问:
Proving that there is a path in set.
Given that $\underline Y$ is a subspace of $\underline ℝ^2$ and defined as follows:

$$I = [0,1]$$

$$X = (\{1\} × I) ∪ \left(I × \left(\{0\} \cup\left\{ \frac{1}{n} \,\Bigg\vert\, n \in \Bbb N \right\}\right)\right)$$

$$Y = X \setminus \{ (0.5, 0) \}$$

How can one prove that there is no path from $(0,0)$ to $(1,0)$ using sequences?

To show the contradiction I started by assuming that such a map exists:

$$f: \underline{I} \rightarrow \underline Y \ \text{continuous map}$$

$$f(0) = (0,0),\ \ f(1) = (1,0)$$

If $f$ is continuous, then it must hold that $f^{-1}[\ [0,0.5)\times \{0\} ] = A$ is a closed set. So $A$ contains its maximum $m$. From here on I'm not sure how to proceed. I wanted to somehow show that there is a sequence that due to $m$ being in $A$ cannot converge to $f(m) = (x,0)$.

答:
You are on the right track.

So $A$ contains its maximum $m\in[0,1)$. On the other hand $f(m)=(t,0)$ for some $0\leq t<0.5$. Now take any sequence $(x_n)\subseteq [0,1]$ converging to $m$ from above, something like $x_n=m+(1-m)/n$. It follows that $f(x_n)\to f(m)$ and so if we write $f(x_n)=(a_n,b_n)$ then $a_n<0.5$ eventually, but since $m$ is the maximum of $A$ then $b_n>0$ (still convergent to $0$) whenever $a_n<0.5$.

It follows that the image of $f$ is not locally connected (around $f(m)$), by considering the open subset $\{(x,y)\in im(f)\ |\ x<1\}$, the projection onto $Y$-axis and preimages of clopen points. But any continuous function $f:[0,1]\to X$ into a Hausdorff space is a quotient map onto image. And being locally connected is preserved under quotients. This contradiction shows that no such continuous $f$ may exist.

Title: Proving that there is a path in set.


问:
How to find the indefinite integral of $\arctan((x^2+1)^{1/3})$?
I have to find the indefinite integral of $\arctan((x^2+1)^{1/3})$. I have tried to do different variable substitutions, then integration by parts, but in the end I always got a another integral that I could not calculate. Please give a hint how to calculate this integral.

答:
Change variables $y^3 = 1+x^2$ to get
$$
\int \arctan(\sqrt[3]{x^2+1}) \, dx = \frac{3}{2}\int\frac{y^2\arctan y}{\sqrt{y^3-1}}\;dy
$$
Integrate by parts to get
$$
{\frac {\arctan \left( y \right)  \left( y-1 \right)  \left( {y}^{2}+y
+1 \right) }{\sqrt {{y}^{3}-1}}}-\int \!{\frac { \left( y-1 \right)
 \left( {y}^{2}+y+1 \right) }{ \left( {y}^{2}+1 \right) \sqrt {{y}^{3}
-1}}}\,{\rm d}y
\\=
{ {\arctan \left( y \right)  \sqrt{ y^3-1 } }}-\int \!{\frac { \sqrt{y^3-1}
}{  {y}^{2}+1
 }}\,{\rm d}y
$$
That is an elliptic integral.  In Maple it is about a page long,
in terms of $F$ and $\Pi$.

ContentLicense: CC BY-SA 4.0


问:
Finding sample space and outcome
If I have 2 red candies, 3 blue candies, 4 pink candies inside a bowl. Candies of the same colour are indistinguishable. Then I randomly select one candy and record its colour. What is the sample space and how many outcomes does it have?

Would the sample space be {red, blue, pink}? Or would it be {red, red, blue,....,pink} having 9 elements in the sample space?

Would there be 3 outcomes? Or 9 outcomes?

答:
The singleton subsets sets $\{R\}, \{B\}, \{K\}$ of sample space $S =\{R,G.K\}$ need not be assigned equal probabilities. Here you could have
$P(\{R\}) = 2/9,$ $P(\{B\}) = 3/9,$ $P(\{K\})= 4/9.$
Then you could find probabilities such as
$P(\mathrm{not\, R}) = 1 - 2/9 = 7/9$ and
$P(\mathrm{Red\,or\,Blue}) = 2/0 + 3/9 = 5/9,$ by
using the axioms of probability.

Alternatively, even though the candies of the same color
are not distinguishable, you could imagine they
are numbered in invisible ink. Then you could have
sample space $S_2 = \{r_1. r_2, b_1,b_2,b_3,k_1,k_2,k_3,k_4\},$ and assign each singleton
event probability $1/9: P(\{b_1\}) = 1/9, \dots,$
$P(\{k_4\}) = 1/9.$

Then $P(\mathrm{Blue}) =1/9 + 1/9 + 1/9 = 3/9 = 1/3.$

If you had an ultraviolet lamp that allows you to read the invisible ink, you could say
$P(\mathrm{Invisible\, number\ 3}) = 2/9.$

Title: Finding sample space and outcome


问:
How to prove that the general solution of a homogeneous linear first order ODE has dimension 1?
I would like to prove it by using Linear Algebra.

I first defined a Linear Application on an open interval $I$

$$L(.) : C^1(I) \to C(I).$$

$$y(x) \to  y(x)'+p(x)y(x)$$

For a Homogeneous Equation $L(y)=0$, The solution is the Kernel of $L(.)$ which is by definition a Subspace but how can I prove that the dimension is 1?

答:
If $P$ is an antiderivative of $p$ then $y(x) = C e^{-P(x)}$ will satisfy
$$y'(x) = -C e^{-P(x)} p(x) = -p(x) y(x),$$ and hence solve the homogeneous differential equation. Different choices of antiderivatives correspond to different choices of $C$ because $e^{a+K} = e^K e^a = \text{another constant} \times e^a$ is a scalar multiple of $e^a$.

If $f$ is another solution to the equation then
$$
(y/f)' = (y'f - f'y)/f^2 = (-pyf + pfy)/f^2 = 0,
$$
implying that $y/f$ is constant and hence that $f$ is a scalar multiple of $y$.

This analysis does not fully treat the possibility that $f$ might sometimes be zero, where the computation of $(y/f)'$ would not be sensical. In fact, it might make more sense to do the computation with $(f/y)'$ and achieve the same result because $y$ by definition is never zero as long as $C$ is nonzero. I hope this helps. It can be phrased in the language of linear algebra, but does involve engaging with the form of the differential equation. Most results in ODE are similar.


问:
How does positivity affect operator norms?
Take a measure space $\Omega$, an exponent $1 \le p < \infty$ and a linear operator $T \colon L^p(\Omega) \to L^p(\Omega)$. Suppose that $T$ is positive, that is $Tf \ge 0$ a.e. if $f \ge 0$ almost everywhere. Then

$$\lVert T \rVert = \sup_{f \ge 0, \int_{\Omega} f^p \le 1} \lVert Tf \rVert_p;$$

that is, it suffices to take the supremum over positive functions.

I was wondering if there is something more general here. For example, does this hold if $T$ is an operator on $C(K)$ for compact $K$? Is this property true in some abstract class of Banach spaces?

答:
I've got my hands over Aliprantis and Burkinshaw's Positive operators (@t.b.: thank you for hints!) and read something on Riesz spaces and Banach lattices. A Banach lattice is a triple $(E, \le , \lVert \cdot \rVert )$, where

$E$ is a real vector space.

$\le$ is an ordering on $E$ compatible with vector space operations and s.t. for every $f, g\in E$ there exist $f \vee   g , f \wedge  g \in E$. We then define $\lvert f \rvert= f \vee (-f)$.

$\lVert \cdot \rVert$ is a Banach norm on $E$ s.t. $\lvert f \rvert \le \lvert g \rvert \Rightarrow \lVert f \rVert \le \lVert g \rVert$.

A positive operator on $E$ is then a linear operator that is order preserving. It is not obvious that every positive linear operator is continuous (cfr. Aliprantis & Burkinshaw, §4.3), hence bounded.

Now let $T$ be a positive operator on $E$. By definition

$$\lVert T \rVert= \sup_{\lVert f\rVert \le 1} \lVert Tf \rVert.$$

We claim that, being $T$ positive, it suffices to take the supremum over positive $f$ (that is, over all $f$ that are greater than the origin):

$$\lVert T \rVert = \sup_{\lVert f \rVert \le 1, f \ge 0} \lVert Tf \rVert.$$

In fact, it is clear that the second supremum is lesser than the first. To prove the reverse inequality take $\lVert f \rVert \le 1$ and observe that $ \lVert \lvert f \rvert \rVert \le 1$ also. We have $Tf \le T\lvert f \rvert$ and so $\lVert Tf \rVert \le \lVert T \lvert f \rvert \rVert$.

ContentLicense: CC BY-SA 3.0


问:
Fitch-Style Proof ¬(A → B) → (A ∧ ¬B)
Good Evening,

I want to proof this statement: ¬(A → B) → (A ∧ ¬B). I have no premises.

I already did other proofs where I have no premises and a implication. I guess I have to start with the antecedent as my premise and try to generate the consequence. But there is not much I can use, so I thought I can proof it with contradiction.

This is my idea:

I used this website: http://proofs.openlogicproject.org/

答:
Observe this fact: $\neg(p\to q)\equiv p \wedge \neg q$ this is true because both propositions have the same truth table.

Now notice that $p\to p$ is a tautology.
Then use the fact, if $P\equiv Q$ then we can "replace" $P$ instead of $Q$ in any statement that includes $Q$.

Now $p\to p$ is a tautology then $\neg(p\to q)\to \neg(p\to q)$ is a tautology, and now
$\neg(p\to q)\to (p \wedge \neg q)$  is a tautology.

Title: Fitch-Style Proof ¬(A → B) → (A ∧ ¬B)


问:
Extension of a prime ideal in a Dedekind domain
Suppose that $A$ is a Dedekind domain and that $P$ is a prime ideal of $A$. Suppose that $K$ is the fraction field of $A$ and that $L$ is some finite separable extension of $K$. If $B$ is the integral closure of $A$ in $L$, is it necessarily the case that $PB\neq B$? If so, why?

I am not sure why this question has been closed. Apparently it has offended the hall monitors of MSE by contradicting some nebulous rule. However, I believe it is a straightforward question (and the comments indicate an answer exists in some book which I do not own and do not have access to!) and I hope the question will be reopened. Perhaps, after all, someone else has the same question!

答: Yes, because of the Lying over property: as $B$ is integral over $A$ and the morphism  $A\longrightarrow B$ is injective, there exists a prime ideal $\mathfrak q\in\operatorname{Spec}B$ such that $\mathfrak q\cap A=\mathfrak p$, so $\mathfrak p B\subset \mathfrak q\varsubsetneq B$.


问:
Complement of zero section homotopy equivalent to $\mathrm{Gr}_{k-1}(\mathbb{C}^{\infty})$
Let $\mathrm{Gr}_{k}(\mathbb{C}^{\infty})$ be the Grassmanian of complex $k$-dimensional subspaces of $\mathbb{C}^{\infty}$, and let $\gamma:E\to \mathrm{Gr}_{k}(\mathbb{C}^{\infty})$ be the tautological $k$-plane bundle, whose fiber over a point $X\in \mathrm{Gr}_{k}(\mathbb{C}^{\infty})$ is the space
$$
\gamma^{-1}(X)=\{v\in \mathbb{C}^{\infty}\mid v\in X\}.
$$
Let $E_0\subset E$ be the complement of the zero section of $\gamma$, i.e.
$$
E_0=\{(X,v)\in \mathrm{Gr}_{k}(\mathbb{C}^{\infty})\times \mathbb{C}^{\infty}\mid  v\in X,\; v\neq 0\}.
$$

Let $f:E_0\to \mathrm{Gr}_{k-1}(\mathbb{C}^{\infty})$ be the map
$$
f(X,v)=X\cap v^{\perp}\in \mathrm{Gr}_{k-1}(\mathbb{C}^{\infty}),
$$
where the orthogonal complement $v^{\perp}\subset \mathbb{C}^{\infty}$ is computed with respect to the standard Hermitian inner product on $\mathbb{C}^{\infty}$,
$
v\cdot w=\sum_{i=1}^{\infty}v_i\bar{w}_i.
$
In their computation of the cohomology ring of $\mathrm{Gr}_{k}(\mathbb{C}^{\infty})$, Milnor-Stasheff show (Theorem 14.5) that $f$ induces an isomorphism of the cohomology rings
$$
f^*:H^*(\mathrm{Gr}_{k-1}(\mathbb{C}^{\infty});\mathbb{Z})\cong H^*(E_0;\mathbb{Z}).
$$

Question: Is $f$ in fact a homotopy equivalence, and if so, how can this be shown?

In the case $k=1$, I am able to show that $E_0$ and $\mathrm{Gr}_{k-1}(\mathbb{C}^{\infty})$ are homotopy equivalent: indeed, $\mathrm{Gr}_{k-1}(\mathbb{C}^{\infty})=\{0\}$ is contractible, and hence homotopy equivalent to the unit sphere $S^{\infty}\subset \mathbb{C}^{\infty}$ (which is also contractible), and the map $E_0\to S^{\infty}$, defined by
$$
(X,v)\mapsto \frac{v}{\|v\|},
$$
is a homotopy equivalence.

答:
A cohomology equivalence of simply connected spaces is a homotopy equivalence. This follows from the universal coefficient theorem and homology Whitehead.

It is well known that complex Grassmannians are simply connected, so what remains to be shown is that the complement of the zero section is simply connected.

This complement is homotopy equivalent to the sphere bundle of the tautological bundle by normalizing, hence it suffices to show the sphere bundle is simply connected. By our earlier remarks, the base is simply connected and the fiber is as well (assuming $k>1$) since it is homotopy equivalent to a sphere of dimension $2k-1$. Hence, the total space is simply connected by the $LES$ of a fiber bundle.

So Milnor and Stasheff's work to show this is a cohomology equivalence implies that it is a homotopy equivalence.


问:
For two odd primes $p<q$, can we deduce positive integers $a,b$ solving $a^2+b^4=pq$ without trial & error (brute force)?
Let us fix two primes $p,q$ with $2<p<q$. How can we find positive integers $a,b$ which solve the equation $a^2+b^4=pq$ without brute force?

Interestingly there exist sometimes two solutions:

$5\cdot13=7^2+2^4=8^2+1^4$

$5\cdot821=3^2+8^4=53^2+6^4$

$17\cdot113=25^2+6^4=36^2+5^4$

In more rare cases I even obtain three solutions, e.g.:

$73\cdot89=49^2+8^4=64^2+7^4=79^2+4^4$

Is there a systematic way to obtain the solutions $a,b$ directly? Or at least, can we establish a relationship between the two primes and these natural solutions? I generated a CSV file containing some more cases, which maybe help in finding patterns or connections between $p,q$ and $a,b$.

Can we anticipate (state in advance) how many positive integer solutions we will get depending on $p,q$?

答:
With kind assistance I have received a useful hint for a direction to investigate, wich I would like to share with you all:

In the case that $p\equiv3\pmod4$ or $q\equiv3\pmod4$ no solution exist.

Let us set $p=r^2+s^2$ and $q=u^2+v^2$. If $p\equiv1\pmod4$ and $q\equiv1\pmod4$ we exactly obtain one solution $r>s>0$ for $p$ and one solution $u>v>0$ for $q$.

If we now have these unique solutions $p=r^2+s^2$ and $q=u^2+v^2$, then the product of both primes is $pq=(r^2+s^2)(u^2+v^2)=(ru+sv)^2+(rv-su)^2=(ru-sv)^2+(rv+su)^2$.

Consider $b^2=c$, other integer solutions for $pq=a^2+c^2=a^2+b^4$ do not exist, unless one of the four integers $ru+sv$, $|rv-su|$, $|ru-sv|$ and $rv+su$ is a perfect square.

To retrace what happens in practice, I generated variuos cases, where three solutions exist (feel free to download the CSV):

$p\cdot q=233\cdot12409=256^2+41^4=1616^2+23^4=1681^2+16^4$

$p\cdot q=89\cdot233=1^2+12^4=129^2+8^4=144^2+1^4$

$p\cdot q=17\cdot4241=81^2+16^4=256^2+9^4=264^2+7^4$

In the first case we have $p=233=8^2+13^2$ and $q=12409=72^2+85^2$.

In the second case we have $p=89=5^2+8^2$ and $q=233=8^2+13^2$.

In the third case we have $p=17=1^2+4^2$ and $q=4241=4^2+65^2$.

Hypothetically, there could be four solutions, although so far I have only found these cases with a maximum of three solutions?


问:
How to minimize a 2D function given a constraint?
How can I minimize
$$
2a^2 + b^2
$$
given the constraint that
$$
a+b = 1
$$

I am able to obtain the answer using Wolfram Alpha, but I'm not sure how to go through the math.

答:
Substitute using $b = 1-a$.

Then we have
$$
f(a) = 2a^2 + (1-a)^2 = 3a^2 - 2a + 1
$$
Then we have
$$
f^{\prime}(a) = 6a-2
$$
and setting to zero and solve we obtain $a=1/3$, so then because $a+b=1$, then $b=2/3$.


答:
You could also use a giant hammer for the tiny nail here with Lagrange Multipliers: $F = 2a^2 + b^2 - \lambda(a+b -1)$ and minimize this with respect to $a$ and $b$ which will result in $F_a = 4a - \lambda = 0$ and $F_b = 2b - \lambda = 0$ thus $2a = b$ and using the constraint: $1 = a + b = 3a$ so $a = 1/3$ and the rest follows.
https://en.wikipedia.org/wiki/Lagrange_multiplier

Title: How to minimize a 2D function given a constraint?

Body:
How can I minimize
$$
2a^2 + b^2
$$
given the constraint that
$$
a+b = 1
$$

I am able to obtain the answer using Wolfram Alpha, but I'm not sure how to go through the math.


答:
Form an object function by combining with a Lagrange multiplier.

Or more easily solve together

$$2a^2+b^2= c^2, a+b=1$$

evaluate $(a,b) $ in terms of $c$ which yields

$$ 2a=b, a= \frac13,b= \frac23$$


问:
Prove $O(x^n) * O(x^m) = O(x^{n+m})$
${x\to 0}$,$m>n>0$ Prove
$O(x^n)$ $*$ $O(x^m)$ = $O(x^{n+m})$

What I did.

$f(x)$ = $O(x^n)$

$g(x)$ = $O(x^m)$

$\rvert\frac{f(x)}{x^n}*\frac{g(x)}{x^m}\lvert\leq\rvert\frac{f(x)*g(x)}{x^{m+n}}\rvert\leq L*M$

From here we can say that prove is complete?This is my first steps to asymptotic analysis proofs.

答:
Just a small detail that you could be missing.

$f(x) = O(x^n)$ means that there is $x_0$ and $L >0$ such that for $x \geq x_0$, $|f(x)| \leq L x^n$.

$g(x) = O(x^m)$ means that there is $x_1$ (which can be different from $x_0$) and $M >0$ such that for $x \geq x_1$, $|g(x)| \leq M x^m$.

With this, you can conclude that for $x \geq \max\{x_0,x_1\}$,

$|f(x) g(x)| \leq L x^n M x^m = LM x^{n+m}$, where $LM >0$.

Thus, $f(x)g(x) = O(x^{n+m})$.

Title: Prove $O(x^n) * O(x^m) = O(x^{n+m})$


问:
When can we embed field extensions into one another?
I was recently informed of the following theorem:

Let $K$ be a field of characterisic $0$, and let $L$, $M$ be field extensions of $K$ such that

i) $L$ is algebraic over $K$,

ii) every irreducible polynomial $p(x) \in K[x]$ with a root in $L$ has a root in $M$,

then $L$ embeds over $K$ into $M$.

I was also told that this can also be shown by model theoretic compactness.

If $L/K$ is a finite extension, this immediately follows from the fact that finite extensions over fields of characteristic $0$ are simple. Could anyone outline an algebraic/ model-theoretic proof of this fact for the case L is not a finite extension of K?

答:
Consider the following two-sorted first order theory.  We have one sort for $L$ and one sort for $M$, and we have constant symbols for each element of $L$ and each element of $M$ and function symbols for the ring operations of $L$ and $M$.  We also have a symbol $f$ for a function from $L$ to $M$ (which will be our embedding).  We have axioms that say the constant symbols are all distinct elements, the ring structure behaves the way it should on all our constant symbols, and that $f$ is a ring homomorphism.  Finally, we have the following axioms.  Given $\alpha\in L$, let $p$ be the minimal polynomial of $\alpha$ over $K$.  Let $\beta_1,\dots,\beta_n$ be all the roots of $p$ in $M$.  We then have an axiom saying $f(\alpha)=\beta_1\vee\dots\vee f(\alpha)=\beta_n$, and we have an axiom like this for each $\alpha\in L$.

Suppose we have a model of this theory.  The final collection of axioms guarantees that $f$ maps all the constant elements from $L$ to constant elements from $M$ and fixes each element of $K$, so we get an embedding of $L$ into $M$ over $K$.

It thus remains to show that this theory has a model.  Suppose we take any finite subset of our axioms.  These axioms will involve only finitely many of our constant symbols from $L$; let $L_0$ be the subfield of $L$ that they generate.  Then using the primitive element theorem, we can embed $L_0$ into $M$.  This then gives a model of our finite subset of our axioms, where we just interpret all the constant symbols for elements of $L\setminus L_0$ as $0$ (which is fine since we don't have any axioms about those constant symbols).  Thus this theory is finitely satisfiable, and thus satisfiable by compactness.

Here's another way to think about it.  Let $S$ be the set of subfields of $L$ that are finite extensions of $K$, which we can consider as a directed set ordered by inclusion.  For each $s\in S$ let $E_s$ be the set of embeddings $s\to M$ over $K$.  Note that each $E_s$ is a finite nonempty set.  Also, if $s\subset t$, we have a restriction map $E_t\to E_s$, and so these finite sets $E_s$ form an inverse system.  Any inverse limit of finite nonempty sets is nonempty, so the inverse limit of this system is nonempty.  But an element of the inverse limit is exactly a compatible family of embeddings $s\to M$ for each $s\in S$, which we can then glue together to get an embedding of all of $L$.


问:
Algebraic problem for satisfying a given equation
I'm trying to solve the following exercise:

My backward equation looks like:
$P_{i,j}'(t) = i\lambda P_{i+1,j}(t) - i\lambda P_{i,j}(t) $

So i started with differentiating $P_{i,j}(t)$:

${j-1 \choose i-1}(e^{-i\lambda t}(1-e^{-\lambda t})^{j-i})'$
By using $(uv)' = u'v + uv'$

${j-1 \choose i-1}-i\lambda (e^{-i\lambda t}(1-e^{-\lambda t})^{j-i}) + {j-1 \choose i-1}(j-i)\lambda (e^{-(i+1)\lambda t}(1-e^{-\lambda t})^{j-i-1}) $

where the first term is cleary $-i\lambda P_{i,j}(t) $.

I don't see how i can turn $ {j-1 \choose i-1}(j-i)$ into  ${j-1 \choose i}i$  so the second term equals  $i\lambda P_{i+1,j}(t) $

答: $$P_{ij}(t)=\binom{j-1}{i-1}e^{-i\lambda t}\left(1-e^{-\lambda t}\right)^{j-i}\\\begin{align*}P'_{ij}(t)&=\binom{j-1}{i-1}\left(-i\lambda e^{-i\lambda t}(1-e^{-\lambda t})^{j-i}+\lambda(j-i)e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{j-i-1}\right)\\&=-i\lambda P_{ij}(t)+\lambda(j-i)\binom{j-1}{i-1}e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{j-(i+1)}\\&=-i\lambda P_{ij}(t)+\lambda\binom{j-1}{(i+1)-1}e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{j-(i+1)}\\&=-i\lambda P_{ij}(t)+\lambda P_{i+1,j}(t)\end{align*}$$... which follows from the fact that $\binom{j-1}{(i+1)-1}=\binom{j-1}i=(j-1)(j-2)\cdots(j-(i-1))(j-i)$</p>


Body:
Here it is an approach without calculus:

\begin{align*}
2a^{2} + b^{2} & = 2a^{2} + (1-a)^{2}\\\\
& = 3a^{2} -2a + 1\\\\
& = 3\left(a^{2} - \frac{2a}{3} + \frac{1}{9}\right) + \frac{2}{3}\\\\
& = 3\left(a-\frac{1}{3}\right)^{2} +\frac{2}{3} \geq \frac{2}{3}
\end{align*}

Hopefully this helps!


问:
How to define hypergeometric function ${}_1 F_1(-n+1;-n+1;z)$ for $n$ positive integer
Consider a truncated Taylor series of the exponential function to approximate $e$:
$$
E(n) = \sum_{k=0}^{n-1} \frac{1}{n!}
$$

I thought of computing this using the hypergeometric finite series $_1 F _1(a;c;z)$ for $a=-n+1$; $c=-n+1$; $z=1$. This is based on the fact that (see the Wikipedia link above)

The series is finite if the first argument, $a$, is a nonpositive integer, in which case the function reduces to a polynomial.

The series is finite in this case because the Pochhammer symbol in the numerator of $_1 F _1$ becomes $0$ from a given index onwards.

However, what if $b=a$? (as happens for the approximation of $e$: $_1 F _1(-n+1;-n+1;1$). In that case the Pochhammer symbol in the denominator of $_1 F _1$ becomes $0$ at the same index as that in the numerator, giving $0/0$. How is the function defined in that case?

According to Wolfram, that $0/0$ appears to be interpreted as $0$, and the series remains finite. See for example here: for arguments $-2$, $-2$, $1$ we get $2.5$, corresponding to $E(3) = 1 + 1 + 1/2$.

Matlab's hypergeometric function, on the other hand, seems to treat that $0/0$ as $1$: the series is infinite and always gives $e$:

```<code>>> hypergeom(-2,-2,1)
ans =
   2.718281828459045
>> hypergeom(-7,-7,1)
ans =
   2.718281828459045
</code>```

The approximation to $e$ was just an example; my actual question is what is the correct definition for $_1 F _1(-n+1;-n+1;z)$? Or more generally, how is $_1F_1(a;c;z)$ defined for $c$ nonpositive integer? Or yet more generally if possible: how is $_mF_n$ defined when one of the denominator coefficients is a nonpositive integer?

答:
In regards to the specific example provided you can avoid the whole issue of negative bottom parameters by using DLMF 16.2.4 to formally write your truncated series as
$$
E(n)=\frac{1}{(n-1)!}{_2F}_0\left({1,1-n\atop -};-1\right),
$$
for $n=1,2,\dots$

You can confirm in MATLAB that the follow code will give you the correct answer for your truncated exponential Taylor series:

```<code>syms n
f(n)=hypergeom([1,1-n],[],-1)/factorial(n-1);
</code>```

It is important to note that this representation for your series only holds for positive integer $n$ as the ${_2F_0}(1,1-n;;1)$ function diverges for $n\notin\Bbb N$. For a continuous generalization one could use
$$
E(n)=\sum_{k=0}^\infty\frac{1}{k!}-\sum_{k=n}^\infty\frac{1}{k!}=e-\frac{{_1 F}_1(1;n+1;1)}{\Gamma(n+1)},
$$
which works for noninteger values of $n$. In fact, according to the methods of Summability Calculus this representation is the unique generalization of your truncated series expansion to $n\in\Bbb C$.

But is $E(n)={_1F}_1(1-n;1-n;1)$ also a valid representation? It comes down to how you define such a quantity. As you have noticed, MATLAB's symbolic package does not know how to handle such expressions but this does not mean such a representation is invalid. In the literature, we typically interpret $(-n)_k/(-n)_k$ in the context of hypergeometric functions as being equal to one when $k\leq n$ and zero otherwise. In other words
$$
{_pF}_q\left({-n,a_2,\cdots,a_p\atop-n,b_2,\dots,b_q};z\right)=\sum_{k=0}^n\frac{(a_2)_k\cdots(a_p)_k}{(b_2)_k\cdots(b_q)_k}\frac{z^k}{k!}.
$$
An example of this is the Lerch sum. Indeed, Mathematica knows how to properly interpret hypergeometric functions for these parameters and will give you the numeric value of your truncated series. In your case, we would write

```<code>f[n_]:=Hypergeometric1F1[1 - n, 1 - n, 1]
</code>```

Body:
In regards to the specific example provided you can avoid the whole issue of negative bottom parameters by using DLMF 16.2.4 to formally write your truncated series as
$$
E(n)=\frac{1}{(n-1)!}{_2F}_0\left({1,1-n\atop -};-1\right),
$$
for $n=1,2,\dots$

You can confirm in MATLAB that the follow code will give you the correct answer for your truncated exponential Taylor series:

syms n
f(n)=hypergeom([1,1-n],[],-1)/factorial(n-1);

It is important to note that this representation for your series only holds for positive integer $n$ as the ${_2F_0}(1,1-n;;1)$ function diverges for $n\notin\Bbb N$. For a continuous generalization one could use
$$
E(n)=\sum_{k=0}^\infty\frac{1}{k!}-\sum_{k=n}^\infty\frac{1}{k!}=e-\frac{{_1 F}_1(1;n+1;1)}{\Gamma(n+1)},
$$
which works for noninteger values of $n$. In fact, according to the methods of Summability Calculus this representation is the unique generalization of your truncated series expansion to $n\in\Bbb C$.

But is $E(n)={_1F}_1(1-n;1-n;1)$ also a valid representation? It comes down to how you define such a quantity. As you have noticed, MATLAB's symbolic package does not know how to handle such expressions but this does not mean such a representation is invalid. In the literature, we typically interpret $(-n)_k/(-n)_k$ in the context of hypergeometric functions as being equal to one when $k\leq n$ and zero otherwise. In other words
$$
{_pF}_q\left({-n,a_2,\cdots,a_p\atop-n,b_2,\dots,b_q};z\right)=\sum_{k=0}^n\frac{(a_2)_k\cdots(a_p)_k}{(b_2)_k\cdots(b_q)_k}\frac{z^k}{k!}.
$$
An example of this is the Lerch sum. Indeed, Mathematica knows how to properly interpret hypergeometric functions for these parameters and will give you the numeric value of your truncated series. In your case, we would write

f[n_]:=Hypergeometric1F1[1 - n, 1 - n, 1]


问:
Almost sure convergence in parameters preserves convergence in distribution
Let $X_n$, $n\in\mathbb{N}$ denote a sequence of real-valued random variables that converges in distribution to the standard normal distribution. In addition, each $X_n(c)$ is a function of a  real-valued parameter $c$. I.e. $X_n(c) \stackrel{d}{\to}\mathcal{N}(0,1)$ as $n\to\infty$.

Now, assume $c$ is unkown but the sequence $c_n$, $n\in\mathbb{N}$ converges almost sure to $c$.

What are sufficient conditions for the functions $X_n(c)$, such that $X_n(c_n) \stackrel{d}{\to}\mathcal{N}(0,1)$ as $n\to\infty$? (As stated above, at the limit point $c$ the sequence $X_n(c)$ also converges in distribution to $\mathcal{N}(0,1)$.)

This statement cannot be true in general. But, I assume it holds if $X_n$ are continuous functions of $c$. Does it?

答:
[EDIT]
Here is a counterexample for $X_n:(0,\infty)\rightarrow\mathbb{R}$ continuous. (This doesn't seem to work for $X_n$ defined over all of $\mathbb{R}$, as you commented).

Let $N\sim \mathcal{N}(0,1)$ and define:
$$
X_n(c):=\begin{cases}
N, & \text{if }n>c^{-1}+1,\\
N(n-c^{-1}), & \text{if }c^{-1}\leq n\leq c^{-1}+1,\\
0, & \text{else.}
\end{cases}
$$

In other words, $c\mapsto X_n(c)$ is $0$ for small $n$ but normally distributed for large $n$. In between, the function linearly interpolates - hence continuity of $c\mapsto X_n(c)$. (You could even find a polynomial interpolation in between with Weierstrass, then you have a smooth counter-example).
Evidently, $X_n(n^{-1})=0$ for all $n\in\mathbb{N}$, which completes the counterexample.

I think these references might be helpful to find sufficient conditions:
this and that.


问:
Show that if $D$ does not contain a directed path of length $m$, then $\chi(D) \le m$
Given a digraph $D$, we define $\chi(D)$ as the chromatic number of the underlying undirected graph. Show that if $D$ does not contain a directed path of length $m$, then $\chi(D) \le m$.

I've tried doing this proof by contradiction.

Assume by contradiction that $\chi(D) > m$. Let $P$ where $|P| = m-1$ be the directed path with maximum length in $D$. Since $\Delta(D) + 1 \ge \chi(D) > m \Rightarrow \Delta(D) > m-1 \Rightarrow \Delta(D) > |P|$. (Show that $\Delta(D)$ can't be greater than the largest path? Or maybe that if $\Delta(D) > |P|$ we can extend $|P|$?)

This is as far as I've gotten, I'm not sure if this path I'm taking is a good one for approaching this proof.

答:
I'll show that if $k$ is the length of the longest directed path in $G=(V, E)$ (measured as the number of edges on that path), then $\chi(G) \leq k+1$, which is equivalent to the statement in question.

Let $k$ be the length of the longest directed path in $G$ and let $G' = (V, E' \subseteq E)$ be a maximal acyclic subgraph of $G$. Graph $G'$ contains all the vertices and a maximal set of edges such that there are no directed cycles in $G'$. That means that adding any edge $e \in (E \setminus E')$ to $G'$ generates a directed cycle.

Let us now colour each vertex $v$ of $G'$ with the length of the longest directed path ending in $v$. As the longest path in $G'$ has length at most $k$, we use at most $k+1$ colours (from $0$ to $k$). Notice that the edge is always oriented from the vertex coloured with a lower number to the vertex coloured with a higher number. Also, two vertices of the same colour cannot be adjacent, as one of them would be lying further down the longest path than the other. That means that $G'$ is properly coloured with at most $k+1$ colours.

Let us now consider the remaining edges of $G$. Let $e = (u, v) \in E \setminus E'$ be any such edge. As $G'$ was a maximal acyclic subgraph of $G$, adding $e$ to $G'$ generates a directed cycle. That means that that there exists a directed path from $v$ to $u$ in $G'$ and, therefore, $u$ and $v$ have different colours. That means that adding $e$ to $G'$ does not break the colouring and also that the colouring is valid for $G$. We have properly coloured $G$ with at most $k+1$ colours, which concludes the proof.


问:
A probability problem on conditional expectation
Consider $n$ multinomial trials, where each trial independently results in outcome $i$ with $p_i$, $\sum_{i=1}^{k}p_i=1$. With $X_i$ equal to the number of trials that result in outcome $i$. How to find $E[X_1 | X_2 =0]$ ?</p>

答:
Use the linearity of (conditional) expectation.

Let $I_i$ be the indicator variable that the $i$th trial shows outcome 1, given that it does't show outcome 2.

Then, $E[I_i] = \frac{p_1} { 1 - p_2}$.

Hence, $E[X_1 | X-2 = 0 ] = E[\sum I_i] = \sum E[I_i] = n \frac{p_1} { 1-p_2} $.


问:
Prove a graph is not planar with angles
In graph theory, I remember the first time proving $K_5$ is non-planar, and I thought it was pretty slick to use Euler's Theorem and some other results to prove this. However, in the back of my mind, I thought there could be another way of doing this that involved angles, vectors, or a constructive argument. There are a lot of results in geometry involving triangulation and other spatial theorems, and I would be curious if there was a way to do it this way.

Has anyone seen or able to prove a particular graph is non-planar by using angles, vectors, or a constructive argument?

答:
We could use geometry to prove that $K_5$ has no straight-line embedding, for example.

To do this, first argue that three of the points $A,B,C$ must form a triangle containing the other two points $D,E$; there's several ways to do this. Then, the line $DE$ can only intersect two sides of $\triangle ABC$ (I mean the line segments); suppose it intersects $AB$ and $AC$. Then $BCDE$ is a convex quadrilateral, so its diagonals intersect.

Proving that $K_5$ is not planar is much harder. For the above argument to be enough, we would need to invoke Fáry's theorem that every planar graph has a straight-line embedding. This comes down to Euler's formula, among other things :)

Geometric arguments are used in some problems related to planar graphs, where this is not an issue. For example, Kurz and Pinchasi's proof that there is no $5$-regular matchstick graph is all about looking at the angles of various faces. (A matchstick graph is a graph with a planar embedding in which all the edges are unit line segments: we could represent it by putting matchsticks together.)

Title: Prove a graph is not planar with angles


问:
Logarithm Question With Square Root
There is a problem that I cannot solve:

$$ \sqrt{ (\log {8})^2 + \left(\log {\frac {1}{16}}\right)^2} $$

$$ A) \sqrt{2} \log {2}$$
$$ B) \log {2}$$
$$ C) 3\log {2}$$
$$ D) 5\log {2}$$
$$ E) 5$$

Rules for logaritm:

$$ \log_a{b^c} = c.\log_a{b}  $$

Therefore inner of the square:

$$ \log({2^3})^2 + \log({2^{-4}})^2 $$
$$ \log({2^6}) + \log({2^{-8}})  $$
$$ 6 \log({2}) -8 \log({2})  $$
$$ -2 \log({2})  $$

Where is my error at the calculation? Thanks in advance.

Mr. Oscar Lenzi has informed me about the solution. Here is the solution: (inner of the square)

$$ (\log{2^3})^2 + (\log{2^{-4}})^2 $$
$$ (3\log{2})^2 + (-4 \log{2})^2 $$
$$ 9 (\log{2})^2 + 16 (\log{2})^2 $$
$$ 25 (\log{2})^2 $$

$$  \sqrt{25 (\log{2})^2} $$
$$ 5 \log{2} $$

Answer is D.

答: You misdid parentheses.  Properly, $(\log 8)^2=(3\log 2)^2=9[(\log 2)^2]$.  Similarly for $[\log (1/16)]^2$.  With careful use of parentheses you should get a positive radicand whose square root matches (the correct) one of the given choices.

Title: Logarithm Question With Square Root


问:
Algebraic proof of an inequality being given different constraints on 2 variables
Let $x,y \in \mathbb R$ and $$(1+x+x^2)(1+y+y^2)=2x^2y^2-1$$ and assume that $2<x<y$. I want to algebraically show that $xy<16$ but  I cannot get anywhere. I have tried substitution and nothing seems to help. Graphically I know that it is try because I can simply graph and check visually but I want an actual proof if any one can offer any hints or proof thank you.

答:
$$(1+x+x^2)(1+y+y^2)=2x^2y^2-1\tag{1}$$  $2<x<y$.

Plug $x=2$ in $(1)$
$$7 \left(y^2+y+1\right)=8 y^2-1\to y=-1;\;y=8$$
$(2,8)$ is a point where the curve has its maximum because
in that point the curve $(1)$ is decreasing
$$y'=\frac{-2 x y^2+2 x y+2 x+y^2+y+1}{2 x^2 y-x^2-2 x y-x-2 y-1}$$
as $y'(2,8)=-\frac{49}{3}<0$. It can be proved that the curve is decreasing in the whole region $2<x<y$.

Thus the maximum value of $xy$ is $16$.

Hope this helps

$$...$$


问:
Functional Expectation
Problem: Let  $X$  be distributed over the set  $\Bbb N$  of non-negative integers, with pmf: $$P(X=i)=\frac{\alpha}{2^i}$$

$\alpha$

$E[X]$

For $Y=X$ mod $3$, find:

$P(Y=1)$

$E[Y]$

I will assume $\alpha=1$ for now to set the stage althought is there a way to find this value rather than assume?

Assuming this is simlar to an infinite expectation that starts with Basel problem: $$\sum_{i=0}^\infty \frac{1}{i^2}=\frac{\pi^2}{6}$$
$$\frac{6}{\pi^2}\sum_{i=0}^\infty \frac{1}{i^2}=1$$
$$p_i = \frac{6}{\pi^2}\cdot\frac{1}{i^2}$$
$$E[X]=\sum_{i=0}^\infty i \cdot p_i=\frac{6}{\pi^2}\sum_{i=0}^\infty \frac{1}{i}=\infty$$
However, just not certain.

For the "$Y=X$ mod $3$" part I don't even know what that means.  Does that mean $Y=X+3$?
Thus I don't know how to answer the rest of them without understanding what "mod $3$" means.

Title: Functional Expectation

Body:
In order to guarantee that $p_i=\frac{\alpha}{2^i}$ is a distribution, we have that:

$$\sum_{i=0}^{+\infty} p_i = 1 \Rightarrow \\\alpha \sum_{i=0}^{+\infty}\frac{1}{2^i} = \alpha \frac{1}{1 - \frac{1}{2}} = 2 \alpha = 1 \Rightarrow \alpha = \frac{1}{2}.$$

1.
$$P(Y=1) = P(X ~\text{mod}~ 3 = 1) = \sum_{i : i ~\text{mod}~ 3 = 1}\frac{\alpha}{2^i}.$$

But all $i$ such that $i ~\text{mod}~ 3 = 1$ can be rewritten as:

$$i = 3j + 1,$$

for $j \geq 0.$ Hence:

$$P(Y=1)  = \sum_{j=0}^{+\infty} \frac{\alpha}{2^{3j + 1}} = \frac{\alpha}{2}\sum_{j=0}^{+\infty}\frac{1}{8^{j} } = \frac{\alpha}{2}\frac{1}{1 - \frac{1}{8}} = \frac{4}{7}\alpha = \frac{2}{7} .$$

2.

$$\mathbb{E}[Y=1] = \sum_{j=0}^{+\infty} (3j + 1)\frac{\alpha}{2^{3j + 1}} = \sum_{j=0}^{+\infty} \frac{3j\alpha}{2^{3j + 1}} + \sum_{j=0}^{+\infty} \frac{\alpha}{2^{3j + 1}} = \\
=\frac{3\alpha}{2}\sum_{j=0}^{+\infty} j\left(\frac{1}{8}\right)^j + \frac{4}{7}\alpha = \frac{3\alpha}{16}\sum_{j=0}^{+\infty} j\left(\frac{1}{8}\right)^{j-1} + \frac{4}{7}\alpha =\\
 = \frac{3\alpha}{16}\frac{1}{\left(1 - \frac{1}{8}\right)^2}  + \frac{4}{7}\alpha   = \frac{3\alpha}{16}\frac{64}{49}  + \frac{4}{7}\alpha = \frac{12\alpha}{49} + \frac{28 \alpha}{49} = \frac{40}{49}\alpha = \frac{20}{49}.
  $$


问:
Partial derivatives of the hypergeometric ${_2F_1}$
Do formulas for the partial derivatives of the hypergeometric function ${_2F_1}$ exist?

I mean I am interested in $$\frac{\partial}{\partial a}\  {_2F_1}(a,b,c,z)$$$$\frac{\partial}{\partial b}\  {_2F_1}(a,b,c,z)$$
$$\frac{\partial}{\partial c} \ {_2F_1}(a,b,c,z)$$

答:
As @Lucian pointed out the derivatives for the hypergeometric function w.r.t. its parameters are given in terms of the differentiated series expansion
$$
\partial_a F(a,b;c;z)=\sum_{k=1}^\infty\frac{(a)_k(b)_k}{(c)_k}(\psi(a+k)-\psi(a))\frac{z^k}{k!},
$$
$$
\partial_b F(a,b;c;z)=\sum_{k=1}^\infty\frac{(a)_k(b)_k}{(c)_k}(\psi(b+k)-\psi(b))\frac{z^k}{k!},
$$
and
$$
\partial_c F(a,b;c;z)=\sum_{k=1}^\infty\frac{(a)_k(b)_k}{(c)_k}(\psi(c)-\psi(c+k))\frac{z^k}{k!}.
$$
For higher order derivatives see here. Only in special cases do these series have closed-forms in terms of elementary functions. Furthermore, by the ratio test one can confirm that these series converge for $|z|<1$ and for certain values of $a$, $b$, and $c$ convergence also takes place at the endpoints $z={-1,1}$.


问:
Are periodic tilings stable against defects outside of some region away from the defect?
Suppose that I have a set of Wang tiles on a 2D infinite grid, and that normally the tiling pattern is periodic. Assume it has period $p$ in both vertical and horizontal direction. Then at fixed points on the lattice, I remove the tile/introduce a defect. Is there any notion of the periodic tiling recovering its structure sufficiently far away from the defects?

There appears to be a similar result for types of aperiodic tilings. Here it's is shown that a Robinson tiling is stable with respect to defects: https://www.mimuw.edu.pl/~miekisz/stablejsp.pdf

答:
As I understand your question, you are asking:

Suppose we have a finite set of Wang tiles, which have a periodic tiling $T$ of the plane. Now, fix the colors on finitely many grid squares, such that the result can still be extended to a tiling of the plane. Can this extension can be chosen such that the tiling is identical to $T$ up to a translation, except for finitely many errors near the defect?

The answer to this question is "no". Consider the eight Wang tiles given by rotations of the following two tiles:

Observe that in any tiling of these, the green squares must be matched up, so we are using domino shapes with certain matching conditions. The resulting dominoes look like this, up to rotation:

where the grey circles indicate a side that can be either red or blue, whichever we choose.

After some thought, it should be clear that the only restriction this imposes on a domino tiling is that no two dominoes can be aligned edge-to-edge, with their union forming a $2\times 2$ square. Note that every arrangement of these Wang tiles uniquely determines a squarefree domino tiling of the plane, and every squarefree domino tiling yields at least one corresponding arrangement of the Wang tiles.

First, note that these tiles have a periodic arrangement:

(I've shown the domino tiling, but it should be obvious how to convert this into a periodic arrangement of the Wang tiles.)

On the other hand, suppose that we start off with a domino configuration like this:

Then it is not too hard to verify that the unique squarefree domino tiling containing this arrangement looks like:

But no translation of this latter tiling will ever line up with our periodic one! So we cannot place Wang tiles beyond those original eight in a way that reverts to our original periodic tiling. (In fact, we only need to fix three of them for this behavior to occur.)


问:
If (condition) is true for all $x \in X$, does there exist an $x$ such that it has such a condition for all $x \in X$?
If $\forall x \in X$, (condition) is true, is it then also true to say $\exists x$ such that (condition holds) $\forall x \in X$?

答:
I don't understand the downvotes, this is actually an interesting question. It boils down to whether you allow vacuous quantifiers (many systems don't), that is quantifiers that do not bind any variable.

Let's say that your condition is $P$ and I'm going to change from the extensive formulation of $x \in X$ to $x$ has property $F$.

So $\forall x \in X, \text{(condition is true)}$ becomes $\forall x (Fx \to Px)$ which is a closed formula, call it $\phi$. What you're asking is whether $\exists x\phi$ is true... it depends, if you allow empty domains then no. If you allow for vacuous quantifiers (without  empty domains) then it makes sense to let $\exists x\phi$ be logically equivalent to $\phi$, and the same with the universal quantifier.

Know though that these formulae do not correspond to any meaningful english sentence, it's just a pure logical shibboleth.


问:
Bounded variation implies Lebesgue integrable?
Let $f: [a,b] \rightarrow \mathbb{R}$ be a bounded variation.

Let $f_a(x) = f(a)+ \int_a^x f' dm$ be the indefinite integral of $f$.

Let $g= f - f_a$.

Show $g$ is singular i.e $g'=0$ a.e.

Attempt:

Since $f \in BV[a,b]$, $f' $ exists a.e $f'$ is integrable.

So we can use just simply use the definition of differentiation and Lebesgue differentiation theorem to show that $f_s'(x)= f'(x)-f'(x)$.

But the problem is that to use the Lebesgue differentiation theorem, we need to have $f \in L^1[a,b]$.

I'm not sure if that is true.

So does $f\in BV[a,b]$ imply $f \in L^1[a,b]$?

Thank you.

答: Write $f = v-(v-f)$ where $v(x)$ is the variation of $f$ from $a$ to $x$.  From this decomposition we see that $f$ is the difference of two increasing functions, $v$ and $v-f$. Monotone functions are measurable so $f$ is measurable, and any monotone functions are bounded on $[a,b]$ by their endpoint values, so in fact $f$ is bounded and hence integrable.


问:
Show that the fiber $f^{-1}(a)$ is finite if $a∈ℝ,a≠0$
Let $f:ℝ→ℝ$ be a real analytic function. If $f$ has infinitely many zeros, then we know that the fiber $f^{-1}(0)$ is an infinite discrete and countable set. Let $a∈ℝ,a≠0$, we know also that the fiber $f^{-1}(a)$ is a discrete set unless $f = a$. Then my question is: Some observations show that the number of the fibers $f^{-1}(a)$ is finite (not the number of elements in each fiber). However, I am not able to prove that. We can define the equivalence relation as: $$xRy⇔f(x)=f(y)$$

The equivalence classes of the relation $R$ are the fibers $f^{-1}(x)$. If the number of those fibers is infinite, then this will create a contradiction with the fact that $f$ is an analytic function.

答:
Summarizing the discussion in comments: the fibers of $f$ are in bijective correspondence with the points in the range of $f$. The range of a nonconstant continuous real function is uncountable.

Side remark: when editing a question, it is advisable to check if the title still matches the question.


问:
Dividing an integral by a variable.
In my lecture notes I got this:
$$\int_u^vdv = \int_0^T(-4v)dt$$
$$\int_u^v\frac {dv}v = -4\int_0^Tdt$$
$$ln{\frac vu} = -4T$$
Is the second step actually valid? I mean, can you divide an integral by upper limit of the integral on LHS?

答: What would be valid is$$\int_u^vdv^\prime=\int_0^T(-4v)dt\implies v-u=-4vT,$$or $$\int_u^v\frac{dv^\prime}{v^\prime}=-4\int_0^Tdt\implies\ln\frac{v}{u}=-4T.$$But in your notes, something has gone wrong due to conflating a free variable $v$ with a bound variable I've denoted $v^\prime$. The latter is a dummy integration variable.

Title: Dividing an integral by a variable.


问:
For which values of $ \alpha$, $e^x+ \alpha \cdot x^3$ is convex in $\mathbb{R}$?
I have to say for which values of $ \alpha$ the following function is convex in  $\mathbb{R}$:
$$ e^x+ \alpha \cdot x^3.$$

Applying the definition, the function is convex if
$$ e^{ \lambda x_1+ (1- \lambda) x_2} + \alpha ( \lambda x_1+ (1- \lambda) x_2) \le \lambda (e^{x_1+ \alpha (x_1)^3} ) + (1- \lambda)(x_2+ \alpha (x_2)^3)   \\\Rightarrow \alpha  \le \frac{\lambda e^{ \lambda x_1}+ (1- \lambda)e^{ \lambda x_2} - e^{ \lambda x_1+ (1- \lambda) x_2}}{\lambda x_1+ (1- \lambda) x_2-\lambda (x_1)^3+ (1- \lambda) (x_2)^3}.$$

But I don't know how to proceed.

答:
$f(x) = e^x + \alpha x^2$ is convex in $\Bbb R$ exactly if
$$
g(x) = f''(x) = e^x +6 \alpha x
$$
is non-negative for all $x \in \Bbb R$.

If $\alpha > 0$ then $\lim_{x \to -\infty} g(x) = -\infty$, so $f$ is not convex in that case.

If $\alpha = 0$ then $g(x) = e^x > 0$, i.e. $f$ is convex.

If $\alpha < 0$ then $g$ has a unique minimum where $g'(x_0) = 0$, that is at $x_0 = \ln(-6\alpha)$. The minimal value of $g$ is
$$
 g(x_0) = -6\alpha + 6 \alpha \ln(-6\alpha) = -6\alpha(1-\ln(-6\alpha))
$$
and that is non-negative if $\ln(-6\alpha) \le 1$ or $\alpha \ge -e/6$.

Combining these results we see that $f$ is convex if $-e/6 \le \alpha \le 0$.


问:
The formula for the integral of the floor function, and ceiling function from $0$ to $x$.
I would like to know whether my observations have led me to the right path. Ok, so suppose we wanted to integrate $\int_{0}^{x}$ $\lfloor x \rfloor dx$. How would one find a formula for such a function?

Well, if you graph this function, you obtain these steps. Now what an integral would do is find the area underneath those steps. (It is like Riemann integration). Also if one graphs this function, you can observe that the area of each rectangle is $n^2$, for any value of $n$ on the $x$-axis. So we have that $\int_{0}^{x}$ $\lfloor x \rfloor dx$ =$\sum_{n=0}^{x} n^2$=$\frac{x(x+1)(2x+1)}{6}-1$, and the $-1$ came from the fact that $\lfloor x \rfloor$=$0$, for $x$$\in$$[0,1)$.

Now using this fact, we can also conclude that the integral of the ceiling function is $\frac{x(x+1)(2x+1)}{6}-1$. (More generally if we are integrating on an interval $[a,b]$, we have the value of the integral of the floor function is $\frac{b(b+1)(2b+1)}{6}-1-(\frac{a(a+1)(2a+1)}{6}-1)$).

答:
$$ \int_0^n \lfloor x \rfloor \ dx = \sum_{j=0}^{n-1} \int_j^{j+1} \lfloor x \rfloor \ dx = \sum_{j=1}^{j-1} j = \frac{j(j-1)}{2}. $$

EDIT: This is for the special case when the $n$ is an integer. For $y \not \in \mathbb N$, you can split the integrand into $0$ to $\lfloor y \rfloor$ and $\lfloor y \rfloor$ to $y$ (see other answers).


问:
Using deductive reasoning to prove a true statement with false premises?
Using the fact that

$$ A \vDash B ~~~~ \text{iff} ~~~~ \vDash (A \implies B) $$

What if I take $A := (1=2)$ and $B := (1=1)$ Then we have that:

$$ \vDash (1=2) \implies (1=1) $$ Which is $$ \vDash \mathbb{T}$$ as $A = \mathbb{F}, B = \mathbb{T}$ thus $\mathbb{F} \implies \mathbb{T} := \mathbb{T}$

And so we get that:

$$ (1=2) \vDash (1=1)$$

What have I done here? this sounds like nonsense... how can you prove $1=1$ starting from $1=2$ (?!)

Thank you!

答: Indeed, what you have done is both correct and nonsense. If you start with a false assumption, you can prove anything. You can prove true statements, false statements, contradictory statements, independent statements, anything. That is why we prefer to avoid false assumptions in our proofs.


问:
Fundamental group of lens space
It's direct to find the fundamental group of lens space by using more advanced tools like covering spaces and theory of group actions.

But is there any way to show it using Van Kampen's theorem?

答:
This probably comes under "more advanced tools", but it does use van Kampen.

The lens space $L(p,q)$ can be realized by attaching two solid tori $D^2\times S^1$ along their boundary, via the homeomorphism that sends a meridian of the boundary torus of one of the solid tori to a curve on the boundary torus of the other that winds $p$ times around the longitude and $q$ times around the meridian. This is called a "Heegaard splitting" of $L(p,q)$ and is detailed nicely in Rolfsen's Knots and Links, Chapter 9B.

From this construction, the fundamental group can be calculated directly using van Kampen's theorem. The detail of how to do so is given in Hempel's 3-Manifolds, Chapter 2.

Title: Fundamental group of lens space


问:
Fitch-Style Proof ¬(A → B) → (A ∧ ¬B)
Good Evening,

I want to proof this statement: ¬(A → B) → (A ∧ ¬B). I have no premises.

I already did other proofs where I have no premises and a implication. I guess I have to start with the antecedent as my premise and try to generate the consequence. But there is not much I can use, so I thought I can proof it with contradiction.

This is my idea:

I used this website: http://proofs.openlogicproject.org/

答:
HINT:

Your proof by contradiction is a good idea, but instead of doing that on the whole $A \land \neg B$ statement, try and prove each of $A$ and $\neg B$ using proof by contradiction.  It's clear how to set that up for the latter (assume $B$, and try and get $\bot$). For the former, you would assume $\neg A$, and if that leads to $\bot$, you get $\neg \neg A$, from which you can then infer $A$.

Title: Fitch-Style Proof ¬(A → B) → (A ∧ ¬B)


答:
The area of rectangle standing over the interval $[n,n+1]$ where $n$ is integer, is $\color{red}{n \times 1 = n}$, not $n^2$.

Hence
$$\int_{0}^{x} \lfloor x \rfloor dx = \int_{0}^{\lfloor x \rfloor} \lfloor x \rfloor dx + \int_{\lfloor x \rfloor}^{x} \lfloor x \rfloor dx$$
$$ = \sum _{n=0}^{\color{blue}{\lfloor x \rfloor-1}} n + \int_{\lfloor x \rfloor}^{x} \lfloor x \rfloor dx$$
$$=\frac{\lfloor x \rfloor(\lfloor x \rfloor -1)}{2} + \lfloor x \rfloor \{x\}$$

where $\{x\} = x - \lfloor x \rfloor$, is the fractional part of $x$.

Can you similarly find the other integral?
$$\int_{0}^{x} \lceil x \rceil dx = \ldots$$

Title: The formula for the integral of the floor function, and ceiling function from $0$ to $x$.

Body:
I would like to know whether my observations have led me to the right path. Ok, so suppose we wanted to integrate $\int_{0}^{x}$ $\lfloor x \rfloor dx$. How would one find a formula for such a function?

Well, if you graph this function, you obtain these steps. Now what an integral would do is find the area underneath those steps. (It is like Riemann integration). Also if one graphs this function, you can observe that the area of each rectangle is $n^2$, for any value of $n$ on the $x$-axis. So we have that $\int_{0}^{x}$ $\lfloor x \rfloor dx$ =$\sum_{n=0}^{x} n^2$=$\frac{x(x+1)(2x+1)}{6}-1$, and the $-1$ came from the fact that $\lfloor x \rfloor$=$0$, for $x$$\in$$[0,1)$.

Now using this fact, we can also conclude that the integral of the ceiling function is $\frac{x(x+1)(2x+1)}{6}-1$. (More generally if we are integrating on an interval $[a,b]$, we have the value of the integral of the floor function is $\frac{b(b+1)(2b+1)}{6}-1-(\frac{a(a+1)(2a+1)}{6}-1)$).


问:
Hypergeometric function ratios: $\frac{_{2}F_{1}(a+1,b;c;x)}{_{2}F_{1}(a,b;c;x)}$?
I need a numerically stable way to compute the following ratio:

$$\frac{_{2}F_{1}(a+1,b;c;x)}{_{2}F_{1}(a,b;c;x)},$$

All the parameters are real numbers, with $a< 0$,$\ $ $b,c > 0$ and $0<x<1$.

Right now I am using GSL's implementation of the hypergeometric function, but I keep getting underflows/overflows.

Is there a simplification that I can use?

Update: Following @Leiucippus comment, is there a continued fraction expansion for $\frac{_{2}F_{1}(a+1,b;c;x)}{_{2}F_{1}(a,b;c;x)}$ and/or $\frac{_{2}F_{1}(a+2,b;c;x)}{_{2}F_{1}(a,b;c;x)}$? Hopefully one that converges quickly so it is numerically useful.

答:
I know this is a little late to the game but you can work with the contigious relations of the hypergeometric function to get the ratio into a more useful form for using continued fraction expansions. Using the notation of the linked paper we may write
$$
\begin{aligned}
F\left({a+1,b\atop c};z\right)%
&=\mathcal A_1 F\left({a,b\atop c};z\right)\\
&=\left(\mathcal I+\frac{bz}{c}\mathcal A_1\mathcal A_2\mathcal A_3\right) F\left({a,b\atop c};z\right).
\end{aligned}
$$
Hence,
$$
\begin{aligned}
\frac{F(a+1,b;c;z)}{F(a,b;c;z)}%
&=1+\frac{b z}{c}\frac{F(a+1,b+1;c+1;z)}{F(a,b;c;z)}\\
&=1+bz\frac{\mathbf F(a+1,b+1;c+1;z)}{\mathbf F(a,b;c;z)}.
\end{aligned}
$$
Then, DLMF $\S 15.7$ provides continued fraction expansions for the remaining ratio of hypergeometric functions for certain values of $z$. If this does not help you can further work with these contiguous relations to turn your expression into a ratio of hypergeometric functions that will match one of the continued fraction expansions.


答:
Well, I see that I simply took the upper limit written in notes as v instead of V.
The above is actually correctly written in the notes as:
$$\int_u^Vdv = \int_0^T(-4v)dt$$
$$\int_u^V\frac {dv}v = -4\int_0^Tdt$$
$$ln{\frac Vu} = -4T$$

Title: Dividing an integral by a variable.

Body:
In my lecture notes I got this:
$$\int_u^vdv = \int_0^T(-4v)dt$$
$$\int_u^v\frac {dv}v = -4\int_0^Tdt$$
$$ln{\frac vu} = -4T$$
Is the second step actually valid? I mean, can you divide an integral by upper limit of the integral on LHS?


问:
Existence of specific non-symplectic manifolds
I am fairly new to symplectic geometry but I started wondering about this :

From the definition of a symplectic manifold we can extract that $M$ has to be even dimensional , orientable , and in the case that it's compact $H^2_{dR}(M)\neq 0$.  Now is there extra conditions that we can impose on even-dimensional compact manifold with  $H^2_{dR}(M)\neq 0$ such that $M$ will be symplectic , other than having a symplectic form of a $2-$form $\omega$ such that $\omega^n$ is a volume form ? And does anybody know if there exists an example of a compact manifold that satisfies all those 3 conditions such that it is not symplectic ? Or if it will automatically be a symplectic manifold?

Thanks in advance.

答: Try $S^2\times S^4$. What happens then? You need $H^2(M)$ to generate $H^{2n}(M)$


问:
Is there a space of all finite models?
Let $\mathcal{L}=\{R_i\}_{i \in I}$ be a relational language. There is a natural construction of a compact space of countable $\mathcal{L}$-structures, namely, the space $$ \prod_{R_i \in \mathcal{L}} 2^{\mathbb{N}^{n_i}}$$
where $n_i$ is the arity of the relation symbol $R_i$.

Here the intuition behind the topology is that the more the relations of two structures agree on finite fragments of the underlying set (which is chosen as $\mathbb{N}$ without loss of generality) the closer these structures are. This idea will clearly break down if one considers finite $\mathcal{L}$-structures since a finite amount of data regarding its relations will enable us to distinguish a structure from all the others, so we should end up getting the discrete topology. Indeed, we do get this if $\mathcal{L}$ is finite and we replace $\mathbb{N}$ by $\{0,1,\dots,n\}$ and take the disjoint union of these spaces as $n$ ranges over $\mathbb{N}$.

Can one construct a topological space of all finite $\mathcal{L}$-structures other than the discrete space? Just to avoid any non-trivial topologies arising from the cardinality of $\mathcal{L}$, let us assume that $\mathcal{L}$ is finite; so we are looking for a countably infinite (preferably Hausdorff) topological space.

Obviously I'd prefer to have some kind of intuition behind the notion of closeness in this space. Given how much the space of countable structures has been studied, surely someone must have tried to construct such a space.

答:
I was able to think of a way to topologize the set of finite $\mathcal{L}$-structures that resembles the idea behind the topology of the space of countable structures, namely, declaring the set of finite structures that have a common finite substructure as a basic open set. However, it produces a bad space in terms of separation axioms and whether or not this can become useful is another issue...

For each $n \in \mathbb{N}$, let $\text{Mod}_n(\mathcal{L})$ denote the set of $\mathcal{L}$-structures with universe $\{0,1,\dots,n\}$ and set $\text{Mod}_{<\omega}(\mathcal{L})=\bigcup_{n \in \mathbb{N}} \text{Mod}_n(\mathcal{L})$. For every $\mathcal{M} \in \text{Mod}_{<\omega}(\mathcal{L})$ set $$U_{\mathcal{M}}=\{\mathcal{N} \in \text{Mod}_{<\omega}(\mathcal{L}):  \mathcal{M} \text{ is a substructure of } \mathcal{N}\}$$
Here, when I say "is a substructure of", I do not mean "contains a copy of" but I actually do want $\mathcal{M}$ to be physically contained in $\mathcal{N}$ as a substructure with its interpretations as the restrictions of those of $\mathcal{N}$.

Given $\mathcal{M}$ and $\mathcal{M}'$ in $\text{Mod}_{<\omega}(\mathcal{L})$ with domains $\{0,1,\dots,n\}$ and $\{0,1,\dots,n'\}$ respectively, if we have $\mathcal{N} \in U_{\mathcal{M}} \cap U_{\mathcal{M}'}$, then the substructure of $\mathcal{N}$ generated by $\{0,1,\dots,n\}$ is $\mathcal{M}$ and the substructure of $\mathcal{N}$ generated by $\{0,1,\dots,n'\}$ is $\mathcal{M}'$. It follows that if $U_{\mathcal{M}} \cap U_{\mathcal{M}'} \neq \emptyset$, then we have $U_{\mathcal{M}} \subseteq U_{\mathcal{M}'}$ or $U_{\mathcal{M}'} \subseteq U_{\mathcal{M}}$ depending on whether $n' \leq n$ or $n \leq n'$.

It follows that the set $\{U_{\mathcal{M}}: \mathcal{M} \in \text{Mod}_{<\omega}(\mathcal{L})\}$ is a basis for a topology. Let us endow $\text{Mod}_{<\omega}(\mathcal{L})$ with the topology generated by this basis. Then $\text{Mod}_{<\omega}(\mathcal{L})$ is a $T_0$-space since if $\mathcal{A} \neq \mathcal{B}$ and $|A| \leq |B|$, then $\mathcal{A} \notin U_{\mathcal{B}}$ and $\mathcal{B} \in U_{\mathcal{B}}$. This space is not $T_1$ since if $\mathcal{A}$ is a substructure of $\mathcal{B}$, then every open set containing $\mathcal{A}$ also contains $\mathcal{B}$.

As Eric Wofsey mentioned in the comments, convergence in this space is a bit unusual. If there is a sequence of finite $\mathcal{L}$-structures, then any finite $\mathcal{L}$-structure which is eventually a substructure of the elements of the sequence is a limit of this sequence. This is simply because every point $\mathcal{M}$ has a smallest open neighborhood, namely, $U_\mathcal{M}$.

The subspaces $\text{Mod}_n(\mathcal{L})$ are discrete since $U_{\mathcal{M}} \cap \text{Mod}_n(\mathcal{L})= \{\mathcal{M}\}$ whenever $\mathcal{M} \in \text{Mod}_n(\mathcal{L})$. But instead of taking the disjoint union of these discrete spaces which would result in a discrete space, we are "gluing" them at each finite structure in a way that I honestly don't understand.

Title: Is there a space of all finite models?


Title: Calculate the limit $\lim\limits_{n \to \infty}\left(\frac{1}{\sqrt[3]{(8n^{3}+2)}^{2}}+\cdots+\frac{1}{\sqrt[3]{(8n^{3}+6n^{2})}^{2}}\right)$

Body:
I have to calculate the limit $$\lim_{n \to \infty}\left(\frac{1}{\sqrt[3]{(8n^{3}+2)}^{2}}+\frac{1}{\sqrt[3]{(8n^{3}+4)}^{2}}+\frac{1}{\sqrt[3]{(8n^{3}+6)}^{2}}+\cdots+\frac{1}{\sqrt[3]{(8n^{3}+6n^{2}-2)}^{2}}+\frac{1}{\sqrt[3]{(8n^{3}+6n^{2})}^{2}}\right)$$

I tried to use Sandwich Theorem like this $$\frac{3n^{2}}{\sqrt[3]{(8n^{3}+6n^{2})}^{2}} \leq \Bigg(\frac{1}{\sqrt[3]{(8n^{3}+2)}^{2}}+\frac{1}{\sqrt[3]{(8n^{3}+4)}^{2}}+\cdots+\frac{1}{\sqrt[3]{(8n^{3}+6n^{2}-2)}^{2}}+\frac{1}{\sqrt[3]{(8n^{3}+6n^{2})}^{2}}\Bigg) \leq \frac{3n^{2}}{\sqrt[3]{(8n^{3}+2)}^{2}}$$

And for result I got that limit is $\frac{3}{4}$

Is this correct?

Body:
Bit of an overkill, but an alternative approach giving a bound on the rate  of convergence as well:

$$
S_{n}:=\sum_{i=1}^{3n^{2}}\frac{1}{\sqrt[3]{\left(8n^{3}+2i\right)^{2}}}=\frac{1}{4n^{2}}\sum_{i=1}^{3n^{2}}\frac{1}{\sqrt[3]{\left(1+\frac{2i}{8n^{3}}\right)^{2}}}
$$

By Taylor expansion, for small enough $x$:

$$
1-\frac{2}{3}x\leq\frac{1}{\left(1+x\right)^{2/3}}\leq1-\frac{2}{3}x+\frac{5}{9}x^{2}
$$

Some double-checking if this really holds:
$$
1-\left(1+x\right)^{2}\left(1-\frac{2}{3}x\right)^{3}=\frac{1}{27}x^{2}\left(8x^{3}-20x^{2}-10x+45\right)
$$

This is positive for $1\geq x\geq0$ as $8x^{3}-20x^{2}-10x+45\geq45-30\geq15$.

$$
1-\left(1+x\right)^{2}\left(1-\frac{2}{3}x+\frac{5}{9}x^{2}\right)^{3}=-\frac{1}{729}x^{3}\left(125x^{5}-200x^{4}+440x^{3}+144x^{2}-270x+1080\right)
$$

This is negative for $1\geq x\geq0$, as $125x^{5}-200x^{4}+440x^{3}+144x^{2}-270x+1080\geq-200-270+1080>0$.

As $6n^2<8n^3$ (maximum of $i$ for a given $n$), these estimations hold for every $n$.
Using these approximations to bound the sum:

$$
S_{n,0}:=\frac{1}{4n^{2}}\cdot\sum_{i=1}^{3n^{2}}1=\frac{3}{4}
$$

$$
S_{n,1}:=\frac{1}{4n^{2}}\cdot\frac{2}{3}\sum_{i=1}^{3n^{2}}\frac{2i}{8n^{3}}=\frac{1}{4n^{2}}\cdot\frac{2}{3}\cdot\frac{3n^{2}+1}{4n}= \mathcal{O}\left( \frac{1}{n}\right)
$$

$$
S_{n,2}:=\frac{1}{4n^{2}}\cdot\frac{5}{9}\sum_{i=1}^{3n^{2}}\left(\frac{2i}{8n^{3}}\right)^{2}=\frac{1}{4n^{2}}\cdot\frac{5}{9}\frac{\left(3n^{2}+1\right)\left(6n^{2}+1\right)}{48n^{4}}= \mathcal{O}\left( \frac{1}{n^2}\right)
$$
Hence:
$$
\frac{3}{4}+S_{n,1}+S_{n,2}\geq S_{n}\geq\frac{3}{4}+S_{n,0}
$$
And the rate of the convergence is $\mathcal{O}\left( \frac{1}{n}\right)$


问:
Prove that $2^{n} s$ can be written as the sum of perfect squares, where $s$ is the sum of perfect squares
I need require assistance in proving that $$2^{n} s$$ can be written as the sum of two perfect squares, where $$s$$ is a sum of two perfect squares. My teacher has told me I'm not allowed to use induction and I'm not really sure how to go about it.

If anyone could nudge me in the night direction, I'd be appreciative.

答:
Hint: If the numbers $x$ and $y$ can each be written as the sum of 2 perfect squares, then so can the number $xy$. This is known as the Brahmagupta-Fibonacci Identity.

You can either prove this statement yourself, or click on the Wikipedia link.


问:
Product over Dirichlet characters mod n
I've recently learned about Dirichlet characters and am trying to develop a better understanding of the structure of characters mod n. Here's a problem that's kept me occupied for a bit:

Expand the following product into a polynomial with integer coefficients.

$\prod_{\chi \mod 9} (x-\chi(2))$.

My first approach was to use brute force (i.e. we know the values of all 6 characters mod 9. We can plug the values of these characters at 2 in and evaluate the polynomial.) But this has proven to be pretty tedious, and I'm wondering if there's some way we can capitalize on the fact that 2 is a primitive root mod 9?

Apologies if this is a trivial question...

答: There are $\phi(N)$ characters $\bmod N$. If $a$ has order $k$ in $\Bbb{Z/NZ}^\times$ then $k|\phi(N)$ and $$\prod_{\chi\bmod N} (x-\chi(a))=\prod_{m=0}^{k-1}(x-e^{2i\pi m/k})^{\phi(N)/k}= (x^k-1)^{\phi(N)/k}$$

Title: Product over Dirichlet characters mod n


问:
When to use simplicial, singular and relative homology
From my understanding, basically, the homology groups of all basic objects (especially those which are fundamental polygon) can be found using the simplicial homology. But specifically, when should I use simplicial, singular, or relative homology? Is there any indication for this?</p>

答:
First, you can use relative homology within either simplicial or singular settings, so it is not a matter of using "simplicial, singular, or relative": "relative" doesn't fit in this group. ("Absolute" is the word to best contrast with "relative" in the world of homology.)

As far as simplicial vs. singular goes, simplicial is easier to compute and singular is better-behaved theoretically. With just the definitions you can compute the simplicial homology of the boundary of a 1-simplex, but you wouldn't want to compute the singular homology of $S^1$ just from the definitions. On the other hand, it's easy to show that singular homology is a topological invariant, but this is far from obvious for simplicial homology.


问:
Nonexistence proof of divisibility
Here is the proof problem:

Prove that:

$$\forall a \in \mathbb{N} \ \ \forall p \in \mathbb{N}\setminus \{0\} \ \exists x \in \left\{ a+k \ | \ k \in 0.. (p-1)\right\}(p|x)$$

First, I need some help translating this. I think I've done it correctly. $x$ would belong to the set cardinality $1$ of $a+k$, which would mean x is either the one element, or the empty set, which doesn't work.

Then, I know that this isnt true of all a and p, plenty of counterexamples.

So, what strategy would I use to tackle this nonexistence proof?

答: HINT: By the division algorithm there are non-negative integers $q$ and $r$ such that $a=pq+r$, and $0\le r\le p-1$. Use this $r$ to find a $k\in\{0,\ldots,p-1\}$ such that $a+k$ is a multiple of $p$.

Title: Nonexistence proof of divisibility


问:
Writing down homeomorphisms in a formula
Hey I am dealing with this question since some months.
How rigorously do I have to prove that two "obviously homeomorphic" spaces are really homeomorphic? Example: Cup homeomorphic to a donut... Is it even possible to write down a homeomorphism in this example?</p>

答:
The comments above summarize nicely, but I thought it should be put in an answer.

To describe the homeomorphisms, you first need to describe a "cup" and a "donut" as topological spaces. As Tesla says, a parametrization of the cup will work fine (although it will be very complicated). The homeomorphism would then probably be a composition of homeomorphisms that could be described in terms of the coordinates: one that "lifts up" the base of the mug, one that "squashes" the now solid mug into its handle, etc. The formula you want will probably look horrible, but it's possible to write one down (because, as you said, it exists!).

Gerry's method also works, provided that the donut is hollow and so is the mug: if both are solid (i.e. the donut is a copy of $D^2\times S^1$, which is intuitively "more like a donut than the torus"), then some more work needs to be done, as they are now handlebodies, not surfaces.


问:
Optimal interest compounding frequency
I am trying to figure out what is the optimal compounding frequency for annual interest $\mathbf r$ and compounding count $\mathbf n$.

I calculated that for $\mathbf n \rightarrow \infty$ the compounded interest approaches $\mathbf e^r$, but compounding infinitely many times is not very practical.

Especially when the compounding must be done manually or there might be a small fixed transaction fee charged for every compounding (in cryptocurrency yield farming for example).

To start I am trying to figure out how many times do I need to compound to reach 90% of the theoretical maximum $\mathbf e^r$ so I figured I need to solve the following equation for $\mathbf n$:

$$\left( 1+\frac{r}{n}\right)^n = 0.9e^r$$

How do I do that? Using natural log does not seem to help much. Also how would I modify the equation to account for a fixed transaction fee (every time compounding is done)?

答:
You can make a quick determination using the following inequality which holds for all $n \in \mathbb{N}$ and $r > -n$,

$$e^r \left(1 - \frac{r^2}{n}\right) \leqslant \left( 1 + \frac{r}{n}\right)^n \leqslant e^r$$

If $r = 5\%$, then $  \left( 1 + \frac{r}{n}\right)^n \geqslant e^r \left(1 - \frac{r^2}{n}\right)> 0.9 e^r$ holds for all $n >0.05^2/0.1 = 0.025$. In other words,  the discretely compounded amount exceeds $0.9e^r$ in the case where $r = 5\%$ for any frequency.

Only with very high interest rates will this condition not hold for all $n \geqslant 1$. For example, if $r = 55\%$, then

$$\left( 1 + \frac{0.55}{1}\right)^1= 1.55 < 0.9e^{0.55}\approx 1.55998 < 1.625625 = \left( 1 + \frac{0.55}{2}\right)^2$$

In this case, we have $e^r \left(1 - \frac{r^2}{n}\right)> 0.9 e^r$ for $n > 0.55^2/0.1 = 3.025$. Because the bound is not extremely tight it determines that $0.9e^r$ is first exceeded when $n \geqslant 4$ when in fact this is true for $n \geqslant 2$. Nevertheless, it provides a reasonable guide.

Proof of inequality

Since $e^x \geqslant 1 +x$ and $e^{-x} \geqslant 1 - x$ for $x \geqslant 0$ we have for $r \geqslant -n$,

$$\left(1+
   \frac{r}{n}\right)^n \leqslant e^{r}\leqslant \left(1- \frac{r}{n}\right)^{-n},$$

and

$$0 \leqslant e^{r} - \left(1+ \frac{r}{n}\right)^n = e^{r}\left[1 - e^{-r}\left(1+ \frac{r}{n}\right)^{n}\right]\\ \leqslant e^{r}\left[1 - \left(1- \frac{r}{n}\right)^{n}\left(1+ \frac{r}{n}\right)^{n}\right]\\= e^{r}\left[1 - \left(1- \frac{r^2}{n^2}\right)^{n}\right]$$

By Bernoulli's inequality, we have

$$\left(1- \frac{r^2}{n^2}\right)^{n}\geqslant 1 - n \cdot \frac{r^2}{n^2} = 1 - \frac{r^2}{n},$$

and it follows that

$$0 \leqslant e^{r} - \left(1+ \frac{r}{n}\right)^n \leqslant e^r \frac{r^2}{n}$$

Title: Optimal interest compounding frequency


问:
Finding a clever solution to a game of chance
This question may smell like my older question A probability game, but this time my intentions are different.

The above problem came from my friend. Or so I thought, it really came from me incorrectly remembering his problem. The problem he actually gave me was:

Suppose you start with 1 dollar and flip a weighted coin. With probability 3/4 (call this heads) you win a dollar and with probability 1/4 (call this tails) you lose a dollar. What is the probability you eventually run out of money? (Suppose the house has as many dollar bills as needed).

I know how to solve this problem in the following way:
First observe the number of tails must equal the number of heads plus one. If a string of Hs and Ts satisfy this property, call it Good. Now we want to count the number of Good strings that don't have Good proper prefixes. For a string of length 2n-1 this is just the nth Catalan number. Now we just take the appropriate summation and we arrive at the probability we eventually run out of money which turns out to be 1/3.

Now when my friend told me this problem, he said he was able to apply a clever trick and solve it almost instantly. The thing is, he doesn't remember how he did it. He says he vaguely remembers using a recurrence relation and then telescoping a sum. (Which sounds very plausible)

So can you come up with a short and sweet solution?

On a side note, I am aware my title is poor at best. If you can think of a better one, please let me know.

答:
If $p$ is the probability that you eventually lose a dollar then we must have that

$$p = \frac{1}{4} + \frac{3}{4}p^{2}$$

(We lose a dollar, or win one and lose two eventually).

Solving gives $p=1$ or $p=\frac{1}{3}$. I suppose we can eliminate $p=1$ somehow and get $p=\frac{1}{3}$ as the required answer.

ContentLicense: CC BY-SA 2.5


问:
Is there a bijection from $[0,1]$ to the Cantor set $C$?
Does there exist a bijective function $\phi$ from the unit interval $[0,1]$ to the Cantor set $C$? If so, how can it be constructed? I could then proceed to build a measure space ($C$, $\mathcal{M}_\phi$, $m_\phi$) where $m_\phi(E)$ = $m(\phi^{-1}(E))$ for $E \subset C$. How would $m_\phi(C) = 1$? </p>

答:
We look only at the less interesting part of the question, that of finding an explicit bijection from $[0,1]$ to the Cantor set $C$.

Certainly Cantor-Bernstein-Schroeder is perfectly constructive, and can be used to give an explicit, albeit messy, bijection.

But the "near-bijection" described by @mixedmath can be tweaked to produce an explicit bijection.

For any real number $x$ in $[0,1]$, let the canonical representation of $x$ be defined as follows.  If $x$ is not a dyadic rational (that is, if $x$ does not have an eventually constant binary representation), the canonical representation of $x$ is the ordinary binary representation.  If $x\ne 1$, and $x$ is a dyadic rational, the canonical representation of $x$ is the one that is eventually $0$'s.  Finally, the canonical representation of $1$ is the representation with all $1$'s.

Now we define the bijection.  If $x$ is not a dyadic rational, map $x$ to the number $y$ whose ternary representation replaces the $1$'s in the binary representation of $x$ by $2$'s, and leaves the $0$'s unchanged.

Suppose now that $x$ is a dyadic rational.  Write down its canonical representation.

Let $x$ be a dyadic rational whose canonical representation begins with $0$.  Remove the $0$, move the remaining bits one to the left. Then replace the $1$'s by $2$'s. Call the number which has this ternary representation $y$, and map $x$ to $y$.

Let $x$ be a dyadic rational whose canonical representation begins with $1$. Represent it in the "non-terminating form" that is, after a certain point all $1$'s.  Remove the initial $1$, move all bits one to the left. Then replace the $1$'s by $2$'s.  Call the number which has this ternary representation $y$, and map $x$ to $y$.


问:
Necessary condition for the notion of thick set in topological semigroup.
Let $T$ be a topological semigroup. A subset $A$ of $T$ is called syndetic if there is compact subset $K$ of $T$ such that $K^{-1}A= T$. This means that for every $t\in T$, $Kt\cap A\neq \emptyset$.
Also a subset $R$ of $T$ is called a thick set if for every compact subset $K$ of $T$, there is $t\in T$ such that $Kt\subseteq R$.
I interest to study relation between these notions.

If for every compact set $K\subseteq T$, we have $S\cap (\bigcap_{k\in K} k^{-1}S)\neq \emptyset$, then $S$ is thick set.

Q.What can say about the converse of inclusion?

Please help me to know it.

答:
Let $T$ be a topological semigroup. Then we show that the following condition are equivalent.

$(1)$ $A\subseteq T$ is a thick set,

$(2)$ for every compact set $K$ there is some $t\in A$  such that $Kt\subseteq A$.

$(3)$ for every compact set $K$ we have

$A\cap_{k\in K}k^{-1}A\neq \emptyset$.

$(1)\Rightarrow (2)$. Let $r\in T$ be arbitrary. For compact set $K\subseteq T$, we have $Kr\cup{r}$ is a compact set in $T$. Since $A$ is thick set, for compact set $Kr\cup{r}$, there is $s\in T$ with $(Kr\cup{r})s\subseteq A$. Now, let $t=rs$, then $t\in A$ and $Kt\subseteq A$.

$(2)\Rightarrow (3)$. Let $K\subseteq T$ be a compact set in $T$. There is  $t\in A$, with $Kt\subseteq A$. This means that $t\in K^{-1}A$. Hence, $t\in A\cap_{k\in K}k^{-1}A$, i.e. $A\cap_{k\in K}k^{-1}A\neq \emptyset$.

$(3)\Rightarrow (1)$ Let $K\subseteq T$ be a compact set. Since    $A\cap_{k\in K}k^{-1}A\neq \emptyset$, hence assume that $t\in A\cap_{k\in K}k^{-1}A$. Then $Kt\subseteq A$.


问:
Study the monotonicity of the following function? No derivatives/graphing
The function is the following:

$f(x) = x^{21} + 3x + 1$

I have to study the monotonicity without graphing (as I don't know how to graph higher than mx+b at the moment) and without derivatives (as I haven't learnt them). Intuitively it is strictly increasing, according to Desmos graphing it is strictly increasing (so I was correct I guess).

My approach would be with the "rate of monotonicity" as we call it here:

We take two variables $x_1 \ne x_2$. We proceed by doing $R(x_1,x_2) = (f(x_1) - f(x_2)/(x_1-x_2)$

If my calculations are correct, $R(x_1,x_2) = (x_1^{21}-x_2^{21})/(x_1-x_2) + 3$

I just have to prove this $R(x_1, x_2)$ is strictly greater than $0$. Any advice?

答:
I will write $x$ and $y$ instead of $x_1$ and $x_2$, just to make typing easier.  We may assume that $x>y$.  Then $$\frac{f(x)-f(y)}{x-y}=\frac{x^{21}-y^{21}}{x-y}+3=3+x^{20}+x^{19}y+x^{18}y^2+\cdots+xy^{19}+y^{20}$$

Now, this is clearly $>0$ if $x$ and $y$ are both $\geq0$ or both $\leq0$, so (since $x>y$,) we have only the case $y<0<x$ left to deal with.  Rather than trying to deal with the above expression, we go back to the original definition of $f$ and see that when $y<0$ we have $f(y)<1$ and when $x>0$ we have $f(x)>1.$


问:
Writing formulas in first order logic

Let $\sigma = \{+, \cdot\}$ be a signature with binary function symbols and let $\mathcal{N} = (\mathbb{N}, +^\mathcal{N}, \cdot^\mathcal{N})$ be a structure with the normal $+$ und $\cdot$ operations. Give for each $\varphi_{i}$ a fitting formula:

$\varphi_0(\mathcal{N}) = \{ 0 \}$.

$\varphi_1(\mathcal{N}) = \{ 1 \}$.

$\varphi_2(\mathcal{N}) = \{ 2k, k \in \mathbb{N} \}$.

$\varphi_3(\mathcal{N}) = \{ (x, y), x<y \}$.

For $\varphi_0$ I have  $\varphi_0(x) := \forall y (y + x = y)$, since the free variable $x$ can only be $0$.

$\varphi_1 (x) := \exists y (y \cdot x = y \land \lnot \varphi_0(y))$.

Here I'm saying that the free variable $x$ has to be $1$ but $y$ cannot be $0$.

For $\varphi_2(x)$ I have first defined $\varphi_2'(x)$ with $\varphi_2'(\mathcal{N}) = \{2 \}$:

$\varphi_2'(x) = \exists y \exists z (y + z = x \land \varphi_1(y) \land \varphi_1(z))$

$\varphi_2(x) = \exists y \forall z (y \cdot z = x \land \varphi'_2(y))$

Here I'm saying that $y$ must be $2$, forcing $x$ to be even.

I haven't been able to come up with a formula for $\varphi_3$, all I know is that it has to have $2$ free variables and would really appreciate any hints.

Can someone please check my work?

答:
Your solutions so far are almost* correct. An easier solution for $\varphi_2(x)$ would be $\exists y(x = y + y)$.

For $\varphi_3(x, y)$ you can use that $x < y$ precisely when there is some nonzero $z$ such that $x + z = y$.

*: I noticed that you have one wrong quantifier in $\varphi_2(x)$. The $\forall z$ should be $\exists z$. Because otherwise that formula reads "$x = 2 \cdot z$ for all $z$", which can of course not happen.

Title: Writing formulas in first order logic


问:
$f: G→G$ defined by $f(x) =x^2$ is a homomorphism if and only if $G$ is abelian.
The function $f: G→G$ defined by $f(x) =x^2$ is a homomorphism if and only if $G$ is abelian.
Can anyone give me any tips how to work on this question?</p>

答:
Hint When is it true that $(ax)^2=a^2x^2$? That is $$axax=aaxx$$

Further hint Multiplying on the left by $a^{-1}$ and right by $x^{-1}$ gives what?


问:
Prove that any two continuous functions from a simply connected, locally path connected space to the circle are homotopic.
I came across a problem I just cannot seem to figure out:

Let $X$ be a simply connected, locally path-connected space, and let $f,g:X \to \mathbb{S}^1 $ (that is how John Lee denotes the unit circle) be two continuous functions.
Prove $f\simeq g $ (i.e. $f$ and $g$ are homotopic).

Is the trick to show that $f$ and $g$ are each homotopic to a constant map?
The most I could think of was that since $X$ is simply connected, the image of any loop based at $x\in X$ under $f$ and $g$ is null path-homotopic. However $f(x)$ and $g(x)$ may not be equal. Also, we are not given that $f$ or $g$ are surjective.

Or are we supposed to create a homotopy between $f$ and $g$?

Would anyone be willing to give a suggestion on how to think about this problem?

答: Any two maps $f,g$ would lift to maps into $\mathbb{R}$ since $\mathbb{R}$ covers $S^1$ by the lifting property and the fact that $X$ is simply connected. Since $\mathbb{R}$ is contractible, any continuous map from any topological space into $\mathbb{R}$ is null homotopic. This finishes the proof.


问:
Proof that inequality involving logarithms and exponentials holds.
Let $a\in(0,1)$ and $x\geq 1$. I want to show that

$$g(a,x)=-\ln(a)\cdot a^x\cdot x -1+a^x\leq 0.$$

My attempt: I can define $G(x)=\frac{1}{x}(1-a^x)$ and then write

$$\int_a^1z^{x-1}\,dz=G(x).$$

Then, as $0\leq z^{x-1}\leq 1$ for every $x\geq 1$ and $z\in(0,1)$, the dominated convergence theorem allows me to exchange integral and derivative to get

$$G'(x)=\int_a^1\log(z)z^{x-1}dz\leq 0,$$

Finally, direct computation yields $G'(x)=\frac{g(a,x)}{x^2}$, so we conclude that $g(a,x)\leq 0$.

My question is if anyone have any ideas to show that $g(a,x)\leq 0$ directly. Thanks!

答: Consider the function $g(x) = -\ln(a)a^xx - 1 + a^x$ where $a\in (0,1)$.  Then we have $$g'(x) = -\ln^{2}(a)a^xx - \ln(a)a^x + \ln(a)a^x = -\ln^{2}(a)a^xx.$$  Now, as $a\in (0,1)$, we have that $g'(x) = 0 \implies x = 0$.  Furthermore, $g'(x) > 0$ for all $x\in (-\infty,0)$ and $g'(x)<0$ for all $x\in (0,\infty)$.  Therefore, we have that $x = 0$ is a global maximizer for $g(x)$, and as $g(0) = 0$ we have $g(x) \leq 0$.


问:
Why is the graph of $\sin(y)=x$ different from the graph of $y=\sin^{-1}(x)$?
Why is the graph of $\sin(y)=x$ different from the graph of $y=\sin^{-1}(x)$? Aren't they just the same thing but in the latter y is written in terms of x?

答:
Note that the explicit function has the range

$$y=\sin^{-1}(x)\in (-\frac\pi2,\frac\pi2)$$

while the implicit function below has quite different range

$$\sin(y)=x,\>\>\>\>\>y\in(-\infty,\infty)
$$


问:
Why do algebraic curves have to involve polynomials?
I.e., why, when we talk about an algebraic curve $f(x, y) = 0$, do we assume that $f(x,y)$ is a polynomial?

Surely the analysis of a curve like $|\cos(z)|^2 = 1$ is not beyond the reach of mathematics -- let alone modern mathematics.  But results like Bézout's Theorem require the curves in question to be polynomial.  Is it that the extension to general analytic functions is too hard?

Or, on the other hand, is it too easy (e.g, do results for more general $f(x,y)$ follow easily from results for polynomial $f(x,y)$)?

答:
You can study any equation you want. There's tons of mathematics for studying all kinds of equations. Some tools work better for this bunch of equations, other tools work better for that other bunch of equations.

It just so happens that there is an incredibly rich and beautiful set of tools for studying solution sets of polynomial equations, and which (for the most part) apply only to solution sets of polynomial equations and not other kinds of equations. This collection of tools is known as algebraic geometry.

As for your title question, what shall we call the solution sets of equations defined by polynomials? As the study of polynomial equations has evolved over the last two millenia or so, a terminology has emerged: nowadays their solution sets are known as algebraic varieties, and in the case of $f(x,y)=0$ as algebraic curves.

So, the real and literal answer to your title questions is: That's the definition of algebraic curves.


答: They aren't the same thing, because $y=\arcsin x$ (or $\sin^{-1}x$, as you write it), is defined, not only by $\sin y=x$, but also by the constraint  $-\frac\pi2\le y\le\frac\pi 2$.

Body: Why is the graph of $\sin(y)=x$ different from the graph of $y=\sin^{-1}(x)$? Aren't they just the same thing but in the latter y is written in terms of x?


问:
Nilpotent operator / Orthogonal projection
If you have a nilpotent operator $A \colon V \to V$, $V$ being $A$-cyclic (meaning that $V$ is generated by a single vector $v$ in $V$), is it true that the minimal polynomial of the Gram operator $P = A^{T} A$ is $p(t) = t^2 - t$, i.e., $P$ is the orthogonal projection?</p>

答:
If you grant Jordan canonical form, then in a suitable basis $A$ looks like
$$\begin{bmatrix}0&1&0&\dots&0 \\ 0&0&1&\dots&0 \\ & & \ddots&\ddots \\ 0&0&\dots&0&1 \\ 0&0&\dots &0&0
\end{bmatrix}\,.$$
Then $$A^\top A = \begin{bmatrix}0 & \\ & 1 & \\ & & \ddots \\ & & & 1\end{bmatrix}\,.$$
So 'twould appear you're right.

But .... Note that once you bring the transpose into the game, things are no longer basis-independent. The basis that brings $A$ into this nice form is not likely to be orthonormal, and so, if you're bringing the dot product into the game, it really is not meaningful to describe this in terms of an orthogonal projection. The characteristic polynomial of $A$ is basis-independent, but the characteristic polynomial of $A^\top A$ is not.


答: You need to look at the definition of $\arcsin(x)$. In particular, the range of $\arcsin(x)$ is the closed interval $[-\pi/2,\pi/2]$. So when you look at $\arcsin(0)$, it is the value $x$ in $[-\pi/2,\pi/2]$ such that $\sin(x) = 0$, namely $x=0$. However, the solution set to $\sin(y)=0$ is every number $y$ such that $\sin(y)=0$, for which there are infinitely many such $y$. This explains why your graph of $\sin(y)=0$ has infinitely many horizontal lines. Note here that I am using the notation $\arcsin(x)$ to denote the inverse sine function which you denote as $\sin^{-1}(x)$.

Title: Why is the graph of $\sin(y)=x$ different from the graph of $y=\sin^{-1}(x)$?


问:
Summation of gcd on the divisors of a number.
Let $x=3^{a_1}5^{a_2}7^{a_3}$ and $y=3^{b_1}5^{b_2}7^{b_3}$ where $a_i,b_i$ are nonnegative integers such that $a_k + b_k \le 5$ for all $k$ find
$$\sum \gcd(x,y).$$
where the sum is over all possible $x,y$.

Progress: I tried small cases on this problem but I didn't find a pattern. This is however equal to having $xy$ a divisor of $3^55^57^5$. I couldn't see what to do with this information, however. Maybe there is a systematic way to do count this? I also tried multiplying geometric series
$$(1+3+9+\cdots)(1+5+25+\cdots)(1+7+49\cdots)$$
for the desired term, but that didn't work either.

答:
HINT

It's the product of three terms of the form $$\sum_{a=0}^5\sum_{b=0}^{5-a}p^{\min(\{a,b\})}$$

for $p=3,5,7$, so you just need to figure out how to compute those.


答:
We have $$g(a,x) = -\ln(a)\cdot a^x\cdot x -1+a^x = x\frac{f(x) - f(0)}{x-0} = xf'(\xi)$$

for $f(x) = -\ln(a)\cdot a^x\cdot x +a^x$ and $\xi \in (0,x)$

With $f'(x) = -\ln^2(x)a^x x$ it follows $$g(a,x) = -x\ln^2(\xi)a^\xi \xi$$ for a $\xi \in (0,x)$ hence $g(a,x) \le 0$

Title: Proof that inequality involving logarithms and exponentials holds.

Body:
Let $a\in(0,1)$ and $x\geq 1$. I want to show that

$$g(a,x)=-\ln(a)\cdot a^x\cdot x -1+a^x\leq 0.$$

My attempt: I can define $G(x)=\frac{1}{x}(1-a^x)$ and then write

$$\int_a^1z^{x-1}\,dz=G(x).$$

Then, as $0\leq z^{x-1}\leq 1$ for every $x\geq 1$ and $z\in(0,1)$, the dominated convergence theorem allows me to exchange integral and derivative to get

$$G'(x)=\int_a^1\log(z)z^{x-1}dz\leq 0,$$

Finally, direct computation yields $G'(x)=\frac{g(a,x)}{x^2}$, so we conclude that $g(a,x)\leq 0$.

My question is if anyone have any ideas to show that $g(a,x)\leq 0$ directly. Thanks!


问:
Linear independence of solutions to Floquet equation
I have a question about solutions to the Floquet equation, i.e.:
$$
\dot{x} = A(t) x
$$
where $x$ is an $N$-dimensional time-dependent column vector and $A(t)$ is a $T$-periodic $N \times N$ matrix.

To solve the equation I first find the eigenvalues $q_\beta$ and eigenvectors $\phi_\beta(t)$ of the equation
$$
( A(t) - \frac{d}{dt} ) \phi_\beta = q_\beta \phi_\beta
$$
where the $\phi_\beta$ live in the space of $T$-periodic column vectors.
With the $q_\beta$ and $\phi_\beta$ we form the Floquet equation solutions $\psi_\beta(t)$ by
$$
\psi_\beta(t) = \phi_\beta(t) e^{q_\beta t}
$$
As I understand it, there can only be $N$ linearly independent solutions $\psi_\beta$ out of the infinite number of $q_\beta, \phi_\beta$ pairs.
This is because the Floquet equation is an ordinary linear differential equation and so possesses an $N$-dimensional space of solutions.
The redundancy of solutions (again, as I understand it) is explained by the fact that for every periodic eigenvector $\phi$ with eigenvalue $q$ we can generate another $\phi'(t) = \phi(t) e^{i 2 \pi n \frac{t}{T}}$ with eigenvalue $q' = q - i \frac{2 \pi n}{T}$.

My question is: if $N$ eigenvectors $\phi_\beta$ are selected so that their associated Floquet solutions $\psi_\beta$ are linearly independent, under what conditions are the solutions orthogonal? For instance, the matrix $A(t)$ that I currently working on is real, normal ($AA^T=A^TA$), and has the property that $A + A^T$ is constant.

答:
I assume that $\|A(t)\|\le C$ for all $t\in [0,T]$, that $A(\cdot)$ is continuously differentiable, and that $A(0) = A(T)$.

Let $M_A$ be the operator of multiplication in $L^2(0,T;\mathbb R^N)$ with the matrix $A(\cdot)$, that is $M_Ax(t) = A(t)x(t)$. It's adjoint is given by $M_A^*=M_{A^T}$. Since $AA^T = A^TA$, it follows that $M_A$ is a bounded normal operator.

Now, let $Dx(t) = x'(t)$, defined on $\mathcal D = \{x\in H^1 : x(0) = x(T)\}$, where $H^1 := H^1(0,T;\mathbb R^n)$. It is not hard to see that its adjoint is then given by $D^* = -D$. Now, consider the operator $T := M_A-D$, defined on $\mathcal D$. Its adjoint is $T^* = M_{A^T}+D$.

Ok, consider the operators $TT^*$ and $T^*T$. We have
$$
dom(TT^*) = \{x\in dom(T^*) : T^*x\in dom(T)\} = \{x\in\mathcal D : M_{A^T}x+x'\in\mathcal D\}
$$
By the assumptions on $A$, $\mathcal D$ is invariant under both $M_A$ and $M_{A^T}$. Thus, $dom(TT^*) = \{x\in\mathcal D : x'\in\mathcal D\}$, that is,
$$
dom(TT^*) = \{x\in H^2 : x(0) = x(T),\;x'(0) = x'(T)\}.
$$
By the same reasoning, we get $\mathcal D' := dom(T^*T) = dom(TT^*)$. Now, for $x\in\mathcal D'$ we have
\begin{align*}
TT^*x &= (M_A-D)(M_{A^T}+D) = M_AM_{A^T}x - DM_{A^T}x + M_Ax' - x''\\
TT^*x &= (M_{A^T}+D)(M_A-D) = M_{A^T}M_Ax + DM_Ax - M_{A^T}x' - x''.
\end{align*}
Furthermore,
\begin{align*}
-DM_{A^T}x + M_Ax' &= -\tfrac d{dt}A^Tx + Ax' = -(A^T)'x-A^Tx' + Ax'\\
DM_Ax - M_{A^T}x'&= \tfrac d{dt}Ax -A^Tx' = A'x + Ax' - A^Tx',
\end{align*}
For these to be the same we need that $A' + (A^T)' = 0$, which means that $A+A^T$ is constant, which you say is the case. So, $T$ is normal.

Now, let us see that the eigenvectors corresponding to different eigenvalues are orthogonal in $L^2$. If $S$ is a normal operator and $x,u\in dom (SS^*)$, then
$$
\langle Sx,Su\rangle = \langle S^*Sx,u\rangle = \langle SS^*x,u\rangle = \langle S^*x,S^*u\rangle.
$$
In particular, if we set $u=x$, then $\|Sx\| = \|S^*x\|$. Now, since for any $\lambda\in\mathbb C$ the operator $T-\lambda I$ is normal, this implies $\ker(T-\lambda I) = \ker(T^* - \overline\lambda I)$. So, let $x,u\in\mathcal D$ such that $Tx = \lambda x$ and $Tu = \mu u$ with $\lambda\neq \mu$. Then $Tx = \lambda x\in\mathcal D$ and so $x,u\in\mathcal D'$. We obtain
$$
\lambda\langle x,u\rangle = \langle\lambda x,u\rangle = \langle Tx,u\rangle = \langle x,T^*u\rangle = \langle x,\overline\mu u\rangle = \mu\langle x,u\rangle,
$$
thus $\langle x,u\rangle=0$.

This is of course the $L^2$-inner product. So, $\langle x,u\rangle = 0$ means that
$$
\int_0^T\langle x(t),u(t)\rangle\,dt = 0.
$$
It does not say anything about the orthogonality between $x(t)$ and $u(t)$ at particular times $t$.


问:
Can these simultaneously be integers?
I am trying to figure out if there exist $k \in \mathbb{N}$ and $\omega \in \mathbb{R}$ ($\omega>0$ and  $\omega$ is not odd) such that
\begin{align}
\frac{k}{1+\omega} +\frac12  \quad \text{and}  \quad  \frac{k}{1+\tfrac{1}{\omega}} +\frac12 \\
\end{align}
are both integers  strictly greater than $1$.

I know I have a possibly strange constraint on $\omega$ but it is relevant to the problem I am trying to work on.

Assuming that $\frac{k}{1+\omega} +\frac12  = n $ for some $n \in \mathbb{N}$ and
$ \frac{k}{1+\tfrac{1}{\omega}} +\frac12 =m $ for some $m \in \mathbb{N}$

I arrive at
\begin{equation}
(2n-1)(1+\omega) = (2m-1)(1+\tfrac{1}{\omega})
\end{equation}

My suspicion is that there is no solution to this problem, and I tried to arrive at a contradiction that $\omega$ would have to be odd.

Is this the right approach? Or is my conjecture false and there is a solution?

I would greatly appreciate any help. Thank you!

答: $2m-1$ is odd while $\frac{\omega+1}{\omega}$ has an even denominator and odd numerator, so product ($\frac{odd}{even})$ of these terms cannot be integer .

Title: Can these simultaneously be integers?


问:
Order Statistics are Sufficient under Minimum requirements
I am having lots of trouble with a question that seems at firs quite elementary.

Let $X_1,X_2,\dots,X_n$ be independent and identically distributed random variables, $X_i:(\Omega,\mathcal{F},P')\rightarrow(\mathbb{R},\mathcal{B}(\mathbb{R}),P_X)$ such that $P'$ determines a continuous cumulative distribution function $F_X(x)=P[X\leq x]=P'[\{\omega\in\Omega:X(\omega)\leq x\}]$ which is not necessarily absolute continuous wrt the Lebesgue measure. Then, given the order statistics $X_{i_1}<X_{i_2}<\dots<X_{i_n}$ the conditional distribution of $X_1,X_2,\dots,X_n$ is discrete, and assigns probability $\frac{1}{n!}$ to each "point" $X_{i_1},X_{i_2},\dots,X_{i_n}$ where $i_1.i_2,\dots,i_n$ is a permutation of $1,2,\dots,n$.

The questions does not seem to refer to the distribution of $X_1,X_2,\dots,X_n$ given a particular realization of the order statistics $X_{i_1}=x_{i_1},X_{i_2}=x_{i_2},\dots,X_{i_n}=x_{i_n}$, but the general conditioning.

I have found rather difficult to compute the conditional probabilities as they do not reduce to the simplest cases of conditional distributions.

Any insights or references?

Best Regards,

JM

答:
This is an obvious statement (esp if you ignore the portion about absolutely continuous).

You should be able to show that $P(X_1 < X_2) = P(X_2 < X_1)$, which is the 2 variable case.

And similar, for any permutation $\sigma \in S_n$, we have

$$P(X_1 < X_2 < \cdots <X_n) = P(X_{\sigma(1)} < X_{\sigma(2)} < \cdots < X_{\sigma(n)})$$

Hence, each of these $n!$ events are equally likely to occur, hence occur with probability $\frac{1}{n!}$ (with the disclaimer that they sum to 1).

Note that you are restricting your cases to when all the order statistics are distinct, which means that all the original RV's have distinct values. This might restrict your probability space, if it is not absolutely continuous.

The exact distribution of $X_i$ does not matter. All that matters is that they are IID, and also that it is not the constant RV, in which case, $P(X_1< X_2) = 0$.


问:
Confused about mathematically (formally) finding a conditional expectation in a double expectation problem versus in words
Textbook example: Losses on an insurance policy are uniformly distributed on [0, 10000]. A policy covering these losses has a 1000 deductible. Calculate the average payment made by this policy for each loss.

Textbook solution: Let $X$ be the loss, $Y = \text{max}\{0, X - 1000\}$ be the payment. Condition on the loss being above or below 1000. Then

$$E[Y] = P[X \leq 1000]E[Y| X \leq 1000] + P[X \geq 1000]E[Y|X \geq 1000].$$

If the loss is below 1000, then the payment is 0, so the mean payment is 0. If the loss is greater than 1000, then the payment is uniform on [0, 9000], so the mean payment is 4500.

$$E[Y] = P[X \leq 1000](0) + P[X \geq 1000](4500)$$

The probability that $X$ is greater than 1000 is $1 - F(1000) = 1 - 1000/10000 = 0.9$. Therefore

$$E[Y] = (0.9)(4500) = 4050.$$

My solution: Now if I was doing this on my own I would have proceeded as follows:

$X$ follows a uniform distribution on [0, 10000] so that the pmf of $X$ is 1/10000.  The payment amount $Y$ is a piecewise function defined in terms of $X$.  It is $0$ if $X \leq 1000$ and it is $X - 1000$ if $X > 1000$.

Using double expectation we have that

$$E[Y] = E_X[E[Y|X]] = P[X \leq 1000]E[Y| X \leq 1000] + P[X > 1000]E[Y|X > 1000].$$

The probabilities $P[X \leq 1000]$ and $P[X > 1000]$ are gotten by integrating 1/10000 over the intervals $[0,1000]$ and $[1000, 10000]$, respectively, to get $1/10$ and $9/10$.

Question: But now I'm confused about how to formally find the remaining conditional expectations.  By my definition of conditional expectation in the continuous case, which is new to me and have never used for inequalities, I would have that

$$E[Y| X \leq 1000] = \int_0^{9000} y \cdot f_{Y|X}(Y|X \leq 1000)\, dy.$$

And for $f_{Y|X}(Y|X \leq 1000)$ I have by my definition of conditional distribution that

$$f_{Y|X}(Y|X \leq 1000) = \frac{f(y,x)}{f_X(x \leq 1000)}.$$

If this is correct then how would I find the numerator?  The denominator is just $\int_0^{1000} 1/10000 \, dx = 1/10$.

答:
As you noted, it is much more cumbersome to compute the conditional expectation through the conditional density of $Y$ given $\{X \le 1000\}$ or given $\{X > 1000\}$. It is easier to directly make use of $Y = \max\{0, X-1000\}$ to get
$$E[Y \mid X \le 1000] = E[\max\{0, X-1000\} \mid X \le 1000] = 0$$
and
$$E[Y \mid X > 1000] = E[\max\{0, X-1000\} \mid X > 1000] = E[X \mid X > 1000] - 1000$$
where you can use the knowledge that the conditional distribution of $X$ given $X > 1000$ is uniform on $[1000, 10000]$. This is essentially the textbook solution.

Finding the conditional distribution of $Y$ given $\{X \le 1000\}$ or given $\{X > 1000\}$ is a bit thorny because of the mixture of continuous and discrete distributions.

For instance, the conditional distribution of $Y$ given $\{X \le 1000\}$ is actually the non-random quantity $0$, which does not even have a density function, so doing an integral or finding the conditional density does not make sense.

For $Y$ given $\{X > 1000\}$, you can show that the conditional distribution is uniform on $[0, 9000]$, but the way to show this would essentially be using the knowledge that $X$ given $X > 1000$ is uniform on $[1000, 10000]$ and shifting it, which is exactly what the textbook solution does anyway.


问:
How to prove if two matrices are similar without explicitly finding the basis matrix?
My understanding is that two matrices A and B are similar if and only if there exists some other matrix M such that A = MBM-1.

Is there a way to prove that two matrices are similar without knowing exactly what the basis matrix is? Because obviously if we did, then the "proof" would be a trivial one.

Specifically I'd like to know how the guy in this YouTube video  comes up with similar matrices off the top of his head without explicitly finding the basis matrix.

答:
For two-by-two matrices $A,B$, if the traces and the same, and if the determinants are the same, then at least the characteristic polynomials are the same. Even then, there is still the risk that one has a non-trivial Jordan block, while the other is diagonalizable, and it's maybe not generally easy to distinguish.

An approach for bigger matrices, which already needs some hand computation, is to verify that the trace of $A^n$ is equal to the trace of $B^n$, for $n=1,2,3,\ldots$ up to the size of the matrices. This proves that the characteristic equations are the same. There is still the possibility that the Jordan blocks are different...