问:
Proof that the sum of a subset of small powers must equal a larger power which full sum exceeds.
What it says in the title.

Just kidding, the title is very poorly explained, sorry about that. Here is a more descriptive question.

How can I prove the following.

Suppose we have some set of coins $C$, each with a value of the form $b^a$ where $a<k$, the base $b$ is constant and the exponent $a$ varies from coin to coin.

Suppose as well that the sum of all the coins in $C$ is greater than or equal to $b^k$. i.e.

$$\sum_{c \in C}\text{value}(c) \geq b^k$$

Then we must show that there exists $D \subseteq C$ such that all the coins in $D$ sum up exactly to $b^k$. i.e.

$$\sum_{d \in D}\text{value}(d) = b^k$$

I am trying to use this as a lemma in a longer proof, and it is the last link I need to complete the proof. I would appreciate any help I can get to point me in the right direction. Thank you.

答:
Consider the collection, $\mathscr{C}$, of all subsets of $C$ which sum to $≥b^k$.  We know that $\mathscr{C}$ is not empty (since it at least contains $C$).  Let $C_0$ be a subset with minimal sum in $\mathscr C$. We claim that the sum of the coins in $C_0$ is exactly $b^k$, which will clearly conclude the proof.  So, suppose otherwise. Suppose $$N=\sum_{c\in C_0}\text {value}(c)>b^k$$

As a special case, to illustrate the idea here, first note that $C_0$ does not contain a coin with value $1$, else we could remove it, contradicting the minimality of $C_0$.

We can easily generalize this argument. Let $b^{a_1}$ be the minimal value of a coin in $C_0$.  Then $b^{a_1}\,|\,N$ so $b^{a_1}\,|\,(N-b^k)$.  But then $N≥b^k+b^{a_1}$ so we can remove a coin of value $b^{a_1}$ and get a subset in $\mathscr{C}$ with smaller sum than $C_0$, contradicting the minimality of $C_0$, so we are done.


问:
Convert a polar equation into parametric
I am getting hung up on the terminology in Calc 2 and making this subject much harder than it needs to be.

Question: Convert $r=7\sin(\theta)-2$ into parametric.

Does that simply mean putting it into the $x, y$ form?

$x=r\cos(\theta)$ --> $$x=(7\sin(\theta)-2)(\cos(\theta))$$
$y=r\sin(\theta)$ --> $$y=(7\sin(\theta)-2)(\sin(\theta))$$

答:
Consider $r = 7 \sin(\theta)$,

\begin{align*}
r^2 &= 7r \sin(\theta)\\
x^2 + y^2 &= 7y\\
x^2 &= 7y - y^2
\end{align*}

Hence $x^2 = t$ and $7y - y^2 = t$. Using quadratic equation, we have
$$x = \pm \sqrt{t} \hspace{30pt} \mbox{and} \hspace{30pt} y = \frac{7 \pm \sqrt{49 - 4t}}{2}$$

ContentLicense: CC BY-SA 4.0


问:
$\sum_{j=1}^{n}x_{j}f_{j}(x_1, x_2, \ldots, x_n) = 0 $ $\forall (x_1, x_2, \ldots, x_n)\in\mathbb{R}^{n}\backslash\{0\}$.
Let $n\geq 2$ be an odd integer. Prove that there is no function $f := (f_{1}, f_{2}, \ldots, f_{n})$ of class $C^{\infty}$ from $\mathbb{R}^{n}\backslash\{0\}$ to $\mathbb{R}^{n}\backslash\{0\}$ such that $\sum_{j=1}^{n}x_{j}f_{j}(x_1, x_2, \ldots, x_n) = 0 $ $\forall (x_1, x_2, \ldots, x_n)\in\mathbb{R}^{n}\backslash\{0\}$.

This question is from concours SMF junior 2020 https://smf.emath.fr/evenements-smf/annonce-concours-smf-junior-2020

I only know the case where $f$ is linear. If $f$ is linear, we represent $f$ by an matrix $A\in M_{n}(\mathbb{R})$. Then $\forall x = (x_1, x_2, \ldots, x_n)\in\mathbb{R}^{n}\backslash\{0\}$, $x^\top Ax = 0$. So $A$ is antisymetric. Since $\det(A) = \det(A^\top) = (-1)^{n}\det(A)$ and $n$ is odd, so $\det(A) = 0$. But $x\in\mathbb{R}^{n}\backslash\{0\} \Rightarrow Ax\in\mathbb{R}^{n}\backslash\{0\}$ means that $\ker(A) = 0$. Impossible.

In general case, we have $df(x)(x)+f(x) = 0$ $\forall x\in\mathbb{R}^{n}\backslash\{0\}$. I'm not sure if this is useful.

答: The given condition is $\langle x,f(x)\rangle = 0$ for all $x \in \Bbb R^n \setminus \{0\}$. This means that the restriction $f|_{\Bbb S^{n-1}}$ defines a vector field on the sphere $\Bbb S^{n-1}$, since $x \in \Bbb S^{n-1}$ and $\langle x,f(x)\rangle = 0$ means that $f(x) \in x^\perp = T_x(\Bbb S^{n-1})$. Since $n$ is odd, $n-1$ is even. So by the hairy-ball theorem, there is $x_0 \in \Bbb S^{n-1}$ with $f(x_0) = 0$. But this contradicts your assumption that $f$ takes values in $\Bbb R^n \setminus \{0\}$.


问:
Proving a Summation using induction
I am trying to prove the following summation inequality via induction:

$$\sum_{j=1}^n\frac{1}{\sqrt{j}}\geq 2\sqrt{n+1}-2$$

I know that first I must check base case, which is $n=1$.
$$\frac{1}{1}=1\geq2\sqrt{2}-2=0.8...$$
which checks out. Next, I assume that the inequality holds for $k$. Thus, for $k+1$:
$$\sum_{j=1}^{k+1}\frac{1}{\sqrt{j}}=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}\geq 2\sqrt{k+1}-2+\frac{1}{\sqrt{k+1}}$$
Here I recognized, that the first $k$ terms correspond to the case that I assume is true. Now I need to manipulate this epxression algebraically to find:
$$\sum_{j=1}^{k+1}\frac{1}{\sqrt{j}}\geq 2\sqrt{k+2}-2$$
However, when I manipulate the expression I got for $k+1$:
$$2\sqrt{k+1}-2+\frac{1}{\sqrt{k+1}}=\frac{2(k+1)+1}{\sqrt{k+1}}-2$$
From this point I have been unable to make progress. I feel like I am very close to the answer, yet I seem to be unable to manipulate the expression in the correct way.
What could I do next?

答:
From your last step, it suffices to show $2(k+1)+1 \ge 2\sqrt{(k+2)(k+1)}$, which will give you the desired result.
To check this inequality, it may help to square both sides.

Title: Proving a Summation using induction


答: Hint: $\sqrt {k+2}+\sqrt {k+1} >2\sqrt {k+1}$ so $\frac 1 {\sqrt {k+1}} >\frac  2 {\sqrt {k+1}+\sqrt {k+2}} =2 (\sqrt {k+2}-\sqrt {k+1}) $. Can you use this to go from $k$ to $k+1$?

Title: Proving a Summation using induction

Body:
I am trying to prove the following summation inequality via induction:

$$\sum_{j=1}^n\frac{1}{\sqrt{j}}\geq 2\sqrt{n+1}-2$$

I know that first I must check base case, which is $n=1$.
$$\frac{1}{1}=1\geq2\sqrt{2}-2=0.8...$$
which checks out. Next, I assume that the inequality holds for $k$. Thus, for $k+1$:
$$\sum_{j=1}^{k+1}\frac{1}{\sqrt{j}}=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}\geq 2\sqrt{k+1}-2+\frac{1}{\sqrt{k+1}}$$
Here I recognized, that the first $k$ terms correspond to the case that I assume is true. Now I need to manipulate this epxression algebraically to find:
$$\sum_{j=1}^{k+1}\frac{1}{\sqrt{j}}\geq 2\sqrt{k+2}-2$$
However, when I manipulate the expression I got for $k+1$:
$$2\sqrt{k+1}-2+\frac{1}{\sqrt{k+1}}=\frac{2(k+1)+1}{\sqrt{k+1}}-2$$
From this point I have been unable to make progress. I feel like I am very close to the answer, yet I seem to be unable to manipulate the expression in the correct way.
What could I do next?


问:
Let $G$ be a group and $a,b \in G$. Prove that for each integer such as $n$, we have $(a b^{-1} a)^n = a b^n a^{-1}$
I tried expanding $n$ parentheses such as $a b^{-1} a$ and then tried to replace the second $a$ with $a^2 a^{-1}$ but didn't know how to proceed further.

Also, is it possible to use commutativity here somehow?

答:
I do it via a simple inductive argument:

The trivial identity

$aba^{-1} = aba^{-1} \tag 1$

provides the $k = 1$ base case; for general $k$ we assume

$(aba^{-1})^k = ab^ka^{-1}; \tag 2$

then right multiplying by $aba^{-1}$:

$(aba^{-1})^{k + 1} = (ab^ka^{-1})aba^{-1} = (ab^k)(a^{-1}a)ba^{-1}$
$= (ab^k)eba^{-1} = ab^kba^{-1} = ab^{k + 1}a^{-1}, \tag 3$

where $e \in G$ is the identity element.  This completes the induction and hence we have

$(aba^{-1})^n = ab^na^{-1}, \tag 4$

for all $n \in \Bbb N$.


问:
n*2 mod p is not always equal to 0 mod 2?
Edit: as a first-timer here, I sincerely do not understand the downvotes to this legitimate question.

With regular integers in Z, n*2 is always equal to 0 mod 2.

In a finite field Zp of order P, n*2 is equal to 0 mod 2 only if n < (P+1)/2 and 1 mod 2 otherwise.

Is it possible to multiply n by another integer in order to obtain a number equal to 0 mod 2 in Zp for any n in Zp? Would appreciate the proper way to reason about this problem.

答:
The accepted answer has very little to do with the question - whether or not the notion of "parity" persists $\!\bmod n.\,$  This is true $\!\iff\! n\,$ is even. Indeed by the linear congruence solvability criterion

$$\,a\,\ \text{is even }\!\bmod n\!\iff\! \exists\:\! x\!:\ 2x \equiv a\!\!\!\pmod{\! n}\! \iff\!\color{#c00}{\gcd(2,n)}\mid a\qquad$$

If $\,n\,$ is odd then $\,\color{#c00}{\gcd(2,n)=1}\,$ divides $\,a\,$ for all $\,a,\,$ so every integer is even (i.e. a multiple of $\:\!2).\,$ Hence there is no useful notion of parity modulo odd $\,n.$

If $\,n\,$ is even then $\,\color{#c00}{\gcd(2,n)=2}\,$ so $\,a\,$ is even $\!\bmod n\!\iff\!\color{#c00}2\mid a\!\iff\! a\,$ is even in $\,\Bbb Z$.

Further, by congruences persist mod factors of the modulus,  congruences $\!\bmod 2k$ persist $\!\bmod 2,\,$ which means that parity arithmetic works fine $\!\bmod 2k\,$ (i.e. $\Bbb Z_{\large 2k}\,$ has $\,\Bbb Z_{\large 2}\,$ as a ring image).

More generally, as explained here, we can extend the notion of parity not only to even moduli but further to any ring having image $\,\Bbb Z_{\large 2} = \Bbb Z\bmod 2 =$ integers $\!\bmod 2,\,$ e.g. the Gaussian integers $\,\Bbb Z[i]\,$ where $\,i\,$ is odd, and $\,\Bbb Z[\sqrt 5]\,$ where $\,\sqrt 5\,$ is odd (here where we use parity in this quadratic number ring to give a simple proof that the integer $\,(9+4\sqrt{5})^n + (9-4\sqrt{5})^n\,$ is even), and here we extend parity to a ring with infinite elements. See also here and here for further background.


问:
Double commutant of an injective subalgebra still injective?
Consider a von Neumann algebra $A_0$ and an injective *-isomorphism $\pi : A_0 \to B(H)$.

Then we have a *-subalgebra $\pi(A) \subset B(H)$, which is abstractly *-isomorphic to the von Neumann algebra $A_0$, but which might not be a "von Neumann subalgebra" of $B(H)$, i.e. we are not guaranteed $\pi(A)''=\pi(A)$.  (E.g. this answer.)

If $A_0$ is injective, can we conclude $\pi(A)''$ injective?  If it helps, I'm interested in the case $A_0$ is the hyperfinite $II_1$ factor.

答: No, it isn't. An irreducible representation of a II$_1$ factor represents faithfully as a dense sot-subalgebra of a non-separable $B(H)$, which then cannot be injective.


问:
Let $G$ be a simple regular graph of odd degree, with $n\geq 4$ vertices. Show that $G$ or its complementary graph $\overline{G}$ is hamiltonian.
Let $G$ be a simple regular graph of odd degree, with $n\geq 4$ vertices. Show that $G$ or its complementary graph $\overline{G}$ is hamiltonian.

First, if we assume that $G$ is $r-$regular we obtain:
Since $G$ is $r-$regular so if $r\geq n/2$ we have that $G$ is Hamiltonian. In the other hand we have that $\overline{G}$ is $(n-r-1)-$regular so if $n-r-1\geq n/2$ then $\overline{G}$ is hamiltonian.

I think that proof of this result is made by contradiction using Ore's Theorem. That is, to assume, absurdly, that $G$ and $\overline{G}$ are both non-hamiltonian. Then we have that $r<n/2$ and $n-r-1<n/2$. I'm not sure, just an idea.

Can anyone help me to prove this?

答:
If $r\ge\frac{n}2$, then

$$\deg_Gu+\deg_Gv=2r\ge n$$

for every pair of vertices of $G$, so $G$ is Hamiltonian by Ore’s theorem.

Let $v$ be any vertex of $G$. There are $n-1$ other vertices, and $G$ is $r$-regular, there are $n-1-r$ vertices that are not adjacent to $v$ in $G$. These are precisely the vertices adjacent to $v$ in $\overline{G}$, so $\deg_{\overline{G}}v=n-1-r$, and $\overline{G}$ is $(n-1-r)$-regular.

Suppose now that $r<\frac{n}2$; then

$$n-1-r>n-1-\frac{n}2=\frac{n}2-1\,.$$

If $n$ is even, $\frac{n}2-1$ is an integer, so $n-1-r\ge\frac{n}2$, and $\overline{G}$ is Hamiltonian by Ore’s theorem.

If $n$ is odd, the smallest integer greater than $\frac{n}2-1$ is $\frac{n}2-\frac12=\frac{n-1}2$, so all that we can conclude is that $n-1-r\ge\frac{n-1}2$. If in fact $n-1-r\ge\frac{n+1}2$, we can apply Ore’s theorem to conclude that $\overline{G}$ is Hamiltonian, but if $n-1-r=\frac{n-1}2$, Ore’s theorem doesn’t apply. In this case, however, we can apply a theorem due to Nash-Williams to conclude that $\overline{G}$ is Hamiltonian. In fact, in this case both $G$ and $\overline{G}$ are $\frac{n-1}2$-regular, so both are Hamiltonian.


问:
Induction proof of $T(n)=T(\lfloor \alpha n \rfloor)+T(\lfloor(1−\alpha)n\rfloor)+1=O(n)$
Im trying to prove the following using induction but I am stuck with a constant in my induction step which I can't rid of:

$T\left(n\right)=T\left(\left\lfloor\alpha n\right\rfloor\right)+T\left(\left\lfloor(1-\alpha)n\right\rfloor\right)+1=O\left(n\right) \ \ \ \ for\ \ \ \alpha<1.$

Target: want to show for a specific constant $c>0$ and "big enough"   $ n $   that $T\left(n\right)\le c\cdot n$

base case:

Assuming $T(0)=1$ and showing for $n=1$:

For every constant $c\geq3 $ , the folloing it true: $T\left(1\right)=T\left(\left\lfloor\alpha\right\rfloor\right)+T\left(\left\lfloor(1-\alpha)\right\rfloor\right)+1\le1+1+1=3\le c\cdot1$

induction step:
$ T\left(n\right)=T\left(\left\lfloor\alpha n\right\rfloor\right)+T\left(\left\lfloor(1-\alpha)n\right\rfloor\right)+1\le\ c\cdot\left\lfloor\alpha n\right\rfloor+\ c\cdot\left\lfloor(1-\alpha)n\right\rfloor+1\le c\cdot an+\ c\cdot(1-\alpha)n+1\le cn+1$

And this is where I get stuck

I would like to say someting like $\left\lfloor a n\right\rfloor+\left\lfloor(1-\alpha)n\right\rfloor\le n-1$  , but well, it is not accurate for exampla in case $\alpha=\frac{1}{3}$ and $n=9$

Any assistance would be appreciated!

答:
Let $f(n) = T(n)+1$; then the recurrence relation for $T(n)$ becomes
$$
    T(n) + 1 = T(\lfloor \alpha n\rfloor) + 1 + T(\lfloor (1-\alpha) n\rfloor) + 1
$$
after we add $1$ to both sides, and this simplifies to
$$
   f(n)  = f(\lfloor \alpha n\rfloor) + f(\lfloor(1-\alpha)n\rfloor).
$$
We can begin by proving $f(n) = O(n)$; now, the approach you're taking will work just fine, because the inequality we need is $\left\lfloor \alpha n\right\rfloor+\left\lfloor(1-\alpha)n\right\rfloor\le n$, and this is always true.

This means $T(n) = O(n)$ because $T(n) \le f(n)$.


问:
Differential Geometry - Computation Help
I'm trying to learn differential geometry through one of MIT's online courses (lecture notes found here:  http://ocw.mit.edu/courses/mathematics/18-950-differential-geometry-fall-2008/lecture-notes/ch1_revised.pdf) and am stuck with what should be an easy question.  The question asks to show the following:  Suppose $c(s)$ is a regular curve ($c'(s) \ne 0$) in the plane with $|c(s)| \le 1$ and suppose there is a point t with $|c(t)| = 1$.  Then $|\kappa(t)| \ge 1$.  Here, $\kappa$ is defined as

$\frac{det(c', c'')}{||c'||^3}.$

It's should be straightforward if we write $c$ as the graph of a function and rotate so that $c(t) = (0,1)$.  However, I feel as if there should be a more elegant solution using only what's found in the first day's notes (found in the link).  I just don't see it.

答:
Let me assume, without loss of generality, that $c$ is parametrized by arclength, so that $|c'|\equiv 1$, and let me also assume $t_0=0$ (where $c(t_0)$ has length one). Furthermore, upon reparametrizing $c$ backwards, we may assume that $\mathcal{B}=(c(0),c'(0))$ is a direct orthonormal basis of $\Bbb R^2$ (this frame is indeed orthonormal, as follows from the fact that $|c|^2$ attains a local maximum at $0$, and the the preceding hypothesis).

The curvature then takes on the form
$$\kappa(0)=\det(c'(0),c''(0))=\det_{\mathcal B}(c'(0),c''(0))=-c(0)\cdot c''(0)$$
Also, the Taylor expansion of $c$ near $0$ is $c(t)=c(0)+tc'(0)+\frac12t^2c''(0)+o(t^2)$, and thus $$\begin{array}{ccl}
|c(t)|^2&=&\underbrace{c(0)\cdot c(0)}_{=1}+2t\underbrace{c(0)\cdot c'(0)}_{=0}+(\underbrace{2\times\frac12 c''(0)\cdot c(0)}_{=-\kappa(0)}+\underbrace{c'(0)\cdot c'(0)}_{=1})t^2+o(t^2)\\
&=&1+(1-\kappa(0))t^2+o(t^2)
\end{array}$$
This expression has to stay $\leq 1$ near $0$, which in turn forces $1-\kappa(0)\leq 0$

ContentLicense: CC BY-SA 3.0


问:
Does the Axiom of Substitution Have Limits?
According to Terence Tao's book Analysis I, the axiom of substitution is one of the four axioms of equality for objects (which are basically eligible set elements). In particular, it says that for any two objects $x$ and $y$ of the same type, if $x$ = $y$, then $f(x)=f(y)$ for all functions and operations $f$, and that furthermore, $P(x)$ and $P(y)$ are logically equivalent propositions (i.e. they imply each other) for all  propositions $P$ whose truth-value depends on $x$ and $y$.

Sometimes, the axiom of substitution makes sense (i.e. is correct), such as when considering sets to be the object in question (they are indeed valid objects according to the first set-theory axiom presented in the book), and considering the operation $f(X) = X \cup C$, where $C$ is some arbitrary set. Then indeed, one may verify that if two sets $A$ and $B$ are equal to each other (i.e. $A=B$), then $f(A)=f(B)$ for the operation $f$ defined above.

But there are times when it starts acting up. Still considering sets to be the object in question, let $P(X)$ be the proposition that the elements of $X$ are ordered from least to greatest (this proposition's truth value clearly depends on what the set $X$ is that you "feed" it so to speak). Evidently however, as Tao himself acknowledges, even if two sets $A$ and $B$ were equal to each other, because sets are unordered collections of elements, there are plenty of counterexamples to give that would thoroughly disprove the statement: $P(A)$ iff $P(B)$. So my question then is: does the axiom of substitution have limits to its applicability, and if so, can someone very explicitly clarify what they are? Thanks.

答: Your statement that $P(X)$ is the proposition that the elements of $X$ are ordered from least to greatest tacitly relies on two assumptions: (1) that $X$ is a sequence of objects and (2) that there is an ordering relation on those objects. When you make those assumptions explicit (as you must if you want to model $P$ in set thoery), then the axiom of substitution holds. You might be interested in the notion of referential transparency.


问:
What's the order of special orthogonal group of finite field $SO(2, \mathbb F_p)$
So I am wondering how to compute $|SO(2, \mathbb F_p)|$ when $p$ is a prime. And by definition $$SO(2, \mathbb F_p)=\left\{A=\begin{pmatrix}a&b\\c&d\end{pmatrix}: a,b,c,d\in\mathbb F_p,\,\,\,A^TA=I\,\,\text{mod}\,\,\,p,\,\,\,\det(A)=1\,\,\text{mod} \,\,p\right\}$$

答:
It is $\pmatrix{a&b\\-b&a}$ with $a^2+b^2=1$. Then it depends on $p$, if $p\equiv 1\bmod 4$ then $i \in \Bbb{F}_p$, $(a+ib)(a-ib)=1$ and $u=a+ib,v=a-ib,a=(u+v)/2,b=(u-v)/(2i)$ let us reduce to $uv=1$ ie. $p-1$ elements.

If $p\equiv 3\bmod 4$ then $\{ \pmatrix{a&b\\-b&a}, a,b\in F_p\}$ is the field with $p^2$ elements. $ \pmatrix{a&b\\-b&a}^p =  \pmatrix{a&-b\\b&a}$ so $ \pmatrix{a&b\\-b&a}\in SO_2$ iff $ \pmatrix{a&b\\-b&a}^{p+1}=I$. The multiplicative group of $F_{p^2}$ is cyclic with $p^2-1$ elements and $p+1$ of them satisfy $u^{p+1}=1$.


问:
(SOLVED) Inverse Z transform of a complicated fractional expression

Problem

Consider the following recurrence equation:
$$y[n+2]+ \frac{1}{4}y[n] = \cos(n \frac{\pi}{3})$$
$$\text{Where }y[0] = y[1] = 0$$
$$\text{What is }y[n]?$$

My attempt

Using $Z$ transform for left and right side of the recurrence equation we get:
$$Z(y[n+2]+ \frac{1}{4}y[n] = \cos(n \frac{\pi}{3}))$$
$$\implies Y(z)(z^2 + \frac{1}{4}) = \frac{z(z-\frac{1}{2})}{(z-\frac{1}{2})^2 + \frac{3}{4}}$$
$$\iff$$
$$Y(z) = \frac{z(z-\frac{1}{2})}{((z-\frac{1}{2})^2 + \frac{3}{4})(z^2 + \frac{1}{4})}$$
And now I have come to the conclusion that there must be another way of solving this, it is practically impossible to manually with a pen and paper solve for $y[n]$.
We don't even get a nice function when using the "InverseZTransform" function in Mathematica.

Help with a new perspective on the problem?

答:
Sorry that it took a while for me to answer, but I did manage to solve it. I think I made it a lot trickier than it was.

Posting the answer before asking another question on this site, seems quite fair to the community.

So we have:
$$Y(z) = \frac{z(z-\frac{1}{2})}{((z-\frac{1}{2})^2 + \frac{3}{4})(z^2 + \frac{1}{4})}$$
$$\implies$$
$$\frac{Y(z)}{z} = \frac{4z-2}{((z-\frac{1}{2})^2+\frac{3}{4})(4z^2+1)}$$
$$\iff$$
$$\frac{Y(z)}{z} = \frac{4z}{(z^2-z+1)(4z^2+1)}-\frac{2}{(z^2-z+1)(4z^2+1)}$$

$$\text{Partial fraction decomposition on both terms gives}$$
$$\frac{Y(z)}{z} = \frac{1}{13}(8\frac{2z-5}{4z^2+1}
-2\frac{2z-7}{((z-\frac{1}{2})^2+\frac{3}{4})})$$
$$\implies$$
$$Y(z) = \frac{1}{13}(8\frac{2z^2-5z}{4z^2+1}
-2\frac{2z^2-7z}{((z-\frac{1}{2})^2+\frac{3}{4})})$$

Divide and conquer, working on both the fractional terms in the parenthesis, let the first term be (1), and second (2)

$$\implies$$
$$(1) \implies \frac{8}{13}\frac{2z^2-5z}{4z^2+1} = \frac{8}{26}(\frac{z(z-0)}{(z-0)^2+(\frac{1}{2})^2}-\frac{5}{2}*2\frac{\frac{1}{2}}{(z-0)^2+(\frac{1}{2})^2})$$

Using the same method as above, we divide and conquer on the second part aswell and look for the z-transform on the table

$$(2) \implies -2(2\frac{z^2}{(z-\frac{1}{2})^2+\frac{3}{4}}-7\frac{z}{(z-(\frac{1}{2})^2)+\frac{3}{4}})$$

As of now my motivation to keep entering these in TeX commands is fading, so I'll just go straight to the point. Everything is now laid perfectly, simplified as much as needed, maybe a little bit too much as well. Now just look at the Z-Transform table and see that you'll get:

$$y[n] = \frac{4}{13}((\frac{1}{2})^n(cos[n\frac{\pi}{2}]-5sin[n\frac{\pi}{2}])-\frac{4}{13}(cos[n\frac{\pi}{3}]-2\sqrt{3}sin[n\frac{\pi}{3}])$$
Where we see the transient and stationary bit of the functions, the first term being the transient bit, and the second is the stationary bit of the function.


问:
Which of these ways is correct to find $a$ and $b$ in $a+b=13$ in which they are in the golden ratio?
I have two quantities $a$ and $b$ which satisfy $a+b=13$. I want to find the values of $a$ and $b$ for which, they are in the golden ratio, i.e. $\frac{a}{b}=\frac{1}{2} \left(\sqrt{5}+1\right)$. My question is :

Should I solve $a+b=13$ and $\frac{a}{b}=\frac{1}{2} \left(\sqrt{5}+1\right)$ to find the values?

or I should solve $\frac{a}{b}=\frac{1}{2} \left(\sqrt{5}+1\right)$ and $\frac{a+b}{a}=\frac{13}{a}=\frac{1}{2} \left(\sqrt{5}+1\right)$?

答:
As usual let $\varphi=\frac12\left(1+\sqrt5\right)$ be the golden ratio. You know (or can easily check) that $\varphi^2=\varphi+1$, so if $a=b\varphi$, then $$a\varphi=b\varphi^2=b\varphi+b=a+b=13\,.$$ Clearly the system

$$\left\{\begin{align*}
&b\varphi=a\\
&a\varphi=13
\end{align*}\right.$$

is a very easy one to solve, first for $a$ and then for $b$.


问:
How to find all ring homomorphisms from $\mathbb{Z}$ to $\mathbb{Z}$
I am trying to find all homomorphisms from $\mathbb{Z}$  to    $\mathbb{Z}$. I am looking at a solution, but i do not understand it. Here it is:

Let  $\Phi\colon \mathbb{Z}\to\mathbb{Z}$  be a ring homomorphism. Because $1^2 = 1$, we see that $\Phi(1)$ must be an integer whose square is itself, namely either $0$ or $1$. If $\Phi(1) = 1$ then $\Phi(n)= \Phi(n · 1)= n$, so $\Phi$ is the identity map of $\mathbb{Z}$ onto itself which is a homomorphism. If $\Phi(1)= 0$, then $\Phi(n)=\Phi(n · 1)0$, so  maps everthing onto $0$, which also yields a homomorphism.

I do not understand here what "Because $1^2= 1$, we see that $\Phi(1)$ must be an integer whose square is itself, namely either $0$ or $1$.". All proof is built upon this and I do not understand what it means. Can someone help me with this?

Thank you

答: By additivity you must have $\phi(n)\phi(1)= \phi(n)=n \phi(1)$ thus if $\phi(1) \neq0$ you get $\phi(n)=n$ and so $\phi$ is the identity on $\mathbb{Z}$, otherwise $\phi=0$.</p>


问:
Compute the limit $\lim\limits_{t \to + \infty} \int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin tx} $
I have been working on this limit for days, but I am not getting it. The question is

Compute the limit $$\lim_{t \to + \infty} \int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (tx)}$$

Note that the integral is well defined and convergent for every $t >0$. Indeed the integrand function is a positive function for every $t >0$ since
$$e^x + \sin tx > e^x-1 > x>0$$
And as $x \to + \infty$ the integrand function behaves like $e^{-x}$.

WHAT I TRIED:

I consider $t=2n \pi$ a multiple of $2 \pi$, and see what happens:
$$\int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)} = \sum_{k=0}^\infty \int_{k /n}^{(k+1) /n} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)}$$
Making the change of variables $u = 2n \pi x$ I get
\begin{align}\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{u/2n \pi}+ \sin (u)} &\ge
\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(2k+2) \pi/2n \pi}+ \sin (u)} \\&=
\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(k+1)/n}+ \sin (u)}\end{align} where I write the lower bound with the minimum of the function at $u=(2k+2) \pi$. Now I use the fact that the integrand function does is integrated over a period of $2 \pi$, and using the result for $C>1$
$$\int_0^{2 \pi} \frac{ \mathrm d u}{C+ \sin (u)} = \frac{2 \pi}{\sqrt{C^2-1}}$$
I get the estimate
\begin{align}\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(k+1)/n}+ \sin (u)} &= \sum_{k=0}^\infty \frac{1}{2n \pi} \frac{2 \pi}{\sqrt{e^{2(k+1)/n} -1 }} \\&= \frac{1}{n} \sum_{k=0}^\infty \frac{1}{\sqrt{e^{2(k+1)/n} -1 }}\end{align}
Summing all up, I got that
$$\int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)} \ge \frac{1}{n} \sum_{k=0}^\infty \frac{1}{\sqrt{e^{2(k+1)/n} -1 }}$$
As $n \to \infty$ the series converges to the Riemann integral
$$\int_0^{+ \infty} \frac{\mathrm d y}{\sqrt{e^{2y}-1}} = \frac{\pi}{2}$$

Hence the limit should be a number larger than $\pi/2$, or $+ \infty$.

Using WA I got for large values of $t$ that the integral is between $1$ and $2$, thus $\pi/2$ could be the actual limit.

Body:
$$
I(t)=\int^{\infty}_{0}\frac{1}{e^x+\sin(tx)}dx=\int^{\infty}_{0}e^{-x}\frac{1}{1+e^{-x}\sin(t x)}dx=
$$
$$
=\int^{\infty}_{0}e^{-x}\sum^{\infty}_{l=0}(-1)^le^{-lx}\sin(tx)dx
=\sum^{\infty}_{l=0}(-1)^l\int^{\infty}_{0}e^{-x(l+1)}\sin^l(tx)dx=
$$
$$
=t^{-1}\sum^{\infty}_{l=0}(-1)^l L\left(\sin^l(x),x,\frac{l+1}{t}\right)=
$$
$$
=t^{-1}\sum^{\infty}_{l=0} L\left(\sin^{2l}(x),x,\frac{2l+1}{t}\right)-t^{-1}\sum^{\infty}_{l=0} L\left(\sin^{2l+1}(x),x,\frac{2l+2}{t}\right).\tag 1
$$
where $L(f,x,w)=\int^{\infty}_{0}f(x)e^{-wx}dx$ is the Laplace transform and
$$
L(\sin^l(x),x,s)=\left\{
\begin{array}{cc}
l!/s\prod^{l/2}_{j=1}((2j)^2+s^2)\textrm{, if }l=even\\
l!/\prod^{(l-1)/2}_{j=0}((2j+1)^2+s^2)\textrm{, if }l=odd
\end{array}\right\}\tag 2
$$
Factoring $(2)$ and using the Pochhammer symbol ($(a)_s=\frac{\Gamma(a+s)}{\Gamma(a)}$), we get
$$
L(\sin^{2l}(x),x,s)=\frac{2^{-l}l!}{s\left(1-\frac{is}{2}\right)_{1/2}\left(1+\frac{is}{2}\right)_{1/2}}\tag 3
$$
and
$$
L(\sin^{2l+1}(x),x,s)=\frac{2^{-l}l!\tanh(\pi s/2)}{s\left(1-\frac{is}{2}\right)_{1/2}\left(1+\frac{is}{2}\right)_{1/2}},\tag 4
$$
Taking now the limit $t\rightarrow+\infty$ in $(1)$ using $(3),(4)$, we arive to
$$
\lim_{t\rightarrow\infty}I(t)=\sum^{\infty}_{l=0}\frac{4^{-l}(2l)!}{(2l+1)l!^2}+0=\frac{\pi}{2}.
$$


问:
$\mathbb E [||X_n|| ^ \alpha] = O(a_n) \implies X_n = O_p (a_n)$. True?
Helloo everyone, just in case, excuse my english i am a french native.

Please i'd like to have a proof, if it's possible, that

$\mathbb E [||X_n|| ^ \alpha] = O(a_n) \implies X_n = O_p (a_n)$

I saw it at

http://jeanalain.monfort.free.fr/Dicostat2005/O/Ordres_de_convergence_en_probabilite.pdf

property (5) (i)

答:
I think there is a typo in your file. So let's say it my way.

" If $\mathbb{E}\left( \Vert X_n \Vert ^{\alpha} \right) = O( a_n^{\alpha})$ then $X_n= O_p(a_n)$ ".

Proof is a direct consequence of Markov's.

$ \mathbb{P} \left( \frac{\Vert X_n \Vert}{a_n} \le M \right) \le \frac{1}{M^{\alpha}}.\mathbb{E}\left( \frac{\Vert X_n \Vert^{\alpha}}{a_n^{\alpha}} \right)= \frac{ O(1)}{M^{\alpha}}$


问:
L'Hospital's rule application
I have to find out for which pair of functions $f(x)$ and $g(x)$ the following limit
$$\lim_{x\to x_{0}}\frac{f(x)}{g(x)}$$
can be solved using L'Hospital's rule, and my options are:

$f(x)=x^2+x-6,\quad g(x)=x^2-3x+2,\quad x_{0}=2$

$f(x)=\sin(x)-x,\quad g(x)=e^x+e^{-x}-2,\quad x_{0}=0$

$f(x)=x^2\cdot\sin\displaystyle\left(\frac{1}{x}\right),\quad g(x)=x,\quad x_{0}=0$

$f(x)=2x+7,\quad g(x)=4x+1,\quad x_{0}=3$

$f(x)=\ln(x),\quad g(x)=\displaystyle\frac{1}{x},\quad x_{0}=0^{+}$

For me, the only item I can't use the rule is the 4th; but my problem is that, in the exercise I'm doing, there is no alternative that the items 1, 2, 3 and 5 are the correct answer.

So, my question is, in the 5th item, since the limit is
$$\frac{+\infty}{-\infty}$$
can I still use L'Hospital's rule here?

答:
L'Hopital's rule can be applied to item 5.  For indeterminate forms, the signs don't matter.

L'Hopital's rule does not resolve item 3.  (Actually calculate the equivalent limit by L'Hopital's rule.)  Item 3 is resolved by the squeeze theorem.

Title: L'Hospital's rule application


问:
Help Understanding Genus of a Plane Algebraic Curve
I have been attempting to teach myself algebraic geometry, and I am unfortunately confused about the geometric genus of an algebraic curve. My understanding is that from the degree of the curve, $d$, one computes $$(d-1)(d-2)/2$$ Then one subtracts $$r(r-1)/2$$ for each node, where $r$ is the multiplicity of that node.

Here is where I get confused. Suppose I have a polynomial $P(x)$ of degree $d$. Then the curve defined by $$Q(x,y) = P(x)-y = 0$$ is also of degree $d$. $Q$ is furthermore rational, because it can be parameterized by
$$x=t$$
$$y=P(t)$$

Further, it seems that I can easily choose $P(x)$ such that $Q(x,y)=0$ has no nodes of multiplicity 2 or more, (am I mistaken?) and hence the genus of $Q$ would be $(d-1)(d-2)/2$.
However, I also understand (incorrectly?) that curves are rational if and only if their geometric genus is $0$.
Can someone help me with what I am missing?

Perhaps it would be useful to deal with a concrete example. What is the genus of this curve? $$y=x^4 + x^3 + x^2 + x + 1$$

答:
The clarification I needed was provided by Ted Shifrin in the comments. The mistake in my thinking was in the notion that:

"it seems that I can easily choose P(x) such that Q(x,y)=0 has no nodes of multiplicity 2 or more."

If one extends the curve $$Q(x,y)=P(x)-y=-y+\sum_{i=0}^dc_ix^i=0$$ to the projective plane, one has $$Q^h(x,y,z)=-yz^{d-1}+\sum_{i=0}^dc_ix^iz^{d-i}=0$$ If one then dehomogenizes onto the $xz$ plane, one gets
$$\hat{Q}(x,z) = -z^{d-1}+\sum_{i=0}^dc_ix^iz^{d-i}=0$$
for which $$\hat{Q}(0,0)=0$$ is clearly a solution, and for which
the lowest degree term is $z^{d-1}$. Thus, $Q^h(x,y,z)=0$ has a singular point at $(0:1:0)$ of multiplicity $d-1$
and so, the curve defined by $Q$ will, of necessity, be of genus $0$.


问:
Exercise 3.11.d of Chapter II in Hartshorne: How to describe the scheme theoretic image of a reduced scheme?
I'm trying to do the final part of Exercise 3.11.d in chapter II of Hartshorne, i.e. that if $(Z,\mathcal{O}_Z)$ is a reduced scheme and $f:(Z,\mathcal{O}_Z)\to(X,\mathcal{O}_X)$ is a scheme morphism, then the scheme theoretic image of $f$ is given by the reduced induced closed subscheme structure on $\overline{f(Z)}$. With the help of this answer by KReiser, I was able to prove that the ideal sheaf $\ker f^{\#}$ defines a closed subscheme $(Y,\mathcal{O}_Y)$ of $X$ (with $Y\subseteq X$), and as $Z$ is reduced, $Y$ carries the reduced induced closed subscheme structure. However, to conclude the solution of the problem, we still need to show that $Y=\overline{f(Z)}$, which one easily reduces to showing that $\overline{f(Z)}\cap U=V(\ker f^{\#}(U))$ for all affine open $U\subseteq X$. And this is where I'm stuck. So how can one show this? Thanks in advance!

答:
Let $U=\operatorname{Spec} A$ be an affine open, and cover $f^{-1}(U)$ by $\operatorname{Spec} B_i$ as $i\in I$ ranges over an index set. Then $\operatorname{Spec} (B_i)_g$ cover $f^{-1}(\operatorname{Spec} A_g)$. If $g\in\ker f^\sharp(U)$, then $(B_i)_g=0$ and so $f^{-1}(\operatorname{Spec} A_g)=\emptyset$ and $\ker f^\sharp(\operatorname{Spec} A_g)=A_g$, so the underlying set of the scheme-theoretic image is a subset of $\overline{f(Z)}$.

On the other hand, the underlying set of the scheme-theoretic image can't be smaller than $\overline{f(Z)}$: if it was, then there exists a point $z$ in $f(Z)$ not in the scheme-theoretic image, and a local function which vanishes on the image but not at $z$ defines a local function on $Z$ which is nonvanishing by pullback. So the underlying sets of $\overline{f(Z)}$ and $\operatorname{im}(f:Z\to X)$ are equal, and both are reduced (check that $\ker f^\sharp(U)$ is radical by the computations in the linked post). By the uniqueness of the reduced induced structure, they must be the same subscheme.


问:
Is there a name for the minimal surface connecting two straight line segments in 3-dim Euclidean space?
In the $2$-dim Euclidean plane, the minimal curve that joins (and necessarily contains) two given points is a straight line.

Here what is being minimized by the curve is the $1$-dim measure of the $1$-dim object in the $2$-dim space that joins the two given $0$-dim objects.

In $3$-dim Euclidean space, what is the minimal surface that connects (and necessarily contains) two given straight line segments (of known finite length)?

Again, here what is being minimized is the surface area ($2$-dim measure) of the $2$-dim object (embedded in $3$-dim space) connecting two given $1$-dim objects.

Is there an umbrella name for this whole category?

By "the whole category" I mean given line segments can be in general configurations and not just special cases like being parallel, or the two segments having the same length. (btw the special case of parallel and same length seems to yield helicoid as the minimal surface)

As pointed out by Rahul in the comment, besides the given two straight line segments, it is necessary that one specifies the "other" boundaries. For simplicity, a reasonable choice seems to be connecting the end points to have a non-planar (skew) quadrilateral frame (for the soap film, so to speak).

Note:
This post seems relevant, but it doesn't solve my problem (of finding the keywords).

I suspect this should be one of the basic stuff in differential geometry that everybody knows or agrees on (just like how people agree on the notion of a straight line). The next section describes my current guesswork, and it would be nice to know if I'm on the right track.

My Naive Guess: the unique ruled surface

Given two straight line segments in the $3$-dim space, it suffice to consider the two skew lines $L_1$ and $L_2$ that contain them respectively. As one can parametrize a segment as part of an infinite line, the desired minimal surface from the given segments is part of the surface "generated" by $L_1$ and $L_2$.

One can find the unique pair of points, $P_1$ on $L_1$ and $P_2$ on $L_2$, that have the smallest distance. Accordingly one obtains $\overset{\longrightarrow}{P_1P_2}$ as the common normal vector between for two skew lines.

Now one can define a ruled surface by parametrizing the "moving" lines $L(t)$, thus making a ruled surface, with parameter $t$: starting with $L(t) = L_1$ at $t = 0$, "move" $L(t)$ towards $L_2$ along $\overset{\longrightarrow}{P_1P_2}$, "anchoring" at $P_1$ and rotate $L(t)$ along the way to make it eventually coincide with $L_2$.

The surface is not yet unique before specifying the amount of rotation (with respect to the distance to $L_2$). I'm guessing the rotation should be done linearly, but I haven't figure out if this indeed minimizes the surface area.

Obviously a standard approach is to formulate the whole thing in terms of calculus of variations.

I've done the almost trivial case of a straight line minimizing the distance, but apparently things get very complicated already for a $2$-dim surface in $3$-dim.

Even in the special cases where the coordinates are setup nicely so that the surface is a function like $z = f(x,y)$, it's an elliptic PDE and the solution seems to vary greatly depending on the boundary conditions. In short, I don't know how to handle the situation.

答:
The surface you’re asking about is called a “ruled surface” in engineering/manufacturing.

In the special case where both curves are straight lines, the surface is actually a hyperbolic paraboloid.

If the line segment end-points are $A,B,C,D$, then the surface can be described by the parametric equations
$$
S(u,v)= (1-u)(1-v)A + (1-u)vB + u(1-v)C + uvD
$$


问:
Integral of a two dimensional function
A function is defined by
$$f(x,y) = \int_{x^2}^{xy} e^{t^2} \,dt$$
We are to decide the partial derivatives of the function.

I am quite unsure whether this is a trick question because we could do an approximation of the integrand with the Taylor series, but I can't see how that would simplify the work.

Could I just get a direction to as how I should approach this problem? I would very much not like to just get the solution.

Note that this is one of the easiest questions on the paper.

答:
Hint: If we let $F(s) = \displaystyle\int_{0}^{s}e^{t^2}\,dt$, then $f(x,y) = F(xy)-F(x^2)$.

Now, try using the chain rule to compute $\dfrac{\partial f}{\partial x} = \dfrac{\partial }{\partial x}\left[F(xy)-F(x^2)\right]$ and $\dfrac{\partial f}{\partial y} = \dfrac{\partial }{\partial y}\left[F(xy)-F(x^2)\right]$. Remember that by construction, $F'(s) = e^{s^2}$.

Title: Integral of a two dimensional function


问:
$\displaystyle\int_C (e^x+\cos(x)+2y)\,dx+(2x-\frac{y^2}{3})\,dy$ in an ellipse
I have to compute $\displaystyle\int_C (e^x+\cos(x)+2y)\,dx+(2x-\frac{y^2}{3})\,dy$ in the ellipsoide $\frac{(x-2)^2}{49}+\frac{(y-3)^2}{4}=1$ using Green's Theorem.

The first thing I did was getting the parametric equation of the ellipse.

$\alpha(t)=(2+7\cos(t),3+2\sin(t)),\; t\in[0,2\pi]$.

Now, I'm not sure if I have to replace in the function every $x$ and $y$ with the result I got, and then using (maybe) polar coordinates to finally compute the result. Is my approach correct?

答: Notice the gradient of $$f(x,y)=e^x+\sin(x)+2xy-\frac{y^3}{9}$$ is the vector field $$\bigg<e^x+\cos(x)+2y,2x- \frac{y^3}{3} \bigg>$$ This means your vector field is conservative, so its integral over any closed curve (like your ellipse, not ellipsoid) is zero. With Green's Theorem, $$\int_{C}\vec{\nabla} f \cdot d\vec{r}=\int \int _{R} 0\cdot dA=0$$ Here $C$ is your ellipse and $R$ is its interior.


问:
How do I find the minimal upper bound of flow that can pass for each edges in this graph?
What is in red is the supply/demand. If it's 0 that means that the node is just there for transfering the flow. If it's negative this means that it's the demand and if it's positive then it's the supply.

Brackets in black on edges are the capacity limit. If there is no [ ] on an edge this simply mean that there is no limit. Otherwise, the numbers on the edges are the cost per unit of transfer.

I'm asked for each edge to find the lowest upper bound of flow that can pass through it. However, I'm not sure how to do this? For the node 1 I can put 10 units on each of its edges. Is this the lowest upper bound? How do I figure this out?

答:
Let $b_i$ be the supply or demand at node $i \in N$.  For each arc $(i,j)\in A$, introduce a nonnegative flow variable $x_{i,j}$ (some with upper bounds).  For each node $i\in N$, impose linear flow-balance constraints:
$$\sum_{(i,j)\in A} x_{i,j} - \sum_{(j,i)\in A} x_{j,i} = b_i \quad \text{for $i\in N$}$$
For each arc $(i,j)$, maximize $x_{i,j}$.  The maximum values turn out to be:
\begin{matrix}
i & j & \max \\
\hline
1 &2 &10 \\
1 &3 &20 \\
1 &4 &20 \\
2 &5 &30 \\
3 &2 &25 \\
3 &4 &30 \\
4 &5 &30 \\
\end{matrix}


问:
Laplace transform of convolution with modified limits
I have an expression such as
$\int_0^{x+l}y(z)g(x-z) dz$
and I want to evaluate its Laplace transform w.r.t $x$ in terms of the Laplace transform of $y(x)$. I know that I can substitute $t=x+l$, and coerce it into the standard from to get the Laplace transform w.r.t $t$, but I need its transform w.r.t $x$.

Motivation:

I would like to solve an integral equation:
$y(x) = f(x) + \int_0^{x+l} y(z) g(x-z) dz$.

If the integral limit had been to $x$, we would have had
$y(x) = f(x) + \int_0^{x} y(z) g(x-z) dz$. This leads to
$Y(s) = \frac{F(s)}{1+K(s)}$.

Body:
Hint:

$y(x)=f(x)+\int_0^{x+l}y(z)g(x-z)~dz$

$y(x)=f(x)+\int_x^{-l}y(x-t)g(t)~d(x-t)$

$y(x)=f(x)-\int_x^{-l}y(x-t)g(t)~dt$

$y(x)=f(x)+\int_{-l}^xy(x-t)g(t)~dt$


问:
What is the number of animals in a farm when the quantities add up?
The problem is as follows:

In a farm there are cows, horses and pigs. Not counting the cows there
are $124$ animals, not counting the horses $36$ animals, not counting
the pigs $28$ animals. Using this information, what is the number of
horses on that farm?

The alternatives in my book are as follows:

$\begin{array}{ll}
1.&\textrm{16 horses}\\
2.&\textrm{18 horses}\\
3.&\textrm{12 horses}\\
4.&\textrm{8 horses}\\
5.&\textrm{16 horses}\\
\end{array}$

What I've attempted to solve this problem was to make a set of equations as follows:

Assuming cows is $c$, horses is $h$ and pigs is $p$ then:

$h+p=124$

$c+p=36$

$c+h=28$

Then adding subtracting the third from the second:

$h-p=-8$

And adding this to the first becomes into:

$2h=116$

$h=58$

Thus the number of horses would make $58$ horses but it doesn't appear in the alternatives. On a second analysis I found that:

$2h+2c+2p=188$

$h+c+p=94$

But since: $h+p=124$ this makes the earlier equation a contradiction. What could I be doing wrong?. Can somoene help me here?. What part did went not right?.

答:
There’s an error in the problem. If you add all three of your original equations together, you find that $2(c+h+p)=188$, which implies that there are altogether $94$ animals. However, this clearly contradicts the statement that there are $124$ animals not counting the cows.

I’ll take a guess that it was supposed to say that there are $24$ animals not counting the cows. In that case we’d have $2(c+h+p)=88$, for a total of $44$ animals, and since $36$ of them are not horses, there would be $8$ horses.


问:
Can generalised continued fractions always be written so that they are simple?
Can every generalised continued fraction be written in simple continued fraction form? If not, how else does one represent a generalised continued fraction using the standard $[a_0;a_1,a_2,\ldots]$ notation when there are $b_1,b_2,\ldots$ to consider?

答:
How does one represent ...

There is this notation
$$
a_0 + \underset{k=1}{\overset{\infty}{\sf K}} \frac{b_k}{a_k} =
a_0 + \cfrac{b_1}{a_1 + \cfrac{b_2}{a_2 + \cfrac{b_3}{a_3 + \cfrac{b_4\;\;}{{\ddots}}}}}.
$$


问:
How do I prove this for all binary trees?
Ok, so here is my question.

For a binary tree B,

count of nodes is nodeCount(B)

count of leaves is leafCount(B)

count of half nodes is halfnodeCount(B)  (Half nodes are nodes that have just one child).

P(B) is given by nodeCount(B) = 2 x leafCount(B) + halfnodeCount(B) - 1

Using mathematical induction, prove that P(B) holds for all binary trees.

I was able to prove the base case, when the tree has just one node, and no leaves or half nodes.

That is, nodeCount(B) = 1, leafCount(B) = 1, halfnodeCount(B) = 0

2 x leafCount(B) + halfnodeCount(B) - 1

= (2 x 1) + 0 - 1

= 2 + 0 - 1

= 1

= nodeCount(B)

Now I need to assume that P(B) is true for a tree B of k nodes, then prove that P(B) also holds for B with k+1 nodes.

Please help, I am not sure how to go about this.

答:
Moving what I put in comments to an answer (answering “how to go about this?” without providing a complete proof):

You can use regular induction for this. You proved that $P(1)$ holds, so you need to prove that $P(k) \implies P(k+1)$ when $k\ge 1$.

Let $B$ be a tree with $k+1$ nodes.

$B$ has at least one leaf, so choose one ($v$) and remove it to obtain $B'$.

Consider these two cases:

Case 1: The removed leaf $v$ was an only child.
Case 2: The removed leaf $v$ had a sibling.

Case 1: In the smaller tree $B'$, $v$’s parent is now a leaf, because $v$ was an only child. Therefore leafCount($B$) = leafCount($B'$). (Removing $v$ removed one leaf but created one new leaf.) Also, in $B$ $v$’s parent was a half-node and is no longer, so halfnodeCount($B'$) = halfnodeCount}($B$) - 1.

Therefore $2\times$leafCount($B$) + halfnodeCount($B$) - 1 = 2 leafCount($B'$) + (halfnodeCount($B'$)+1) - 1 = $k+1$ (which is what was to be shown).

Case 2: See if you can do this similarly.


问:
Reduced costs of a linear programming problem
If I have a linear programming problem in standard form, how can I prove that:

If $x^*$ is the unique optimal solution of the LP problem, then the reduced costs of all nonbasic variables are strictly positive ??

答: I suggest proving the contrapositive.  Suppose some nonbasic variable has nonpositive reduced cost.  What happens if you pivot?</p>


问:
APR Calculation
I'm hoping someone can clarify this for me.

The model/example is this:

We lend an amount of 1498.50 (loan amount). Other fees total 39.95.
The term of the loan is for 12 months. There is no interest per month, per se, but a 10.00 fee is charged for each of the 12 months.

Using federally-approved software that calculates APR (to comply with Regulation Z, Appendix J), the calculation returns the APR to be 21.488%.

The monthly payment is calculated as follows:

(Loan Amount + (10.00 x 12 months)) / 12 = 134.87.

According to the amortization schedule, none of the interest for any particular month exceeds 14%.  How can the APR for the entire term be ~ 21% when none of the effective monthly interest rates even come close to 21%?

My intuition tells me that there is some averaging going on, but to achieve an average of 21%, some months would have to exceed 21% by quite a bit, while being offset by smaller percentages on the fringes of the curve.

If you are familiar with Reg Z, Appendix J, and the given example above, would you say that the formula being used is not appropriate?  (There are a few others that the feds provide; maybe we/the software applied the wrong one)

Additionally, playing around with the software, keeping all numbers the same, but changing the term (in months), I find that the APR also changes.  I find this confusing and counter-intuitive.

For example:
3 months, APR is 39.126%
6 months, APR is 27.393%
12 months, APR is 21.488%.

Can someone please explain this?

*All numbers above are the input and output values of approved calculation software.

Huge thanks in advance!

PS: Apologies for the seemingly uninformative tag(s). I couldn't find one that was more meaningful.

答: First, if the other fees are paid as part of the loan, I find a payment of 138.1625.  If they are paid in advance, I believe they still count as interest.  Then the effective interest is 159.95.  The average balance is about half of the amount borrowed, as you start off owing 1498.5 and end at 0.  So the effective annual interest rate is about 159.95/(1498.5/2)=23.15%.  I think they get a lower value because they do an amortization, which keeps the balance higher for longer, but it is not far off.  For the shorter terms, the interest rate goes higher because the fixed 39.95 fee gets spread over fewer months.

Title: APR Calculation


问:
Arc length of logarithm function
I need to find the length of $y = \ln(x)$ (natural logarithm) from $x=\sqrt3$ to $x=\sqrt8$.

So, if I am not mistake, the length should be $$\int^\sqrt8_\sqrt3\sqrt{1+\frac{1}{x^2}}dx$$
I am having trouble calculating the integral. I tried to do substitution, but I still fail to think of a way to integrate it. This is what I have done so far:
$$\sqrt{1+\frac{1}{x^2}}=u-\frac{1}{x}$$
$$x=\frac{2u}{u-1}$$
$$dx=\frac{2}{(u-1)^2}du$$
$$\sqrt{1+\frac{1}{x^2}}=u-\frac{1}{x}=u-\frac{1}{2}+\frac{1}{2u}$$
$$\int\sqrt{1+\frac{1}{x^2}}dx=2\int\frac{u-\frac{1}{2}+\frac{1}{2u}}{(u-1)^2}du$$
And I am stuck.

Title: Arc length of logarithm function


问:
Let $u(x)$ - smooth function $u(x)\in C^{\infty }(\mathbb{R})$
Let $u(x)$ - smooth function $u(x)\in C^{\infty }(\mathbb{R})$ and $\int_{-\infty}^{\infty}|u(x)|^2 (\frac{e^x+e^{-x}}{2})dx< \infty$.

Proof that  $\int_{-\infty}^{\infty}|\frac{\mathrm{d}^2 u(x) }{\mathrm{d} x^2}|^2 (\frac{e^x+e^{-x}}{2})dx< \infty$

I tried to use integration by parts - $\int_{-\infty}^{\infty}|u(x)|^2ch(x)dx=u^2(x)sh(x)|_{-\infty}^{\infty}-2\int_{-\infty}^{\infty}|u(x)u'(x)|sh(x)dx$

But what to do next?

答:
You can consider the behavior at $\pm \infty$ separately.
Reduce to the problem: the behavior of $|u(x)|^2 e^{x}$ vs $|u''(x)|^2 e^x$.  But we can clearly see counterexamples: $u^2 e^x$ integrable, but $(u'')^2 e^x$ is not, for instance take $u(x) =e^{-x} \sin e^x$.

Maybe your problem is in the opposite direction?  This somehow makes sense, since if $\frac{u''(x)}{e^x} \to \alpha$, then $\frac{u(x)}{e^x} \to \alpha$ (l'Hospital), but not conversely.


问:
Using differentiation to find the power series for $x/(1+2x)^2$
So I am not sure how to go about this problem: finding the series representation of

$$f(x) = \frac{x}{(1+2x)^2}$$

I used $1/(1+2x)$ and created the power series

$$\sum_{n=0}^\infty (-2x)^n = \frac{1}{1+2x}$$

and differentiated both sides. My derivative was

$$\sum_{n=0}^\infty (n+1)(-1)^{n+1}(2x)^n = \frac{-2}{(1+2x)^2}$$

I found that I needed to multiply my power series by $-x/2$ in order for the power series to be equivalent to $x/(1+2x)^2$. What I don't understand is why this is wrong. Can someone help point me in the right direction?

答:
First, how I would approach this:

You'll want to start the geometric series, $|x| < 1$,

$$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$

Differentiate both sides with respect to $x$, and then multiply both sides by $x$. You will obtain

$$\sum_{n=0}^\infty nx^n = \frac{x}{(1-x)^2}$$

Replace $x$ with $-2x$:

$$\sum_{n=0}^\infty n(-2x)^n = \frac{-2x}{(1+2x)^2}$$

Multiply both sides by $-1/2$:

$$- \frac 1 2 \sum_{n=0}^\infty n(-2x)^n = \frac{x}{(1+2x)^2}$$

You can bring the $-1/2$ into the summation since $(-2x)^n = (-1)^n 2^n x^n$, causing some simplification:

$$\sum_{n=0}^\infty n(-1)^{n+1} 2^{n-1} x^n= \frac{x}{(1+2x)^2}$$

Of course, $(-1)^{n+1} = (-1)^{n-1}$, so for a slightly cleaner-looking final answer, you can use that fact to recombine the $-1$ and $2$:

$$\sum_{n=0}^\infty n(-2)^{n-1} x^n= \frac{x}{(1+2x)^2}$$

Your work, and where your error seems to arise:

You begin by claiming, via the geometric series,

$$\sum_{n=0}^\infty (-2x)^n = \frac{1}{1+2x}$$

Differentiating with respect to $x$ on both sides, you should get

$$\sum_{n=0}^\infty (-2)n(-2x)^{n-1} = \frac{-2}{(1+2x)^2}$$

The summation simplifies a bit to

$$\sum_{n=0}^\infty (-2)n(-2x)^{n-1} = \sum_{n=0}^\infty n(-2)^n x^{n-1}$$

which doesn't at all seem like the result you claimed of

$$\sum_{n=0}^\infty (n+1)(-1)^{n+1} (2x)^n$$

So I assume a calculation error arises here.


问:
Trivial cup product in a (cohomology) subring
I was going through a proof that $S^n\times S^n$ and $S^n\vee S^n \vee S^{2n}$ are not homotopy equivalent spaces. Both spaces have the same cohomology groups in degree $n$ and $2n$, namely

$$H^k(\ \cdot\ ;\Bbb{Z}) = \begin{cases} \mathbb{Z}\oplus \Bbb{Z} & k = n \\ \mathbb{Z} & k = 2n \end{cases}$$

Now the proof states the following

$H^*(S^n\vee S^n\vee S^{2n};\mathbb{Z})$ is a subring of
$H^*(S^n;\mathbb{Z})\times H^*(S^n;\mathbb{Z}) \times
 H^*(S^{2n};\mathbb{Z})$. Thus the product of any two degree $n$
elements is trivial.

Could someone please elaborate why the subring property implies the product of any two degree $n$ elements is trivial? It would be really helpfull if someone could demonstrate how the product in this case looks like.

And what precisely is meant by trivial cup product anyway? I was under the impression that the cup product is called trivial whenever the product vanishes i.e. if $a\smile b = 0$, but we are also calling products $ab$ trivial if one factor is the unit, don't we? What is meant in the above statement?

Thanks for any help!

答:
As what Nick L has pointed out, there is a more direct argument, and you can even triangulate the space and use the definition to conclude that cup products are trivial (i.e. vanish).

The subring property is given by Mayer-Vietoris sequence (I think). It induces isomorphisms
$$\eta:H^p(S^n\vee S^n\vee S^{2n})\to H^p(S^n)\oplus H^p(S^n)\oplus H^p(S^{2n})$$
for all $p>0$, so the direct sum (the following) corresponds to the positive dimension of $H^*(S^n\vee S^n\vee S^{2n})$.
$$H^*(S^n\vee S^n\vee S^{2n})\supset\bigoplus_{p>0}(H^p(S^n)\oplus H^p(S^n)\oplus H^p(S^{2n}))$$
Actually, this is already enough for deducing the triviality of cup products.

Edit:

As requested by the OP, I'll elaborate more on this statement. Because we've decomposed the positive dimension part of $H^*(S^n\vee S^n\vee S^{2n})$ into $H^*(S^n)\oplus H^*(S^n)\oplus H^*(S^{2n})$ (the direct sum of three graded polynomial rings), it should be noted that in the definition of graded ring, we have
$$\operatorname{im}(H^k(X)\times H^l(X)\overset{\smile}{\to}H^{k+l}(X))\subset H^{k+l}(X)$$
Now it becomes immediate that the cup product of any two degree $n$ elements vanishes, since $H^{2n}(S^n)\cong 0$ and $H^{2n}(S^n\vee S^n)\cong 0$.

But the idea of subring comes in dimension $0$, the MV sequence for wedge products gives
$$0\to H^0(X\vee Y)\overset{a}{\to}H^0(X)\oplus H^0(Y)\overset{i^*-j^*}{\to}\Bbb{Z}\to\ldots$$
(Apply this twice to your space) So, $H^*(S^n\vee S^n\vee S^{2n})$ contains triads $(\alpha,\beta,\gamma)\in H^0(S^n)\oplus H^0(S^n)\oplus H^0(S^{2n})$ s.t. their restrictions to the $0$-th cocycle are equal, i.e., their images under the pullback of the inclusion $\{*\}\hookrightarrow S^n$ (resp. $S^{2n}$) are equal. And I think this is probably what the proof mean by "subring".

Supplement (Something that I recall from Evan Chen's Napkin project), :)

It can also be concluded from "the cohomology pseudo-ring" consisting of the direct sum of  reduced cohomology groups. Now these two graded pseudo-rings
$$\tilde{H}^*(X\vee Y)\cong \tilde{H}^*(X)\oplus \tilde{H}^*(Y)$$
are indeed isomoprhic. Since we also know that the reduced cohomology groups are isomorphic to absolute ones in positive dimensions, this can be passed to the argument of cohomology rings to conclude the triviality of cup products. These two argument actually utilize the same idea to express (part of) the cohomology ring in terms of what we're familiar with.


问:
A noetherian ring whose ideals are idempotent is artinian
I have to prove the folowing:

If $R$ is a Noetherian ring, and for every ideal $I$ of $R$ we have $I = I^{2}$, then $R$ is Artinian.

My first thought was to try to prove that the nilradical of $R$ is equal to the Jacobson, but I got nothing with that. Then I tried to prove that any prime ideal of $R$ is maximal, but I couldn't, so I got stuck.

Thank you for any help!

答:
If $I^2=I$ and $I$ is finitely generated, it is generated by an idempotent $e$, and thus it is a direct summand of $R$ (since $R=eR\oplus (1-e)R$).

Since $R$ is Noetherian, all ideals are finitely generated and hence by the last sentence they are summands of $R$. Consequently $R$ is semisimple, hence Artinian.

Rings for which $I^2=I$ for all ideals are sometimes called "fully semiprime rings," and probably a more popular name I'm unaware of. Another way to summarize the above argument is that commutative fully semiprime rings are von Neumann regular and a Noetherian von Neumann regular ring is semisimple.

(Bonus unrelated discussion: I started learning about fully semiprime rings when learning about noncommutative rings whose ideals are all prime ideals. If I remember correctly, all ideals are prime iff $R$ is fully semiprime and the ideals are linearly ordered.)


答:
$\frac{1}{1+2x} = \sum _0^{inf} \left(-2x \right )^n$

$-\frac{2}{\left ( 1+2x \right )^2} = \sum _1^{inf}{n \times \left( -2\right )^n \times x^{n-1}}=\sum _0^{inf}{\left( n+1\right ) \times \left( -2\right )^{n+1} \times x^{n}}$

Title: Using differentiation to find the power series for $x/(1+2x)^2$

Body:
So I am not sure how to go about this problem: finding the series representation of

$$f(x) = \frac{x}{(1+2x)^2}$$

I used $1/(1+2x)$ and created the power series

$$\sum_{n=0}^\infty (-2x)^n = \frac{1}{1+2x}$$

and differentiated both sides. My derivative was

$$\sum_{n=0}^\infty (n+1)(-1)^{n+1}(2x)^n = \frac{-2}{(1+2x)^2}$$

I found that I needed to multiply my power series by $-x/2$ in order for the power series to be equivalent to $x/(1+2x)^2$. What I don't understand is why this is wrong. Can someone help point me in the right direction?


问:
Convert a polar equation into parametric
I am getting hung up on the terminology in Calc 2 and making this subject much harder than it needs to be.

Question: Convert $r=7\sin(\theta)-2$ into parametric.

Does that simply mean putting it into the $x, y$ form?

$x=r\cos(\theta)$ --> $$x=(7\sin(\theta)-2)(\cos(\theta))$$
$y=r\sin(\theta)$ --> $$y=(7\sin(\theta)-2)(\sin(\theta))$$

答:
We can solve this problem with Mathematica:

```<code>CoordinateTransform[
     "Polar" -> "Cartesian", {7 Sin[θ] - 2, θ}]
</code>```

Body:
We can solve this problem with Mathematica:

CoordinateTransform[
     "Polar" -> "Cartesian", {7 Sin[θ] - 2, θ}]


问:
Solving an integral equation: $y(x) = 16+\int_0^x 2t \sqrt{y(t)} dt$
How do I solve the following integral equation?

$$y(x) = 16+\int_0^x 2t \sqrt{y(t)} dt$$

答:
We have
$$y(x) = 16+\int_0^x 2t \sqrt{y(t)} dt \tag{$\star$}$$
From fundamental theorem of calculus,
$$y'(x) = 2x \sqrt{y(x)} \implies \dfrac{dy}{y^{1/2}} = 2x dx \implies 2y^{1/2} = x^2 + \text{ constant}$$
Hence, we get that
$$y = \left(\dfrac{x^2+ c}2 \right)^2$$
Now plug this into $(\star)$ to determine the constant $c$.</p>


问:
Proving two relations are the same with set difference
I'd just like to know where to start with this problem, where $R$ and $S$ are equivalence relations on non-empty set $A$, and suppose that $A-R = A-S$. I am looking to show that equivalence relations $R = S$.

If I say, let an example pair be an element of $A - R$. But with the exclusion of the set, how would you then make the steps to say that $R$ = $S$? Would it involve arbitrating something being equal to what's not removed?

答:
As it was pointed out in the comments, equivalence relations are subsets of the cartesian product of the set, since they are formed via pairing elements with another elements. Then your statement has to be proven from $A\times A\setminus R=A\times A\setminus S$.

The statement is true, but not because $R$ and $S$ are equivalence relations, it is true because you're taking the set difference from the total set. What I mean is: If $X$ is a set and $A,B\subset X$ are subsets, then $A=B\Leftrightarrow X\setminus A=X\setminus B$. Since $X\setminus A$ is the complement of $A$ in $X$, let's denote it by $A^\mathsf c$ (same for $B$), this is the same as saying that $(A^\mathsf c)^\mathsf c=A$.

The $\Rightarrow$ part is obvious. Let's see the $\Leftarrow$ part: $x\in A\Leftrightarrow x\notin X\setminus A\Leftrightarrow^\text{by hypothesis} x\notin X\setminus B\Leftrightarrow x\in B$, so $A=B$.


问:
How do I define Injective/Surjective functions in terms of sets and not the elements within them?
I am in a Proof writing class and we are currently on functions. I have a good understanding of what it means to be injective or surjective in terms of elements in the set but I am having trouble coming up with a formal definition that just generalizes the definitions in terms of sets using an iff statement.

So if we have a function  $f: A\rightarrow B$

Would my statement be considered a proper theorem?

If $G\subseteq B$ then  $f: A\rightarrow B$  is surjective if and only if  $G=B$ and $f(f^{-1}( B)) = B$.

My reasoning here being that if our function is surjective everything in B has to have some pre-image in A. I am not sure If I need to say anything about A.

For Injective I came up with a similar theorem:

If $H\subseteq A$ then $f: A\rightarrow B$ is injective if and only if $A=H$ and $f(f^{-1}(A))=A$

My reasoning again being we need everything in our domain to have some image in B. So our domain subset must take on the whole set A.

Your help and knowledge would be extremely appreciated!

答:
Here is a pair of categorical definitions:

A function $m:Y \to Z$ is injective, iff for all functions $a,b$ with codomain $Y$,

$$m \circ a = m \circ b \implies a = b$$

A function $e:X \to Y$ is surjective, iff for all functions $f,g$ with domain $Y$,

$$f \circ e = g \circ e \implies f=g$$


问:
Factorial simplification help; how did wolfram get this?
in this problem n is assumed to be odd:
Here's what I am trying to simplify:
$$ \frac {(n-2)!( \frac {n-1}{2})!} { (n-1)! (\frac {n-3} {2}) !}
$$

Wolfram is telling me that this should simplify to $$\frac {1}{2} $$ but I am very confused about how it gets there. I understand that the two factorial terms combine, but then I have $$ (n-2) (\frac {n-1} {2}) $$ as my final simplification and I really have no idea how to move from that to $$\frac{1}{2}$$ Any help would be appreciated.

答:
First, note that

$$\frac{(n-2)!}{(n-1)!} = \frac{(n-2)(n-3)(n-4) \cdots (2)(1)}{(n-1)(n-2)(n-3)(n-4) \cdots (2)(1)} = \frac{1}{n-1}$$

Similarly, notice that

$$\dfrac{ \left( \dfrac{n-1}{2} \right)! }{\left( \dfrac{n-3}{2} \right)!} = \dfrac{ \left( \dfrac{n-1}{2} \right)\left( \dfrac{n-3}{2} \right)\left( \dfrac{n-5}{2} \right)\cdots(2)(1) }{\left( \dfrac{n-3}{2} \right)\left( \dfrac{n-5}{2} \right)\cdots(2)(1) } = \frac{n-1}{2}$$

Multiply the two together for the desired result.


问:
Can someone help me understand this "family of sets"?
I have been trying to understand what this "family of sets" really is all day and I just can't wrap my head around it. it goes like this:
$\{A_n\}_{n\in{\mathbb{N}}}$ such that $A_n \subseteq{\mathbb{N}}$ for every $n\in{\mathbb{N}}$.
At first, I thought that this "family of sets" is practically equal to $P(\mathbb{N})$ but then I also think it might be just a random subset of $\mathbb{N}$ for every different $n$ like as long as $A_n \subseteq{\mathbb{N}}$ then it is counted as part of the family of sets. Can someone give me an idea of what this family of sets represents?
for more context, the question continued to define $D=\{n\in{\mathbb{N}} : n\notin{A_n}\}$ and then continues to ask multiple questions such as: define the family of sets $\{A_n\}_{n\in{\mathbb{N}}}$ such that $D=\emptyset$

答:
For starters, this

it might be just a random subset of $\mathbb{N}$ for every different
$n$ like as long as $A_n \subseteq \mathbb{N}$

is correct.

Suppose you have such a sequence. For definitiness, imagine that it starts as
$$
A_1 = \{1,2,3\}, A_2 = \text{the odd numbers}, A_3= \text{primes}, A_4 = \{3\}, \ldots
$$

Then the set $D$ for this particular sequence starts with
$$
D = \{2, 4, \ldots \}
$$
since $2 \not \in A_2$ and $4 \not \in A_4$ while $1   \in A_1$ and $3  \in A_3$.

Now you might be able to continue with the "multiple questions" that follow.

Note: I suspect that this exercise is leading up to an argument showing that the list $\{A_n\}$ cannot be all of $\mathcal{P}(\mathbb{N})$.


问:
Symmetric difference of sets, Ring
The symmetric difference between $2$ sets $A,B$ is $A \Delta B=(A \setminus B)\cup (B \setminus A)$. Let $Ω$ be a set. The characteristic function for $A \subset Ω$ is $\mathcal X_A:Ω\rightarrow \mathbb{Z}$ with $\mathcal X_A(x)=0$ for $x \notin A$ and $\mathcal X_A(x)=1$ for $x \in A$. Let $\overline{\mathcal X}_A:Ω \rightarrow \mathbb{Z}/2\mathbb{Z}$ be the function with $x$ modulo $2$.

Show that $\overline{\mathcal X}_A+\overline{\mathcal X}_B=\overline{\mathcal X}_{A \Delta B}$ for $A,B \subset Ω$.

Show that $(A \Delta B)\Delta C=A\Delta(B\Delta C)$ for $A,B,C \subset Ω$.

I already showed 2. but I don't know how to do 1..

答:
We have to show that $\overline{\chi}_A(x)+\overline{\chi}_B(x)=\overline{\chi}_{A\Delta B}(x)$, for every $x\in\Omega$ and $A,B\subset\Omega$.

So let $x\in\Omega$.
There are three cases to consider.
$x\in A$ and $x\notin B$.
$x\in A$ and $x\in B$.
$x\notin A$ and $x\notin B$.

The first case would play out like this:

We have that $x\in A$ and $x\notin B$. Therefore $x\in A\Delta B$. (Why?)

Hence

$\underbrace{\overline{\chi}_A(x)}_{=1}+\underbrace{\overline{\chi}_B(x)}_{=0}=\underbrace{\overline{\chi}_{A\Delta B}(x)}_{=1}$

So $1+0=1$, which is correct.

I leave the other cases to you.

Title: Symmetric difference of sets, Ring


问:
Hermitian Matrix and its Eigenvalues
Suppose B is an N x N Hermitian matrix with $\lambda _i$ eigenvalues. if  $w_k$ are any orthonormal real-valued vectors. I need to prove that
\begin{equation}
\sum _{k=1}^N\left(w_k^TBw_k\right)\:=\:\sum _{k=1}^N\left(\lambda _k\right)
\end{equation}
Here's what I know so far

Since $B$ is Hermitian, then $B = Q\Lambda Q^H$ Where $Q$ is the eigenvector matrix of $B$ and $\Lambda$ is the diagonal matrix of eigenvalues of $B$

now $B$ can be represented as $B = \sum _{j=1}^N\left(\lambda _jq_jq^H_j\right)$

If we substitute that in the original question, we get
\begin{equation}
\sum _{k=1}^N\left(w_i^T\sum _{j=1}^N\left(\lambda _jq_jq_j^H\right)w_i\right)\
\end{equation}

But am confused on what to do from here. Please help me

答: In terms of matrices, the sum is just $\operatorname{tr}(W^TBW)$. Since $W$ is orthogonal, $W^TBW$ is similar to $B$. Therefore $\operatorname{tr}(W^TBW)$ is equal to $\operatorname{tr}(B)$, which in turn is equal to the sum of all eigenvalues of $B$.

Title: Hermitian Matrix and its Eigenvalues


问:
Proving a sequence diverges to infinity using definition of divergence
Here is the question I am working on:
Suppose {$s_n$} is a sequence with the property that $\lim\limits_{x \to \infty} t_n$ $=c$, where $c$ is some positive real number and $t_n = \frac{s_n}{n}$.  Show that {$s_n$} diverges to infinity.

This is what I have. We know $t_n$ converges, so by the definition of convergence, for all $\epsilon$, there exists an $N$ such that $|t_n - c| < \epsilon$. By substitution, $|\frac{s_n}{n} - c| < \epsilon$. I'm skipping a bunch of algebra here, but I created a neighborhood around $s_n$ and wound up with this: $nc-n\epsilon < s_n < n\epsilon+nc$. I know that I need to select an M so that $nc-n\epsilon \ge M$ always. I feel like this might depend on an $N$ that I would have to choose for $t_n$ as well, but that's where I've run out of steam. Is there any advice on how to define these? I would appreciate any assistance. Thanks!

(I apologize if the title is poor.)

答: First we show that for some $N_{0} \ge 1$, if $n \ge N_{0}$ then $t_n > 0$. Indeed, for $\epsilon = \dfrac{c}{2} > 0$, there exists a $N_{0} \ge 1$ such that if $n \ge N_{0} $ then $|t_n - c| < \dfrac{c}{2} \implies \left|\dfrac{s_n}{n} - c\right| < \dfrac{c}{2}\implies \dfrac{s_n}{n} - c > -\dfrac{c}{2}\implies t_n = \dfrac{s_n}{n}> \dfrac{c}{2} > 0$. Thus for $n \ge N_{0}$, we have: $s_n = nt_n = n|t_n| =n|t_n-c+c|\ge n(|c|-|t_n -c|) > n(c-\dfrac{c}{2})= \dfrac{nc}{2}\implies \displaystyle \lim_{n \to \infty} s_n = +\infty$.


答:
$\overline{\mathcal X}_A(x)+\overline{\mathcal X}_B(x)=0\Leftrightarrow{\mathcal X}_A(x)+{\mathcal X}_B(x)=0,2\Leftrightarrow (x\notin A\land x\notin B)\lor(x\in A\land x\in B)\Leftrightarrow (x\in\Omega\setminus A\land x\in\Omega\setminus B)\lor(x\in A\cap B)\Leftrightarrow x\in (\Omega\setminus A)\cap(\Omega\setminus B)=\big((\Omega\setminus A)^\mathsf c\cup(\Omega\setminus B)^\mathsf c\big)^\mathsf c=\big(A\cup B)^\mathsf c\lor (x\in A\cap B)\Leftrightarrow x\in(A\Delta B)^\mathsf c\Leftrightarrow{\mathcal X}_{A \Delta B}=0\Leftrightarrow \overline{\mathcal X}_{A \Delta B}=0$

Can you do the other case?

Title: Symmetric difference of sets, Ring

Body:
The symmetric difference between $2$ sets $A,B$ is $A \Delta B=(A \setminus B)\cup (B \setminus A)$. Let $Ω$ be a set. The characteristic function for $A \subset Ω$ is $\mathcal X_A:Ω\rightarrow \mathbb{Z}$ with $\mathcal X_A(x)=0$ for $x \notin A$ and $\mathcal X_A(x)=1$ for $x \in A$. Let $\overline{\mathcal X}_A:Ω \rightarrow \mathbb{Z}/2\mathbb{Z}$ be the function with $x$ modulo $2$.

Show that $\overline{\mathcal X}_A+\overline{\mathcal X}_B=\overline{\mathcal X}_{A \Delta B}$ for $A,B \subset Ω$.

Show that $(A \Delta B)\Delta C=A\Delta(B\Delta C)$ for $A,B,C \subset Ω$.

I already showed 2. but I don't know how to do 1..


问:
How to calculate Laplace inverse of $\frac{1}{(s^{2}+1)^{2}}$?
Laplace inverse of $\frac{1}{(s^{2}+1)^{2}}$.

I have beaten my head about this but have not got a clue as to how to do it! Can someone help? I tried to express this as some form of a derivative, but I am really just not sure what to do since a derivative of $\frac{1}{s^{2}+1}$ would involve a 2s in the numerator. Can someone please help?

答:
$$F(s)=\frac{1}{(s^{2}+1)^{2}}=\frac{2s}{(s^{2}+1)^{2}} \times \dfrac 1 {2s}$$
$$F(s)=- \dfrac {d}{ds}\left  (\frac{1}{(s^{2}+1)} \right) \times \dfrac 1 {2s}$$
$$F(s)=G(s) H(s)$$
And their inverse Laplace Transform are:
$$g(t)=t \sin t \,\, , h(t)=\dfrac 12$$
Then use the convolution theorem. Evaluate this integral:
$$f(t)=\int_0^t g(\tau) h(t-\tau)  d \tau$$
$$f(t)=\int_0^t  \tau \sin (\tau) \times \dfrac 12 d \tau$$

For the Convolution Theorem look here formula 33
here

You should find:
$$f(t)=\dfrac 12 (\sin t - t\cos t)$$


答:
An alternative argument to the one presented in the other answer: we have
$$
\begin{align}
\sum _{k=1}^N\left(w_k^TBw_k\right) &=
\sum _{k=1}^N\left(\sum _{j=1}^Nw_k^H\left(\lambda _jq_jq_j^H\right)w_k\right)
\\ & =
\sum_{j=1}^N \sum_{k=1}^N  \lambda_j \,(w_k^Hq_j)( w_k q_j^H)
\\ & =
\sum_{j=1}^N \lambda_j \left(\sum_{k=1}^N (w_k^Hq_j)( w_k q_j^H)\right)
\\ & =
\sum_{j=1}^N \lambda_j \left(\sum_{k=1}^N |w_k^Hq_j|^2\right)
= \sum_{j=1}^N \lambda_j \cdot 1
\end{align}
$$

Title: Hermitian Matrix and its Eigenvalues

Body:
Suppose B is an N x N Hermitian matrix with $\lambda _i$ eigenvalues. if  $w_k$ are any orthonormal real-valued vectors. I need to prove that
\begin{equation}
\sum _{k=1}^N\left(w_k^TBw_k\right)\:=\:\sum _{k=1}^N\left(\lambda _k\right)
\end{equation}
Here's what I know so far

Since $B$ is Hermitian, then $B = Q\Lambda Q^H$ Where $Q$ is the eigenvector matrix of $B$ and $\Lambda$ is the diagonal matrix of eigenvalues of $B$

now $B$ can be represented as $B = \sum _{j=1}^N\left(\lambda _jq_jq^H_j\right)$

If we substitute that in the original question, we get
\begin{equation}
\sum _{k=1}^N\left(w_i^T\sum _{j=1}^N\left(\lambda _jq_jq_j^H\right)w_i\right)\
\end{equation}

But am confused on what to do from here. Please help me


问:
Proving problem about relative extrema
Prove that if $f$ is increasing on $[a, b]$ and $g$ is increasing on $[f(a),f(b)]$ then if $g \circ f$ exists on $[a,b]$, $g \circ f$ is increasing on $[a,b]$.

It's about the relative extrema of functions, and I don't know how to do this.

Thanks for the answers!

答:
Take some $x,y\in[a,b]$ such that $x\le y$. Then $a\le x\le y\le b$, and since $f$ is increasing on $[a,b]$ we have that $f(a)\le f(x)\le f(y)\le f(b)$, so $f(x),f(y)\in[f(a),f(b)]$ and $f(x)\le f(y)$. Therefore, since $g$ is increasing on $[f(a),f(b)]$ (try to figure it out first!):

 $g(f(x))\le g(f(y))$, which is the same as $(g\circ f)(x)\le (g\circ f)(y)$. So we took $x,y\in[a,b]$ such that $x\le y$ and we got that then $(g\circ f)(x)\le (g\circ f)(y)$, hence we can conclude

$g\circ f$ is increasing on $[a,b]$

Title: Proving problem about relative extrema


问:
Open balls in $\mathbb{R}^d$ are Jordan Measurable
I'm trying to solve the following question from Terrence Tao's An Introduction to Measure Theory.

Show that an open Euclidean
  ball $B(x, r) := \{y \in \mathbb{R}^d
: |y − x| < r\}$ in $\mathbb{R}^d$ is Jordan measurable, with Jordan
  measure $c_d r^d$
  for some constant $c_d > 0$ depending only on
  $d$.

Is there an elementary way to approach this problem?

答:
HINT:

Cover the ball $B(x, (1-\epsilon) r)$ with finitely many cubes of diameter $<\epsilon r$. This Jordan cover $C_1$ will be contained in $B(x,r)$.  Do a homothety of center $x$ and ratio $\frac{1}{1-\epsilon}\ $. We get a Jordan cover $C_2$ of $B(x,r)$ contained in $B(x,\frac{1}{1-\epsilon} r)$, so contained in $B(x,2r)$ (assume $\epsilon < \frac{1}{2})$. Therefore,
$$C_1\subset B(x,r) \subset C_2$$
and
$$\mu(C_2)- \mu(C_1) = (1- (1-\epsilon)^n)
\cdot \mu(C_2) \le \\
 \le (1- (1-\epsilon)^n) \mu^{\star}( B(x,2r))\underset{\epsilon \to 0}{\rightarrow} 0$$

Note: remember how the Greeks approximated the circle with inscribed and circumscribed regular polygons.


问:
Find the domain of $f(x, y) = \sqrt{1-x^2} - \sqrt{25-y^2}$
Sorry in advance for posting this simple question. This is kinda important and I just need to quickly double-check on this.

Question: Find the domain of the following function: $$f(x, y) = \sqrt{1-x^2} - \sqrt{25-y^2}$$

My attempt: I was able to graph the function out with GeoGebra but I couldn't exactly figure out if the domain is $[-1, 1]$ or $(1, 1)$. From the looks of the function, there is no restriction for the domain at $x =-1$ and $1$ so I'd assume that the domain is $[-1, 1]$. Is there a mathematical way of solving this and proving what the domain is?

答:
Yes there is a simple algebraic way to analyze this. Unless you want complex numbers, $\sqrt{1 - x^2}$ is only valid as long as $1 - x^2 \geq 0$ or $x^2 \leq 1$ or $|x| \leq \sqrt{1} = 1$. In other words you need $-1 \leq x \leq 1$.

Similarly $\sqrt{25 - y^2}$ is only valid as long as $25 - y^2 \geq 0$ or $y^2 \leq 25$ or $|y| \leq \sqrt{25} = 5$. In other words you need $-5 \leq y \leq 5$.

Thus, the domain is $\{(x, y) \in \mathbb{R}^2 : -1 \leq x \leq 1, -5 \leq y \leq 5\}$ which is just $[-1, 1] \times [-5, 5]$.


问:
If $f$ is continuous on $(a,b)$ , $\lim_{x\to a+}f(x)$ and $\lim_{x\to b-}f(x)$ exist then $f$ is bounded.
It's quite intuitive if you draw the graph-- there are no asymptotes so it must be bounded-- but how do you prove it? proving there exists a maximum and a minimum? Using the extreme value theorem? But it's not a closed interval! Any help would be appreciated! Thanks!</p>

答:
Suppose that $f$ is not bounded, so there is a sequence of points $(x_n)$ so that $f(x_n)>n$ but $(x_n)\in [a,b]$ so there is a subsequence $(x_{n_j})$ that convergers on $[a,b]$ the limite point cannot belong to $(a,b)$ because $f$ is continuos on this interval. So the limit point is $=a$ or $=b$, without loss generality, suppose that $x_{n_j}\rightarrow a$ but by hypothesis the limit below exist
$$
\lim_{j}f(x_{n_j})
$$

but by constructer $\lim_jf(x_{n_j})=\infty$ an contradiction.


问:
Notation for a region $R=[0,2]×[0,4$] when calculating double integrals
Can someone explain this notation.

I understand that in the $x$ direction it is a closed interval from $0$ to $2$ and in the $y$ direction it is a closed interval from $0$ to $4$.

But why do we take the cross product?

答:
When we integrate a univariate function $f(x)$ over a region $R=[0,2]$, this means to calculate:
$$\int_0^2 f(x) dx$$

Likewise, when we  integrate a bivariate function $f(x,y)$ over a region $R=[0,2] \times [0,4]$, this means to calculate:
$$\int_0^4\int_0^2 f(x,y) dx dy$$

In other words, you integrate $dx$ with respect to $x$'s range, then integrate that $dy$ with respect to $y$'s range. (Or vice versa, doing it the other way is fine too)


问:
Proof that sequence converges
How can I prove that $ (a_n) = \frac{n^3 -1}{2n^3-n} $ converges?

I've calculated the limit and got a result of a 1/2.

Now I need to prove that this limit exists. So, I tried to use the definition and find an $M$ that $n > M \implies |a_n - L| < \epsilon$ for $ \epsilon > 0 $, but I couldn't reach in a result.

Is there any strategy to prove that this sequence converges?

答:
HINT: You can use the sandwich theorem here as an alternative. Note that
$$\frac{n^3-1}{2n^3}\leq\frac{n^3-1}{2n^3-n}\leq\frac{n^3}{2n^3-n}$$
Now, $\frac{n^3-1}{2n^3}=\frac{1}{2}-\frac{1}{2n^3}$ and you can use the result that if $a_n$ converges to a and $b_n$ converges to b, then $a_n+b_n$ converges to a+b to find it's limit.

For the $\frac{n^3}{2n^3-n}$, you can use the $\epsilon-N$ definition of convergence.

 Choose any $\epsilon>0$ and thus, $\exists N\in\mathbb{N}$, such that $N>\sqrt{\frac{1}{4\epsilon}+\frac{1}{2}}$ by the Archimedean property.

Can you do the rest?

Title: Proof that sequence converges


答:
Define a new function:
$$
f^\star(x) = \begin{cases} \lim_{x \to a+} f(x) &\text{ if } x = a\\
f(x) &\text{ if } a < x < b\\
\lim_{x \to b-} f(x) &\text{ if } x = b
\end{cases}
$$ This new function is defined on the closed interval $[a, b]$ and it is continuous on $[a, b]$, by construction. So now you can use the extreme value theorem on $f^\star$ instead to conclude that $f^\star$ is bounded. Since $f$ is just $f^\star$ restricted to $(a, b)$, it is must also be bounded.

Title: If $f$ is continuous on $(a,b)$ , $\lim_{x\to a+}f(x)$ and $\lim_{x\to b-}f(x)$ exist then $f$ is bounded.

Body: It's quite intuitive if you draw the graph-- there are no asymptotes so it must be bounded-- but how do you prove it? proving there exists a maximum and a minimum? Using the extreme value theorem? But it's not a closed interval! Any help would be appreciated! Thanks!</p>


问:
Ideal sets of polynomials

The set I of all polynomials of even degree in Z(x) is not ideal because $1 \in I$ and $I \ne Z(x)$. Or consider $1 \in I$, $1x=x \notin I$.

I understand the second reasoning since x is in Z(x) so by definitiion of ideals, 1x or x1 has to be in I but it is not so I is not ideal--am i right?

Also, I am totally lost for the first reasoning--why $I \ne Z(x)$ mean I is not ideal? And Is the above also true in Q(x) setting? (I would assume so.)

答: Suppose that $I$ is an ideal of $\mathbb Z[x]$ and that $1 \in I$. Then, for any $f(x) \in \mathbb Z[x]$ we have $f(x) = 1f(x) \in I$, so, $\mathbb Z[x] \subseteq I$ and then $I = \mathbb Z[x]$.

Title: Ideal sets of polynomials


问:
Show that $x^4+x+1$ is a prime polynomial over $F_2$
To prove this, the only idea I had was to divide $x^4+x+1$ by all polynomials over $F_2$ and show that it cannot be evenly divided.

This seems tedious and I am sure there is a more elegant way to prove this. Can anyone give hints or guide me towards a proof?

答:
There is no root, and we need a constant term $1.$ We may simply calculate
$$ \left(x^2 + 1 \right) \left(x^2 + 1 \right) $$
$$ \left(x^2 +x+ 1 \right) \left(x^2 +x+ 1 \right) $$
$$ \left(x^2 + 1 \right) \left(x^2 +x+ 1 \right) $$


问:
Lower bound existence of positive semidefinite quadratic function
Let $f(x)=x^TQx+2b^Tx$, where $Q\succeq 0$. When does this function have a lower bound and when not?

I did some survey (especially from Prof. Lieven Vandenberghe’s slides on convex optimization) and the answer seems to be, the lower bound exists iff $b\in \mathcal R(Q)$, i.e. $b$ lies in the vector space spanned by $Q$'s row(column) vectors.

Intuitively I believe this is a constraint that guarentees that we can convert the function in a way like Completing the square, and I can complete this arithmetic when $Q\succ 0$ and in that case $f(x)=\|Q^{1/2}x+Q^{-1/2}b\|_2^2-b^TQ^{-1}b$, and it does have a lower bound. However, in general cases where $Q$ is singular, I don't know what to do. Hope somebody can help me, thanks!

Btw, this problem arises from a nonconvex problem with strong duality: $\min x^TAx+2b^Tx$ s.t. $x^Tx\le 1$, where the dual function is bounded when $b\in \mathcal R(A+\lambda I)$.

答:
Proof:

When $b\not\in\mathcal R(Q)$, we can decompose $b$ into $b=b_1 + b_2$ where $b_1\in\mathcal R(Q)$ and $b_2\perp\mathcal R(Q)$. The direction given by $-b_2$ will lead the function to $-\infty$, as $f(-td_2)=t^2d_2^TQd_2-2tb^Td_2=-2tb^Td_2\to -\infty$ when $t\to\infty$.

when $b\in\mathcal R(Q)$, $\nabla f(x)=2Q^Tx+2b=0$ has a solution denoted by $x^*$. As $f(x)$ is convex, $x^*$ must be global minimum point.


问:
Why did the range change to $(0,4)$ for density of $X^2$ with $[-1,2]$ uniform distribution?
The question is: Suppose $X$ has uniform $[-1,2]$ distribution. Find the density of $X^2$.

The range is $(0, 4)$, but it's $g(-1) = 1$ and $g(2) = 4$, so I don't get why it's $(0, 4)$

Also, the solution says this:

For (0,1), $\frac{1/3+1/3}{2\sqrt{y}}$

For (1,4), $\frac{1/3+0}{2\sqrt{y}}$

Why is it split at 1, and how come $(0,1)$ is $1/3+1/3$ and (1,4) is $1/3+0$?

答:
The range of $g$ is $(\min g(x), \max g(x))$, not $(g(\min x), g(\max x))$. In other words, while $g(-1)=(-1)^2=1$, $g(0)=0^2=0<1$ is the minimum of $g(x)=x^2$ on the range $(-1,2)$.

The density function is split in two parts for the same reason, $\forall a\in (0,1)\implies a^2\in (0,1), X=a$ and $X=-a$ both map to $X^2=a^2$, which means the density of $X^2$ is doubled on $(0,1)$


问:
Is $\bigcup_{k=1}^\infty (\frac{1}{k}, 2)$ a cover of $(0, 1)$?
I was just wondering whether $\bigcup_{k=1}^\infty (\frac{1}{k}, 2)$ is a cover of $(0, 1)$. Intuitively I think that it is because $\frac{1}{k}\rightarrow 0$ as $k\rightarrow\infty$ so I'll start catching all the numbers near zero as I open up the interval. Please let me know if I'm making a mistake, thank you!

答:
We want to show that $(0,1)\subseteq\bigcup_{k=1}^{\infty}(\frac{1}{k},2)$.

Fix $z\in(0,1)$. By the Archimedean property, there exists an $N\in\mathbb{N}$ such that $\frac{1}{N}<z$. Then $z\in(\frac{1}{N},2)$. It follows that $z\in\bigcup_{k=1}^{\infty}(\frac{1}{k},2)$, so $(0,1)\subseteq\bigcup_{k=1}^{\infty}(\frac{1}{k},2)$.

We are able to make use of the Archimedean property because $z>0$ and $z\neq 0$. If $z=0$, there would not exist and $N\in\mathbb{N}$ such that $\frac{1}{N}<0$, so $0$ is never in any set in the union. Zero is however a limit point of the union.


问:
Let $G$ be a group and $a,b \in G$. Prove that for each integer such as $n$, we have $(a b^{-1} a)^n = a b^n a^{-1}$
I tried expanding $n$ parentheses such as $a b^{-1} a$ and then tried to replace the second $a$ with $a^2 a^{-1}$ but didn't know how to proceed further.

Also, is it possible to use commutativity here somehow?

答: The inner adjacent $aa^{-1}$ terms cancel out, and this is fairly trivial:  $(aba^{-1})^n=aba^{-1}aba^{-1}\dots aba^{-1}=ab^na^{-1}$.


问:
Find the vector perpendicular to the curve $f(x, y) = x^2y+y^3$
Question: Which given vector is perpendicular to the following curve at the point (1, 2).$$f(x, y) = x^2y+y^3 = 10$$

Given vectors: $<4, 3>, <-13, 4>, <2, 17>$ or non of these choices are correct.

My attempt: By taking the partial derivatives of this function, I was able to find that the gradient of the function to be $(4, 13)$. I was then able to find the tangent line of the function at the given point to be: $$x = \frac{{13(y-2)}}{4}+1$$
However, I'm not sure what to do next on out from here to evaluate the vector. Thanks in advance!

答: When you have a vector $(a,b)$, an orthogonal vector is just $(-b,a)$, since $(a,b)·(-b,a)=-ab+ab=0$ (remember two vectors are orthogonal if and only if their dot product is zero). Since your gradient is $(4,13)$, which option do you think is the answer?


答:
For all $\epsilon > 0$, by the Archimedean Property, there exists $N \in \mathbb{N}$ such that $N >  \frac{1}{\sqrt[3]{\epsilon + \frac{1}{2}}}$, then for all $n \geq N$, we have $\frac{1}{n^3} < \epsilon + \frac{1}{2}$ and

\begin{align*}
\left| a_n - \frac{1}{2} \right| &= \left| \frac{n^3 - 1}{2n^3 - n} - \frac{1}{2} \right|\\
&= \left| \frac{1 - \frac{1}{n^3}}{2 - \frac{1}{n^2}} - \frac{1}{2} \right|\\
&\leq \left| 1 - \frac{1}{n^3} - \frac{1}{2} \right| \hspace{20pt} \mbox{(since $2 - \frac{1}{n^2} \geq 1$})\\
&= \left| \frac{1}{n^3} - \frac{1}{2} \right| \\
&< \epsilon
\end{align*}

Title: Proof that sequence converges

Body:
How can I prove that $ (a_n) = \frac{n^3 -1}{2n^3-n} $ converges?

I've calculated the limit and got a result of a 1/2.

Now I need to prove that this limit exists. So, I tried to use the definition and find an $M$ that $n > M \implies |a_n - L| < \epsilon$ for $ \epsilon > 0 $, but I couldn't reach in a result.

Is there any strategy to prove that this sequence converges?


问:
Eigenvalues and eigenvectors of integral operator
I am trying  to find the eigenvalues and eigenvectors for the following integral operator in $L^{2}[0,1]$:
$T:H \rightarrow H$ is defined as follows:

$$ T[f(x)]= \int_{1-x}^{1} f(t) dt $$.

Here is my procedure until now:

Let $g(x)$ be an eigenvector associated with the eigenvalue $\lambda$, then
$$T[g(x)]  = \lambda g(x) $$.

Solving this problem is equivalent to solving the following integral equation
$$\int_{1-x}^{1} g(t) dt - \lambda g(x) = 0 $$

Differentiating twice the previous equation, we get

$$ g(1-x)=\lambda \frac{dg(x) }{dx}  $$

But i couldn't find any way to solve the last differential equation. Is the previous procedure correct? And is there any way to solve the last equation?

答: By differentiating of equation $g(1-x)=\lambda g'(x)$, we get $-g'(1-x)=\lambda g''(x)$, from where $\lambda^2 g''(x)+g(x)=0$ (it's obvious that $\lambda\ne0$). By adding boundary conditions $g(0)=g'(1)=0$ and solving the obtained Sturm-Liouville problem we get $g_n(x)=\sin\mu_nx$, $\mu_n=\dfrac{1}{\lambda_n}$, $\mu_n=\dfrac{\pi}{2}+\pi n$, $n\in\mathbb{Z}$.


答: If you have proved the algebraic limit laws for sequences, then this isn't too bad. $$\begin{align*}\lim_{n\to\infty}\frac{n^3-1}{2n^3-n}&=\lim_{n\to\infty}\frac{1-\frac{1}{n^3}}{2-\frac{1}{n^2}}\\&=\frac{\lim\limits_{n\to\infty}(1-\frac{1}{n^3})}{\lim\limits_{n\to\infty}(2-\frac{1}{n^2})}\\&=\frac{1}{2}\end{align*}$$ This works because both the top and bottom limits exist and the bottom limit is non-zero for all $n\in\mathbb{N}$.

Title: Proof that sequence converges


问:
Second order inhomogeneous equation: $y''-2xy'-11y=e^{-ax}$
My question relates to an a second order inhomogeneous equation:

$$y''-2xy'-11y=e^{-ax}$$

First I need to investigate the homogeneous equation:

$$y''-2xy'-11y=0$$

$$y''-2xy'=11y$$

Forms Hermite's Equation where $\lambda = 11$

So I need a general solution of the homogeneous equation
$Ly(x) = 0$.

To do this I need to linearly independent solutions, $y_1(x)$ and $y_2(x)$ say, and
then the general solution of $Ly(x) = 0$.

becomes:

$$Ay_1(x) + By_2(x)$$

where $A$ and $B$ are arbitrary constants.

I am struggling to find two independent solutions of the hermite equation above:

My attempt

If I take a solution to be if the form $$y=\sum_{n=0}^{\infty}a_{n}x^n$$
Then putting this into the ODE I get the following:

$$\sum_{n=2}^{\infty}n(n-1)a_{n}x^{n-2}-2\sum_{n=1}^{\infty}na_{n}x^{n}-11\sum_{n=0}^{\infty}a_{n}x^{n}=0$$

Which I can reduce to:

$$\sum_{n=0}^{\infty}[(n+2)(n+1)a_{n+2}-2(n+\frac{11}{2})a_{n}]x^{n}=0$$

And I have found that rearranging gives:

$$a_{n+2} = \frac{-2(n+\frac{11}{2})a_{n}}{(n+2)(n+1)}$$

This can be used to develop a recurrence relation...

How can I use this to find my two independent solutions $y_1(x)$ and $y_2(x)$ which I need in order to calculate the inhomogeneous solution?

Maybe there are more efficient ways of calculating this ODE that I am unaware of.

Edit

I displayed the equation incorrectly and have since edited it.

Should my solution be in the form of an infinite sum rather than a finite polynomial because I believe that the solution could only be constructed as a series solution which terminates if and only if $\lambda = −2n$ where $n \in \mathbb N$.

In my case $\lambda$ isn't of the above form hence an infinte series is required.

答:
We will solve the equation
\begin{equation}
(a_1x+b_1)f''(x)+(a_2x+b_2)f'(x)+(a_3x+b_3)f(x)=g(x)\tag 1
\end{equation}
where $f$, $g\in L_2(\bf R\rm)$ and $a_1$, $a_2$, $a_3$, $b_1$, $b_2$, $b_3$ are constants in $\bf R\rm$.

Let the Fourier Transform of a function $f$ of $L_2(\bf R\rm)$ is
$$\widehat{f}(\gamma)=\int^{\infty}_{-\infty}f(t)e^{-i t\gamma}dx$$
the Inverse Fourier Transform is
$$f(x)=\frac{1}{2\pi}\int^{\infty}_{-\infty}\widehat{f}(\gamma)e^{i \gamma x}d\gamma$$
Then it is known (integration by parts)
\begin{equation}
\int^{\infty}_{-\infty}f(x)x^ne^{-ix\gamma}dx=i^n(\widehat{f})^{(n)}(\gamma) .
\end{equation}
\begin{equation}
\widehat{(f^{(n)})}(\gamma)=(i\gamma)^n\widehat{f}(\gamma) .
\end{equation}
$$
\int^{\infty}_{-\infty}f'(x)A(x)e^{-i x\gamma}dx=
$$
\begin{equation}
=\int^{\infty}_{-\infty}f(x)A'(x)e^{-ix \gamma}dx+(-i\gamma)\int^{\infty}_{-\infty}f(x)A(x)e^{-ix\gamma}dx .
\end{equation}
$$
\int^{\infty}_{-\infty}f''(x)A(x)e^{-ix\gamma}=\int^{\infty}_{-\infty}f(x)A''(x)e^{-ix\gamma}dx+
$$
\begin{equation}
+2(-i\gamma)\int^{\infty}_{-\infty}f(x)A'(x)e^{-ix\gamma}dx+(-i\gamma)^2\int^{\infty}_{-\infty}f(x)A(x)e^{-ix\gamma}dx .
\end{equation}

Theorem.
When $f$, $g\in L_2(\bf R\rm)$ and $\lim_{|x|\rightarrow \infty}|f(x)x^{2+\epsilon}|=0$, $\epsilon>0$, equation $(1)$ can reduced to
\begin{equation}
(-ia_1\gamma^2+a_2\gamma+ia_3)\frac{\widehat{f}(\gamma)}{d\gamma}+(-b_1\gamma^2-2ia_1\gamma+ib_2\gamma+a_2+b_3)\widehat{f}(\gamma)=\widehat{g}(\gamma)
\end{equation}
which is solvable.

Using the above theorem in your equation we have
$$
y''-2xy'-11y=X_{[0,\infty)}(x)e^{-a x}\tag{eq}
$$
We get
$$
-(13+s^2)Y(s)-2sY'(s)=g(s)\textrm{, }g(s)=\frac{1}{a-is}
$$
Solving this equation we get
$$
Y(s)=-e^{1/2(-s^2/2-13\log s)}\int^{s}_{1}\frac{e^{1/2(t^2/2+13\log t)}}{2(a-it)t}dt
$$
Hence
$$
y(x)=-\frac{1}{2\pi}\int^{\infty}_{-\infty}e^{1/2(-s^2/2-13\log s)}\left(\int^{s}_{1}\frac{e^{1/2(t^2/2+13\log t)}}{2(a-it)t}dt\right)e^{isx}ds
$$
The equation
$$
y''-2xy'-11y=0
$$
Have general solution
$$
y(x)=C_1H_{-11/2}(x)+C_2\cdot {}_1F_1\left(\frac{11}{2};\frac{1}{2};x^2\right)
$$
Hence equation $(eq)$ have general solution
$$
y(x)=C_1H_{-11/2}(x)+C_2\cdot  {}_1F_1\left(\frac{11}{2};\frac{1}{2};x^2\right)-
$$
$$
-\frac{1}{2\pi}\int^{\infty}_{-\infty}e^{1/2(-s^2/2-13\log s)}\left(\int^{s}_{1}\frac{e^{1/2(t^2/2+13\log t)}}{2(a-it)t}dt\right)e^{isx}ds,
$$
where $H_n(x)$ is the $n-$th Hermite function and ${}_1F_1(a;b;x)$ is the ${}_1F_{1}$ hypergeometric function.

For more details in this kind of equations see here .

GENERAL NOTES

The degree of the term $x^my^{(n)}$ is $\nu=m-n$. We gather together all the terms of the DE of $\nu$ degree. In this way we can split a differential equation $A(x)y''+B(x)y'(x)+C(x)=0:(DE)$ into $N$ distinct groups of terms with $\nu_i$, $i=1,2,\ldots,n$ degree. The number $N$ is called degree of the DE.

If the degree $N$ is 2, we call the DE 2-degree. i.e
$$
(1-x^2)y''-2xy'+\lambda y=0\textrm{ (Legendre) }
$$
$$
y''-2xy'''+\lambda  y=0\textrm{ (Hermite) }
$$
$$
x^2y''+xy'+(x^2-\nu^2)y=0\textrm{ (Bessel) }
$$
Every 2-degree DE have two parts: The part with the largest degree ($L_{max}$ part degree) and the part of the smaller degree ($L_{min}$ part-degree).

i) If the part of largest degree have the term $y''$ we call it DE of the first kind.

ii) If the part of largest degree have the term $y'$ we call it DE of the second kind.

iii) If the part of the largest degree have the term $y$ we call it DE of the third kind.

Step $l$ of a 2-degree (DE) is the difference of the degree of the larger part minus the degree of the smaler part.

I) If the 2-degree (DE) is of the first kind, then its solution is
$$
y(x)=x^{\mu}\Phi(a,b;c;\lambda x^{l}),
$$
where
$$
\Phi(a,b,c;x)=c_1\cdot {}_2F_1(a,b;c;x)+c_2\cdot x^{1-c}{}_2F_1(a+1-c,b+1-c;2-c;x),
$$
where ${}_2F_1(a,b;c;,x)$ is the well known Gauss hypergeometric series.

The asymptotic behavior $y=x^s$ of (DE) around $x=0$ lead us to the starting powers $s_1,s_2$. Then $x^{s_1}=x^{\mu}(x^l)^0\Rightarrow s_1=\mu$, $x^{s_2}=x^{\mu}(x^l)^{1-c}\Rightarrow s_2=\mu+l(1-c)$.

The asymptotic behavior in infinty $y=x^{k}$, lead us to $x^{k_1}=x^{\mu}(x^l)^{-a}\Rightarrow k_1=\mu-la$, $x^{k_2}=x^{\mu}(x^l)^{-b}\Rightarrow k_2=\mu-lb$

The parameter $\lambda$ is evaluated demmanding that change of variable $t=\lambda x^l$ leaves the finite singular points of (DE) at $t=1$.

II) If the 2-degree (DE) is of the second kind, then its solution is
$$
y(x)=x^{\mu}\Phi(a;c;\lambda x^{l}),
$$
where
$$
\Phi(a;c;x)=c_1\cdot {}_1F_1(a;c;x)+c_2\cdot x^{1-c}\cdot {}_1F_1(a+1-c;2-c;x).
$$
In $x=0$, we have $x^{s_1}=x^{\mu}(x^l)^0\Rightarrow s_1=\mu$ and $x^{s_2}=x^{\mu}(x^l)^{1-c}\Rightarrow s_2=\mu+l(1-c)$ and in $x=\infty$, we have $x^{k_1}=x^{\mu}(x^l)^{-a}\Rightarrow k_1=\mu-la$. The $\lambda$ is evaluated from the asymptotic subtitution $y_{\infty}(x)\approx e^{\lambda x^l}$ in the (DE).

III) The  last case is when the 2-degree (DE) is if third kind. Then
$$
y(x)=x^{\mu}Z_{k}(\lambda x^{l/2}),
$$
where
$$
Z_{k}=c_1J_{k}(x)+c_2Y_{k}(x)\textrm{, where }J_k(x)\textrm{  and }Y_k(x)\textrm{ are the Bessel... }
$$
The the asymptotic behavior in $x=0$ give us $x^{s_1}=x^{\mu}(x^{l/2})^k\Rightarrow s_1=\mu+\frac{lk}{2}$, $x^{s_2}=x^{\mu}(x^{l/2})^{-k}\Rightarrow s_2=\mu-\frac{lk}{2}$. The $\lambda$ is evaluated from the asyptotic behavior in $x=\infty$, $y_{\infty}(x)\approx \exp\left(\pm i\lambda  x^{l/2}\right)$

Example 1.
$$
y''-xy=0\tag 1
$$
Obviously (1) is of 2-degree with step $l=1-(-2)=3$. We have
$$
L_{mim}=D^2\textrm{ and }L_{max}=-x.
$$
Since the largest term has no differentials this equation is a 2-degree of the third kind with step $l=3$. Hence its general solution is
$$
y(x)=x^{\mu}Z_{k}(\lambda x^{3/2}).\tag 2
$$
Setting $y=x^{s}$ in (1), we get $s(s-1)x^{s-2}=0\Leftrightarrow s_1=0$, $s_2=1$, we get from (2): $x^1=x^{\mu}(x^{3/2})^{-k}\Rightarrow 1=\mu+\frac{3}{2}k$. Also $x^{0}=x^{\mu}(x^{3/2})^{-k}\Rightarrow 0=\mu-\frac{3}{2}k$. Hence $\mu=1/2$, $k=1/3$. For the evaluation of $\lambda$, we set $y_{\infty}(x)\approx \exp(\pm i \lambda x^{3/2})$, (using the asymptotic formula: $(e^{S})''\approx (S')^2e^{S}$, where $S=\lambda x^{\rho}$, $\rho>0$, $x>>1$),
we get after inserting this into (1): $\lambda=\pm i\frac{2}{3}$. Hence the ecxact solution of (1) (Airy equation) is
$$
y(x)=x^{1/2}Z_{1/3}\left(i\frac{2}{3}x^{3/2}\right)
$$

Example 2.
$$
xy''+(2-x^2)y'-2xy=0\tag 2
$$
We rewrite (2) in the form $(xy''+2y')+(-x^2y'-2xy)=0$. Hence $L_{min}=xD^2+2D$, $L_{max}=-x^2D-2x$. Hence (2) is a two degree DE of second kind with step $l=2$. Hence the solution is of the form
$$
y(x)=x^{\mu}\Phi(a;c;\lambda x^{2})
$$
The asymptotic behavior at $x=0$ is: The starting powers are (set $y=x^s$ in $L_{min}y=0$ to get) $s_1=0$, $s_2=-1$.  $x^0=x^{\mu}(x^2)^{0}\Rightarrow \mu=0$ and $x^{-1}=x^{\mu}(x^2)x^{1-c}\Rightarrow c=\frac{3}{2}$.

The asymptotic behavior at $x=\infty$ is: Solve $L_{max}x^s=0\Leftrightarrow -x^{2}sx^{s-1}-2xx^s=0\Leftrightarrow s=-2$. Hence $k_1=-2$. Hence $x^{-2}=x^{\mu}(x^2)^{-a}\Leftrightarrow a=1$. Also if we set $y_{\infty}(x)\approx\exp(\lambda x^2)$ in (2) we get $\left(e^{\lambda x^2}\right)'=2\lambda x e^{\lambda x^2}$, $\left(e^{\lambda x^2}\right)''=4\lambda^2 x^2 e^{\lambda x^2}$. Hence seting these in (2), we get $4\lambda^2x^3e^{\lambda x^2}-2\lambda x^3e^{\lambda x^2}=0\Rightarrow 4\lambda^2=2\lambda\Rightarrow \lambda=1/2$. Hence the ecxact solution of (2) is
$$
y(x)=\Phi\left(1;\frac{3}{2};\frac{x^2}{2}\right)
$$

Hence
$$
y(x)=c_1\cdot {}_1F_1\left(1;\frac{3}{2};\frac{x^2}{2}\right)+c_2 \cdot x^{-1/2}\cdot{}_1F_1\left(\frac{1}{2};\frac{1}{2};\frac{x^2}{2}\right)
$$

Example 3. Solve
$$
(1-x^4)y''+\left(n(n+1)x^2-\frac{m(m+1)}{x^2}\right)y=0
$$
Answer
$$
y(x)=x^{m+1}\Phi\left(\frac{m-n}{4},\frac{m+n+1}{4};\frac{2m+5}{4};x^4\right)
$$

Example 4. The Hermite equation is
$$
y''-2xy'+2\nu y=0
$$
This equation is 2-degree of the second kind and step $l=0-(-2)=2$...etc
$$
y(x)=\Phi\left(\frac{-\nu}{2};\frac{1}{2};x^2\right)
$$


问:
Setting a function to power series with the $\sum_{n=0}^\infty\ a_{n} \ x^n $ method
Let's say I have this function below

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

How do I represent it as a power series ?
The only solution that I know is when using the geometric series :
$\sum_{k=1}^\infty q^k = \frac{1}{1-q}$ but this is not possible in our case .

I asked recently this question with a different function by mistake
$$f(x) =  \frac{(x-1)}{x^2+1}$$
This is actually simpler and can be done with $\sum_{k=1}^\infty q^k = \frac{1}{1-q}$

I tried to integrate the function ($f(x) =  \frac{(x-1)}{x^2+2}$)  and then differentiated it but I stopped the process when I started finding out some bad result .

I checked some documentation and most of these functions (that don't work with a geometric series) should  work using this method   $\sum_{n=0}^\infty\ a_{n} \ . x^n $ . The problem here is I don't know how to apply it.

答:
Via partial fraction decomposition and geometric series:
\begin{align}
\frac{x-1}{x^2+1}
&= \frac{x-1}{(1+i x)(1-i x)} \\
&= \frac{(1-i)/2}{1+i x} - \frac{(1-i)/2}{1-i x} \\
&= \frac{1-i}{2} \sum_{n\ge 0} (-i x)^n - \frac{1-i}{2}\sum_{n\ge 0} (i x)^n \\
&= \sum_{n\ge 0} \frac{(1-i)i^n((-1)^n - 1)}{2} x^n
\end{align}


问:
Reversing the order of integration in $\int_{-1}^{3}\int_{x^2}^{2x+3} x \, dy \, dx$
$$\int_{-1}^{3}\int_{x^2}^{2x+3} x \, dy \, dx$$

I am trying to reverse the order of integration for this problem. I drew the region bounded by the functions of $x$ but I've hit a block.

答:
If you drew the region, you see that the low $y$-value is $0$ and the high $y$-value is $9$. So the outer integral will be $\int_{y=0}^{y=9}$.

As your eyes move up from $0$ to $9$, for a time the left edge and right edge are both parts of a parabola. Then at $y=1$ the left edge transitions to a line. So the outer integral needs to be broken up: $\int_{y=0}^{y=1}+\int_{y=1}^{y=9}$.

For the lower section, the left curve is $x=-\sqrt{y}$ and the right curve is $x=\sqrt{y}$.

Then at $y=1$ the left curve transitions to $x=\frac{y-3}{2}$.

So you have
$$\int_{y=0}^{y=1}\int_{x=-\sqrt{y}}^{x=\sqrt{y}}+\int_{y=1}^{y=9}\int_{x=\frac{y-3}{2}}^{x=\sqrt{y}}$$


问:
If $f:[a,b]\rightarrow \mathbb{R}$ is continuous, for any $x_{1},...,x_{n}\in [a,b]$ exist $c\in [a,b]$ s.t $f(c)=\frac{f(x_{1})+...+f(x_{n})}{n}$
My issue is that I don't understand this problem, so I don't know how to attack it. Could you give me some hints on how to approach this?</p>

答:
I don't really understand what you don't understand, so if you can try to explain more about your problem, it'll be better for both.

As for the solution to this problem, this is a direct result from the continuity of $f$.

WLOG, $f(x_n) \ge f(x_{n-1}) \ge... \ge f(x_1)$

$\Rightarrow f(x_1) \le y:=\frac{ f(x_1)+...+f(x_n)}{n} \le f(x_{n})$

So by the continuity of $f$, there is at least a $c$ between $x_1$ and $x_n$ such that : $f(c)=y$


答:
Since $f$ is continuous and $[a,b]$ is compact, by Weierstrass theorem $f$ attains a maximum and a minimum on $[a,b]$. Let's call them $m$ and $M$ respectively. We can say that

$\dfrac{f(x_1)+\ldots+f(x_n)}{n}\le \dfrac{\overbrace{M+\ldots+M}^{n\text{ times}}}{n}=M$ and

$\dfrac{f(x_1)+\ldots+f(x_n)}{n}\ge \dfrac{\overbrace{m+\ldots+m}^{n \text{ times}}}{n}=m$

Can you take it from here? Hint:

 It involves the intermediate value theorem.

Title: If $f:[a,b]\rightarrow \mathbb{R}$ is continuous, for any $x_{1},...,x_{n}\in [a,b]$ exist $c\in [a,b]$ s.t $f(c)=\frac{f(x_{1})+...+f(x_{n})}{n}$

Body: My issue is that I don't understand this problem, so I don't know how to attack it. Could you give me some hints on how to approach this?</p>


答:
Observe that $f$ is continuous on a closed and bounded interval, so there exists $m,M \in \mathbb{R}$ such that $m\leq M$ and $f([a,b])=[m,M]$. Also observe $\frac{f(x_1)+ \dots + f(x_n)}{n}$ is between $m$ and $M$ for any $x_1,\ldots ,x_n\in [a,b]$ so we can say

 by IVT the result follows.