问:
How do I show that $\sum\limits^\infty_{n=n_0+1}(\frac{n_0}{n})^s$ vanishes as $s\to\infty$?
This is my last step of the proof but I'm not sure if this is true nor prove it. Could you give me any suggestions?</p>

答:
In your problem, the convergence "as $s \to \infty$" really should be "as ${\rm Re}(s) \to \infty$" to force the point $s$ to move to the right.
You don't want $s \to \infty$ along a vertical line, for example.

You write in the comments about knowing uniform convergence of the series only in compact subsets.  Dirichlet series actually converge uniformly on wider regions than compact subsets. If a Dirichlet series converges at a point $s_0$ (I don't believe boundedness of its partial sums at $s_0$  suffices for what I say next) then it converges uniformly on subsets of the right half-plane ${\rm Re}(s) > {\rm Re}(s_0)$ that form an angular sector opening to the right out of $s_0$ at less than a $180$-degree angle. This looks like the region at $0$ opening up to the right between the lines $y = mx$ and $y = -mx$ for $x > 0$ and some (finite) $m$.

Such an angular region is called a Stolz angle.  Having $s \to \infty$ within a Stolz angle is more generous than $s \to \infty$ along a horizontal line (${\rm Re}(s) \to \infty$ while ${\rm Im}(s)$ is fixed).  What Stolz angles avoid is having $s \to \infty$ in a way that allows $|s|$ to grow a lot faster than $|{\rm Re}(s)|$ (e.g., it avoids  $s \to \infty$ along a vertical line).

You can find pictures of Stolz angles by a Google image search on "Stolz angle", but the images you'll see are Stolz angles opening up into a disc from a point on the boundary. Stolz (in 1875) introduced "his" regions in order to generalize Abel's theorem for power series from radial convergence at a boundary point of the disc of convergence of a power series to larger subsets of that disc.  Later on, this concept was applied to Dirichlet series. For a Dirichlet series, the significance of Stolz angles at a point $s_0$ is that inside such a region the ratio $|s - s_0|/{\rm Re}(s-s_0)$ is bounded above.

A book that proves a Dirichlet series converges uniformly inside a Stolz angle (opening up to the right) is Serre's Course in Arithmetic. See Proposition 6 of Chapter VI.  The Stolz angle condition occurs in the proof where Serre writes  "$|z|/R(z) \leq k$".  His $R(z)$ means ${\rm Re}(z)$.

This business with Stolz angles isn't too important.  If a Dirichlet series converges at $s_0$ then it converges absolutely for ${\rm Re}(s) > {\rm Re}(s_0) + 1$, so if you are interested in behavior of a Dirichlet series as ${\rm Re}(s) \to \infty$ then you can focus on a half-plane of absolute convergence, in which case ${\rm Im}(s)$ is irrelevant to convergence issues: a Dirichlet series that converges absolutely at a point $s_0$ converges uniformly on the entire half-plane ${\rm Re}(s) \geq {\rm Re}(s_0)$.


问:
Independent and Identically Distributed Random Variables, with 0.01 and 0.99
good day

Can anybody help me? thanks

Let ${X_i}$ Independent and Identically Distributed Random Variables as normals with mean $μ$ and standard deviation 2, What is the minimum value of n such that $P(|\dfrac{S_n}{n}-μ|<.01) \geq .99$

Thanks

Independent and Identically Distributed Random Variables

答:
It gives you $X\sim N(\mu, 4)$, then $S\sim N(n\mu, 4n)$.

So $$\begin{split}P(|\frac {S-n\mu}{n}|<.01)&=P(|\frac{S-n\mu}{2\sqrt n}|<.005\sqrt n)\\
&=P(|Z|<.005\sqrt n)\ge .99\end{split}$$

Thus you need to find the quantile function $\Phi^{-1}.995$ to find a lower bound for $.005\sqrt n$, say you get z*. Then solve $.005\sqrt n\ge z^*$ to get the minimum value of n. Don’t forget to round up to an integer.

ContentLicense: CC BY-SA 4.0


问:
Is a class of sets a set if its union is a set?
I am wondering if a class of sets X is a set given that $ \cup X $ is a set.

Going in the other direction is trivial -- if X is a set then $\cup X$ is a set by the axiom of union. However, how would one prove it in this direction?

My initial thought is that if $\cup X$ is a set, we can construct its power set $P(\cup X)$. Then, to show X itself (the class of sets) is a set, it would suffice if it were a member of $P(\cup X)$. I'm not sure if this is true or if it is, how I would show that it is?

答: $X$ is not necessarily a member of $P(\bigcup X)$ but rather is a subclass of it.  Indeed, if $x\in X$, then $x$ is a subset of $\bigcup X$ (since $\bigcup X$ is the union of all the elements of $X$), so $x\in P(\bigcup X)$.  So $X$ is a subclass of $P(\bigcup X)$ and hence is a set by Separation.


问:
$|A \cup B| \leq |A| + |B|$
Let $|A| = n$ and $|B| = m$. Wlog, we can assume that $n \leq m$. I proved the following result $|A \cup \{x\} | \leq n + 1$.
I will use this result. Since B is non-empty we have element $y_1$ in $B$. Then it follows that $|A \cup \{y_1\}| = n + 1$. Consider the set $A_1 \leq A \cup \{y_1\}$. Applying the above argument gives us $A_2$. After m steps we have $A_{m}$ such that $A_m = A \cup B$ such that $|A_m| \leq n + 1$.

答:
Let $A$ and $B$ finite sets, then $$|A\cup B|=|A|+|B|-|A\cap B| $$
Notice that the set $A\cup B$ is formed by the $n$ elements of $A$ plus the $m$ elements of $B$. But some of the elements of $B$ have already been accounted for, this is precisely the case of $|A\cap B|$ elements of $A\cap B$, that's why we must subtract them. Since $-|A\cap B|\leq 0$ we have $$|A\cup B|\leq |A|+|B| $$ Maybe in future this can be helpful: if $f:A\to B$ is inyective then $|A|\leq |B|$. Here I will provide a different argument:
$A\cup B$ can be expressed as a disjoint union $$A\cup B = (A\backslash B)\cup (B\backslash A)\cup (A\cap B)  $$ therefore we have $$|A\cup B|=|A\backslash B|+|B\backslash A|+|A\cap B| $$ Since $A$ is the disjoint union of $A\backslash B$ and $A\cap B$ we have $$|A|=|A\backslash B|+|A\cap B| \Rightarrow |A \backslash B|=|A|-|A\cap B|$$ similarly $$|B\backslash A|=|B|-|A\cap B| $$ Replacing we have $$|A\cup B|=|A|+|B|-|A\cap B| $$ from which your result follows.

Title: $|A \cup B| \leq |A| + |B|$


问:
Show that $f(t)=(a_{1}t^{2}+b_{1}t+c_{1}, a_{2}t^{2}+b_{2}t+c_{2}, a_{3}t^{2}+b_{3}t+c_{3})$, $t \in \mathbb{R}$ is a plane curve.
A curve $f:I \subseteq \mathbb{R} \rightarrow \mathbb{R^n}$ is said to be flat, if there is a plane $\beta=\{(x,y,z)\in\mathbb{R^3}:ax+by+cz=d\}$ such that $f(t) \in \beta$, for all $t \in I$. Show that $f(t)=(a_{1}t^{2}+b_{1}t+c_{1}, a_{2}t^{2}+b_{2}t+c_{2}, a_{3}t^{2}+b_{3}t+c_{3})$,  $t \in \mathbb{R}$ is a plane curve.

答:
Torsion for a general parametric curve can be computed using a determinant of the curve's first, second and third derivatives:
$$\tau=\frac{\det(r',r'',r''')}{\|r'×r''\|^2}$$
Since all components of this curve are quadratic, the third derivative $r'''$ is the zero vector, so the determinant and hence the torsion are zero.


问:
Why is there an element with infinite spectrum in a commutative Banach algebra with infinitely many characters?
Let $A$ be a commutative Banach algebra such that set of all characters is infinite. I want to prove that there exists an element in $A$ such that its spectrum is infinite.</p>

答:
Let $(\chi_{n})_{n \in \mathbb{N}}$ be a sequence of pairwise distinct characters.

For $m \neq n$ let $U_{m,n} = \{a \in A \mid \chi_{n}(a) \neq \chi_{m}(a)\}$. Since $U_{m,n}$ is the complement of the closed hyperplane $\ker{(\chi_{m} - \chi_{n})}$, the set $U_{m,n}$ is open and dense in $A$ whenever $m \neq n$. By the Baire category theorem, the countable intersection
$$D = \bigcap_{m \neq n} U_{m,n}$$
is dense and of second category in $A$, so it is non-empty. Every $a \in D$ has infinite spectrum because
$$\sigma(a) \supseteq \{\chi_{n}(a) \mid n \in \mathbb{N}\}$$
and because the set $\{\chi_{n}(a) \mid n \in \mathbb{N}\}$ is infinite by definition of $D$.

ContentLicense: CC BY-SA 3.0


答:
You have $\vec{f}=\vec{c}+t\,\vec{b}+t^2 \, \vec{a}$ with constant vectors $\vec{a},\vec{b},\vec{c}.$  Take $\vec{k} = \vec{a} \times \vec{b}$ to be the cross product of $\vec{b},\vec{a},$  so that $\vec{k} \cdot \vec{a} = 0$ and
$\vec{k} \cdot \vec{b} = 0.$  Then
$$(\vec{f}-\vec{c}) \cdot \vec{k} = 0  $$ is your plane

Body: A curve $f:I \subseteq \mathbb{R} \rightarrow \mathbb{R^n}$ is said to be flat, if there is a plane $\beta=\{(x,y,z)\in\mathbb{R^3}:ax+by+cz=d\}$ such that $f(t) \in \beta$, for all $t \in I$. Show that $f(t)=(a_{1}t^{2}+b_{1}t+c_{1}, a_{2}t^{2}+b_{2}t+c_{2}, a_{3}t^{2}+b_{3}t+c_{3})$,  $t \in \mathbb{R}$ is a plane curve.


问:
How do I define a 3D line that does not leave the horizontal plane?
If I want to define the line y=2x when discussing 3D lines this is actually a plane. We should be able to turn it into a line with cartesian equation x=2y=z/0 but that would give us something undefined. How do you resolve that problem?</p>

答:
Typically you would describe it parametrically:

$$(x,y,z) = (t, 2t, 0), \quad t \in \Bbb R$$

To elaborate on this, two pieces of information that describe a line are a starting point (which you can choose anywhere on the line) as well as a direction vector (which you can scale by any non-zero number):

$$(x,y,z) = (p_x, p_y, p_z) + (v_xt, v_yt, v_zt), \quad t \in \Bbb R$$

Since this notation starts to get cumbersome, we can simplify it by writing
$$\mathbf{x} = \mathbf{p} + \mathbf{v}t, \quad t \in \Bbb R$$
where $\mathbf{x}$ represents $(x,y,z)$ and similarly for $\mathbf{p}$ and $\mathbf{v}$.


问:
Norm of a linear operator on $l^1$
$f:l^1 \to l^1$ be a linear operator given by $f((y_1,y_2, ...))=(x_1y_1,x_2y_2, ...), \forall y=(y_1,y_2,....) \in l^1$ , where $x=(x_1,x_2,....) \in l^{\infty}$ is a fixed element. We have to show that $||f|| = ||x||_{\infty}$. I have shown that -

$||f(y)|| = \sum_{i=1}^{\infty} |x_iy_i| \leq ||x||_{\infty} \sum_{i=1}^{\infty} |y_i| = ||x||_{\infty} ||y||_1$, $\forall y \in l^1$ $\Rightarrow ||f|| \leq ||x||_{\infty}$

But I am unable to find any suitable element in $l^1$ for proving the other part of the inequality. Please give some hints. Thank you.

答: $f(e_n)=x_ne_n$ so $\|f(e_n)\|=\|x_n\|$ By definition of norm of an operator this implies $\|f\| \geq |x_n|$. This is true for all $n$, so $\|f\| \geq \|x\|_{\infty}$. [$e_n$ is the sequence with $1$ in the $n-$th place and $0$ elsewhere].

Title: Norm of a linear operator on $l^1$


问:
Find the value of angle x (Langley's Adventitious Angles variant)
This problem had came to a Facebook post. Given that the left angles are given (with only 40° angle on the upper right triangle,find the angle x . What I did us to make equilateral triangles but it seems I still don't get two angles to get the vale of x . Can anyone give me a solution ???

答:
We can easily solve this by making some clever constructions of some points $X$, and $Y$ in the labelled diagram below;

Note that in the diagram, the triangle you gave is $\triangle ABC,$ and $X$ is produced by reflecting $D$ over $BC$, and $Y$ is constructed so that $BX=BY=BD,$ and $\triangle BXY$ is equilateral.

Note that since $\angle ABX=\angle ABY,$ thus $X$ is the reflection of $Y$ over $AB.$ We know that by Angle Chasing that $\angle BAX=90^\circ-\angle ABC=70^\circ,$ thus by properties of reflection, we know that $\angle BAY=70^\circ$ as well.

But then we have;

$$\angle YAC=\angle YAB+\angle BAX+\angle XAC=70^\circ+70^\circ+40^\circ=180^\circ.$$

Hence, $Y$ lies on $AC.$ We know that $\angle YCB=50^\circ=\angle YBC,$ thus $\triangle YBC$ is isosceles, and thus $YC=YB=YX.$

Hence, $\triangle XYC$ is also isosceles. We will now show that the side length of the lateral sides is equal to the side length of the lateral sides of the isosceles triangle $\triangle XBD,$ note that $XY=XB,$ so the claim is proven.

Now we will prove that $\triangle XYC\cong\triangle XBD$ by showing that $\angle XBD=\angle XYC.$ Note that $\angle XBD=20^\circ$ by the construct, and we have;
$$\angle XYC=90^\circ-\angle BAY=20^\circ.$$

Thus, the claim is proven and the two triangles are congruent by SAS.

We know by Congruence that $XD=XC.$ But since $CD=CX,$ the $\triangle CDX$ is equilateral, and the result follows by noting that;
$$\angle BCD=90^\circ-CDX=30^\circ.$$


问:
Show $\sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k} = \binom{x+y}{n}$.
For $n\ge 0$, show $\sum_{k=0}^{n}  \binom{x}{k}  \binom{y}{n-k} = \binom{x+y}{n}$.

Over a well ordered set of non-negative integers, induction can be used.

Base case: $n=0$

$\binom{x}{0}  \binom{y}{0} = 1.\frac{(y)!}{(y)! \,\,(0)!}=1$.

Taking the rhs, get :
$\binom{x+y}{0} = 1$.

Induction hypothesis: it is true for natural number $n$.

So, $\sum_{k=0}^{n}  \binom{x}{k} \binom{y}{n-k} = \binom{x+y}{n}$

Inductive step: need show for next natural $n+1$

$\sum_{k=0}^{n+1} \binom{x}{k} \binom{y}{n-k+1} = \binom{x+y}{n+1}$

This should lead to :

$\binom{x+y}{n} + \binom{x}{n+1}\binom{y}{0}= \binom{x+y}{n+1}$

get on lhs:

$\binom{x+y}{n} + 1.\binom{x}{n+1}$

$\implies \frac{(x+y)!}{n!(x+y-n)!} + \frac{x!}{(n+1)!(x-n-1)!}$

Please help to proceed further, or suggest alternative approach.

Body:
$$S=\sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k}$$
$$S=\sum_{k=0}^{n}[t^k]~ (1+t)^x~ [t^{n-k}]~ (1+t)^y$$
$$S=[t^n] (1+t)^{x+y}={x+y \choose n}.$$
Here $[t^j]f(t)$ means co-efficient of $t^j$ in $f(t).$


答:
The general rule-of-thumb is that in $n$-dimensional space $\Bbb R^n$, every equation you add cuts the dimension by $1$.

In the plane, $y=x$ gives a line. A system of $2$ equations like $x=y$ and $x=3$ determines a point.

In $\Bbb R^3$, $y=x$ gives a plane. The system $x=y$ and $y=z$ defines a line, and something like $x=y$, $y=z$, and $z=3$ gives just a point.

In $7$-dimensional space, $y=x$ gives a $6$-dimensional hyperplane, you need $6$ equations to describe a line, etc.

All you need to do here is write $x=2y$ and $z=0$. I suppose with some elbow grease you could force this into one equation like $(x-2y)^2+z^2=0$, but I'm not sure this is an improvement.

Title: How do I define a 3D line that does not leave the horizontal plane?

Body: If I want to define the line y=2x when discussing 3D lines this is actually a plane. We should be able to turn it into a line with cartesian equation x=2y=z/0 but that would give us something undefined. How do you resolve that problem?</p>


答: There are much better proof strategies in my opinion, but here's an inductive proof of the Vandermonde convolution identity.</p>

Title: Show $\sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k} = \binom{x+y}{n}$.

Body:
For $n\ge 0$, show $\sum_{k=0}^{n}  \binom{x}{k}  \binom{y}{n-k} = \binom{x+y}{n}$.

Over a well ordered set of non-negative integers, induction can be used.

Base case: $n=0$

$\binom{x}{0}  \binom{y}{0} = 1.\frac{(y)!}{(y)! \,\,(0)!}=1$.

Taking the rhs, get :
$\binom{x+y}{0} = 1$.

Induction hypothesis: it is true for natural number $n$.

So, $\sum_{k=0}^{n}  \binom{x}{k} \binom{y}{n-k} = \binom{x+y}{n}$

Inductive step: need show for next natural $n+1$

$\sum_{k=0}^{n+1} \binom{x}{k} \binom{y}{n-k+1} = \binom{x+y}{n+1}$

This should lead to :

$\binom{x+y}{n} + \binom{x}{n+1}\binom{y}{0}= \binom{x+y}{n+1}$

get on lhs:

$\binom{x+y}{n} + 1.\binom{x}{n+1}$

$\implies \frac{(x+y)!}{n!(x+y-n)!} + \frac{x!}{(n+1)!(x-n-1)!}$

Please help to proceed further, or suggest alternative approach.


问:
How to calculate the probability of 1 dimensional symmetric random walk to escape the ball?
I'm struggling to calculate escaping probability of symmetric random walk with 1 dimension.
How to calculate the probability of symmetric random walk to exit the ball with radius n^{x} centered 0 after n steps?
Is there a paper related to this one?
I'm sorry for my poor English.</p>

答:
Let $S_{n}=\sum_{j=1}^{n} X_{j}$ be the position of the random walk on the number line.

The number of different walks of n steps where each step is +1 or −1 is $2^{n}$. For the simple random walk, each of these walks is equally likely. In order for $S_{n}$ to be equal to a number k it is necessary and sufficient that the number of +1 in the walk exceeds those of −1 by k. It follows +1 must appear $\frac{n + k}{2}$ times among n steps of a walk, hence the number of walks which satisfy $S_{n} = k$ equals the number of ways of choosing (n + k)/2 elements from an n element set. Notice that n+k must be even since n=j+k+j where j is the number of -1 steps, so $n+k = 2j + 2k$ and we get the formula.

$$P(S_{n}=k) = \frac{n!}{(n-\frac{n+k}{2})!(\frac{n+k}{2})!2^{n}}$$

Using a symmetry we get $P(S_{n}=k)=P(S_{n}=-k)$.

So for n and k even:

$$P(|S_{n}|\leq k) = \sum^{k/2}_{j=0}2 P(S_{n} = 2j) - P(S_{n} = 0) = - \frac{n!}{2^{n}(\frac{n}{2}!)^{2}} + 2\sum^{k/2}_{j=0} \frac{n!}{(n-\frac{n+2j}{2})!(\frac{n+2j}{2})!2^{n}}$$


问:
Is there a connection between the Whitney embedding theorem and Cayley's theorem?
Background: I've been working through Guillemin and Pollack's "Differential Topology." They take the approach of defining smooth manifolds as "concrete" submanifolds of some ambient $\mathbb R^N$, as opposed to "abstract" topological spaces with charts and atlases. In my opinion, this approach leads to better intuition about the subject. Moreover, even in the standard "abstract" approach we can establish the Whitney embedding theorem, meaning that we aren't losing generality by working "concretely."

With this viewpoint of "abstract" vs "concrete" definitions, I realized that the Whitney embedding theorem with its ambient $\mathbb R^N$, and Cayley's theorem with its ambient symmetric group $S_N$, are of the same flavor. I know that there are many other representation theorems in math like these two, and that many can be viewed as generalization of the Yoneda embedding.

However, I can't find any sources that include the geometric representation theorems like Whitney and Nash in this Yoneda or categorical framework. So, I'm asking:

Question: Can the Yoneda lemma or other category theory be used to unify the Whitney embedding theorem with Cayley's theorem and/or other representation theorems in math?

My background consists of the category theory that showed up in the first year grad school, e.g. in an algebraic topology sequence and commutative algebra. However, any comments on this circle of ideas is greatly appreciated. Thank you in advanced!

答:
I do not see any relationship between these ideas at all.

Cayley's theorem exploits a structural idea about groups. Groups acting on sets give homomorphisms to the symmetric group. And a group's action on itself is a particularly natural choice of action. $S_n$ is the correct place to think about group actions.

Whitney's theorem is proved by hacking together a map based on charts and partitions of unity. It does not exploit any clever structural ideas about manifolds, nor teach you new ways of thinking about them. There is nothing special about $\Bbb R^N$ except that it is the domain of a chart and it is a vector space.

so... the proofs are different. The reason these objects are special are different. the way these teach you to think about the objects are different. The only relationship seems to be the word "embedding". What could possibly be the relationship? (Of course, "is there a connection" is not something that can be answered negatively because it is not a precise mathematical question which I can disprove.)


问:
Given n mod 2 = 1 and n mod 3 =1. Find n mod 5. How to solve algebraically knowing n = 7
I know that n = 7, so I should expect a remainder of 2 when divided by 5. However, finding the solution algebraically gives me a wrong remainder of 6. Please help me find my errors, or even find a better way to approach this problem. My work is shown below. $$$$
I know that a general solution involves finding n in terms of the equations:
$$ n = 2k + 1$$
$$n = 3m +1$$
I try to make both equations have a divisor of 5 like this:
$$\frac{5}{2}n = 5k + \frac{5}{2}$$ and
$$\frac{5}{3}n = 5m + \frac{5}{3}$$
I try to find the value of n by noticing $n = \frac{5}{2}n -\frac{9}{10} \cdot(\frac{5}{3}n)$.
$$n = (5k + \frac{5}{2}) - \frac{9}{10}\cdot(5m + \frac{5}{3})$$
This becomes $$n = (5k - 5m\cdot \frac{9}{10})+ (\frac{5}{2} -\frac{5}{3}\cdot\frac{9}
{10})$$
When I simplify this I get $$n = 5(k-\frac{9m}{10})+1$$
I note that $1 = -5 + 6$, therefore
$$n = 5(k-\frac{9m}{10}) -5 +6$$
So in the end $$n = 5(k-\frac{9m}{10} -1) +6$$
I get a remainder of 6 but I was expecting a remainder of 2. Thank you for your time.

Body: $n$ could be $1,7,13,19$ or $25$, so $n\pmod5$ could be $1,2,3,4$ or $0$.


问:
If $A$ is positive definite, then $B^TA^{-1}B$ is also positive/negative (semi) definite?
probably this is a basic question, but I can not see it clearly.

If $A \succ 0$, then $B^TA^{-1}B$ is also positive/negative (semi) definite? or in general is undefined?

In addition, you can assume that $B$ is full column rank.

答:
Which definition of $A>0$ do you use?
Let $V$ be arbitrary inner product space. We say a linear map $A$ is pos.definite, if $A=A^*$ and for $x\in V$, $x\neq 0$ we have $\langle x,Ax\rangle>0$. Since $(AB)^*=B^*A^*$, $I=A^*(A^{-1})^*=A(A^{-1})^*$, so the unicity of the inverse yields $(A^{-1})^*=A^{-1}$. Furthermore,
$$
\langle x, A^{-1} x\rangle=\langle Az,A^{-1} Az\rangle=\langle Az,z\rangle=\langle z,Az\rangle >0,
$$
if $x\neq 0$. Thus $A^{-1}$ is also pos.definite.

Using this definition we obtain
$$
\langle x,B^*A^{-1} B x\rangle=\langle Bx,A^{-1} B x\rangle=\langle z,A^{-1}z\rangle.
$$
Thus $B^*A^{-1}B$ is pos.semidefinite, if $B$ is invertable then pos.definite.


答: So I'm a bit confused, $6k+1$ is a solution of $n mod2 =1$ and $n mod3 =1$. So the solution isn't unique unless I'm missing something. Once you tell me that it is 7 then the earlier hypotheses are unnecessary.

Title: Given n mod 2 = 1 and n mod 3 =1. Find n mod 5. How to solve algebraically knowing n = 7

Body:
I know that n = 7, so I should expect a remainder of 2 when divided by 5. However, finding the solution algebraically gives me a wrong remainder of 6. Please help me find my errors, or even find a better way to approach this problem. My work is shown below. $$$$
I know that a general solution involves finding n in terms of the equations:
$$ n = 2k + 1$$
$$n = 3m +1$$
I try to make both equations have a divisor of 5 like this:
$$\frac{5}{2}n = 5k + \frac{5}{2}$$ and
$$\frac{5}{3}n = 5m + \frac{5}{3}$$
I try to find the value of n by noticing $n = \frac{5}{2}n -\frac{9}{10} \cdot(\frac{5}{3}n)$.
$$n = (5k + \frac{5}{2}) - \frac{9}{10}\cdot(5m + \frac{5}{3})$$
This becomes $$n = (5k - 5m\cdot \frac{9}{10})+ (\frac{5}{2} -\frac{5}{3}\cdot\frac{9}
{10})$$
When I simplify this I get $$n = 5(k-\frac{9m}{10})+1$$
I note that $1 = -5 + 6$, therefore
$$n = 5(k-\frac{9m}{10}) -5 +6$$
So in the end $$n = 5(k-\frac{9m}{10} -1) +6$$
I get a remainder of 6 but I was expecting a remainder of 2. Thank you for your time.


答:
$$\sum\limits^\infty_{n=n_0+1}\left(\frac{n_0}{n}\right)^s=\sum\limits^\infty_{n=1}\left(\frac{n_0}{n}\right)^s-\sum\limits^{n_0}_{n=1}\left(\frac{n_0}{n}\right)^s=\zeta (s)-1$$

For large values of $s$
$$\zeta (s)-1=\sum_{k=2}^\infty k^{-s}\sim 2^{-s}$$

Title: How do I show that $\sum\limits^\infty_{n=n_0+1}(\frac{n_0}{n})^s$ vanishes as $s\to\infty$?

Body: This is my last step of the proof but I'm not sure if this is true nor prove it. Could you give me any suggestions?</p>


问:
statement about continuous function - false, why?
Let $f$ be a function defined on the closed interval $[0,1]$, such that $f(0)<f(1)$.

If for every $f(0)<b<f(1)$ there exists a $0<c<1$ such that $f(c)=b$

then $f(x)$ is continuous on the closed interval $[0,1]$.

I'm having trouble understanding why this is statement isn't true.
Any idea to help me visualize it in some form will help a lot.

Thanks!

答:
First example:
$$f(x)=\begin{cases}
2x\quad\quad\ \ \ \text{ if }\quad 0\le x\lt\frac12,\\
2x-1\quad\text{ if }\quad\frac12\le x\le1.\\
\end{cases}$$
Now $f(0)=0\lt1=f(1)$.

If $f(0)\lt b\lt f(1)$, then $0\lt\frac b2\lt\frac12\lt\frac{b+1}2\lt1$ and $f\left(\frac b2\right)=f\left(\frac{b+1}2\right)=b$,

but $f$ is discontinuous at $x=\frac12$.

Second example:

$f\left(\frac13\right)=\frac23$ and $f\left(\frac23\right)=\frac13$, while $f(x)=x$ in all other cases.

Third example:

$f(0)=0$ and $f(1)=1$, while $f(x)=1-x$ for $0\lt x\lt1$.

On the other hand, if $f$ is a weakly increasing function on $[0,1]$ (that is, $0\le x\lt y\le1$ implies $f(x)\le f(y)$), and if $f$ maps $[0,1]$ onto $[f(0),f(1)]$, then $f$ must indeed be continuous on $[0,1]$. This is probably what you were thinking of; it's true because monotone functions can have only jump discontinuities.


问:
Change of variables $x=u+v$, $y=u-v$
I need to calculate $ \displaystyle \int _0 ^ 1 \int_y^ 1 (x ^ 2 + y ^ 2) dx \;  dy $ making the changes of variables $x = u + v, y = u-v$

I know $|Jg(x)|=2$ and $f(g)=2(u^2+v^2)$

I'm stuck, I don't know how to raise the new integral

答:
$x = u + v, y = u-v$

$u = \frac{x+y}{2}, v = \frac{x-y}{2}$

Given the original region, note that  $ \ 0 \leq x-y \leq 1$

i.e $ \ 0 \leq v \leq \frac{1}{2}$

For any value of $v$, the limts of $u$ will be,

$v \leq u \leq 1-v$

So the new integral is

$\displaystyle \int _0^{1/2} \int_{v}^{1-v} 2 (u^2 + v^2) \ |J| \ du \ dv$

Title: Change of variables $x=u+v$, $y=u-v$


问:
Prove that $(x+1)^x \over x^x $ $=e$ as $x$ approaches infinity.
I graphed this function on desmos and noticed that its value got close to $2.7$. However, the graph soon cut off because the numbers were too big for desmos to calculate. So, here is my attempt. We have $\lim \limits_{x \to \infty} {(x+1)^x \over x^x}$ and by Lhopital's rule, we have $\lim \limits_{x \to \infty} {(x+1)^x\ln(x+1)+1 \over x^x\ln(x)}$. I soon realized, using Wolfram that this won't work.

I start at the beginning with $\lim \limits_{x \to \infty} {(x+1)^x \over x^x}$. Let this limit be $L$. So, we have $L = \lim \limits_{x \to \infty} {(x+1)^x \over x^x}$. Taking natural log of both sides, we get $\ln(L) = \lim \limits_{x \to \infty} x\ln({(x+1) \over x})$. I know that $x+1 \over x$ will get close to $1$ as $x$ gets very big, so $\ln({x+1 \over x})$ will approach $0$ when $x$ gets very big. Thus this limit $\ln(L) = \lim \limits_{x \to \infty} x\ln({(x+1) \over x})$ is a $0*\infty$ case, so I rewrite the limit as $\ln(L) = \lim \limits_{x \to \infty} {x \over \frac1{\ln({(x+1) \over x})}}$. I apply Lhopitals rule, but I end up with another complicated expression on the bottom using Wolfram, that is also an undefined limit. I have tried LHopital's rule twice and a "clever" manipulation, but the problem still won't yield, could someone please help? Thanks.

答: $$\lim_{x \to \infty}\frac{(x+1)^x}{x^x} = \lim_{x \to \infty}\frac{x^x(1+\frac{1}{x})^x}{x^x} = \lim_{x \to \infty}\Big(1+\frac{1}{x}\Big)^x = e$$


答:
Here you are
$$\lim_{x\rightarrow \infty}\frac{\ln\left ( 1+ \frac{1}{x} \right )}{\frac{1}{x}}= 1\Rightarrow \left ( 1+ \frac{1}{x} \right )^{x}\rightarrow e\,{\rm as}\,x\rightarrow \infty$$

Title: Prove that $(x+1)^x \over x^x $ $=e$ as $x$ approaches infinity.

Body:
I graphed this function on desmos and noticed that its value got close to $2.7$. However, the graph soon cut off because the numbers were too big for desmos to calculate. So, here is my attempt. We have $\lim \limits_{x \to \infty} {(x+1)^x \over x^x}$ and by Lhopital's rule, we have $\lim \limits_{x \to \infty} {(x+1)^x\ln(x+1)+1 \over x^x\ln(x)}$. I soon realized, using Wolfram that this won't work.

I start at the beginning with $\lim \limits_{x \to \infty} {(x+1)^x \over x^x}$. Let this limit be $L$. So, we have $L = \lim \limits_{x \to \infty} {(x+1)^x \over x^x}$. Taking natural log of both sides, we get $\ln(L) = \lim \limits_{x \to \infty} x\ln({(x+1) \over x})$. I know that $x+1 \over x$ will get close to $1$ as $x$ gets very big, so $\ln({x+1 \over x})$ will approach $0$ when $x$ gets very big. Thus this limit $\ln(L) = \lim \limits_{x \to \infty} x\ln({(x+1) \over x})$ is a $0*\infty$ case, so I rewrite the limit as $\ln(L) = \lim \limits_{x \to \infty} {x \over \frac1{\ln({(x+1) \over x})}}$. I apply Lhopitals rule, but I end up with another complicated expression on the bottom using Wolfram, that is also an undefined limit. I have tried LHopital's rule twice and a "clever" manipulation, but the problem still won't yield, could someone please help? Thanks.


答:
You will want to simplify the fraction $\frac{(x+1)^{x}}{x^{x}}$ like so:

$$\frac{(x+1)^{x}}{x^{x}}= (\frac{x+1}{x})^{x} = (1+\frac{1}{x})^{x}$$

Now it should look more familiar to the definition of $e$ that you have probably.


问:
The volume of the image of a map with vanishing Jacobian is zero
Let $\Omega \subseteq \mathbb{R}^n$ be a nice domain with smooth boundary (say a ball), and let
$f:\Omega \to \mathbb{R}^n$ be smooth. Set $\Omega_0=\{ x \in \Omega \, | \, \det df_x =0 \} $

Is there an elementary way to prove that $m(f(\Omega_0))=0$? ($m$ is the Lebesgue measure).

I know that this follows from the (co)area formula or Sard's theorem*, but is there a way that avoids them?

Edit:

In this answer, zhw proves that if $x \in \Omega_0$, and $B(r)$ is an Euclidean ball of radius $r$ centered at $x$, then $\frac{m(f(B(r))}{m(B(r))} \to 0$.

Since $\Omega_0$ is compact, we can proceed as follows:

Cover $\Omega_0$ by finitely many balls $B_i$ with radius $r$, centered around points that belong $\Omega_0$. (The centers depend on $r$)Then,

$$ m(f(\Omega_0)) \le \sum_i m(f(B_i))=\sum_i \frac{m(f(B_i))}{m(B_i)}m(B_i).$$

If we could prove that $\frac{m(f(B_i))}{m(B_i)} \to 0$  when $r \to 0$ uniformly in $i$, then we could get
  $$
m(f(\Omega_0)) \le o(1) \sum_i m(B_i)=o(1) \sum_i m(B(r)).
$$
  Since we can cover $\Omega_0$ by $N(r)$ balls of radius $r$ with centers in $\Omega_0$, where $N(r) \le c \frac{1}{m(B(r))}$, we could conclude that
  $m(f(\Omega_0)) \le o(1)$, so it must be zero.

The problem is that I am not sure if $\frac{m(f(B_{x_i(r)}(r)))}{m(B_{x_i(r)}(r))} $ converges to zero independently of $i$. I asked about this separately here.

*If I am not mistaken, then Sard's theorem implies that almost every $y \in \mathbb R^n$ is a regular value of $f$- so it doesn't have a preimage in $\Omega_0$.

**I don't know an elementary proof even in the case where $\Omega_0=\Omega$. That is, even if we assume the the domain where the Jacobian vanishes is "open and nice" I don't know if its trivial. In general, $\Omega_0$ is an arbitrary closed set, which might be pretty complicated.

答:
This is quite well-known. You do not need to have proven area (or coarea) formula in order to prove this. The result follows from basic covering arguments. It will be included in a soon to be publicized paper of mine with Piotr Hajlasz. I am using Hausdorff measure below, but that should not be any issue.

Lemma: Let $f:R^n \to R^N$ be $L$-Lipschitz. Let
$$
E_k=\{ x\in R^n:\, \text{$Df(x)$ exists and $rank \, Df(x)=k$.}\},
\quad
0\leq k\leq n.
$$
Then every $x\in E_k$ has the following property:
For every integer $m\geq 1$, there is $r=r(x,m)>0$ such that $f(B(x,r))$ can be covered by $m^k$ balls, each of radius $\frac{2\sqrt{k}Lr}{m}$.

Proof sketch: Given $m$, there is a threshold $\delta>0$ such that if $r < \delta$ then
$$
 \|f(y)-f(x) - Df(x)(y-x)\| \lesssim \frac{1}{m} \|y-x\| \leq \frac{r}{m}  \, \quad \text{for all $y \in B(x,r)$.}
$$
where $\lesssim$ means up to a constant that depends of $k$ and the Lipschitz constant of $f$. This says that the image $f(B(x,r)$ is within $\lesssim r/m$ thin strip around the image $Df(x)(B(x,r))$. Now the image of the derivative is contained in a $k$-dimensional affine space. So, we may cover this image by $\lesssim m^k$ many balls of diameters $\lesssim r/m$. As a result $f(x,r)$ is also covered by those same balls except after doubling/trippling their radii.

Theorem (Sard-type): Suppose $f\colon R^n \to R^N$, is Lipschitz. Let $E$ be the set of all $x$ where $Df(x)$ exists and $rank \, Df(x) <n$. Then $\mathcal{H}^n(f(E))=0$.

Proof sketch: If we use the lemma then we can estimate the Hausdorff content as follows (the whole point is we compute $n$-dim measure and $k<n$):
$$
\mathcal{H}^n_\infty(f(B(x,r))) \lesssim \sum_{covering\, of \, f(B(x,r)) \, by \, C} (diam \, C)^n \lesssim m^k \times (r/m)^n = r^n m^{k-n}
$$
Now by a Vitali covering we find covering of $E$ with $B(x,r_i)$ such that $B(x,r_i/5)$ are disjoint. Then summing the inequality above over this collection produces (notice that $\sum_i r_i^n$ is bounded by $5^n$ times Lebesgue measure of $E$.)
$$
\mathcal{H}^n_\infty(f(E)) \lesssim  m^{k-n} m(E)
$$
But $m$ is arbitrary and $k<n$. $\mathcal{H}^n_\infty(f(E)) = 0$ from which as a general rule it follows that $\mathcal{H}^n(f(E))=0$.


问:
Prove: If a linear function mapping ${\rm I\!R}^{K}$ into itself is onto then it is also one-to-one.
I have been trying to prove this statement but it is escaping me.

If T is a linear function from ${\rm I\!R}^{K}$ to ${\rm I\!R}^{K}$, then if T is onto the T is one-to-one.

I can see why if I think of the K=1 case, with a straight line passing through the origin, but I can't find the way to build the proof.

Thank you!

答:
From Axler's Linear Algebra Done Right:

3.2 Proposition: Let $T \in \mathcal{L}(V, W)$. Then $T$ is injective if and only if null $T=\{0\}$.

PROOF: First suppose that $T$ is injective. We want to prove that null $T=\{0\}$. We already know that {0}$\subset$ null $T$ (by 3.1). To prove the inclusion in the other direction, suppose $v \in$ null $T$. Then
$$
T(v)=0=T(0)
$$
Because $T$ is injective, the equation above implies that $v=0$. Thus null $T=\{0\},$ as desired.
To prove the implication in the other direction, now suppose that null $T=\{0\}$. We want to prove that $T$ is injective. To do this, suppose $u, v \in V$ and $T u=T v .$ Then
$$
0=T u-T v=T(u-v)
$$
Thus $u-v$ is in $\operatorname{null} T,$ which equals $\{0\} .$ Hence $u-v=0,$ which implies that $u=v$. Hence $T$ is injective, as desired.

If $Tv=0$ for some nonzero $v \in \mathbb R^k$ (i.e., if $T$ is not one-to-one, by Proposition 3.2), writing $v = v_1e_1 + \cdots + v_ke_k$ ($e_j$ is the vector whose only nonzero entry is a $1$ in the $j$-th place) we have that $$0 = v_1Te_1 + \cdots + v_kTe_k,$$ that is, the vectors $Te_1,\dots,Te_k$ are linearly dependent. Thus $Te_1,\dots,Te_k$ cannot span $\mathbb R^k$ (why?), meaning that $T$ cannot be onto.


问:
Probability of words
The question is as follows:

A word of $6$ letters is formed from a set of $16$ different letters of English alphabet (with replacement). Find out the probability that exactly two letters are repeated.

What I think is there is a word that has six letters among which there are there are two that have repetitions i.e. among the 6 only 4 are different, rest are repetitions.

Based on this, I first found out the total number of 6 letter words possible from the 16 letters. It is, of course $16^6$. Now to find out the required no. of words, I have the following expression (E is event space),
$$ n(E) = {16\choose 4} \times \frac{6!}{2! \times 2!}$$

That's because, I need to choose only 4 letters from 16 and then the arrangements of 6 letters in the word among which 2 and 2 are same, e.g., fghkgk. So the probability comes out to be,

$$ P(E)= \frac{{16\choose 4} \times \frac{6!}{2! \times 2!}}{16^6}$$
However the answer given is,
$$ P(E)=\frac {18080}{16^6}$$
which is much less than the answer I am getting. I would like others to explain to me what is wrong in the logic that I have used.

答:
There are 16 different letters and we have to form words with 2 letters repeated.

Six different letters can be selected by 16C6 ways. Further, from these 6 different letters, 6! ways words can be formed.

Hence, the Exhaustive number of words = 16C6 x 6!

= 8008 x 720 words can be formed.

In the selection of 6 letters, we want exactly 2 letters (2 x 2) to be repeated. This amounts to selection of only 4 letters form the words with given condition.

Favorable cases = 16C4 x 6!/2!x 2! = 1820 x 720/4 = 1820 x 180

So, the required probability = 1820 x 180/8008 x 720 = 0.227

Title: Probability of words


问:
how to study the stability of the fixed point
I wonder how to study the stability of the fixed point of the difference equation $x^{‘}=x\exp{r(1-x)}$

答:
Since you say "fixed point", I am going to assume it is a difference equation. That is, you really mean that you have instead:

$$x_{n+1}=x_ne^{r(1-x_n)} \hspace{5mm}(1)$$

Note how this is different from, say:

$$x'=xe^{r(1-x)} \hspace{5mm}(2)$$

Put simply, $(1)$ is a "discretized" version of $(2)$. It should be then clear that we need different criteria for stability (due to the nature of the equation).

My assumption is based on the fact that you say "fixed points", and technically speaking, you find fixed points of a "map", which is a term given to the R.H.S. of equation $(1)$.

When speaking of differential equations, i.e. $(2)$, you find equilibrium points instead. But let's disregard the technicalities for a minute.

Here are my hints for the first equation $(1)$:

Let $f(x):= xe^{r(1-x)}$, i.e the R.H.S. of the equation. Find the points $x \in \mathbb{R}$ such that $f(x)=x$. These are your "fixed points". There should be two of them.

Evaluate the derivative $f'(x)$ at these fixed points. The criterion for a stable fixed point $x_*$(of a map) is:
$$ |f'(x_*)| <1$$

You will then get a criterion from 2. for the stability of each fixed point in terms of $r$.

Try this and let me know if you get stuck.

By the way, if you meant $(2)$, then the criterion is similar but slightly modified. Here are my hints in that case:

Let $f(x):= xe^{r(1-x)}$, i.e the R.H.S. of the equation. Find the points $x \in \mathbb{R}$ such that $f(x)=0$. These are your "fixed points". This should be just $x=0$.

Evaluate the derivative $f'(x)$ at these fixed points. The criterion for a stable fixed point $x_*$ (for an ODE) is:
$$ f'(x_*) <0$$

You will then get a criterion from 2. for the stability of each fixed point in terms of $r$.


问:
Expectation of Chi square distribution
Given that $Y$ is a degree $8$ chi square distribution i.e $Y \sim \chi_{8}^2$

I want to find $E[Y^2]$ and $Var[Y^2]$

I only know that $E[Y]$ of it should be $8$, but I don't know how to find $E[Y^2]$ from it?

Thank you

答:
$$\chi_{8}^2=Gamma(4;1/2)$$

Where $1/2$ is the rate parameter. Now it is not difficult to calculate any moment you want solving the integrals

Alternative method:

Given that $Y\sim \chi_8^2$,

$$Y=Z_1^2+\dots +Z_8^2$$

Where $Z_i$ are iid Standard gaussian. Moments of the standard gaussian are well known thus...

$$\mathbb{E}[Y]=\underbrace{1+1+\dots +1}_{\text{8 times}}=8$$

$$\mathbb{V}[Y]=8\Bigg[\frac{4!}{2^2\cdot 2!}-1\Bigg]=16$$

Title: Expectation of Chi square distribution


问:
Question on the proof of the uniqueness of Levy triplets in the Levy-Khintchine formula
I am reading the proof on the uniqueness of the Levy triplets in the Levy-Khintchine formula from Ken Iti Sato's Levy Processes. However, there are two questions I cannot answer from the proof below.

First, why is the nonnegative-definite matrix $A$ uniquely determined by $\mu$ given that we have
$$s^{-2} \log \hat \mu(sz) \to -\frac{1}{2} \langle z, Az \rangle \; \text{as} \; s \to \infty?$$

Second, why is $\rho$ uniquely determined by $\psi$ given that we have (8.10), i.e. the Fourier transform of $\rho$ is given by $\int_c (\psi(z)-\psi(z+w))dw$?

I cannot think of a formal argument or other theorems guaranteeing the uniqueness from these forms. I would greatly appreciate any explanation.

答:
Concerning your $\textbf{first question}$: Let us assume the existence of another Lévy triplet $(A',\nu',\gamma')$. By repeating the steps that were done to get $$\lim_{s \to \infty} \frac{\log \widehat{\mu}(sz)}{s^{2}} = -\frac{1}{2} \langle Az,z \rangle$$ we will then get $$\lim_{s \to \infty} \frac{\log \widehat{\mu}(sz)}{s^{2}} = -\frac{1}{2} \langle A'z,z \rangle.$$ The left-hand side is independent of the Lévy triplets (since $\log \widehat{\mu}$ is already uniquely determined by $\widehat{\mu}$), hence we can conclude that the equation $$\begin{equation} -\frac{1}{2} \langle Az,z \rangle = -\frac{1}{2} \langle A'z,z \rangle \quad \text{or} \quad \langle Az,z \rangle = \langle A'z,z \rangle \tag{$1$} \end{equation}$$ holds for every $z \in \mathbb{R}^{d}$, especially for the standard unit vectors $e_{1}, \dots, e_{d}$. This leads us to $$\begin{equation} a_{i,i} = \langle Ae_{i},e_{i} \rangle = \langle A'e_{i},e_{i} \rangle = a'_{i,i} \tag{$2$} \end{equation}$$ for all $i$, assuring equal main diagonals of $A$ and $A'$. Moreover, $$\begin{align} \langle A(e_{i} + e_{j}),e_{i}+e_{j} \rangle &= \langle Ae_{i},e_{i} \rangle + \langle Ae_{i},e_{j} \rangle + \langle Ae_{j},e_{i} \rangle + \langle Ae_{j},e_{j} \rangle \\ &= a_{i,i} + a_{j,i} + a_{i,j} + a_{j,j} \\ &= a_{i,i} + 2a_{i,j} + a_{j,j} \end{align}$$ and $$\begin{align} \langle A'(e_{i} + e_{j}),e_{i}+e_{j} \rangle &= \langle A'e_{i},e_{i} \rangle + \langle A'e_{i},e_{j} \rangle + \langle A'e_{j},e_{i} \rangle + \langle A'e_{j},e_{j} \rangle \\ &= a'_{i,i} + a'_{j,i} + a'_{i,j} + a'_{j,j} \\ &= a'_{i,i} + 2a'_{i,j} + a'_{j,j} \end{align}$$ hold, as $A$ and $A'$ both are symmetric. Due to $(1)$, the left-hand sides are equal, resulting in $$a_{i,i} + 2a_{i,j} + a_{j,j} = a'_{i,i} + 2a'_{i,j} + a'_{j,j}.$$ Due to $(2)$, this is equivalent to $a_{i,j} = a'_{i,j}$ for all $i$ and $j$, proving that $A = A'$.

Concerning your $\textbf{second question}$: Because of the uniqueness of $\log \widehat{\mu}$ and $A$, the function $\psi$ is likewise uniquely determined by $\widehat{\mu}$. This means that the Fourier transform of $\rho$ is uniquely determined by $\widehat{\mu}$, as we can compute it with $\psi$. Knowing this, $\rho$ is also uniquely determined by $\widehat{\mu}$. This conclusion is possible, since $\rho$ is a finite measure. (Actually, $\rho$ can be treated as a probability measure and $\widehat{\rho}$ can be treated as its characteristic function in this context. With that, the step is maybe easier to understand.)


问:
Changing variables double integral
I want to know if there's a way to know the best variables change for an integral calculation. For example, if we consider the integral

$$ \iint_{R}\left(x+y\right)dx\,dy $$

Where $ R $ is bounded by the line $ y=x,\thinspace\thinspace\thinspace y=2x,\thinspace\thinspace\thinspace\thinspace y+x=2 $.

This integral is very easy to calculate directly, without variables change. But I wonder how simple it can get if one knows how to change the variables in an efficient way.

I'll give another example;

$$ \iint_{R}\left(x-y\right)e^{x^{2}-y^{2}}dx\,dy $$

Where $ R $ is bounded by the curves $ x^{2}-y^{2}=1,\thinspace\thinspace\thinspace\thinspace x^{2}-y^{2}=-1,\thinspace\thinspace\thinspace\thinspace x+y=1,\thinspace\thinspace\thinspace\thinspace x+y=3 $.

In this example it is pretty obvious that we should change the variables in the following way:

$$ \left(x-y\right)=u,\thinspace\thinspace\thinspace\thinspace\left(x+y\right)=v $$

But how can I find a convenient variables change when its not very clear and implied, like in the first example I gave

Thanks in advance.

答:
Note that the integration boundaries are  $\frac yx \in (1,2)$ and $x+y\in(0,2)$, which naturally leads to the variable changes
$$u=x+y,\>\>\>\>\>v = \frac yx$$ and the corresponding Jacobian $\frac u{(1+v)^2}$. Thus, the integral becomes
$$\int_S(x+y)dxdy =\int_1^2 \int_0^2 \frac{u^2}{(1+v)^2}dudv =\frac49
$$

Title: Changing variables double integral


答:
Every odd integer is of the form $2k+1$.

When you say that $n = 3k+1$, you mean that $n$ is an odd integer which gives remainder $1$ when divided by $3$.

Thus, you mean that $n$ is of the form $6k+1$.

This means that:

$$n = 5k + k + 1$$
$$n = k+1(mod 5)$$

Thus the expected set of remainders must be:$$remainder = {0,1,2,3,4}$$


问:
$\lim_{x \to -\infty} \sqrt{4x^2-nx}+2x = \frac{n}{4}$

Prove that $$\lim_{x \to -\infty} \sqrt{4x^2-nx}+2x = \frac{n}{4}$$

My attempt: I simplified $\sqrt{4x^2-nx}+2x$ by conjugation $$\sqrt{4x^2-nx}+2x \left(\frac{\sqrt{4x^2-nx}-2x}{\sqrt{4x^2-nx}-2x}\right) = -\frac{nx}{\sqrt{4x^2-nx}-2x} $$ I thought about substituting $x$ for $\frac{1}{y}$ to get $$\lim_{x \to -\infty} \sqrt{4x^2-nx}+2x =  \lim_{y \to 0} \frac{\sqrt{4-ny}+2}{y}$$ But this means that the limit doesn't exist! Did I go wrong somewhere in the simplification or substitution? (I noticed the $\frac{n}{4}$ part in Wolfram Alpha)

答:
$$\lim_{x\rightarrow-\infty}-\frac{nx}{\sqrt{4x^2-nx}-2x} = \lim_{y\rightarrow  \infty}\frac{ny}{\sqrt{4y^2+ny}+2y} = \lim_{y\rightarrow  \infty}\frac{n}{\sqrt{4+n/y}+2} = \frac{n}{4}$$

Alternatively,
$\lim_{x \to -\infty} \sqrt{4x^2-nx}+2x = \lim_{y \to 0^-} \frac{-\sqrt{4-ny}+2}{y}$

Note the negative sign as $y$ is a negative quantity.


问:
Proof of one of the most beautiful equations in Math.
$\cfrac{1}{1^2} + \cfrac{1}{2^2} + \cfrac{1}{3^2} + ...... \infty = \cfrac{{\pi}^2}{6}$

Often termed as one of the most beautiful equations in math:

I tried devising a proof for the above formula.

But I got stuck just as soon as I begun. Addition of any two or more rational numbers always results in a rational number. The $LHS$ is rational but the $RHS$ is irrational. How is this possible?

答:
At least as far as your question is concerned, take your favorite irrational number. For instance we can take
$$\pi(=3.1415...)$$
and write it as an infinite convergent sum
$$
\pi=3+\frac{1}{10}+\frac{4}{100}+\frac{1}{1000}+\frac{5}{10000}+\cdots
$$
So actually, every irrational can be viewed as the sum of an infinite series consisting of rational numbers.


问:
A general question about constructing subsequences.
I am currently covering sequences and the Bolzano-Weierstrass Theorem.  My question is regarding construction of subsequences. This material is still new to me and I am in a class that is designed for people heading into Real Analysis.

An example:

"A sequence that does not contain 0 or 1 as a term but contains subsequences converging to each of these values."

It took me a while but I was able to construct a sequence using a piece-wise argument.

$$
x_n=\begin{cases}
  1+\dfrac1n,&\text{n is odd}\\
  \dfrac1n,&\text{n is even}
\end{cases}
$$

Since $\left\{1+\dfrac1n\right\}\rightarrow 1$ and $\left\{\dfrac1n\right\}\rightarrow 0$.

But even this feels rather cheap because I have no way of expressing it as a single subsequence say $(x_{n_k})\color{red}{\text{ (correct me if I am wrong!)}}$. As in, I cannot manipulate the subscript to give me the desired equation. This worries me because I do not have a closed form of a subsequence $-$ as example, if then I am asked to show said subsequences converge, I would not know what to do.

For example we know $(x_n)\rightarrow 0$ where $x_n=\dfrac1n$

then we can show the subsequence $(x_{2n})\rightarrow 0$ where $x_{2n}=\dfrac1{2n}$

and the same follows with $(x_{2n+1})$, etc.

which is fairly easy to show using theorems of convergence, but these do not have the restrictions as the example.

I am wondering if there is an easier method or an approach to finding such subsequences rather than trial and error? I suppose this one was easy as I was able to think of plenty of sequences that converge to these values, yet I still feel as if it is incorrect with how it was constructed.

答:
There is absolutely nothing wrong with the sequence that you constructed; it is one of the simplest that will do what you want.

You do have a closed form for the terms of the sequence: you’ve described $x_n$ directly in terms of $n$. Yes, it’s a multipart definition, but there’s nothing wrong with that. As it happens, it could be rewritten with a single formula, but the result would be less easily understood:

$$x_n=\frac{1+(-1)^n}2+\frac1n\,.\tag{1}$$

Sometimes, however, there is no nice formula that covers all indices; it really doesn’t matter as long as the terms of the sequence can be defined somehow.

And it appears to me that you’ve already seen that the relevant subsequences are easily described: one is

$$\langle x_{2n}:n\in\Bbb N\rangle=\left\langle\frac1{2n}:n\in\Bbb N\right\rangle\,,$$

and the other is

$$\langle x_{2n+1}:n\in\Bbb N\rangle=\left\langle\frac1{2n+1}:n\in\Bbb N\right\rangle\,.$$

I don’t know what you mean by ‘these do not have the restrictions as the example’: they are very straightforwardly subsequences of the sequence $\langle x_n:n\in\Bbb N\rangle$ defined by your two-part definition or, equivalently, by the more opaque $(1)$.


问:
please help me solve the differential equation
I'm having a slight issue trying to solve this diffrential equation. I would greatly appreciate any help or tips to solve this problem.

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

答:
This is nearly a homogeneous DE. Set $f(x)-\frac12=(x-\frac12)u(x)$, then you should get a separable DE.
$$
(x-\tfrac12)u'(x)+u(x)=\frac{u(x)-1}{u(x)+1}.
$$

In general you can insert arbitrary shifts in $x$ and $f(x)$ and solve the linear system that reduces the sum of the constants to zero, so that the resulting equation becomes properly homogeneous (in the sense of a fraction of equal-degree expressions)


问:
Proving that a formula is not a WFF
So we know that the basic properties of a well formed formula is one where the number of left and right brackets are equal, every initial proper segment is contains more left brackets than right brackets and that the WFF is either an atomic formula or starts with a left bracket. But I'm having trouble proving that this formula is not a WFF:

$$(((p \rightarrow q)) \wedge p)$$

答:
One way of defining a well-formed formula goes that it comes as possible to construct a sequence of symbols where every step $\alpha$ gets built up according to the following :

$\alpha$ is a propositional variable.

$\alpha$ got obtained from a well-formed formula $\beta$ by prefixing a '$\lnot$' to $\beta$.

$\alpha$ got obtained from two well-formed formulas '$\beta$' and '$\gamma$' such that it has the following form ($\beta$$\land$$\gamma$).

$\alpha$ got obtained from two well-formed formulas '$\beta$' and '$\gamma$' such that if has the following form ($\beta$$\rightarrow$$\gamma$).

(((p→q))∧p) can't get obtained that way, since there is no sequence of well-formed formulas that yields (((p→q))∧p) as a well-formed formula.

Also, sometimes the author using the term 'well-formed formula' writes something like:

"Nothing else is a well-formed formula."

after writing the other parts of the inductive definition of a well-formed formula.

(((p→q))∧p) falls into that nothing else category, as follows from the fact that the left conjunct ((p→q)) can't get constructed, since (p$\rightarrow$q) has to exist in some construction sequence first, and there is no rule which allows one to add more parentheses.

In summary:

A sequence of symbols is a well-formed formula on the condition that it can get constructed according to the formation rules.  If a sequence of symbols can't get constructed according to the formation rules, it's not a well-formed formula.

Edit: One might point out that if a well-formed formula is well-formed in some infix language, then it has some unique counterpart in a prefix (or postfix) language and conversely.  But, were it the case that ((p$\rightarrow$q)) were well-formed, then $\rightarrow$pq (or pq$\rightarrow$) wouldn't have a unique infix language counterpart for the infix language at hand, since presumably (p$\rightarrow$q) and ((p$\rightarrow$q)) have the same meaning.

Title: Proving that a formula is not a WFF


答:
$X\sim \chi_{n}^2$

$E(X)=n,V(X)=2n$

$E(X^2)=V(X)+E(X)^2$

Above mentioned stuff is sufficient to solve this problem

Title: Expectation of Chi square distribution

Body:
Given that $Y$ is a degree $8$ chi square distribution i.e $Y \sim \chi_{8}^2$

I want to find $E[Y^2]$ and $Var[Y^2]$

I only know that $E[Y]$ of it should be $8$, but I don't know how to find $E[Y^2]$ from it?

Thank you


问:
Convergence of sub-optimal $k$-means algorithm?
Does the $k$-means algorithm converge to a local minimum if we alter the cluster assignment step? More specifically, if instead of assigning a point to the cluster whose centroid is closest to it, we assign it based on a different closest metric, will the algorithm still converge?

Surely the $SSE$ will no longer monotonically decrease as I can see in my experiments also but the algorithm does converge after a few steps. Is there any way to prove this?

答:
No. You must put some more restrictions. As a counterexample, take a 2 observation space with an expected $2$ clusters I believe if they take the second closest centroid, the assignments will be periodic with period 2. Then again you should perhaps make more precise what should converge, the centroids themselves or the centroid assignments and centroids. Centroid asignments are usually important to have as consistent.

Also many folks here may not know about k-means as it is an algorithm and may fall outside the topic. If you will post here you should at least repose it as a math question by defining all steps of the algorithm and defining 'convergence' in terms of sequences and topology.


问:
Show that the point $P$ of the line $f(t)=(at+\alpha, bt+\beta, ct+\gamma)$, $t \in \mathbb{R}$
Show that the point $P$ of the line $f(t)=(at+\alpha, bt+\beta, ct+\gamma)$, $t \in \mathbb{R}$, closest to the origin is that $P=f(t_{0})$ with $t_{0}=-\frac{a\alpha+b\beta+c\gamma}{a^{2}+b^{2}+c^{2}}$.

答: It suffices to show the vector from the origin to $P$ is orthogonal to the given line. This vector is given by $$\frac{1}{a^2+b^2+c^2}(-a b B - a c C + A b^2 + A c^2, a^2 B - a A b - b c C + B c^2, a^2 C - a A c + b^2 C - b B c).$$ Taking the dot product of this with the direction numbers $(a,b,c)$ for the line yields $0$ as desired.


问:
Determine whether the given set as follows is a vector space or not.
Determine whether the set $V=\Bbb R^2$ with vector addition and scalar multiplication defined as:
$(x,y)+(x_1,y_1)=(x+x_1,y+y_1+1)$ and
$r \cdot (x,y) = (rx,ry+r-1)$ for all $(x,y),(x_1,y_1) \in V$ and for all $r \in \Bbb R$
is a vector space or not.

Attempt:
Let $(x,y),(x_1,y_1),(a,b),(c,d),(e,f) \in V$ and $r \in \Bbb R$. Then,

$(x,y)+(x_1,y_1) \in V$ (by definition of $+$).

$r\cdot (x,y) \in V$ (by definition of $\cdot$).

Since
\begin{equation*}
(0,-1) + (x,y) = (0+x,-1+y+1) = (x,y),
\end{equation*}
then $(0,-1)$ is the additive identity in $V$.

Since
\begin{equation*}
(-x,-y-2) + (x,y) = (-x+x,-y-2+y+1) = (0,-1),
\end{equation*}
then $(-x,-y-2)$ is the additive invers in $V$.

$(a,b)+((c,d)+(e,f)) = ((a,b)+(c,d))+(e,f)$.

$r \cdot ((a,b)+(c,d)) = r\cdot (a,b) + r \cdot (c,d)$.

$(r+s)\cdot (a,b) = r \cdot (a,b) + s \cdot (a,b)$.

$1 \cdot (a,b) = (a,b)$, where $1$ is the multiplicative identity in $\Bbb R$.

$(rs)\cdot (a,b) = r\cdot (s \cdot (a,b))$.

$(a,b)+(c,d)=(c,d)+(a,b)$.

Hence, $V$ is a vector space under $+$ and $\cdot$ defined as above.

Am I true?

Body:
This is my prove of the other 6 axioms above (axiom 5,6,7,8,9,10).

For 5:
\begin{align*}
(a,b)+((c,d)+(e,f)) &= (a,b)+(c+e,d+f+1) \\
&= (a+(c+e),b+(d+f+1)+1) \\
&= ((a+c)+e,f+(b+d+1)+1) \\
&= (a+c,b+d+1)+(e,f) \\
&= ((a,b)+(c,d))+(e,f).
\end{align*}

For 6:
\begin{align*}
r \cdot ((a,b)+(c,d)) &=
r \cdot (a+c,b+d+1) \\
&= (r(a+c),r(b+d+1)+r-1) \\
&= (ra+rc,rb+rd+r+r-1) \\
&= (ra+rc,rb+r-1+rd+r-1+1) \\
&= (ra,rb+r-1)+(rb,rd+r-1) \\
&= r\cdot (a,b) + r \cdot (c,d).
\end{align*}

For 7:
\begin{align*}
(r+s)\cdot (a,b) &= ((r+s)a,(r+s)b+(r+s)-1) \\
&= (ra+sa,rb+sb+r+s-1-1+1) \\
&= (ra,rb+r-1)+(sa,sb+s-1) \\
&= r\cdot (a,b) + s\cdot (a,b).
\end{align*}

For 8:
\begin{equation*}
1 \cdot (a,b) = (1\cdot a, 1\cdot b + 1-1) = (a,b).
\end{equation*}

For 9:
\begin{align*}
(rs)\cdot (a,b) &= ((rs)a,(rs)b+(rs)-1) \\
&= (r(sa),r(sb)+rs-r+r-1) \\
&= r \cdot (sa,sb+s-1) \\
&= r \cdot (s \cdot (a,b)).
\end{align*}

For 10:
\begin{align*}
(a,b)+(c,d) &= (a+c,b+d+1) \\
&= (c+a,d+b+1) \\
&= (c,d)+(a,b).
\end{align*}


答:
Let $P = f(t)$ be the point on the given line closest to the origin.

The distance $D$ between $P$ and the origin is given by

$$D^2 = (at+\alpha)^2 + (bt+\beta)^2 + (ct+\gamma)^2$$

Or,
$$2D \frac{dD}{dt} = 2a(at+\alpha) + 2b(bt+\beta) + 2c(ct+\gamma)$$

To minimize $D$, let us set $$\frac{dD}{dt} = 0$$

Or, $$a(at+\alpha) + b(bt+\beta) + c(ct+\gamma) = 0$$

Or,  $$t = -\frac{a\alpha+b\beta+c\gamma}{a^{2}+b^{2}+c^{2}}$$

Body: Show that the point $P$ of the line $f(t)=(at+\alpha, bt+\beta, ct+\gamma)$, $t \in \mathbb{R}$, closest to the origin is that $P=f(t_{0})$ with $t_{0}=-\frac{a\alpha+b\beta+c\gamma}{a^{2}+b^{2}+c^{2}}$.


问:
Show $\sum_{k=0}^{n} \binom{2k}{k} \binom{2(n-k)}{n-k} = 4^n$.
For $n\ge 0$, show $\sum_{k=0}^{n}  \binom{2k}{k}  \binom{2(n-k)}{n-k} = 4^n$.

Want to first clarify the logic of the problem.

It is not only having variable $k$ (in number of items to be chosen in each category) as variable, but also having the quantity available in each category as variable.

I mean if instead of taking variable quantity : $2k, 2(n-k)$, have fixed qty. (say, $x,y$), then can define a problem as choosing a total of $k$ items from two categories having fixed quantities $x,y$.

But, now the two input quantities are variable and this makes it difficult to find a problem that can be represented by the equality.

Now, try to prove it using MI.

Over a well ordered set of non-negative integers, induction can be used.

Base case: $n=0$

$\binom{0}{0}  \binom{0}{0} = 1$.

Taking the rhs, get :
$4^0= 1$.

Induction hypothesis: it is true for natural number $n$.

So, $\sum_{k=0}^{n}  \binom{2k}{k} \binom{2(n-k)}{n-k} = 4^n$

Inductive step: need show for next natural $n+1$

$\sum_{k=0}^{n+1} \binom{2k}{k} \binom{2(n+1-k)}{n+1-k} = 4^{n+1}$

This should lead to :

$4^n + \binom{2(n+1)}{(n+1)} \binom{2((n+1)-(n+1))}{((n+1)-(n+1))}= 4^{n+1}$

$\implies \binom{2(n+1)}{(n+1)} = 4^{n}.(3)$

Please help to proceed further, or suggest alternative approach.

答: Take squares on both sides of the expansion $$(1-4x)^{-1/2}=\sum_{n=0}^{\infty}{2n \choose n}x^n.$$


问:
How to find eigenfuctions of shifts operator and dilation operators
I am trying to find eigenfunction of shift operators and dilation operators. Shift operators $S_t$ maps function $f(x)$ to $f(x+t)$. I am trying to find continuous function $f$ such that $(S_tf)(x) = \lambda_t f(x)$ for all $t\in\mathbb{R}$, i.e. $f(x+t)=\lambda_tf(x)$ for all $t$. I know the answer is $ce^{ax}$ for some $c,a$. It's easy to verify it, but how to prove it? In other words, how to prove that other continuous function can not be the eigenvector function?

I've shown that $f(0)$ is not 0 and I think we could also prove $f(x)=a^x$ for some a.

Similarly, I try to find eigenfunction of dilation operator which is defined as $(D_rf)(x)=f(rx)$ for $r>0$. I think they are similar, is there any way to solve them?

答:
Suppose that $f: \mathbb{R} \to \mathbb{C}$ is a continuous eigenfunction for shift operators as you say, i.e. for every $t \in \mathbb{R}$ there is a $\lambda_t \in \mathbb{C}$ such that $(S_t f)(x) = f(x + t) = \lambda_t f(x)$ for all $x$. Also assume that $f(0) \neq 0$, as otherwise it's fairly quick to derive that $f \equiv 0$. Then the mapping $t \mapsto \lambda_t$ must be a continuous homomorphism from the reals under addition to the (nonzero) complex numbers under multiplication.

The homomorphism part (setting aside continuity) comes from noting that for any $s, t \in \mathbb{R}$,
$$
\lambda_{s+t} f(0)
= (S_{s+t} f) (0)
= (S_s S_t f) (0)
= \lambda_s \lambda_t f(0).
$$
(Technically we should also show that no $\lambda_t$ can be equal to $0$, but I'll leave that as an exercise.) Being continuous comes from the continuity of $f$ together with the fact that $f(t) = \lambda_t f(0)$, i.e. $\lambda_t = f(t) / f(0)$.

It now follows (see here) that there is a complex number $z$ such that $\lambda_t = e^{zt}$. Therefore $f$ has the form $f(t) = f(0) e^{zt}$.

The question for dilations is fairly similar, though with a couple additional considerations. Here we begin by assuming that at least one of $f(1)$ and $f(-1)$ is nonzero, as otherwise we would obtain $f \equiv 0$. Using the notation $(D_r f)(x) = \mu_r f(x)$, $r > 0$, we can derive that $\mu_{rs} = \mu_r \mu_s$; and $f(r) = \mu_r f(1)$ and $f(-r) = \mu_r f(-1)$ for $r > 0$. So now the map $r \mapsto \mu_r$ is a continuous homomorphism from the positive reals under multiplication to the nonzero complex numbers under multiplication. But further, by the continuity of $f$ at $0$ we can conclude that the limit $\lim_{r \to 0} \mu_r$ must exist and be finite; and moreover that $f(1) = f(-1)$.

Via a logarithmic change of variable, we can convert to a continuous homomorphism from the reals under addition to the nonzero complex numbers under multiplication. The new homomorphism is given by $t \mapsto \mu_{e^t}$. As with the first part above for shift operators, we obtain a $z \in \mathbb{C}$ so that $\mu_{e^t} = e^{zt}$. Using the fact that $\lim_{r \to 0} \mu_r = \lim_{t \to -\infty} \mu_{e^t}$ exists and is finite, we conclude that $z$ must be a nonnegative real number. So now, denoting $e^z$ by $a$ (so that $a \geq 1$), we have $f(r) = f(1) a^{\log |r|}$ for $r \neq 0$, and $f(0)$ is either $f(1)$ (if $a = 1$) or $0$ (if $a > 1$).


问:
Given $n=36$, $ p=20/36$, and $p_0=0.5$; calculate $p$-value
I know so far that:

$H_0:\ p= 0.5,$

$H_1:\ p> 0.5.$

And that $z$-score: $0.672$

I did $1-0.44 = 0.56$ for $p$-value but got it wrong, the correct answer is $0.2514$

答:
The correct z- score is

$$z=\Bigg(\frac{5}{9}-\frac{1}{2}\Bigg)12=\frac{2}{3}$$

Thus p-value is (using paper table)

$$\mathbb{P}[Z>0.67]=1-0.7486=0.2514$$

EDIT : I understood your error: you misread z table because you confused quantiles and probabilities; the numbers in the body of the table are probabilities while quantiles are in row+ column


问:
Find the special function form of $\int_0^{+\infty} \left(\sqrt{1+x^4} - x^2\right) dx$
I'd like to have an expression of the following integral:
$$\int_0^{+\infty} \left(\sqrt{1+x^4} - x^2\right) dx$$
in terms of some special functions (but not in the form given by Wolfram Alpha).</p>

答:
The simplified answer seems to be :
$$\frac 23K\left(\frac 1{\sqrt{2}}\right)$$
This means $\frac 23K\left(\frac 12\right)$ using Alpha's convention with a rather interesting alternate form :

\begin{align}
\frac{\Gamma\bigl(\frac 14\bigr)^2}{6\;\sqrt{\pi}}&=\frac{\Gamma\bigl(\frac 14\bigr)^2}{6\;\Gamma\bigl(\frac 12\bigr)}\\
&=\frac 16 B\left(\frac 14,\frac 14\right)\\
&=\frac 13 \int_0^\infty \frac {t^{1/4-1}}{\sqrt{1+t}}\;dt\\
\end{align}
with $B(x,y)$ the Euler integral of the first kind ('beta function') and using the DLMF classical relation $(5.12.3)$. Setting $x:=t^{1/4}$ should bring you nearer to the initial question...

As noted by O.L. the rewriting of the elliptic integral $K(z)$ in terms of $\Gamma$ function is due to $z=\frac 1{\sqrt{2}}$ being one of the singular values of the elliptic integral. It is in fact the first one (see equation $(11)$ here). Your specific case appears in page $3$ of the original article of Chowla and Selberg.


问:
Solving limit without using l'hopitals rule
Find the limit as $h \to 0$ for $\frac{f(x)-f(x-h)}{h}$ at $x=0$, where $f(x) = x + |x|$ without using L'hopital rule.

Once I plug in the numbers and try to work it down I get brought to $\frac{-(-h + |-h|)}{h} = \frac{h-h}{h}$ which is $\frac{0}{h}$ and then if go to plug in $0$ for $h$ then I would get $0/0$ which is indeterminate. How should I approach this problem to get the answer?

答:
In general,

$$
\lvert h\lvert=
\begin{cases}
h &,& h\geq0 \\
-h &,& h<0
\end{cases}
$$

If we want to find $\lim_{h\to 0}\lvert h\lvert$, we need to consider what happens as $h$ approaches $0$ from the right and the left.

If $h$ is approaching $0$ from the right, then $h>0$; that is, we are in the first case of $\lvert h\lvert$, and $$\lim_{h\to 0^+}\lvert h\lvert = \lim_{h\to 0^+}h.$$

If $h$ is approaching $0$ from the left, then $h<0$; that is, we are in the second case of $\lvert h\lvert$, and $$\lim_{h\to 0^-}\lvert h\lvert = \lim_{h\to 0^-}(-h).$$

Now, to your question. We want to find
$$\lim_{h\to 0}\frac{f(0)-f(0-h)}{h}=\lim_{h\to 0}\frac{h-\lvert h\lvert}{h},$$
after some simplifying.

For the right limit,
\begin{align*}
\lim_{h\to 0^+}\frac{h-\lvert h\lvert}{h}&=\lim_{h\to 0^+}\frac{h-h}{h}=\lim_{h\to 0^+}\frac{0h}{h}=\lim_{h\to 0^+}0=0
\end{align*}

For the left limit,
\begin{align*}
\lim_{h\to 0^-}\frac{h-\lvert h\lvert}{h}&=\lim_{h\to 0^-}\frac{h+h}{h}=\lim_{h\to 0^-}\frac{2h}{h}=\lim_{h\to 0^-}2=2
\end{align*}

Since the left and right limits don't agree, the limit does not exist.


问:
Proof of an inverse Matrix
I am very raw at proofs, this is only my first semester learning them and I am having trouble with this problem. How would I approach this ?

If $$ is an $ × $ matrix that satisfies the equation $ $ is $A^3 -4A^2 +3A -5I_n = 0$, Find the $A^{-1}$

Title: Proof of an inverse Matrix

Body:
$$A^3 -4A^2 +3A -5I_n = 0$$

Or,
$$5I_n = A^3 -4A^2 +3A$$

Pre-multiplying both sides by $A^{-1}$ and dividing by $5$,
$$A^{-1} = \frac{1}{5}(A^2 -4A +3I_n)$$


问:
ODE involving matrix exponentiation
Exercise 5.14 (Hunter and Nachtergaele chapter 5). Suppose that $A$ is an $n \times n$ matrix. For $t \in \mathbb{R}$ we define $f(t)=$ $\operatorname{det} e^{t A}$

(a) Show that
$$
\lim _{t \rightarrow 0} \frac{f(t)-1}{t}=\operatorname{tr} A
$$
where $\operatorname{tr} A$ is the trace of the matrix $A,$ that is the sum of its diagonal elements.

(b) Deduce that $f: \mathbb{R} \rightarrow \mathbb{R}$ is differentiable, and is a solution of the ODE $\dot{f}=(\operatorname{tr} A) f$

(c) Show that
$$
\operatorname{det} e^{A}=e^{\operatorname{tr} A}
$$

I have used that $A$ is similar to its Jordan form. But not able to proceed. Any help is appreciated.

答:
Let $J$ be the Jordan form of $A$. Then $A$ and $J$, as well as $e^{tA}$ and $e^{tJ}$ are similar. So
$$
f(t)=\det e^{tA} = \det e^{tJ} = \prod_{i=1}^n e^{t\lambda_i},
$$
where $\lambda_i$ are the $n$ eigenvalues of $A$. Now the three task should be very simple. Note that in (a) and (b) the matrix $A$ is fixed, and $t$ is the parameter to do differentiation. You do not have to take derivatives with respect to the eigenvalues.

Title: ODE involving matrix exponentiation


问:
A curious equality: Where do these numbers come from?
This identity, which was shared on math.stackexchange and seem curious at first sight, caught my attention. Here is the equality:

$$\color {red} {\dfrac{1646-736\sqrt{5}} {2641-1181\sqrt{5}} =\color{blue}{\dfrac{17+15\sqrt{5}} {7+15\sqrt{5}}}}$$

The truth is that even the almost all students can easily prove this.

But, what I'm wondering about isn't the proof that actually uses a cross multiplication.

Goal : Knowing where these numbers come from

Maybe, here something is hiding from us. Has a complicated fraction been simplified to an easy fraction?

I have reasons not to believe this.

I will try to show that the numbers are not randomly chosen and moreover, where they come from.

Let, $$m,n,r,s \in\mathbb {Z}.$$

We have:

$$\frac{m-n\sqrt{5}} {r-s\sqrt{5}} =\frac{17+15\sqrt{5}} {7+15\sqrt{5}}$$

$$\implies ( r-s\sqrt{5})(17+15\sqrt{5})=(m-n\sqrt{5})(7+15\sqrt {5})$$

$$\implies (17r-75s)+\sqrt {5}(15r-17s)=(7m-75n)+\sqrt {5}(15m-7n)$$

Then we have,

$$\begin{cases} 15r-17s=15m-7n \\ 17r-75s=7m-75n \end{cases}$$

There are probably good methods for solving such equations in number theory.  But since I don't know, I usually solve such equations by methods I know myself.

$$\begin{cases} 255r-289s= 255m-119n \\ 255 r -1125s =105 m - 1125 n \\ 1125r-1275s=1125 m - 525 n \\ 289r-1275s= 119 m - 1275 n \end{cases}$$

$$ \implies \begin{cases} r=\dfrac {503m+375n}{418} \in\mathbb Z \\ s=\dfrac {75m+503n}{418} \in\mathbb Z \end{cases}$$

First restriction

$$\begin {align} r=\dfrac {503m+375n}{418} \in\mathbb Z \Longrightarrow r=\dfrac {418m+85m+418n-43n}{418} \Longrightarrow \dfrac{85m-43n}{418}=k_1 \Longrightarrow m=\dfrac{418k_1+43n}{85}=\dfrac{425k_1-7k_1+43n}{85} \Longrightarrow \dfrac {43n-7k_1}{85}=k_2 \Longrightarrow k_1=\dfrac{42n-84k_2+(n-k_2)}{7} \Longrightarrow \dfrac {n-k_2}{7}=k_3 \Longrightarrow n=7k_3+k_2 \end {align} $$

$$\implies k_1=\dfrac{43(7k_3+k_2)-85k_2}{7}=43k_3-6k_2 $$

$$\implies m= \dfrac { 418(43k_3-6k_2)+43(7k_3+k_2)}{85}=215k_3-29k_2$$

$$\implies n=7k_3+k_2$$

$$\implies r=\dfrac {503m+375n}{418} =\dfrac {503(215k_3-29k_2)+375(7k_3+k_2)}{418}=265k_3-34k_2, ~ \text {where} ~ k_2, k_3 \in\mathbb Z.$$

$$\color {gold}{\boxed {\color{black} {r=265k_3-34k_2}}}$$

Second restriction

$$\begin{align} s=\dfrac {75m+503n}{418} \in\mathbb Z \Longrightarrow s=\dfrac {75(215k_3-29k_2)+503(7k_3+k_2)}{418}=47k_3-4k_2, ~ \text {where} ~ k_2, k_3 \in\mathbb Z. \end{align}$$

$$\color {gold}{\boxed {\color{black} {s=47k_3-4k_2}}}$$

Thus, no new restrictions are required.

Here for simplicity, I am changing $k_2 = v, k_3 = u$, where $u,v\in\mathbb Z$. Then we have,

$$\color {gold}{\boxed {\color{black} {\begin {align} r=265u-34v\\ s=47u-4v \\m=215u-29v \\n=7u+v \end{align}}}}$$

Now we have to solve any of the equations we obtain. Therefore, solving the first equation will be sufficient:

$$r=265u-34v.$$

$$\begin{align} v=\dfrac{265u-r}{34}=8-\dfrac{7u+r}{34}\Longrightarrow \dfrac {7u+r}{34}=k_4 \Longrightarrow u=5k_4-\dfrac{k_4+r}{7} \Longrightarrow k_4+r=7k_5 \Longrightarrow k_4=7k_5-r, ~ \text{where} ~ k_4, k_5 \in\mathbb Z \end{align}$$

$$\implies u=34k_5-5r$$

$$\implies v=\dfrac {265(34k_5-5r)-r}{34}=265k_5-39r$$

For simplicity, I accept $k_5 \longmapsto k$. So, we have

$$\begin{align} u=34k-5r \\ v=265k-39r \end{align}$$

Finally we have,

$$\color {gold}{\boxed {\color{black} {\begin{align} \color{blue}{r=r} \\ \color {red}{s=-79 r+538k} \\ \color{green} {m=56 r- 375 k} \\ \color{blueviolet}{n=- 74 r+503k} \end{align}}}}$$

Let's go back to the original problem.

If $r=2641$, then we have

$$\begin{align} r=2641 \\ s=-208639+538k \\ m=147896- 375 k \\ n=−195434+503k \end{align}$$

Now we can choose $k$ as we wish.

Let, $k=390$, then we have

$$\begin {align} m=1646 \\ n=736 \\ r=2641 \\ s=1181.\end{align}$$

Thus, we obtain the required equality:

$$\color {red}{\boxed {\color{black} {\dfrac{1646-736\sqrt{5}} {2641-1181\sqrt{5}} =\dfrac{17+15\sqrt{5}} {7+15\sqrt{5}}}}}$$

And now here we can make the numbers as difficult as we want.  Let's try one example.

If $m,n,r,s \in\mathbb {Z^{+}}$, then we must solve the following system of equations:

$$\begin{cases} r>0 \\ -79 r+538k>0 \\ 56 r- 375 k>0 \\- 74 r+503k >0 \end{cases} \Longrightarrow k>0, \dfrac {375k}{56}<r<\dfrac {503k}{74}$$

Let, $k=17639735$ and $r=118123483$ ( a prime number), then we have

$$\begin{align} r=118123483 \\ s = 158422273 \\m=14423 \\ n=131648963 \end{align}$$

Finally, our new "magical" identity will be as follows:

$$\color {gold}{\boxed {\color{blue} {\dfrac{14423-131648963\sqrt{5}} {118123483-158422273\sqrt{5}} \color{red}{=\dfrac{17+15\sqrt{5}} {7+15\sqrt{5}}}}}}$$

Short proof

$$\dfrac{1646-736\sqrt{5}} {2641-1181\sqrt{5}} =\dfrac{17+15\sqrt{5}} {7+15\sqrt{5}}$$

$$\Longleftrightarrow \dfrac{445\sqrt 5-995} {1646-736\sqrt{5}} =\dfrac{10} {17+15\sqrt{5}}$$

$$\Longleftrightarrow \dfrac{89\sqrt 5-199} {1646-736\sqrt{5}} =\dfrac{2} {17+15\sqrt{5}}$$

$$ \Longleftrightarrow \dfrac{1845-825\sqrt 5} {1646-736\sqrt{5}} =\dfrac{15+15\sqrt{5}} {17+15\sqrt{5}}$$

$$\Longleftrightarrow \dfrac{123-55\sqrt 5} {1646-736\sqrt{5}} =\dfrac{1+\sqrt 5} {17+15\sqrt{5}}$$

$$\Longleftrightarrow \dfrac  {1646-736\sqrt{5}} {123-55\sqrt 5}=\dfrac  {17+15\sqrt{5}}{1+\sqrt 5}=15+\dfrac{2}{1+\sqrt 5}$$

$$\Longleftrightarrow \dfrac{89\sqrt 5-199} {123-55\sqrt{5}} =\dfrac{2} {1+\sqrt{5}}$$

$$\Longleftrightarrow \dfrac {246-110\sqrt 5}{89\sqrt 5-199} =1+\sqrt 5$$

$$\Longleftrightarrow  \dfrac {246-110\sqrt 5-89\times 5+199 \sqrt 5}{89\sqrt 5-199}=\dfrac {89\sqrt 5-199}{89\sqrt 5-199}=1.$$

Observation

We always have the chance to choose at least one of the numbers $m, n, r, s$ as a prime number. According our formulas, we have

$$\color {red}{\boxed {\color{black} {\begin{align} r=r \\ s=-79 r+538k \\ m=56 r- 375 k \\ n=- 74 r+503k \end{align}}}}$$

So, we can choose the number $\color{blue}{r}$ always as a prime.

Let's look at the following equality, which is completely randomly selected.

$$\color {red} {\dfrac{179277361-8263\sqrt{5}} {9273317-5193\sqrt{5}} =\color{blue}{\dfrac{a+b\sqrt{5}} {c+d\sqrt{5}}}}$$

We have no good reason for solving this equation that "small" values of $ a, b, c, d$ exist.

Therefore, I came to the following conclusion.

The expression $\color {red}{\dfrac{17+15\sqrt{5}} {7+15\sqrt{5}}}$ is not taken from the expression $\color{blue}{\dfrac{1646-736\sqrt{5}} {2641-1181\sqrt{5}}}.$  On the contrary, the expression $\color{blue}{\dfrac{1646-736\sqrt{5}} {2641-1181\sqrt{5}}}$  is taken from the expression $\color{red}{\dfrac{17+15\sqrt{5}} {7+15\sqrt{5}}}.$

Generalization:

For any $p\in\mathbb Z^{+}$ and small $a,b,c,d$, there exist infinitely many enough large pozitive integers $m,n,r,s$ (we can choose at least one as a prime number) such that,

$$\frac{m-n\sqrt{p}} {r-s\sqrt{p}} =\frac{a+b\sqrt{p}} {c+d\sqrt{p}}$$

Questions

Could these numbers have come from a completely different place that has nothing to do with the method I use?

Does mathematics allow for the techniques I have applied above? Are the things I've done correct?

Thank you.

答: We go from the blue fraction to the red one by multiplying the numerator & denominator by $\tfrac{-199+89\sqrt{5}}{2}$. More generally $\tfrac{17+15\sqrt{5}}{7+15\sqrt{5}}=\tfrac{(17+15\sqrt{5})(a+b\sqrt{5})}{(7+15\sqrt{5})(a+b\sqrt{5})}=\tfrac{17a+75b+(15a+17b)\sqrt{5}}{7a+75b+(15a+7b)\sqrt{5}}$. Now take $a=-\tfrac{199}{2},\,b=\tfrac{89}{2}$. These satisfy $a^2-5b^2=-1$, but we can do the above with arbitrary $a,\,b$.


问:
Show that eigenvalues are negative
I have to consider the eigenvalue problem:
$$ L[u] := \frac{d^2 u}{dx^2}= λu,x \in (0,1)\quad  u(0)-\frac{du}{dx}(0)=0, u(1)=0.$$
I need to show that the eigenvalues are negative. </p>

答:
Suppose we have $\lambda > 0$, then $u'' = \lambda u$ gives us $$u(t) = \alpha\exp(-\sqrt\lambda t) + \beta\exp(\sqrt\lambda t)$$
with $\alpha, \beta \in \mathbb R$. Now
\begin{align*}
  u(0) - u'(0) &= \alpha + \beta -\sqrt \lambda \alpha + \sqrt \lambda \beta\\
     &= \alpha(1 - \sqrt \lambda) + \beta(1 + \sqrt\lambda)\\
\text{and } u(1)&= \alpha\exp(-\sqrt \lambda) + \beta\exp(\sqrt\lambda)
\end{align*}
So we must have by the boundary conditions
$$\beta = -\alpha \exp(2\sqrt\lambda), \quad \beta = -\alpha\cdot\frac{1-\sqrt \lambda}{1+\sqrt \lambda} $$
that is $\alpha = \beta = 0$ or
$$ \exp(2\sqrt \lambda) = \frac{1-\sqrt\lambda}{1+ \sqrt\lambda} $$
But this is impossible for $\lambda > 0$ as $\lambda \mapsto \exp(2\sqrt\lambda)$ is strictly increasing and $\lambda \mapsto \frac{1-\sqrt\lambda}{1+\sqrt \lambda}$ is decreasing and they are both equal at $0$. So $\alpha = \beta = 0$ and no $\lambda > 0$ is an eigenvalue.</p>

Title: Show that eigenvalues are negative


问:
How to determine if this Differential Equation has a singular solution?
I found this Differential Equation $y'^2=16x^2$ and the text says it has a singular solution...

first they make an implicity derivation with respect to $y'$ then they have $2y'=0$. Therefore $y=constant$ and it proves that the D.E. has a singular solution but I can't figure out why it is correct.

答:
Singular Solution is the Envelope of a Family of Curves.

For an equation to be Singular Solution, it has to be common in both p-discriminant and c-discriminant and also has to satisfy the Differential Equation.

Here,

Differential Equation is $ f(x, y, y’) = y'^2 - 16x^2 = 0 $ and

General Solution is $ \phi(x, y, c) = (y-c)^2 - 4x^4 = 0 $

For finding p-discriminant, we find $ \frac {\partial f} {\partial y’} = 0 $ which gives $ 2y’ = 0 $

Now using the above, we need to eliminate $ y’ $ from $ f(x, y, y’) $ which gives $ x^2 = 0 $ which is a possible singular solution.

In this case, when we check whether $ x^2 = 0 $ satisfies the above given Differential Equation, we can see that it doesn’t ( $ \because y’ $ for this curve is tending to $\infty $ )

Graphically also, we can see that $ x^2 = 0 $ is not the envelope of the given family of curves. So, it’s not a singular solution.


问:
Filters in Topology (Converge)
Let $X$ be an infinite set, $\mathscr{F}$ the filter on $X$ generated by the filter base consisting of all complements of finite sets. To which points does $\mathscr{F}$ converge if $X$ is given the cofinite topology?

I would suspect the answer to be every point in $X$. My reasoning, though may be wrong, is given below.

Let $\mathscr{C}$ denote the filter base, i.e. $\mathscr{C}:=\{A^c: A\subset X\text{ is finite}\}$ and $\mathscr{U}_x$ denote the set of all nhbds of $x$. The filter $\mathscr{F}$ converges to $x\in X$ iff each $U\in \mathscr{U}_x$ contains some $C\in\mathscr{C}$.

Notice,

$$C\subseteq U\iff U^c\subseteq C^c.$$

By definition of the topology, the set $U^c$ is finite, whereas by definition of the filter base, the set $C^c$ is infinite. Therefore, each $U\in \mathscr{U}_x$ contains some $C\in\mathscr{C}$.

答: Yes, if $x \in X$, then any open neighbourhood of $x$ is of the form $F^\complement$, where $F$ is finite. By definition $F^\complement \in \mathcal{F}$ (It has finite complement so is in the filterbase), and so all (open) neighbourhoods of $x$ are in the filter and we have convergence $\mathcal{F}\to x$, for any $x$.

Title: Filters in Topology (Converge)


问:
Changing variables double integral
I want to know if there's a way to know the best variables change for an integral calculation. For example, if we consider the integral

$$ \iint_{R}\left(x+y\right)dx\,dy $$

Where $ R $ is bounded by the line $ y=x,\thinspace\thinspace\thinspace y=2x,\thinspace\thinspace\thinspace\thinspace y+x=2 $.

This integral is very easy to calculate directly, without variables change. But I wonder how simple it can get if one knows how to change the variables in an efficient way.

I'll give another example;

$$ \iint_{R}\left(x-y\right)e^{x^{2}-y^{2}}dx\,dy $$

Where $ R $ is bounded by the curves $ x^{2}-y^{2}=1,\thinspace\thinspace\thinspace\thinspace x^{2}-y^{2}=-1,\thinspace\thinspace\thinspace\thinspace x+y=1,\thinspace\thinspace\thinspace\thinspace x+y=3 $.

In this example it is pretty obvious that we should change the variables in the following way:

$$ \left(x-y\right)=u,\thinspace\thinspace\thinspace\thinspace\left(x+y\right)=v $$

But how can I find a convenient variables change when its not very clear and implied, like in the first example I gave

Thanks in advance.

答:
Sometimes the region in an integral is such that you cannot compute without splitting the region into $2$ or more sub-regions, and you have to found individual bounds. The idea of finding a change of variable is such that the region becomes simpler and you do not have to break the integral into multiple. But when you simplify the region, the integrand may transform such that the integral becomes difficult. So the change of variables needs to be chosen keeping both in mind.

The same is true when the integral is difficult and we are trying to simplify that by using change of variable - we need to keep in mind what it does to the region.

In your first case,

$\displaystyle \iint_{R}\left(x+y\right)dx\,dy$

Where $R: \ y = x, y = 2x, x+y=2$

For the given region, the integral needs to be split into two. So if the idea is to transform this into a region such that we can compute it with one integral,

Writing the bounds as $1 \leq \frac{y}{x} \leq 2, 0 \leq x + y \leq 2$ makes it obvious that $u = \frac{y}{x}, v = x + y$ will transform the region into a rectanle with $1 \leq u \leq 2, 0 \leq v \leq 2$.

$|J| = \frac{v}{(1+u)^2}$

The integral becomes $\displaystyle \int_0^2 \int_1^2 \frac{v^2}{(1+u)^2} \ du \ dv \ $ which is still not difficult to compute.

Your second case -

$\displaystyle \iint_{R}\left(x-y\right)e^{x^{2}-y^{2}}dx\,dy$

Where $ \ R: \ -1 \leq x^{2}-y^{2} \leq 1, 1 \leq x+y \leq 3$

The integral in fact needs to be split into $3$. So to simplify the region, the natural choice that I see is

$1 \leq  u (= x+y) \leq 3, -1 \leq v ( = x^2-y^2) \leq 1$, which transforms the region into a rectangle. $|J| = \frac{1}{2u}$. So integral becomes,

$\displaystyle \int_{-1}^{1} \int_{1}^{3}  \frac{v e^v} {2u^2} \ du \ dv = \frac{2}{3e}$

Title: Changing variables double integral


问:
Are topological spaces $X,Y$ homeomorphic if each is homeomorphic to a subspace of the other?
The Schroeder-Bernstein theorem states that

If $A$ and $B$ are two sets and there exists injective maps $f:A \to B$ and $g:B \to A$, then there exists a bijection $h:A \to B$.

The statement above is equivalent to the following

Suppose $A_1 \subset A$ and $B_1 \subset B$ are sets such that there exists a bijective map $f:A \to B_1$ and $g:B \to A_1$, then there exists a bijection $h:A \to B$.

So I was thinking about the problem below. I couldn't prove it, but not getting a counterexample either. Any hints will be appreciated.

Now suppose $A \subset X$, $B \subset Y$ are topological spaces, and suppose that there exists a homeomorphism $\alpha:A \to Y$ and $\beta: B \to X$, then can we conclude that $X$ and $Y$ are homeomorphic?

答: For example, consider $X=[0,1]$ and $Y=(0,1)$. $X$ is not homeomorphic to $Y.$ but $X$ is homeomorphic to $[1/4,3/4] \subset Y=(0,1).$ and Y is homeomorphic to $(0,1)\subset X.$


问:
Product of variables

Suppose we have a set of variables $\{a_i,b_i| i=1,2,3\}$ which can take values $\pm 1$ according to some probabilities.

If there's a constraint that $a_1b_2b_3=a_2b_1b_3=a_3b_1b_2$ must equal to $1$, why then must the value of $$a_1a_2a_3=1$$?

I thought of saying that $a_1b_2b_3a_2b_1b_3a_3b_1b_2=a_1a_2a_3(b_1b_2b_3)^2=1$ therefore the result follows, but I don't think this argument is valid, since the first $b_1$ may differ from the second. -- they are variables and since there is a non-zero probability for both $\pm 1$...

Also, I don't understand why we couldn't have $$a_1=a_2=+1, a_3=-1$$ and have the $\{b_i\}$'s take appropriate values -- since there is a probability, it's possible, right?

Or have I totally misunderstood the question?

答:
Your argument involving $a_1a_2a_3(b_1b_2b_3)^2$ is correct.

The vector $(a_1,a_2,a_3, b_1,b_2,b_3)$ can take $2^6=64$ possible values. ($\pm 1$ for each variable).  The constraint limits the set of values that we are allowed to those that satisfy $a_1b_2b_3=a_2b_1b_3=a_3b_1b_2 = 1$. Sp there are only $8$ possible values that satisfy this constraint. (we are free to choose $b_i$ however we like, but this fixes the $a-i$.)

Your probability distribution then assigns a probability to each of these $8$ vectors, and $0$ to the remaining $56$.

So all you have to show is that $a_1a_2a_3 = 1$ for all $8$ possible values.  Which you've already done.

Title: Product of variables


问:
How do you apply finite geometric series in order to determine the distance a bouncing ball travels up and down at the 10th bounce?
My question is very similar to this one, except I would like to determine the distance traveled immediately after the 10th bounce.

Assume the ball is let go from 1 meter above the ground and each successive bounce is 2/3 the height of the previous bounce.

So the series is 1 + 2 * (2/3 + 4/9 + 8/27 + ...) = 1 + 2 (  2   ) = 5  (by applying infinite geometric)

So obviously the final answer for the distance the ball travels by the 10th bounce must be less than 5.

At the 10th bounce the distance the ball travels should be:

1 + $2\sum_{k=0}^{9}(\frac{2}{3})^{k+1} = 1 + 4(1-(\frac{2}{3})^{10}) = 5 - 4(\frac{2}{3})^{10}\approx 4.93$ meters

However the solution to this problem says after the 10th bounce the ball has traveled $6 (\frac{2}{3})^{10} \approx .104 $ meters.

What am I doing wrong?

答:
I think you  have misunderstood the question and calculated how far the ball travels before rather than after the 10th bounce.

After the 10th bounce the distance travelled is $\sum_{k=10}^{\infty} 2\cdot(\frac{2}{3})^k$ which is the required $6\cdot (\frac{2}{3})^{10}$.


问:
Convergence of a filter defined in a subspace
I am reading about convergence of filters in topological spaces and cluster points. My doubt is the following: let $X$ be a topological space, $A\subset X$ and $\mathcal F$ a filter on $A$. Given $x \in \overline A$, what is the meaning of $\mathcal F \longrightarrow x$?

Bourbaki defines the convergence of a filter (page 68 - topology 1) which is defined in the whole space $X$. Of course that is not the case of $\mathcal F$ if $A\ne X$. But in Proposition 6 of page 70 they use this kind of convergence. I am assuming that this means that  $\mathcal F$ is finer than $\mathcal B(x)_A$, which is the trace of the filter of the neighborhoods of $x$. Is that correct? I found it very strange since Bourbaki's book (as far as I know) does not make this kind of imprecision (to work with a not given definition).

In Dugundji's book only the definition of convergence of a filterbase is given (Def. 2.1 of Chapter 10). So his Theorem 4.1 (that is the cited proposition of Bourbaki's book) is coherent.

答:
If $\mathcal{F}$ is a filter on $X$ that contains $A \subseteq X$, then $\mathcal{F}_A:= \{B \subseteq A\mid B \in \mathcal{F}\}=\mathscr{P}(A) \cap \mathcal{F}$ is a filter on $A$ (it's clear that $\emptyset \notin \mathcal{F}_A \subseteq \mathcal{F}$, and $\mathcal{F}_A$ is also clearly closed under intersections and enlargements "within $A$"). And clearly any filter on $A$ contains $A$ and is a filterbase for a filter $\mathcal{F}^\uparrow$ (still containing $A$) on $X$ that obeys $\mathcal{F}^\uparrow_A = \mathcal{F}$ again.
So there is a natural correspondence between them.

It's also easy to see that for such a filter $\mathcal{F}$ on $X$ that contains $A$, and any $x \in A$, $\mathcal{F} \to x$ in $X$ iff $\mathcal{F}_A \to x$ in the subspace topology on $A$.

If $\mathcal{F}$ is a filter on $A$ then $\mathcal{F} \to x$ is defined as the convergence of a filterbase. So as $\mathcal{F}^\uparrow \to x$ if you want to restrict yourself to convergence of (full) filters on $X$. But IIRC Bourbaki defines convergence for filter bases in general anyway.

"Every point of $x$ of $\overline{A}$ is a limit point of a filter on $A$" is true as we can take that filter to be $\mathcal{F}:=\{N \cap A: N \in \mathcal{N}_x\}$, which is a filter on $A$ and a filterbase on $X$ for which $\mathcal F \to x$ is true and makes sense.


问:
statement about continuous function - false, why?
Let $f$ be a function defined on the closed interval $[0,1]$, such that $f(0)<f(1)$.

If for every $f(0)<b<f(1)$ there exists a $0<c<1$ such that $f(c)=b$

then $f(x)$ is continuous on the closed interval $[0,1]$.

I'm having trouble understanding why this is statement isn't true.
Any idea to help me visualize it in some form will help a lot.

Thanks!

答:
It just has to "hit" all values. But it can jump about as much as it likes in doing so.

Suppose for instance $f(x) = \begin{cases} f(\frac 12) = \frac 14\\f(\frac 14) = \frac 12\\f(x)= x&\text{if }x\ne \frac 12;x\ne \frac 14\end{cases}$.

Then every condition is satisfied $f(0) = 0 < f(1) = 1$ and for every $0 < b < 1$ we have a $c: 0< c < 1$ so that $f(c) = b$.  (If $b = \frac 12$ then $f(\frac 14)=b$.  If $b =\frac 14$ then $f(\frac 12) =b$. And if $b \ne \frac 14$ and $b\ne \frac 12$ then $f(b) = b$.

But $f$ is not continuous at $\frac 12$ or at $\frac 14$.

If we want we can totally scramble them up.  Let $f(x) =\begin{cases}x& \text{if }x\in \mathbb Q\\ 1-x &\text{if} x\not\in \mathbb Q\end{cases}$.


问:
State an antiderivative $F: D \longrightarrow \mathbb{C}$ or explain why an antiderivative does not exist.
State an antiderivative $F: D \longrightarrow \mathbb{C}$ or explain why an antiderivative does not exist.

a) $f(z) = z^2 , D = \mathbb{C}$

b) $f(z) = z \sin z,  D = \mathbb{C}$

c) $f(z) = |z|^2 , D = \mathbb{C}$

d) $f(z) = \overline{z} , D = \mathbb{C}$

For a) and b) I just performed simple integration for the equations

a)$ z^3/3 + C$ and b) $-z\cos z + \sin z +C$

but I am stuck on c) and d). It seems like an antiderivative does not exist but how do i explain that?

答:
If $F: D \to \Bbb C$ is complex differentiable at all points of $D$ then it is a holomorphic function in $D$, and its derivative $F'$ is holomorphic in $D$ as well. (That is a consequence of, e.g., Cauchy's differentiation formula, and implies that $F$ is infinitely often complex differentiable.)

You can verify (e.g. with the Cauchy-Riemann equations) that the functions $f$ in (c) and (d) are not holomorphic, and therefore not the derivative of a holomorphic function $F$.

The restriction $D = \Bbb C$ is irrelevant in these cases, $f(z) = |z|^2$ and $f(z) = \overline z$ are not holomophic in any open subset $D \subset \Bbb C$.


问:
Question about the gradient of a function
We consider a function of class $C^1(\mathbb{R}^2)$, by searching a max or min between the points of $S=\{(x,y) \in \mathbb{R}^2 \;\; | \;\; x^2+y^2=1\}$, if $(x_0,y_0)$ is one of them, is it true that there exists $\lambda \in \mathbb{R}$ such that $\nabla f(x_0,y_0)=2\lambda (x_0,y_0)?$ Why?

答: Let $g(x,y)=x^2+y^2$. Then $S=\{(x,y)\in\Bbb R^1\mid g(x,y)=1\}$ and the method of Lagrange multipliers tells us that if $f$ has a local extreme at some point $(x_0,y_0)\in S$, then $\nabla f(x_0,y_0)=\lambda\nabla g(x_0,y_0)$, for some $\lambda\in\Bbb R$. And $\nabla g(x_0,y_0)=(2x_0,2y_0)$.


答: If an antiderivative exists, then $f$ must be holomorphic. But your functions in c) and d) are not because they do not satisfy the Cauchy-Riemann equations.

Body:
State an antiderivative $F: D \longrightarrow \mathbb{C}$ or explain why an antiderivative does not exist.

a) $f(z) = z^2 , D = \mathbb{C}$

b) $f(z) = z \sin z,  D = \mathbb{C}$

c) $f(z) = |z|^2 , D = \mathbb{C}$

d) $f(z) = \overline{z} , D = \mathbb{C}$

For a) and b) I just performed simple integration for the equations

a)$ z^3/3 + C$ and b) $-z\cos z + \sin z +C$

but I am stuck on c) and d). It seems like an antiderivative does not exist but how do i explain that?


问:
Find all prime polynomials
Problem : Find all polynomials $f(x)\in \mathbb{Z}[x]$ such that there exists a natural $N$, such that for any prime $p>N$, $|f(p)|$ is also a prime.

What I found so far : It is obvious that $f(x)$ is irreducible. Few such polynomials are,

$f(x)\equiv \pm p$ for some prime $p$.

$f(x)=\pm x$

But after this I am unable to prove or disprove the existent of any other such polynomials.

答:
All constant solutions are among your findings, obviously. So assume $f$ is not constant.
If $f$ is a solution, then so is $-f$; hence let us assume $f(x)\to+\infty$ as $x\to+\infty$. We may make $N$ larger such that all (real) roots of $f(x)$ as well as $f'(x)$ are $<N$. Then $f$ is positive and injective on $[N,\infty)$.

If $f(p)=p$ for all primes $p>N$, we must have $f(x)=x$ (because $f(x)-x$ has infinitely many  roots), which you already found.
So let $p>N$ be a prime with $q=f(p)\ne p$. Note that $f(x)\bmod q$ depends only on $x\bmod q$. Hence for every prime $p'$ with $p'>N$ and $p'\equiv p\pmod q$, we have $q\mid f(p')$, but equality only for $p'=p$. By Dirichlet, infinitely many primes $p'\equiv p\pmod q$ exist (this is where we use $p\ne q$), and some of these infinitely many are $>N$, thus showing $f(p')$ is not prime and $f$ does not have the property.

We conclude that the solutions you found are all solutions.

Alternate, somewhat more direct approach:

Let $f$ and $N$ as in the problem statement.
Consider a prime $p>N$ and let $q=|f(p)|$.

If $q\ne p$, they are co-prime and by Dirichlet, there exist infinitely many primes $p'\equiv p\pmod q$. For those (still infinitely many) such $p'$ that are $>N$, we conclude $f(p')\equiv f(p)\equiv 0\pmod q$ and hence $f(p')=\pm q$. Then one of the two polynomials $f(x)-q$, $f(x)+q$ has infinitely many roots, hence is identically $0$, and ultimately
$$ f(x)=\pm q.$$
Remains the case that $|f(p)|=p$ for all primes $p>N$. Then one of the two polynomials $f(x)-x$, $f(x)+x$ has infinitely many roots, hence is identically $0$, and ultimately
$$ f(x)=\pm x.$$

Title: Find all prime polynomials


问:
Solving for x in a specific range
I am trying to find the minimum value, analytically, of a somewhat complicated function, and I've been struggling with it all day. It's complicated (to me) as the function is a sum of a periodic function (in $x$) and a function that is not periodic:

$$ f =\frac{\alpha}{L^2}\left[\cos^2 \left( \frac{\pi x}{L}-\frac{\pi}{2} \right) \right]^{-1} + \frac{\beta}{L^{3/2}}x^{-1/2} .$$

My approach has simply been taking the derivative and setting it to zero,

$$ \frac{df}{dx} = 0 $$

and then solving for $x$. First I considered each part individually as a sanity check. The right-hand side of the expression does not have a minimum (which makes sense), and the left-hand side does not have a global minimum, but rather periodic local minima with solutions,

$$ x = L \left( n - \frac{1}{2} \right) $$

where $n$ is the periodicity (I'm actually not sure how to solve for this, but it makes sense -- Thanks WolframAlpha). $f$ should have local minima though, and this is where I am running into some issues.

For the entire function, I can do the derivative,

$$ - \frac{\alpha}{L^3} \cot\left( \frac{\pi x}{L} \right) \csc^2\left( \frac{\pi x}{L} \right) - \frac{\beta}{L^{3/2}} \frac{1}{2x^{3/2}} = 0$$

but I am not able to solve for $x$. This makes sense as the minimum would now be different for each period, so taking the derivative of $f$ for an unspecified range would not make a ton of sense. I've been trying to think of a "work around".

I am only really interested in the range from $0$ to $L$ and there should be a minimum within that range. Just considering the range from $0$ to $L$, then the solution to the left-hand side should be,

$$ x = \frac{L}{2} $$

(this is intuitive) but I don't know how to actually solve $df/dx=0$ for $x$ in a specific range.

Any pointers would be greatly appreciated.

Jason

答:
Simplifying the cosine and making $\frac{\pi  x}{L}=t$, we end with
$$g(t)=\frac 1{L^2}\left(\alpha  \csc ^2(t)+ \frac{\beta\sqrt{\pi } }{\sqrt{t}} \right)$$
Differentiating with respect to $t$ what is inside the parentheses, we then need to solve
$$\frac{\sqrt{\pi } \beta }{2 t^{3/2}}+2 \alpha  \cot (t) \csc ^2(t)=0$$ that is to say
$$t^{3/2} \cot (t) \csc ^2(t)=k \qquad \text{with} \qquad k=-\frac{\sqrt{\pi } \beta }{4 \alpha }$$ and, according to what you wrote, you are only concerned by the range $0 \leq t \leq \pi$.

If $k >0$ and large, using Taylor expansion around $t=0$, we have
$$t^{3/2} \cot (t) \csc ^2(t)=\frac{1}{t^{3/2}}-\frac{t^{5/2}}{15}+O\left(t^{9/2}\right)$$ So, a very first approximation could be
$$ t\sim k^{-2/3}$$ Trying for $k=12.345$, the approximation would give $t \sim 0.187214$ while the solution given by Newton method is $0.187204$

If $k$ is small, using Taylor expansion around $t=\frac \pi 2$, we have
$$t^{3/2} \cot (t) \csc ^2(t)=-\frac{\pi ^{3/2} \left(t-\frac{\pi }{2}\right)}{2 \sqrt{2}}-\frac{3}{2}
   \sqrt{\frac{\pi }{2}} \left(t-\frac{\pi }{2}\right)^2+O\left(\left(t-\frac{\pi}{2}\right)^3\right)$$ and using the first term only, the estimate
$$t \sim \frac{\pi }{2}-\frac{2 \sqrt{2} k}{\pi ^{3/2}}$$
Trying for $k=1.2345$, the approximation would give $t \sim 0.943733$ while the solution given by Newton method is $0.844008$

Trying for $k=-1.2345$, the approximation would give $t \sim 2.19786$ while the solution given by Newton method is $1.94590$

Now, if $k <0$ and large, using Taylor expansion around $t=\pi$, we have
$$t^{3/2} \cot (t) \csc ^2(t)=\frac{\pi ^{3/2}}{(t-\pi )^3}+\frac{3 \sqrt{\pi }}{2 (t-\pi )^2}+\cdots$$ and then an estimate
$$t\sim \frac{\sqrt{\pi }}{\sqrt[3]{k}}+\pi$$
Trying for $k=-12.345$, the approximation would give $t \sim 2.37468$ while the solution given by Newton method is $2.46583$.

As approximations, this does not look too bad and, using them as initial guesses, Newton method would converge quite fast.

Title: Solving for x in a specific range


问:
Block identity theorem?
If $C$ and $C'$ represent the matrix of a linear transformation with respect distinct bases then we know $C$ and $C'$ are equivalent; in particular we can find invertible matrices $A$ and $B$ such $C'=BCA$. Can we also conclude that $C'$ can be taken to have a partial identity block form? that is an identity block (consisting of only ones) of some size and $0's$ in the remaining entries? Why?</p>

答: Hint: Use elementary row operations to bring $C$ to its reduced row echelon form $R$, then use elementary column operations to bring $R$ to its reduced column echelon form $C'$.

Title: Block identity theorem?


问:
Proving A is B if and only if C
If a statement says " A is B if and only if C," does that mean the forward direction is: Given A, if B then C and the other direction is Given A, if C then B?

Or does it mean, forward direction is If A is B, then C and backward direction is If C, then A is B?

A more specific example would be: a set with property A has property B if and only if it has property C.

For the forward direction, do I assume we are given a set with property A then suppose it has property B and prove it has property C?

And backward direction: I assume we are given a set with property A and suppose it has property C and prove it has property B?

答:
In my opinion this depends on what type of objects $A,B$ and $C$ are, but most likely it is the second interpretation. In particular, if $A$ and $B$ are sets and $C$ is a logical statement, then your second interpretation holds; you'd have $(A = B) \Leftrightarrow C$.

If all three of $A,B,C$ are logical formulas, then it is plausible to interpret the statement as $A \Leftrightarrow (B \Leftrightarrow C)$. However, this seems an unlikely interpretation because this would mean that "is" and "if and only if" are both used to mean the equivalence of logical formulas.

Title: Proving A is B if and only if C


问:
How can I solve the following differential equation by reducing it to first order?
I'm trying to solve the following differential equation

$ \displaystyle y''+ \left( 1+ \frac{1}{y} \right)y'^{2} =0 \tag*{} $

The book in which I found this question says that It can be solved by reducing it to a first order differential equation but I can't seem to find a way to do so.

Body:
$$\displaystyle y''+ \left( 1+ \frac{1}{y} \right)y'^{2} =0 $$
Divie by $y'$ then:
$$\dfrac {y''}{y'}+  y'+ \dfrac{y'}{y}  =0 $$
$$(\ln y')'+(\ln y)' +y'=0$$
$$(\ln y'+\ln y +y)'=0$$

Integrate.


问:
the nessesary condition of branch of logarithm
Let $U$ be an open subset of $\mathbb{C}$.Let $z_0$ be a point in U,and suppose that f is a meromorphic function on U with a pole at $z_0$.Prove that there is no holomorphic function g:U$\setminus${$z_0$}$\rightarrow$$\mathbb{C}$,such that $e^{g(z)}$=f(z)for all z$\in$U$\setminus${$z_0$}

my idea:as we know that if a simply connected domain which does not contain the origin,it can be chosen as a branch of logarithm,however,this is just sufficient condition,and we don't know what the image of f is like,thus how to deal with?this really confuses me.thanks!

答: If $n$ is the order of the pole then $f(z)=(z-z_0)^{-n} h(z)$ for some holomotprhic function $h$ with no zeros in some disk $B(z_0,r)$. Since this disk is simply connected and $h$ has no zeros we can write $h(z)=e^{\phi (z)}$ for some holomorpic function $\phi$. But then $(z-z_0)^{n}$ is itself an exponential in that disk with $z_0$ removed. But $(z-z_0)^{n}$ does not have a holomorpic logarithm in any disk around $z_0$ (with $z_0$ removed)$.