spoonful (that is, the spoonful of the nonuniform mixture that is transported from the cup to the barrel). Let $s$ be the volume of a spoonful and $c$ be the volume of a cup. The quantity of wine in this second spoonful is $\frac{s}{s+c}\cdot s$ and the quantity of tea in this spoonful is $\frac{c}{s+c}\cdot s$. Then the quantity of wine left in the cup is $$s-\frac{s^2}{s+c}=\frac{sc}{s+c}$$ and the quantity of tea in the barrel now is also $\frac{cs}{s+c}.$ So the quantities that we are asked to compare are the same.

However, Arnol'd also says

Children five to six years old like them very much and are able to solve them, but they
may be too difficult for university graduates, who are spoiled by formal mathematical training.

Given the simple nature of the solution, I'm going to guess that there is a trick to it. How would a six year old solve this problem? My university education is interfering with my thinking.

$\color{red}{\star}\quad$ S. H. Lui, An interview with Vladimir Arnol′d, Notices of the AMS, April 1997.

答:
Argument by symmetry

One way to approach the problem is to recognize the importance of the fact that you're expected to find a solution under the assumption that the tea-wine mixture in the teacup is nonuniform. In other words, it's impossible to know whether you're transferring a spoonful of tea back into the barrel, or a spoonful of wine, or some mixture of the two. What this implies is that the relative sizes of the teacup and wine barrel, as well as the proportion of wine that you scoop out of the teacup, are completely irrelevant.

With this knowledge, we can see that we should get the same answer whether we scoop a spoon of wine into the teacup and then go back, or if we scoop a spoon of tea in the wine barrel and then go back. Without knowing (or needing to know) anything about the relative sizes of the containers, we can just fill the barrel with tea and the cup with wine to do the "reverse" experiment. Transferring the wine into the tea will be no different if we have a teacup-sized wine barrel and a barrel-sized teacup - after all, there's nothing to indicate that this isn't the situation being described!

By symmetry, the only logical conclusion is that there is exactly as much tea in the wine barrel as there is wine in the teacup, regardless of whether the teacup or barrel is filled with tea or with wine to begin with. Otherwise, we would arrive at contradictory results when doing both experiments - we can't find more tea in the wine barrel when going one way, and more wine in the teacup when going the other, which is particularly obvious when we just switch the vessels that the liquids are in to begin with.


问:
Subgroup of index $2$ is Normal
Please excuse the selfishness of the following question:

Let $G$ be a group and $H \le G$ such that $|G:H|=2$. Show that $H$ is normal.

Proof:

Because $|G:H|=2$, $G = H \cup aH$ for some $a \in G \setminus H$.

Let $x\in G$. Then $x \in H$ or $x \in aH$.

Suppose $x \in H$. Then $xhx^{-1} \in H$ because $H$ is a group.

Suppose $x \in aH$. Then $ahH(ah)^{-1} = ahHh^{-1}a^{-1} = a(hHh^{-1})a^{-1}$ ***1

***1: Can I say now that $x \in H$, based on the that $hHh^{-1} \in H$ and that $aa^{-1} = e$ ?

Thank you for the time you spent reading my doubts.

答: If $[G:H]=2$, then necessarily $Ha=aH$ for every $a\in G\setminus H$.

Title: Subgroup of index $2$ is Normal


答:
Hints: $$(x^2+4y^2)\frac{dx}{dy}=xy$$

This is a homogeneous differential equation.
Put $y=vx$.then,
$$(x^2+4v^2x^2)=x^2v(v+x\frac{dv}{dx})$$
Now you can proceed. Good luck!

Title: Help needed with the solution about Differential Equations.

Body:
$$(x^2 + 4y^2)\dfrac {dx}{dy} = xy$$

In my opinion, I should separate dx and dy into opposite sites then look for mx = my or not.

Is this true? Or is it that the constant C only influences F(x) and so that makes it an important only to that part of the solution?

Thanks for help, I appreciated.


问:
The proof of inequality
How to prove that $$\sum\limits_{k=1}^n (\frac{a_1+a_2+\dots+a_k}{k})-\sum\limits_{k=1}^n a_k^2\leq \frac{n}{2}-\sum\limits_{k=1}^n\frac{1}{4k}?$$
In my book it is said that this comes from Cauchy-Schwartz inequality and  from that $$\sum\limits_{k=1}^n (\frac{a_1+a_2+\dots+a_k}{k})\leq (2n-\sum_{k=1}^n\frac{1}{k})^\frac{1}{2}(\sum_{k=1}^n a_k^2)^\frac{1}{2},$$ but I don't know how.

Any help is welcome. Thanks in advance.

答:
Put $A=\left(\sum_{k=1}^n a_k^2\right)^\tfrac{1}{2}$ and $M=\left(2n-\sum_{k=1}^n\frac{1}{k}\right)^\tfrac{1}{2}$. Since we are given
$$\sum\limits_{k=1}^n \left(\frac{a_1+a_2+\dots+a_k}{k}\right)\le MA,$$
it suffices to show
$$MA-A^2\le  \frac{n}{2}-\sum\limits_{k=1}^n\frac{1}{4k}=\frac{M^2}4,$$
which is equivalent to an inequality
$$0\le\left(\frac {M}2-A\right)^2.$$

Title: The proof of inequality


问:
Angle between vectors $a-b$ and $-j-k$
I am given vectors $a and b$ and asked to "measure the angle between $a - b$ and $-j - k$".

How do I obtain $-j-k$?

答:
$a = (3,4,-2)$  and $b=(-0,2,1)$ $\Rightarrow$ $a-b=(3,2,-3)$

$\cos\theta=\dfrac{\langle(a-b)(-j-k)\rangle}{ \|a-b||\|-j-k\|}$ $\Rightarrow$ $\theta=\arccos\biggl(\dfrac{\langle(a-b)(-j-k)\rangle}{\|a-b\|\|-j-k\|}\biggr)$

$\cos\theta=\dfrac{1}{2\sqrt{11}}$ which implies $\theta=\arccos\dfrac{\sqrt{11}}{22}$

Title: Angle between vectors $a-b$ and $-j-k$


问:
A proof that $\omega^\alpha$ is countable for every countable ordinal $\alpha$?
I am working on a problem that requires this statement but I am struggling to prove it, I have tried transfinite induction but I am stuck. Any help?</p>

答:
As you noted, there are two cases; one is when $\alpha$ is a successor ordinal, say $\alpha=\beta+1$, and one is when $\alpha$ is a limit ordinal. In the former case, you have $\omega^\alpha=\omega\cdot\omega^\beta$ and you should be able to 'compose' the map $\omega^\beta\mapsto\omega$ that you have with a map $\omega\times\omega\mapsto\omega$ to get the requisite mapping.

In the limit case, because $\alpha$ is a countable ordinal, you have an $\omega$-sequence for it, $\alpha=\bigcup_{n\in\omega}\alpha_n$. In this case $\omega^\alpha=\bigcup_{n\in\omega}\omega^{\alpha_n}$; can you see how to combine your mappings $\omega^{\alpha_n}\mapsto\omega$ (which you 'know' you have, by induction) to get the requisite $\omega^\alpha\mapsto\omega$? (Hint: there are $\omega$ of them. More specifically, you can think of the rank of any element $a$ of $\alpha$, which is to say the smallest $n$ for which $a\in\alpha_n$. Consider how to find an $m\in\omega$ for each $a\in \omega^\alpha$ by looking at the pair $\langle n,f_n(a)\rangle$, where $n$ is the rank of $a$ and $f_n$ is the map from $\omega^{\alpha_n}\mapsto\omega$.)