问:
Find a rotation matrix with two constraints: Aligns two vectors AND a third vector is perpendicular to a fourth vector when transformed
I have four vectors $ d,w,c,n \in \mathscr{R}^3$.  I want to find a rotation matrix $R$ that satisfies these constraints: $w$ is aligned with $d$ after rotation  AND $c$ is perpendicular to $n$ after rotation. You can imagine it in two steps: after a first rotation that aligns $w$ with $d$, make a second rotation around $d$ so that $c$ becomes perpendicular to $n$. In a particular problem I am working on $d=[1,1,1]^T$ and $n=[0,0,1]^T$.

I believe the constraints can be expressed with these two equations

$$ d \times Rw=0$$
$$ n^T  Rc=0$$

In addition, we must impose a third condition for $R$ being a rotation matrix
$$ RR^T=I$$

Is there a nice way to solve for R using matrix algebra? Or how to formulate the two steps above?

I tried to play with Kronecker products to form a vectorized version of $R$, but did not reach a solution. With $S=skew(d)$ being the skew vector for cross product:
$$d \times Rw=SRw=0 \rightarrow (w^T \oplus S)vec(R)=0$$
$$n^T  Rc=0 \rightarrow(c^T \oplus n^T)vec(R)=0$$
but could not impose the $RR^T=I$

答:
I arrived at an alternative formulation which also correctly solved the problem in a few numerical trials.
It uses Rodrigues' formula for the rotation matrix given a unit vector $\mathbf{v}$ along the desired axis of rotation, and the rotation angle $\theta$ :
$$ R=I +\sin(\theta)\ \mathbf {K_{v}} +(1-\cos(\theta))\ \mathbf {K_{v}} ^{2} $$
where $\mathbf {K_{v}}$ is the skew symmetric matrix which represents a cross-product $a\times b=\mathbf{K_{a}}b$

We find the final rotation matrix as a combination of two rotation matrices $R_{1}$, $R_{2}$

Starting with unit vectors $ d,w,c,n $ we first obtain $R_{1}$ which rotates along the axis $w \times d$ by the angle $\theta_{1}$ between $w$ and $d$ using the Rodrigues' formula. Note that $$cos(\theta_{1})=w \cdot d \ , \ \vert sin(\theta_{1})\vert= \Vert w \times d \Vert$$
$$ R_{1}=I +\sin(\theta_{1})\ \mathbf {K_{\frac{w \times d}{\Vert w \times d \Vert}}} +(1-\cos(\theta_{1}))\ \mathbf {K^{2}_{\frac{w \times d}{\Vert w \times d \Vert}}}  $$

In the second step we define $g=R_{1}c$ and seek a rotation along axis $d$ by angle $\theta_{2}$ that makes $g$ perpendicular to $n$. Because $R_{1}w=d$ this second rotation will not affect $w$ .

Writing the constraint as $n^TR_{2}g=0$ and substituting Rodrigues' formula for $R_{2}$ we find
$$ A\sin(\theta_{2})\ +\ B\cos(\theta_{2})=C $$
$$ A=n^T \ \mathbf{K_{d}} \ g$$
$$ B=\ -n^T \ \mathbf{K_{d}^2} \ g$$
$$ C= \ -n^Tg \ -n^T \ \mathbf{K_{d}^2} \ g$$
I used the symbolic toolbox of Octave (which uses Sympy) to find the solution to this trigonometric equation as
$$\theta_{2} = 2\ tan^{-1}(\frac{A \pm \sqrt{A^2+B^2-C^2}}{B+C}) $$
$$ R_{2}=I +\sin(\theta_{2})\ \mathbf {K_{d}} +(1-\cos(\theta_{2}))\ \mathbf {K_{d}} ^{2} $$

and the final solution is $ R = R_{2}R_{1}$.

Note that if $A^2+B^2-C^2 >0$ there are two solutions, if $A^2+B^2-C^2 =0$ there is one solution, and if $A^2+B^2-C^2 <0$ there is no (real) solution.


问:
Solving A Linear Congruential System
I need to find a prime p which makes:
$p\equiv {\pm1}\text{ (mod }8)$ and
$p\equiv {\pm1}\text{ (mod }12)$

How could I find such $p$? Is there any specified method I can use?
I'd be grateful if anyone could point me to the solution. Thanks in advance!

答:
If $\ \color{#c00}{4\mid m,n}\ $ then $\ x\equiv \pm1\pmod m\,$ and $\!\!\pmod n$ $\!\iff\! x\equiv \pm1\pmod{\ell},\,\ \ell = {\rm lcm}(m,n).$

Indeed, $\ m,n\mid x\!-\!1 \iff {\rm lcm}(m,n)\mid x\!-\!1,\ $ and similarly for $\ x\!+\!1.$

Else $\ m\mid x\!-\!1,\,\ n\mid x\!+\!1\,\smash[t]{\stackrel{\large\color{#c00}{ 4\,\mid\, m,n}}\Rightarrow}\, 4\mid x\!-\!1,x\!+\!1\,\Rightarrow\, 4\mid x\!+\!1\!-\!(x\!-\!1) = 2,\,$ contradiction, or

else $\,\ m\mid x\!+\!1,\,\ n\mid x\!-\!1,\ $ which, similarly, yields a contradiction.

Thus in your case $\,m,n = 8,12\,$ case we get $\ p = x\equiv \pm1\pmod{24},\,$ by $\,24 = {\rm lcm}(8,12).$

ContentLicense: CC BY-SA 3.0

Title: Solving A Linear Congruential System


问:
Hint to show $\frac{1}{n^2-n+1} \leq \frac{2}{n^2}$
Looking for a hint as to show the following inequality. I know it should be relatively simple but I just don't see it. I apologise as I'm aware it's rather basic, but I'm getting frustrated.

I'm reading an example concerning the proof of convergence of the following series
$$ \sum_{n=1}^\infty \frac{1}{n^2-n+1} $$
using a comparison with a series I know converges, naimley $\sum_{n=1}^\infty \frac{2}{n^2}$.

For this to work i need to show
$$ \frac{1}{n^2-n+1} \leq \frac{2}{n^2} $$

Algebraic manipulation hasn't gotten me anywhere so far. The subtraction of $n$ in the denominator is what's giving me issues I think. Many thanks.

答:
Here a direct way for $n>1$:

$$\frac 1{n^2-n+1}\leq \frac 1{n(n-1)}=\frac n{n-1}\cdot\frac 1{n^2}=\left(1+\frac 1{n-1}\right)\frac 1{n^2}\leq \frac 2{n^2}$$

For $n=1$ the inequality $\frac 1{n^2-n+1}\leq\frac 2{n^2}$ holds obviously, as well.

ContentLicense: CC BY-SA 4.0


问:
How to solve $\int_0^\infty \frac{\sin(2x)}{e^x} dx$?
I am stuck at this integral. Can anybody help or give me a hint what to do? I have tried to use integration by parts but keep getting stuck.

$$\int_0^\infty \frac{\sin(2x)}{e^x}dx$$

答:
$$\frac{\sin(2x)}{e^x}$$ is the imaginary part of

$$e^{(-1+2i)x},$$

that integrates as

$$\left.\frac{e^{(-1+2i)x}}{-1+2i}\right|_0^\infty=-\frac1{-1+2i}=\frac{1+2i}5.$$


问:
Packing density of 2143 and crossing number of complete graph
I've observed a numerical coincidence, and I'd like to know if there's something deeper here, or if I'm just observing a numerical coincidence due to Richard Guy's Strong Law of Small Numbers. I've been guilty of that before.

I've been looking at permutations $\sigma \in S_n$ and computing the greatest number of occurrences of the pattern $\sigma$ over all permutations in $S_m$. (This is only interesting when $m > n$.)

I asked about one aspect of this on MSE a few days ago with my question Greatest number of occurrences of the pattern 4213 in a permutation. In an answer, Misha Lavrov pointed me at a 2002 paper called On Packing Densities of Permutations, which addresses this idea.

The Observation

Using code from Code Golf Stack Exchange Users Noodle9 and Arnauld, it appears that the greatest number of occurrences of the pattern 2143 over all permutations in $S_n$ is equal to $A000241(n+1)$, the crossing number of complete graph with $n+1$ nodes—that is, the fewest number of edge-intersections of the complete graph $K_{n+1}$ when drawn in the plane.

This is true for all known crossing numbers of $K_n$: $A000241(1)$ through $A000241(12)$. The largest crossing number, $A000241(12)$, was computed back in 2007, and if bigger values have been computed since, I don't know of it. This means that without computing more crossing numbers (which is hard, presumably), you can't prove that these are unequal by a numerical comparison.

Question

I suspect that these numbers are eventually different—although permutation patterns show up in surprising places.

Is there some correspondence which shows that these values are equal in general? Or can you use some asymptotic argument (or other argument) to show that they're different eventually?

答:
The packing density of the pattern $2143$ is much easier to find. It is $\frac38$, and achieved by a permutation of the form $$\left\lfloor \frac n2 \right\rfloor, \left\lfloor \frac n2 \right\rfloor
-1, \dots, 1, n, n-1, \dots, \left\lfloor \frac n2 \right\rfloor
+1.$$
The number of occurrences of $2143$ in this permutation is $\binom{\lfloor n/2\rfloor}{2} \binom{\lceil n/2\rceil}{2}$.

This also happens to be the conjectured crossing number of $K_{n+1}$. A construction for this number of crossing is described in Relations Between Crossing Numbers of Complete and Complete Bipartite Graphs by Richter and Thomassen. So we think that actually, these sequences are equal for all $n$ - but we don't have a proof.

In other words, the maximum number of $2143$ patterns is known to be found in the sequence A028723. The crossing number is defined to be the sequence A000241, and it is conjectured that these sequences are equal.

As an aside: the pattern $2143$ is much easier to understand not by coincidence, but because it is layered. A layered permutation of $[n]$ is one obtained from the permutation $123\dots n$ by partitioning it into several consecutive blocks, then reversing each block. (For $2143$, we split $1234$ into $12\;34$, then reverse both to get $21\;43$.) A theorem of Stromquist says that the optimal way to pack a layered permutation pattern is to use a layered permutation.

With $2143$, if we have blocks of size $n_1 + n_2 + \dots + n_r = n$ in a layered permutation, the number of occurrences of the pattern is $\sum_{i<j} \binom{n_i}{2} \binom{n_j}{2}$, and we can prove that this is maximized by making $r=2$ and by making $n_1, n_2$ as equal as possible. This gives the permutation at the beginning of this answer.


问:
Why do we use divisors in algebraic geometry?
In algebraic geometry there is a correspondence between Weil divisors, Cartier divisors and line bundles. Over an integral separated locally factorial noetherian scheme, the group of isomorphism classes of Weil divisors and the group of isomorphism classes of Cartier divisors are isomorphic. Moreover, in a general scheme, there is a bijection between Cartier divisors and invertible subsheaves of the sheaf of total quotient rings.

In my (admittedly little) experience, it seems that in "nice" situations (when dealing with smooth varieties, for example) we prefer to use Weil divisors as it is more geometric and that we resort to Cartier divisors otherwise.

I wonder why don't we always use line bundles (= invertible sheaves). Let me explain!

Contrarily to Cartier divisors, line bundles are indeed very geometric objects. Perhaps as geometric as Weil divisors.

As with Cartier divisors, we don't have to deal with the trouble of using orders to define principal divisors.

The pullback works at the level of line bundles, where for Cartier divisors we have to use the "moving lemma" to define the pullback on the Picard group.

What do we lose, or becomes harder, when using line bundles?

答:
Question: "I wonder why don't we always use line bundles (= invertible sheaves)?"

Answer: If $k$ is a field and $X$ is a regular scheme of finite type over $k$ there is the Chern character isomorphism

$$ch_i: \operatorname{K}_0(X)_{\mathbb{Q}}^{(i)} \cong \operatorname{CH}^i(X)_{\mathbb{Q}}$$

where $\operatorname{K}_0(X)_{\mathbb{Q}}^{(i)}$ is the $i$'th eigenspace of the Grothendieck group of finite rank vector bundles on $X$ with respect to the Adams operations, and $\operatorname{CH}^i(X)_{\mathbb{Q}}$ is the $i$'th Chow group (with rational coefficients). This isomorphism sends the class of a linebundle $[L]$ to its Weil divisor $D(L)$:
It induce an isomorphism

$$Pic(X) \cong Cl(X)$$

where $Pic(X)$ is the Picard group of $X$ and $Cl(X)$ is the group of Weil divisors (or Cartier divisors) (HH.Prop.II.6.16). Hence if you work with rational coefficients you may identify the Grothendieck group with the Chow group. The Grothendieck group is defined in terms of the "category of finite rank vector bundles on $X$" and the Chow group is defined in terms of the "set of closed integral subschemes of $X$". Hence when working with the Grothendieck (and higher algebraic K-groups) group your proofs and constructions become "more category theoretical". When working with the Chow group (and higher Chow groups) you work with the "set of closed integral subschemes" of $X$, hence there are "fewer" set theoretical complications.  The Chern character exists for higher algebraic K-groups and higher Chow groups

$$ ch^m_i: \operatorname{K}_m(X)_{\mathbb{Q}}^{(i)} \cong \operatorname{CH}^i(X,m)_{\mathbb{Q}}$$

and higher algebraic K-theory is defined in terms of the category $Vect(X)$ of finite rank vector bundles on $X$, and the higher Chow groups are defined in terms of an explicit complex defined in terms of subschemes of $X$. Again algebraic K-theory is "more category theoretical" and Chow theory is "less category theoretical". When $X$ is regular there is an intersection product on the Chow group and this differs from the multiplication on the Grothendieck group. The intersection product is "more geometric", the tensor product is "more general".

Question: "I wonder why don't we always use line bundles (= invertible sheaves)?"

Answer: Algebraic "K-theory is about vector bundles" and "Chow theory is about cycles and subschemes" and in some cases these two theories are equivalent - the equivalence is given by the chern character. But the theories do not exclude each other in general.

Note: There is an exercise in Hartshorne (Ex II.6.11) where they calculate the Grothendieck group of a non-singular curve $C$. They prove using Proposition II.6.16 that  there is an isomorphism

$$f:\operatorname{K}_0(C) \cong \mathbb{Z} \oplus Cl(C)$$

where $Cl(C)$ is the group of Weil divisors  on $C$. By definition (Exercise HH:II.6.10) the Grothendieck group $\operatorname{K}_0(C)$ is the free abelian group on isomorphism classes $[E]$ where $E$ is a coherent sheaf on $C$, modulo exact sequences. If $E$ is a finite rank locally free sheaf on $C$ it follows

$$f([E]):=(rk(E), D(\wedge^e E))$$

where $e:=rk(E)$ and $D(\wedge^e E)$ is the Weil divisor of $\wedge^e E$.
Since $rk(E)=e$ it follows $\wedge^e E$ is a linebundle and you may take its corresponding Weil divisor.
This correspondence between algebraic cycles/divisors and vector bundles is a much studied correspondence, and in the case of a curve $C$ it follows

$$\operatorname{CH}^*(C) \cong \mathbb{Z} \oplus Cl(C)$$

is an isomorphism. In this decomposition you should view the $\mathbb{Z}$-summand as $\mathbb{Z}[C]$ where $[C]:=1$ is the multiplicative unit in the Chow group. Similarly for the Grothendieck group: In the direct sum decomposition $\operatorname{K}_0(C)\cong \mathbb{Z}\oplus Pic(C)$ you should view the $\mathbb{Z}$-summand as $\mathbb{Z}[\mathcal{O}_C]$ where $[\mathcal{O}_C]:=1$ is the multiplicative unit. The structure sheaf $\mathcal{O}_C$ has the property that for any class $[E]\in \operatorname{K}_0(C)$ we get

$$ [\mathcal{O}_C][E]:=[\mathcal{O}_C \otimes E]=[E].$$

Hence via the exercise in HH you may view Proposition HH.II.6.16 as a "special case" of the Chern character construction above.


答:
Write it as

$$\int_0^{+\infty} e^{-x}\sin(2x)\ \text{d}x$$

Integrate by parts getting

$$-\dfrac{1}{2}e^{-x}\cos(2x)\bigg|_0^{+\infty} - \dfrac{1}{2}\int_0^{+\infty} e^{-x}\cos(2)\ \text{d}x$$

The first numerical term goest to $-\dfrac{1}{2}$

Integrate by parts again the second integral to get

$$\dfrac{1}{2}\left(\dfrac{1}{2}e^{-x}\sin(2x) - \int_0^{+\infty} \dfrac{1}{2}e^{-x}\sin(2x)\ \text{d}x\right)$$

The numerical term goest to zero and you are left with

$$\int_0^{+\infty} e^{-x}\sin(2x)\ \text{d}x = \dfrac{1}{2} - \dfrac{1}{4} \int_0^{+\infty} e^{-x}\sin(2x)\ \text{d}x $$

The two integrals are identical but opposite signs. Arrange and you will get:

$$\int_0^{+\infty} e^{-x}\sin(2x)\ \text{d}x = \dfrac{2}{5}$$

Title: How to solve $\int_0^\infty \frac{\sin(2x)}{e^x} dx$?

Body:
I am stuck at this integral. Can anybody help or give me a hint what to do? I have tried to use integration by parts but keep getting stuck.

$$\int_0^\infty \frac{\sin(2x)}{e^x}dx$$


问:
Finding $p$ and $t$ such that the system $px+y+z=1$, $x+2y+4z=t$, $x+4y+10z=t^2$ has one, infinitely-many, or no solutions

Consider the following equations:

$$\begin{align}
px+\phantom{2}y+\phantom{10}z &=1 \\
x+2y+\phantom{1}4z &=t \\
x+4y+10z &=t^2
\end{align}$$

Now find the values of $p$ and $t$ for which

(a) there is a unique solution

(b) there are infinitely many solutions

(c) there is no solution

My attempt: For (a) we require the determinant $\begin{vmatrix}p&1&1\\1&2&4\\1&4&10\end{vmatrix}\neq 0$. Hence $p\neq 1$.

For (b) we need two of the equations to represent the same line - I tried subbing in the difference between the second and third equation into the first but got nowhere. Similarly for (c).

答: If $p=1$, your system becomes$$\left\{\begin{array}{l}x+y+z=1\\x+2y+4z=t\\x+4y+10z=t^2.\end{array}\right.$$If you subtract the first equation from both the second one and the third one, you will get$$\left\{\begin{array}{l}x+y+z=1\\y+3z=t-1\\3y+9z=t^2-1.\end{array}\right.$$Since $3y+9z=t^2-1\iff y+3z=\frac{t^2-1}3$, the system has no solution if $t-1\ne\frac{t^2-1}3$ (that is, if $t$ is neither $1$ nor $2$) and it has infinitely many solutions otherwise.


问:
Is there a characterization of injective $C(X)$-modules analogous to Serre-Swan?
The Serre-Swan theorem in topology says that if $X$ is compact Hausdorff and $C(X)$ the ring of continuous functions on $X$, then the category of finitely generated projective $C(X)$-modules is equivalent to the category of vector bundles over $X$.  Is there an analogous theorem for the dual notion of injective modules?</p>

答:
This is not really an answer.  Rather, it deals with a different question, namely
the analogue of your question in algebraic geometry.  Perhaps it is still
of some interest.

In algebraic geometry, one has an analogue of Serre--Swan, in which f.g. projective modules over a ring (commutative, with 1) $A$ correspond to finite rank locally free sheaves on Spec $A$.

If I recall correctly, under some assumptions, one can obtain injective $A$-modules as local cohomology sheaves of the structure sheaf supported at (not necessarily closed) points of Spec $A$.

E.g. if $A = k[x]$ (with $k$ an algebraically closed field), then the field of rational functions $k(x)$ is obtained as the local cohomology sheaf $\mathcal H^0_{\eta}(\mathcal O)$, where $\eta$ is the generic point of Spec $A$, while for a closed point $(x-a)$, the injective module
$k[x,1/(x-a)]/k[x]$ is obtained as the local cohomology sheaf $\mathcal H^1_{(x-a)}(\mathcal O).$

This construction is due to Grothendieck, I think, and is discussed in Hartshorne's book Residues and duality, as a way of constructing canonical
injective resolutions.

E.g. when $A = k[x]$, we have the injective resolution
$$0 \to k[x] \rightarrow k(x) \to \bigoplus_{a \in k} k[x,1/(x-a)]/k[x] \to 0,$$
which can be rewritten in terms of sheaves on $X = $ Spec $A$ as
$$0 \to \mathcal O_X \to \mathcal H^0_{\eta}(X,\mathcal O_X) \to \bigoplus_{a \in k}
\mathcal H^1_{(x-a)}(X,\mathcal O_X) \to 0;$$
so the structure sheaf has an injective resolution whose $p$th term involves
a sum over local cohomology sheaves supported at codimension $p$ points.

I don't have much of a feeling as to whether one can carry this over in any useful way to the topological setting, since the geometry of closed subsets in algebraic geometry is much more rigid than in general topology.


问:
Infer the convergence of the coefficients of a power series based on the radius of convergence?
I'm taking a course in complex analysis and have been given a homework assignment. Part of the assignment includes True/False type questions, where we are given a mathematical statement and are required to show whether it is true or false. Here is one question from the assignment:

If the radius of convergence of a power series $z\mapsto\sum_{n=0}^\infty a_nz^n$ is $2$, then the series $\sum_{n=0}^\infty |a_n|$ converges.

Some of my classmates were struggling with this question and asked the professor for a hint. She (the professor) said that the statement was false, but obviously didn't provide a proof, as that's the homework problem. This however baffled me, since I was able to construct a very simple argument that the statement was true. Here's what I came up with:

If the series
\begin{equation*}
z\mapsto \sum ^{\infty }_{n=0} a_{n} z^{n}
\end{equation*}
Has a radius of convergence of $\displaystyle 2$, then the series
\begin{equation*}
z\mapsto \sum ^{\infty }_{n=0} a_{n}( 2z)^{n} =\sum ^{\infty }_{n=0} 2^{n} a_{n} z^{n}
\end{equation*}
Has a radius of convergence of $\displaystyle 1$. Having a look at the ratio test for this series,
\begin{gather*}
\lim _{n\rightarrow \infty }\left| \frac{2^{n+1} a_{n+1} z^{n+1}}{2^{n} a_{n} z^{n}}\right| =|2z|\lim _{n\rightarrow \infty }\left| \frac{a_{n+1}}{a_{n}}\right| < 1\\
\lim _{n\rightarrow \infty }\left| \frac{a_{n+1}}{a_{n}}\right| < \frac{1}{|2z|}
\end{gather*}
Since the radius of convergence is $\displaystyle 1$, we can take e.g $\displaystyle z=3/4$,
\begin{equation*}
\lim _{n\rightarrow \infty }\left| \frac{a_{n+1}}{a_{n}}\right| < \frac{2}{3} < 1
\end{equation*}
Hence by the ratio test the sum of the $\{\displaystyle a_{n}\}$ converges absolutely, i.e, $\sum_{n=0}^\infty |a_n|$ converges.

Either my teacher misspoke, or there is an error in my argument, but I cannot find it. The only thing that I don't have a proof for is the assertion that if a series $\sum_n a_nz^n$ has a radius of convergence of $R$, then the series $\sum_n a_n (\lambda z)^n$ has a R.O.C of $R/\lambda$. It seems pretty obvious and I can't think of an instance where this would be false.

Can someone help please? Is my proof right?

答:
In fact everything the OP said here was correct, including most of the proof. I have this pet peeve - it bugs me when people automatically turn to that $|a_n|^{1/n}$ thing any time the words "radius of convergence" appear. As is often the case we get a much simpler proof starting with the definition of "radius of convergence":

Saying that the radius of convergence is $2$ means the series converges for every $z$ with $|z|<2$ and diverges for every $z$ with $|z|>2$. In particular $\sum a_n(3/2)^n$ converges. So $|a_n|(3/2)^n$ is bounded, so $|a_n|\le c(2/3)^n$, so $\sum|a_n|<\infty$.

To my way of thinking this shows that the answer is obviously yes, as opposed to an argument that diddles with $|a_n|^{1/n}$...


答:
Since the integral consists of two transcendental functions $\sin(2x)$ and $e^{-x}$, you may use integration by parts. Set $u = \sin(2x)$ and $dv = e^{-x}dx$ and we get
$$
\begin{align*}
\int_{0}^{\infty} \sin(2x)e^{-x}dx &= \left[-\sin(2x)e^{-x} \right|_{0}^{\infty} + 2\int_{0}^{\infty}\cos(2x)e^{-x}dx \\
&= 2\int_{0}^{\infty}\cos(2x)e^{-x}dx
\end{align*}
$$
Doing integration by parts again with $u = \cos(2x)$ and $dv = e^{-x}dx$ we have
$$
\begin{align*}
\int_{0}^{\infty} \cos(2x)e^{-x}dx &= \left[-\cos(2x)e^{-x} \right|_{0}^{\infty} - 2\int_{0}^{\infty}\sin(2x)e^{-x}dx \\
&= 1-2\int_{0}^{\infty}\sin(2x)e^{-x}dx.
\end{align*}
$$
Putting everything together we obtain
$$\int_{0}^{\infty} \sin(2x)e^{-x}dx = 2 - 4 \int_{0}^{\infty} \sin(2x)e^{-x}dx$$
Setting $y = \int_{0}^{\infty} \sin(2x)e^{-x}dx$ we then need to solve
$$y = 2 - 4y \Rightarrow 5y = 2 \Rightarrow y = \int_{0}^{\infty} \sin(2x)e^{-x}dx = 2/5.$$


问:
joint probability distribution function to find P(X+Y<1)
The question is to find $P(X+Y\leq1)$ given a joint probability distribution function of $f(x,y)=10xy^2{I}_{(0<x<y<1)}$.

My attempt is

$\int_{-\infty}^{\infty}\int_{-\infty}^{1-y}10xy^2I_{(0<x<y<1)}dxdy=\int_{0}^{1}\int_{0}^{\min(1-y,y)}10xy^2dxdy=\int_{0}^{1}5{(\min(1-y,y))}^2y^2dy=\frac{1}{32}+\frac{1}{12}=\frac{11}{96}$

Then I tried calculating $P(X+Y\geq1)$ in the similar way, expecting the sum of $P(X+Y\leq1)$ and $P(X+Y\geq1)$ would be $1$. However, it gave the result of

$\int_{-\infty}^{\infty}\int_{1-y}^{\infty}10xy^2I_{(0<x<y<1)}dxdy=\int_{0}^{1}\int_{\max(1-y,0)}^{y}10xy^2dxdy=\int_{0}^{1}5[y^2-{(\max(1-y,y))}^2]y^2dy=-\frac{5}{96}+\frac{85}{96}=\frac{5}{6}$

which seemed definitely wrong. What have I done wrong?

答:
You have to integrate over a triangle

$$\mathbb{P}(X+Y<1)=\int_0^{0.5}10x\left[\int_x^{1-x}y^2dy\right]dx=\frac{11}{96}$$

Calculating the complement, as you tried, I find

$$\mathbb{P}(X+Y>1)=\int_{0.5}^{1}5y^2\left[\int_{1-y}^{y}2xdx \right]dy=\dots =\frac{85}{96}$$


问:
A good combination of an Olympiad inequality and its solution

Given three side-lengths $a, b, c$ of a triangle. Prove that
$$a^{2}b\left ( a- b \right )+ b^{2}c\left ( b- c \right )+ c^{2}a\left ( c- a \right )\geq\left | \left ( b+ c- a \right )\left ( a- b \right )\left ( b- c \right )c \right |$$
Source: StackMath/@haidangel

buffalo-way kills it, however, I'm looking forward to another solution that dealed the inverse type of Bernhard Leeb's formula with $b:={\rm med}\left ( a, b, c \right )$
$$a^{2}b\left ( a- b \right )+ b^{2}c\left ( b- c \right )+ c^{2}a\left ( c- a \right ):= \left ( c+ a- b \right )\left ( c- a \right )^{2}b- \left ( b+ c- a \right )\left ( a- b \right )\left ( b- c \right )c$$

答:
In the book Over and Over Again by C. Chang and T. W. Sederberg,
page 15 ends with this

PROBLEM 3.2 [IMO 1983/6]. Let $a,b$, and $c$ be the lengths of the sides
of a triangle. Prove that
$$ (3.7) \qquad \qquad  a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0. $$

Continuing later on page 16 is this

We mention here an interesting story of the IMO in 1983.
Bernhard Leeb, a team member of the Federal Republic of Germany,
was awarded the special prize of that year for his "one line proof"
of (3.7). Leeb pointed out that $(3.7)$ has an equivalent form
$$ (3.10) \quad a(b-c)^2(b+c-a)+b(a-b)(a-c)(a+b-c)\ge 0. $$
Note that the first term in (3.10) is always nonnegative and that
a cyclic permutation of $\,(a,b,c)\,$ leaves the given quantity
(3.7) unchanged, so we can assume without loss of generality that
$\,a\,$ is the maximal side length. Thus the second term in (3.10)
is nonnegative too. Equality holds only if $\,a=b=c.$

Use the triangle inequality to define new values
$$
 x \!:=\! b\!+\!c\!-\!a\!>\!0, \quad
 y \!:=\! c\!+\!a\!-\!b\!>\!0, \quad
 z \!:=\! a\!+\!b\!-\!c\!>\!0. \tag{1} $$

Define the quantity we are interested in
$$ Q :=a^2b(a-b)+ b^2c(b-c)+ c^2a(c-a). \tag{2} $$
The Bernhard Leeb
identity states that
$$ Q = a(b-c)^2(b+c-a) + b(a-b)(a-c)(a+b-c). \tag{3} $$
The quantity $\,Q\,$ is invariant if $\,a,b,c\,$ is permuted cyclically.
Thus, equation $(3)$ implies
$$ Q = (c+a-b)(c-a)^2b - (b+c-a)(a-b)(b-c)c. \tag{4} $$
Apply the definitions in $(1)$ to equation $(4)$ to get
$$ Q = y(c-a)^2b - x(a-b)(b-c)c. \tag{5} $$
Suppose the side lengths are ordered so that $\, c>a>b>0. \,$
This implies $\,a-b>0,\,b-c<0.\,$ Combined with $\,x>0,y>0\,$
applied to equation $(5)$ we get
$$ Q > y(c-a)^2b > 0\quad \text{ and }\quad Q > |x(a-b)(b-c)c| > 0. \tag{6} $$

Note that if we name the largest side as $\,c\,$ and if we name the
other two sides so that $\,c>b>a>0\,$ instead, then the second inequality
in equation $(6)$ need not hold.

The simplest example of such a reverse inequality is
$$ a\!=\!2,b\!=\!3,c\!=\!4 \;\; \text{ where }\;\;
 Q\!=\!16 \!<\! x(a-b)(b-c)c \!=\!20. \tag{7} $$


问:
My son's Sum of Some is beautiful! But what is the proof or explanation?
My youngest son is in $6$th grade. He likes to play with numbers. Today, he showed me his latest finding. I call it his "Sum of Some" because he adds up some selected numbers from a series of numbers, and the sum equals a later number in that same series. I have translated his finding into the following equation:
$$(100\times2^n)+(10\times2^{n+1})+2^{n+3}=2^{n+7}.$$

Why is this so? What is the proof or explanation? Is it true for any $n$?

His own presentation of his finding:

Every one of these numbers is two times the number before it.
  $1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$.

  I pick any one of them, times $100$. Then I add the next one, times $10$. Then I skip the next one. Then I add the one after that.

  If I then skip three ones and read the fourth, that one equals my sum!

答:
Nice discovery, truly impressive for a boy.
The answers have already show the justification, I just wanted to display three examples of the property, probably this corresponds to how it was discovered:</p>

```<code>    1 . .        4 . .          1 6 . .
  +   2 .      +   8 .        +   3 2 .
        8          3 2            1 2 8
    -----       ------         --------
    1 2 8        5 1 2          2 0 4 8
</code>```

Body:
Nice discovery, truly impressive for a boy.
The answers have already show the justification, I just wanted to display three examples of the property, probably this corresponds to how it was discovered:

    1 . .        4 . .          1 6 . .
  +   2 .      +   8 .        +   3 2 .
        8          3 2            1 2 8
    -----       ------         --------
    1 2 8        5 1 2          2 0 4 8


问:
Signature of self-intersection submanifold
Let $S\subset M$ be a (complex) codimension $1$ submanifold. $S\cdot S$ denotes the self-intersection manifold of $S$ in $M$. Then it seems that $\text{Sign}(S\cdot S)$ (the signature of $S\cdot S$) the self-intersection number of the divisor $S$? If so, I wonder why it is true?

I think the manifold $S\cdot S$ is given by $S\cap S'$, where $S'$ is gotten by bringing $S$ in transversal position to $S$ via Transversality Homotopy Theorem.

This is from p.78 of this paper.

答:
You have failed to mention the dimension of $M$. This, of course, matters. If $\dim_{\Bbb C} M = 2$ then your $S \cdot S$ is a finite collection of points (and it makes sense to talk about its intersection number: the signed count of those points.)

The signature of a connected oriented $4k$-manifold $M$ is the signature of the pairing $H^{2k}(M;\Bbb Z) \otimes H^{2k}(M;\Bbb Z) \to H^{4k}(M;\Bbb Z) \cong \Bbb Z$, where the isomorphism uses the orientation of $M$.

When $k = 0$ (so you have an oriented point), there is a canonical isomorphism (no orientations in sight yet) $H^0(M;\Bbb Z) \cong \Bbb Z$, and your pairing $H^0 \otimes H^0 \to H^0$ takes the form $\Bbb Z \otimes \Bbb Z \to \Bbb Z$, with $(m,n) \mapsto mn$. The matrix defining this bilinear form is $\begin{pmatrix} 1 \end{pmatrix}.$

However, we have not followed the orientation isomorphism. The orientation isomorphism $H^0(M;\Bbb Z) \to \Bbb Z$ is identified with $\Bbb Z \to \Bbb Z$, with this map being $+1$ if the point is signed positively and $-1$ if the point is signed negatively. Therefore the bilinear form that signature is, well, the signature of is $$\begin{pmatrix} \pm 1 \end{pmatrix},$$ where the sign indicates whether the point is positively oriented or not.

By definition to get signature of a disconnected oriented manifold you take the sum of the signatures of each connected component. Thus if $M$ is an oriented 0-manifold $\text{Sign}(M) = \# M$, where the last expression means the signed count of points.

In particular you are correct: when $\dim_{\Bbb C} M = 2$, the signature $\text{Sign}(S \cdot S)$ is just the intersection number of $S$ with itself.

In general $$\dim_{\Bbb C} (S \cdot S) = \dim_{\Bbb C} M - 2.$$ When $\dim_{\Bbb C} M > 2$ there is no such thing as the intersection number of $S$ with itself. Instead you get another even-dimensional complex manifold which you can take the signature of.


问:
How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$?
How can I evaluate
$$\sum_{n=1}^\infty\frac{2n}{3^{n+1}}$$?
I know the answer thanks to Wolfram Alpha, but I'm more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before. So I feel that there must be a simpler method.

In general, how can I evaluate $$\sum_{n=0}^\infty (n+1)x^n?$$

答:
What about the Cauchy product?
$$
\sum_{n=0}^\infty(n+1)x^n=\sum_{n=0}^\infty\sum_{\ell=0}^nx^n=\left(\sum_{n=0}^\infty x^n\right)\left(\sum_{m=0}^\infty x^m\right)=\frac{1}{(1-x)^2}.
$$


答:
Thanks to José Carlos Santos who has pointed a flaw in my reasoning.

For the cases (b) and (c), we agree that the range of the system's matrix is a vector space with dimension 2, i.e., a plane. Intuitively, the skew
curve with parametric equations $(x=1,y=t,z=t^2)$ intersects a plane in at most 2 points. Therefore we can have "hope" for the cases where $t$ takes one of these values. (see figure 1).

Fig. 1: Case $p=1$ : The space curve $(1,t,t^2)$ (looking here like a parabola) intersects the 2D range of the LHS of the equation.

let us prove it in an algebraic way.

We know that, necessarily $p=1$.

The system of equations being expressible under the form:

$$xV_1+yV_2+zV_3=V_4$$

More explicitly:

$$x\begin{pmatrix}1\\1\\1\end{pmatrix}+y\begin{pmatrix}1\\2\\4\end{pmatrix}+z\begin{pmatrix}1\\4\\10\end{pmatrix}=\begin{pmatrix}1\\t\\t^2\end{pmatrix}\tag{0}$$

As the 3 columns on the LHS belong to the range of the system's matrix and are dependent (the range has dimension 2), we can take a simple basis of this range, out of which we can "sweep" all this range:

System (0) can be written under the form:

$$u\underbrace{\begin{pmatrix} \ \ 1\\ \ \ 0\\-2\end{pmatrix}}_U+v\underbrace{\begin{pmatrix}0\\1\\3\end{pmatrix}}_V=\begin{pmatrix}1\\t\\t^2\end{pmatrix}$$

Before going further, check that

$$U+V=V_1, \ \ U+2V=V_2, \ \ U+4V=V_3$$

$$\begin{cases}u&=&1\\v&=&t\\-2u+3v&=&t^2\end{cases}\tag{1}$$

which will be fulfilled iff:

$$-2+3t=t^2\tag{2}$$

i.e., iff $t$ is a root of the quadratic equation (2),

$$t=1 \ \ \text{or} \ \ t=2.$$

In these two cases, there exists an infinite number of solutions of systems obtained from (0) by replacing $t$ by $1$ or by $2$:

$$\text{case} \ t=1 : \ \begin{cases}
x+\phantom{2}y+\phantom{10}z &=1 \\
x+2y+\phantom{1}4z &=1 \\
x+4y+10z &=1
\end{cases} \text{and }  \text{case} \ t=2 : \ \begin{cases}
x+\phantom{2}y+\phantom{10}z &=1 \\
x+2y+\phantom{1}4z &=2 \\
x+4y+10z &=4
\end{cases}$$

Indeed, in both systems, the 3rd line $L_3$ can be cancelled, being a combination of the 2 others $L_3=-2L_1+3L_2$, giving the intersection  of two affine planes, i.e., an affine line of solutions in both cases:

$$\text{case} \ t=1 : \ \begin{cases}
x&=&1+2a \\ y&=&-3a \\ z&=&a
\end{cases} , \ \  \text{cases} \ t=2 : \ \begin{cases}
x&=&2a \\ y&=&1-3a \\ z&=&a
\end{cases}$$

Otherwise, if $t\ne 1,2$ there will be no solution.

Title: Finding $p$ and $t$ such that the system $px+y+z=1$, $x+2y+4z=t$, $x+4y+10z=t^2$ has one, infinitely-many, or no solutions

Body:
Consider the following equations:

$$\begin{align}
px+\phantom{2}y+\phantom{10}z &=1 \\
x+2y+\phantom{1}4z &=t \\
x+4y+10z &=t^2
\end{align}$$

Now find the values of $p$ and $t$ for which

(a) there is a unique solution

(b) there are infinitely many solutions

(c) there is no solution

My attempt: For (a) we require the determinant $\begin{vmatrix}p&1&1\\1&2&4\\1&4&10\end{vmatrix}\neq 0$. Hence $p\neq 1$.

For (b) we need two of the equations to represent the same line - I tried subbing in the difference between the second and third equation into the first but got nowhere. Similarly for (c).


问:
Prove the existence of a sequence, spectral theorem??
$T \in \mathcal{B} (H)$ is normal and $\lambda \in \sigma (T)$. Show that we can find a sequence $(x_n)$ in $H$ such that $\lVert x_n \rVert =1$ and $Tx_n- \lambda x_n \rightarrow 0$ where $\lambda$ is an approximate eigenvalue.

Can anyone offer any hints or show me the solution? I am new at Functional Analysis and I am having troubles using in practice the propositions related to the Spectral Theorem.

答:
The role of $\lambda$ in the problem is not entirely clear. Does the clause "where $\lambda$ is an approximate eigenvalue" mean that $\lambda$ is assumed to be an approximate eigenvalue, or is it a part of the statement to be proved?

Anyway, every point of the spectrum of a normal operator is an approximate eigenvalue, see, e.g., Proposition 2.2 here. This gives you the desired sequence (and as Zach L. said, it can be normalized).


问:
On the Continuity of Sobolev Functions
The statement is simple. Further discussion follows. Does there exist a proof of the following lemma which doesn't make use of the fact that Schwartz functions or test functions are dense in the Sobolev space?

Lemma: Let $d$ be a positive integer, $\Omega$ a measurable subset of $\mathbb{R}^d$, and $p \geq 1$ a real number. If $f \in W^{k,p}(\Omega)$ for every $k = 0,1,\ldots,$ then $f$ is almost everywhere equal to a continuous function on $\Omega$.

If needed, $\Omega$ can be whatever you want it to be, but I would prefer to not infringe upon $\Omega$'s free will, if possible.

Apparently, this fact is at least true for $\Omega = \mathbb{R}^d$. According to Exercise 33 in Terrence Tao's blog post (link), we can find $k$ great enough that $W^{k,p}(\mathbb{R}^d)$ embeds into $W^{1,2d}(\mathbb{R}^d)$. Then, by Hölder-Sobolev embedding, $W^{1,2d}(\mathbb{R}^d)$ embeds into $C^{0,\alpha}(\mathbb{R}^d)$ for some $\alpha$, the exact value of which is not consequential. Overall,
$
    W^{k,p}(\mathbb{R}^d)
\hookrightarrow
    W^{1,2d}(\mathbb{R}^d)
\hookrightarrow
    C^{0,\alpha}(\mathbb{R}^d)
$
for some $k$ which has to be chosen arbitrarily large given arbitrarily large $d$. In fact, if I'm not mistaken, this reasoning generalizes to show that $W^{\infty,p}(\mathbb{R}^d) \hookrightarrow C^\infty(\mathbb{R}^d)$ where $W^{\infty,p}(\Omega) := \bigcap_{k=0}^\infty W^{k,p}(\Omega).$

I admit that I, myself, still can't articulate the proof of either the Sobolev embedding theorem or the aforementioned Exercise. However, assuming I haven't misstated them, that isn't the present concern. In summary, those embedding theorems both feel like they might be overkill, don't they? Is there not a direct proof?

References are never a bad thing.

Thank you.

答:
If you're interested in avoiding density arguments as much as possible, you might be interested in the following line of reasoning. It somehow reduces the dependence on density.
$\newcommand{\R}{\mathbb{R}}$
$\newcommand{\vp}{\varphi}$

Lemma 1. If $u \in W^{1,1}((a,b))$, then $u$ has a continuous representative.

Proof. Consider the distributional derivative $u' \in L^1((a,b))$ and the auxiliary function
$$
v(x) = \int_a^x u'(t) dt.
$$
It is easily checked that $v'=u'$ in the distributional sense. Indeed, for any $\vp \in C_c^\infty((a,b))$ we have
\begin{align*}
\int_a^b \vp'(x) v(x) dx
& = \int_a^b \int_a^x \vp'(x) u'(t) dt dx \\
& = \int_a^b \int_a^b \vp'(x) u'(t) \chi_{t<x} dt dx \\
& = \int_a^b \int_t^b \vp'(x) u'(t) dx dt \\
& = \int_a^b (\vp(b)-\vp(t)) u'(t) dt \\
& = - \int_a^b \vp(t) u'(t) dt
\end{align*}
by an application of Fubini's theorem. Hence, the function $u-v$ has zero distributional derivative. Let us use the following lemma as a black box:

Lemma 2. If $w \in L^1((a,b))$ and $w' = 0$ in the distributional sense, then $w$ is constant a.e.

Using this, we infer that $u-v$ is constant a.e. and hence
$$
u(x) = v(x)+c \quad \text{for a.e. } x \in (a,b),
$$
where $v$ is a continuous function as an integral of $u' \in L^1$.

Remark 1. I was not able to justify Lemma 2 without some kind of density argument. For example:

One can consider a sequence of mollifications $w_\varepsilon = w * \vp_\varepsilon$. The distributional derivative of $w_\varepsilon$ is $w_\varepsilon' = w' * \vp_\varepsilon = 0$, but since $w_\varepsilon$ is smooth, then also $w_\varepsilon' = 0$ in the classical sense, so $w_\varepsilon$ is constant. At the same time, $w_\varepsilon \to w$ in $L^1$, so $w$ also needs to be constant. We're not using the density of $C_c^\infty$ in $W^{1,1}$ here, but we're almost proving it.

Alternatively, one could fix $x,y \in (a,b)$ and exploit the condition $\int w \vp' = 0$ for smooth functions $\vp$ for which $\vp' \approx \delta_x - \delta_y$. Applying Lebesgue's differentiation theorem, we infer that $w(x) = w(y)$ for all Lebesgue points $x,y$. But again, some density argument is hidden in the proof of Lebesgue's differentiation theorem.

Remark 2. With some care, one should be able to generalize this reasoning to $\R^n$, as in ktoi's answer.

Title: On the Continuity of Sobolev Functions


问:
How to state a recurrence equation for running time of a divide-and-conquer algorithm
I'm working on my homework and I saw this problem:

A divide-and-conquer algorithm $ X $ divides any given problem instance into exactly two smaller problem instances and solve them recursively. Furthermore, whenever $ X $ divides a given problem instance of size $ n $ into smaller instances with sizes $ a $ and $ b $, then $ a + b = n $ and $ \frac 1 4 n \le a , b \le \frac 3 4 n $.

If the splitting and merging procedures require $ 2 n $ steps for a problem instance of size $ n $, derive the asymptotic running time $ T ( n ) $ (in $ \Theta ( ) $).

If the splitting and merging procedures require $ n ^ 2 $ steps for a problem instance of size $ n $, derive the asymptotic running time $ T ( n ) $ (in $ \Theta ( ) $).

I know that I have to use the master theorem after getting the recurrence equation, but I have no idea how to state the form of $ T ( n ) $. In other examples I've seen something like "it splits in three parts" which is understandable, but the $ a + b $ part makes me confused. Can someone help me please? Thanks!!

答:
Let's first formulate the problem in terms of an explicit recurrence relation.

Let $ a _ 1 = a _ 2 = 1 $ and $ b _ 1 = b _ 2 = \frac 1 2 $, and consider the functions $ T , h _ 1 , h _ 2 , g $ and with the property that
$$ T ( x ) = g ( x ) + a _ 1 T \big( b _ 1 x + h _ 1 ( x ) \big) + a _ 2 T \big( b _ 2 x + h _ 2 ( x ) \big) \text . $$
Also assume that $ | h _ 1 ( x ) | , | h _ 2 ( x ) | \le \frac x 4 $. Find asymptotic bounds for $ T $, when $ g ( x ) = 2 x $, and when $ g ( x ) = x ^ 2 $.

I Intentionally formulated the problem as above, so that it looks similar to the form appearing in Akra-Bazzi theorem, which is one of the best-known generalizations of the master theorem. As you can see in Leighton's note on Akra-Bazzi theorem, the given asymptotic bound by the theorem are only guarantied to work when there is some positive constant $ \epsilon $ such that we have $ | h _ 1 ( x ) | , | h _ 2 ( x ) | \le \frac x { ( \log x ) ^ { 1 + \epsilon } } $ for all large enough $ x $ (together with other requirements which hold in our problem). Unfortunately, this is not the case in our problem, and you can't use the master theorem or its well-known generalizations as you expected. But fortunately, the growth order given by the theorem works in our problem, for another simple reason. So we have a good guess using the generalized master theorem, and only need to provide a proof by ourselves. As $ p = 1 $ is the unique real number satisfying $ a _ 1 b _ 1 ^ p + a _ 2 b _ 2 ^ p = 1 $ (the critical power), the given growth order $ \Theta \bigg( x ^ p \left( 1 + \int _ 1 ^ x \frac { g ( u ) } { u ^ { p + 1 } } \ \mathrm d u \right) \bigg) $ becomes $ \Theta ( x \log x ) $ in case $ g ( x ) = 2 x $, and $ \Theta \left( x ^ 2 \right) $ in case $ g ( x ) = x ^ 2 $.

To see why the bounds work, note that both $ x \mapsto x \log x $ and $ x \mapsto x ^ 2 $ are differentiable convex functions (on our domain of interest, say $ x \ge 1 $). Thus their derivatives, $ x \mapsto 1 + \log x $ and $ x \mapsto 2 x $, are strictly increasing, and hence injective. Consider a differentiable real valued function $ f $ defined on an interval $ I = \left[ \frac n 2 - m , \frac n 2 + m \right] $ for some positive $ n $ and $ m $, such that its derivative $ f ' $ is strictly increasing. Define $ F : I \to \mathbb R $ with $ F ( a ) = f ( a ) + f ( n - a ) $. Then $ F $ is differentiable and $ F ' ( a ) = f ' ( a ) - f ' ( n - a ) $. As $ f ' $ is injective, the value of $ F ' $ can become zero only when $ a = n - a $, i.e. $ a = \frac n 2 $. Thus, as the extrema of $ F $ can only happen at the end-points of $ I $ or where the value of $ F ' $ becomes zero, we can check that for all $ a \in I $,
$$ F \left( \frac n 2 \right) \le F ( a ) \le F \left( \frac n 2 - m \right) = F \left( \frac n 2 + m \right) \text , $$
or equivalently
$$ 2 f \left( \frac n 2 \right) \le f ( a ) + f ( n - a ) \le f \left( \frac n 2 - m \right) + f \left( \frac n 2 + m \right) \text , $$
by determining the sign of $ F ' $. This simple observation lets us prove that our bounds  work.

Case $ T ( n ) = T \big( a ( n ) \big) + T \big( b ( n ) \big) + 2 n $

Choose positive constants $ c _ 1 \le 2 $ and $ c _ 2 \ge \frac 1 { 1 - \frac 3 8 \log _ 2 3 } $ such that we have
$$ c _ 1 n \log _ 2 n \le T ( n ) \le c _ 2 n \log _ 2 n \tag 0 \label 0 $$
for sufficiently many values of $ n $. We use (strong) induction to prove that from there on, \eqref{0} holds for all $ n $. As the induction hypothesis, assume that we have \eqref{0} for all $ n $ with $ \frac N 4 \le n \le \frac { 3 N } 4 $. Then we have
$$ T ( N ) = T \big( a ( N ) \big) + T \big( b ( N ) \big) + 2 N \\
\ge c _ 1 a ( N ) \log _ 2 a ( N ) + c _ 1 b ( N ) \log _ 2 b ( N ) + 2 N \\
= c _ 1 a ( N ) \log _ 2 a ( N ) + c _ 1 \big( N - a ( N ) \big) \log _ 2 \big( N - a ( N ) \big) + 2 N \\
\ge 2 c _ 1 \left( \frac N 2 \right) \log _ 2 \left( \frac N 2 \right) + 2 N = c _ 1 N \log _ 2 N + ( 2 - c _ 1 ) N \ge c _ 1 N \log _ 2 N \text , $$
and
$$ T ( N ) = T \big( a ( N ) \big) + T \big( b ( N ) \big) + 2 N \\
\le c _ 2 a ( N ) \log _ 2 a ( N ) + c _ 2 b ( N ) \log _ 2 b ( N ) + 2 N \\
= c _ 2 a ( N ) \log _ 2 a ( N ) + c _ 2 \big( N - a ( N ) \big) \log _ 2 \big( N - a ( N ) \big) + 2 N \\
\le c _ 2 \left( \frac N 4 \right) \log _ 2 \left( \frac N 4 \right) + c _ 2 \left( \frac { 3 N } 4 \right) \log _ 2 \left( \frac { 3 N } 4 \right) + 2 N \\
= c _ 2 N \log _ 2 N + \Bigg( 2 - \left( 2 - \frac 3 4 \log _ 2 3 \right) c _ 2 \Bigg) N \le c _ 2 N \log _ 2 N \text , $$
which completes the induction step.

Case $ T ( n ) = T \big( a ( n ) \big) + T \big( b ( n ) \big) + n ^ 2 $

Choose positive constants $ c _ 1 \le 2 $ and $ c _ 2 \ge \frac 8 3 $ such that we have
$$ c _ 1 n ^ 2 \le T ( n ) \le c _ 2 n ^ 2 \tag 1 \label 1 $$
for sufficiently many values of $ n $. We use (strong) induction to prove that from there on, \eqref{1} holds for all $ n $. As the induction hypothesis, assume that we have \eqref{1} for all $ n $ with $ \frac N 4 \le n \le \frac { 3 N } 4 $. Then we have
$$ T ( N ) = T \big( a ( N ) \big) + T \big( b ( N ) \big) + N ^ 2 \\
\ge c _ 1 a ( N ) ^ 2 + c _ 1 b ( N ) ^ 2 + N ^ 2 \\
= c _ 1 a ( N ) ^ 2 + c _ 1 \big( N - a ( N ) \big)  ^ 2 + N ^ 2 \\
\ge 2 c _ 1 \left( \frac N 2 \right) ^ 2 + N ^ 2 = \left( \frac { c _ 1 } 2 + 1 \right) N ^ 2 \ge c _ 1 N ^ 2 \text , $$
and
$$ T ( N ) = T \big( a ( N ) \big) + T \big( b ( N ) \big) + N ^ 2 \\
\le c _ 2 a ( N ) ^ 2 + c _ 2 b ( N ) ^ 2 + N ^ 2 \\
= c _ 2 a ( N ) ^ 2 + c _ 2 \big( N - a ( N ) \big)  ^ 2 + N ^ 2 \\
\le c _ 2 \left( \frac N 4 \right) ^ 2 + c _ 2 \left( \frac { 3 N } 4 \right) ^ 2 + N ^ 2 = \left( \frac { 5 c _ 2 } 8 + 1 \right) N ^ 2 \le c _ 2 N ^ 2 \text , $$
which completes the induction step.


问:
Sum of Kloosterman sums bound by $O(n^{\epsilon})$
I've been studying the book "Topics in Classical Automorphic forms" by Henryk Iwaniec and I am stuck at a bound.
For even $k$ and any $3/4 < \sigma < 1$ the book claims the following for the Kloosterman sum $S(n,n,c)$ where $\epsilon>0 $ is fixed, as $n$ gets large. (page 73)

$$\sum_{c>0} c^{-2\sigma}|S(n,n,c)| \ll n^{\epsilon} $$

I've been trying to prove this using the Weyl bound $|S(m,n,c)| \le \gcd(m,n,c)^{1/2} c^{1/2}\tau(c)$ but I haven't managed to do so. Maybe I am missing something basic but in order to create a convergent series the bounds for the $\tau$ function seems to exceed any hopes I have for that.

答:
Since $|S(n,n;c)| \leq (n,c)^{1/2} c^{1/2} \tau(c)$ and $\tau(c) \ll_{\varepsilon} c^{\varepsilon}$, this sum is
$$\ll_{\varepsilon} \sum_{c = 1}^{\infty} \frac{(n,c)^{1/2}}{c^{2\sigma - 1/2 - \varepsilon}}.$$
Now write this sum over $c \in \mathbb{N}$ as a double sum over $m \mid n$ and over $c \in \mathbb{N}$ for which $(c,n) = m$ (that is, break up the sum over $c$ based on the $\gcd$ of $c$ and $n$). Having done this, make the change of variables $c \mapsto mc$, since $(c,n) = m$ implies that $m \mid c$. We get
$$\sum_{m \mid n} \frac{1}{m^{2\sigma - 1 - \varepsilon}} \sum_{\substack{c = 1 \\ (c,\frac{n}{m}) = 1}}^{\infty} \frac{1}{c^{2\sigma - 1/2 - \varepsilon}}.$$
The inner sum over $c$ converges absolutely since $\sigma > 3/4$. The outer sum is $\ll_{\varepsilon} n^{\epsilon}$.


问:
Van der Waerden Number Proof

Hi, my professor posted exactly one slide about Van der Waerden numbers and I simply don't have enough information to solve this. I tried looking online for help but none of the lower bounds seem to even closely resemble this one. Seeing as the professor gave us nothing to work with I assume we have to find some expression, e, and prove that 2^(k/2) >= e and then prove that W(k) >= e, but again, I am totally out of my element here. Could anyone please help me with this?

Edit: It's for two colours. So W(k, 2).

答:
Color the interval $\{1,\ldots , N \} =[N]$ randomly uniformly in two colors. There are not more than $\binom{N}{2} < N^2/2$ arithmetic progressions of length $k$ in this interval. To see this, note that we can uniquely fix a $k$-AP by specifying its first two elements for instance. The probability that such a progression is monochromatic is $2 \cdot 2^{-k} = 2^{-k+1}$, as either all of the elements have color 1, or they all have color 2. So we expect less than $ N^2 2^{-k}$ monochromatic $k$-APs. If $N < 2^{k/2}$, then
$$ N^2 2^{-k}<1. $$
So if $N<2^{k/2}$, we expect less than $1$ monochromatic $k$-APs. If every coloring of $[N]$ produces a monochromatic $k$-AP, then the expected value would be at least $1$. Therefore, there is a coloring of $[N]$ with no monochromatic $k$-AP.

Title: Van der Waerden Number Proof


问:
Maximize the sum of cubes
$x_n=\{x:x \in[-1,2]\}$

Find the maximum value of $\ \ \ \ \sum_{i=1}^{100}x^3_i \ \ \ \ \ $ if $\ \ \ \  \sum_{i=1}^{100}x_i=2$

My attempt:

For the "cube sum" to be maximum the $x_i$ has to take the maximum value i.e. 2 but due to the constraint of the linear sum, we have to subtract some terms to balance it out. Hence $x_i \in \{-1,2\}$.

So, $\sum_{i=1}^{100}x_i=2a-b=2$ where $a+b=100$.

On solving the the linear equation, we get $a=34$ and $b=66$

Using this newly found information: $\sum_{i=1}^{100}x^3_i=8a-b=8\cdot34-66=206$ which is the maximum value of the function. But I am not sure how to prove this rigorously.

Please provide hints to solve it rigorously.

答:
Hint.

Make $x_i = a(\sin u_i +b)$ and forming the lagrangian

$$
L = \sum_i^{100} a^3(\sin u_i +b)^3-\lambda\left(\sum_i^{100}a(\sin u_i +b)-2\right)
$$

The stationary points are the solutions for

$$
\cases{
3a^3(\sin u_i+b)^2\cos u_i-\lambda a\cos u_i = 0,\ \ i=1,\cdots, 100\\
\sum_i^{100}a(\sin u_i +b)-2=0}
$$

then we have

$$
\cos u_i = 0\\
a^2(\sin u_i+b)^2 = \frac{\lambda}{3}
$$

and then $100\sqrt{\frac{\lambda}{3}} = 2$ etc.

NOTE

$$
\cases{
a = \frac 32\\
b = \frac 13
}
$$

From $\cos u_i = 0$ we have $u_i = \pm \frac{\pi}{2}$ and then the maximum is attained when

$$
\cases{
-n+2m=2\\
n+m=100
}
$$

associated to this condition we have $\max \sum_{i=1}^{100}x_i^3 = 66(-1)+34(2^3)$ and now we can follow analyzing  the stationary points associated to $\cases{a^2(\sin u_i+b)^2 = \frac{\lambda}{3}\\ 100\sqrt{\frac{\lambda}{3}} = 2}$ etc.

Title: Maximize the sum of cubes


问:
MLE for exponential distribution
lets say $\ Y $ distributed exponential with the following pdf:

$$\ f_{\theta, \tau} = \theta \cdot e^{-\theta(y-\tau)}\mathbb I \{y \ge \tau\} , \theta > 0 $$

and I'm trying to find MLE when both $\ \theta, \tau $ are unknowns.

If I did everything correctly then the log likelihood function is $$\ l(\theta, \tau ; y) = n\cdot ln(\theta) + \theta \cdot \tau \cdot n - \theta \sum y_i $$
derivative w.r.t $\ \theta $  is $$\ \frac{\partial l}{\partial \theta} = \frac{n}{\theta} + \tau \cdot n - \sum y_i  \Rightarrow \theta = \frac{n}{\sum y_i - \tau \cdot n}$$

and derivative w.r.t to $\ \tau $ is just $\ \frac{\partial l}{\partial \tau} = n  \Rightarrow n = 0 $ so I did something wrong here but can't figure out what ?

Thanks.

答:
$\mathbb{I}(\cdot)$ is an indicator function I presume. In that case, your distribution does not have two parameters! It is only $\theta>0.$ Your distribution can actually be written in the following way to make thing clear :

$f_\theta(y)=\theta\cdot e^{-\theta(y-\tau)},$ $y\geq\tau,$ $\theta>0.$

Title: MLE for exponential distribution


答:
System of equations give the below matrix:

$\left(\begin{array}{rrr|r}
p & 1 & 1 & 1\\
1 & 2 & 4 & t\\
1 & 4 & 10 & t^2
\end{array}\right)$

We set third row as $(R_3 - 2 \cdot R_2)$ and second row as $(R_2 - 2 \cdot R_1)$.

$\left(\begin{array}{rrr|r}
p & 1 & 1 & 1\\
1-2p & 0 & 2 & t-2\\
-1 & 0 & 2 & t^2-2t
\end{array}\right)$

If $p = 1$, $t$ must be either $1$ or $2$ for row $2$ and $3$ to be consistent. Also note that it leads to only two equations in $3$ variables.

So we have no solutions, if $p = 1, t \notin (1,2)$

If $p = 1, t \in (1,2)$, we have infinite solutions

If $p \ne 1$, we have unique solution.


问:
How to go from the more intuitive Present Value expression to the more general and concise one?
I am studying finance and I come across some formulas that I've been trying to get some intuition on. Here we have an equality of (simplified and applied to 3 periods without the Cash flow part) PV expressions:

$$\dfrac{1}{1+x}+\dfrac{1}{\left( 1+x\right) ^{2}}+\dfrac{1}{\left( 1+x\right) ^{3}}=\dfrac{1}{x}\left( 1-\dfrac{1}{\left( 1+x\right) ^{3}}\right)$$

I know that they are equal, but after several hours trying to go from the left expression to the right one I have not managed to do it.

Can someone walk me through (or point to) the steps needed to arrive at the right expression?

答:
If you let $r=\frac1{1+x}$, the LHS is $r+r^2+r^3$.

Notice that $1-r=1-\frac1{1+x}=\frac{x}{1+x}=rx$.

Hence the right hand side should be $$r\cdot \frac{1-r^3}{1-r}=r\cdot \frac{1-r^3}{rx}=\frac{1-r^3}{x}$$


问:
Is there a difference between rejecting a null hypothesis and not accepting a null hypothesis?
I understand that we never say that the null hypothesis is accepted, instead we say that the null hypothesis is not rejected since we can never prove an effect does not exist through empirical evidence.

But does the opposite hold? Is there a difference between rejecting the null hypothesis and not accepting the null hypothesis? In the same sense, how can we prove that an effect does exist? Surely we can only show that there is a low probability, say at 1% significance level, that an effect does not exist and thus we can only not accept it?

答:
As a practical matter, there is no difference.  It's "just" a semantic game.

On the other hand, for purposes of deeper understanding, the important philosophical idea is that we never "accept" anything.  All of our knowledge is provisional, and has a degree of uncertainty.

We can certainly reject a hypothesis, as in, "well, that doesn't seem to have worked, let's try something else."  But we don't want to accept something as true because it's a slippery slope to forgetting to doubt it or that it might not be true after all.


问:
If A is an n×n invertible symmetric matrix, then (A^(-1) )^k is symmetric, for any positive integer k.
i have this t/f question with prove required.

If $A$ is an $n×n$ invertible symmetric matrix, then $(A^-$$^1 )^k$ is symmetric, for any positive integer $k$.

答:
Firstly, note that the inverse of a symmetric matrix is also symmetric. I.e. if $A$ is an invertible symmetric matrix, then $A^{-1} = {A^{-1}}^T$. (Is the inverse of a symmetric matrix also symmetric?)

Thus, the question can be simplified to:

If a matrix $A$ is invertible and symmetric, are all of its exponents $A^k$ also symmetric?

The answer to this question is yes as well. You can see this through induction:

For the base case, we have $AA = A^T A^T = (AA)^T$

Now, assume $A^k = (A^k)^T$.

Then $A^{k+1} = A^k A = (A^k)^T A^T = (A A^k)^T = (A^{k+1})^T$

Q.E.D

Thus, the statement

If A is an n×n invertible symmetric matrix, then (A−1)k is symmetric, for any positive integer k.

is True.


问:
Proving that for all numbers $a, b\in\mathbb{R}$, $\min \{a, b\} \le (a+b)/2$
Can anyone help me solve the two cases that are derived from this problem?</p>

答:
$$
\begin{eqnarray}
\text{min}(a, b)&\le& a\\
\text{min}(a, b)&\le& b\\
2 \text{min}(a, b)&\le& a + b\\
\text{min}(a, b)&\le& \frac{a + b}{2}.
\end{eqnarray}
$$</p>


问:
Proving Sard's theorem
Theorem: (Sard) Let $f:U\to \mathbb R^p$ be a smooth map with $U$ open in $\mathbb R^n$, and let $C$ be the set of all critical points of $f$. Then $m([f(C)]=0$ where $m$ is the Lebesgue measure of $\mathbb R^p$.

A proof goes by induction on $n\leq 0$, while $p>0$. The base case is obvious. Let $C_k$ denote the elements of $U$ for which the partial derivatives of $f$ up to the $k$-th order vanish at the said element. We may brake the proof in three steps. We will be focusing on the first step.

1) $m([f(C-C_1)])=0$

2) $m([f(C_k-C_{k+1})])=0, \quad k>0$

3) $m([f(C_k)])=0, \quad\forall i>n_0$, for some natural number $n_0$.

Step 1) Assume $p>1$, since $C=C_1$ when $p=1$. For each $y\in C-C_1$ we will find a an open neighborhood $V$ containing $y$ for which $m([f(V\cap C)])=0$.

Up to an isometry, $y\not\in C_1\Rightarrow \frac{\partial f_1}{\partial x_1}(y)\not=0$. Consider $h(x):=(f_1(x),x_2,x_3,\dots,x_n)$. Since $dh_y$ is nonsingular, $h$ maps some neighborhood of $y$ diffeomorphicaly onto an open set $V'$. The composition $g=f\circ h^{-1}$ maps $V'$ to $\mathbb R^p$, etc.

Questions: How can we see that $g(t,x_2,\dots,x_n)\in t\times\mathbb R^{p-1}$, for each $(t,x_2,\dots,x_n)\in V'$? How can we conclude from the preceding that $g$ caries hyperplanes to hyperplanes?

Comment: Obviously, the first component of $g(t,x_2,\dots,x_n)$ is $t$, but there are many ways to include $\mathbb R^{p-1}$ in $\mathbb R^p$. I guess that the meaning of the statement is that some suitable inclusion exists but it varies with $t$.

Continuing: A point is critical for $g^t:=g|_{t\times\mathbb R^{n-1}\cap V'}$ if and only if it is critical for $g$. Therefore, the set $C'$ of critical values of $g$ intersects intersects each hyperplane $t\times\mathbb R^{p-1}$ in a set of measure zero. By a theorem of Fubini, $m([g(C')])=0$ but $f(V\cap C)=g(C')$. Hence, $m([f(C-C_1)])=0$ because it is a countable union of sets of measure zero. q.e.d.

Questions: Again many ways exist to restict $g$ to a hyperplane with a fixed first coordinate. Which restriction may we mean? (I think that we must mean every possible restriction.)

Moreover, the theorem of Fubini states that if a set $A\subset\mathbb R^n=\mathbb R\times\mathbb R^{n-1}$ intersects each hyperplane $\text{const}\times\mathbb R^{n-1}$ in a set of measure zero, then $A$ must also have measure zero. It seems that we have not satisfied the conditions of the theorem. In order to satisfy them we need to argue about every possible hyperplane.

答:
In John Milnor's book, "Topology from the Differentiable Viewpoint", there is a proof of Sard's theorem.

We have a map $g:f\circ h^{-1}$

Where $f:U \to R^p$ where U is an open set in $R^n$

$h:U\to R^n$ where $h(x)=(f_1(x),x_2,...,x_n)$

$h$ is a diffeomorphism which means its a bijection

For each point $(t,x_2,...x_n)\in V'$,

Let $h^{-1}:V' \to R^n$ where $h^{-1} = (t,x_2,...,x_n)$

Then $f(t,x_2,...,x_n)=(t,x_2,...,x_{p-1})$ which is a hyperplane

Therefore the composition $g:f\circ h^{-1}$ maps hyperplanes to hyperplanes

Title: Proving Sard's theorem


问:
Probability of stopping smoking
thank you for taking the time to read my question.

Suppose a smoker decides to stop smoking. We want to model using a recursive relation, the probability of him smoking on the $n$-th day, given $p_1$, probability of him smoking the first day.

If he doesn't smoke on the $n$-th day, the next day the probability of him not smoking is $90\%$

If he smokes on the $n$-th day, probability of him not smoking the next day is $20\%$

Questions:

What is the correct way to model this "game", and is it correct to assume it's a Markov chain with 2 states?

What I tried that netted different answers: (don't know which one is the correct one)

I got $
T=\left[\begin{array}{ll}
0.9 & 0.1 \\
0.2 & 0.8
\end{array}\right]
$ as a transition matrix, $Q$ is the characteristic polynomial.

$$Q(T) = T² - 1.7T +0.7I_2$$

therefore if we multiply by $P_n-2: P_n = 1.7P_{n-1} - 0.7P_{n-2}$

Total probability theorem:
$$p_{n+1} = 0.8p_n + 0.1(1-p_n)\implies p_{n+1}=.7p_n+.1$$

答:
I get the Transition Matrix as following,
$$
P=\left[\begin{array}{ll}
0.9 & 0.1 \\
0.2 & 0.8
\end{array}\right]
$$
I think we can get the Stationary distribution
$$
\pi \mathbf{P}=\pi
$$
By comparing this definition with that of an eigenvector we see that the two concepts are related and that
$$
\pi=\frac{e}{\sum_{i} e_{i}}
$$
is a normalized$\left(\sum_{i} \pi_{i}=1\right) $ multiple of a left eigenvector $e$ of the transition matrix  $\mathbf{P} \text { with an eigenvalue of } 1 $.
Then we get $\pi = [\frac 23,\,\frac 13]$

i.e. The probability of successful quitting is $\frac 23$ , the contrary is $\frac 13$ after sufficent large $n$ days aprroximately

Title: Probability of stopping smoking


问:
How many five digit numbers can be formed using digits $1,1,2,3,3,4,4$
With digits $1,1,2,3,3,4,4$, how many five digit numbers we can form?

$1)\frac34\times5!\qquad\qquad2)\frac94\times5!\qquad\qquad3)4\times5!\qquad\qquad4)\frac52\times6!$

Ok so the digits $1,3,4$ appears twice and $2$ appears once. I tried to count different cases:

first: having all $1,2,3,4$ digits and choose another digit from $1,3,4$:
$${3\choose1}\times\frac{5!}{2!}$$
second: having two pair of same digit and choose another digit:
$${3\choose2}\times2\times\frac{5!}{2!2!}$$
Summing them we have $3\times\dfrac{5!}{2}+3\times\dfrac{5!}{2}=5!\times3$ but I don't have this in the options.

答:
We have two multiplicity patterns possible for numbers given the digits available:

$(2,1,1,1) \approx wwxyz$ (A) and $(2,2,1)\approx wwxxy$ (B).

Pattern A can be filled in $\binom 31 \binom 33 = 3$ ways and permuted in $\binom{5}{2,1,1,1}$ $ = 5!/2! = 60$ ways

Pattern B can be filled in $\binom 32 \binom 21 = 6$ ways and permuted in $\binom{5}{2,2,1} = 5!/(2!2!) = 30$ ways

So $180+180 = 360$ options altogether which matches your revised result but again doesn't match any given options.


问:
Proof Involving Fourier Basis Functions

Given the equations $$u(x,t) = \sum_{k \in \mathbb{Z}} = \hat{u}(k,t) \exp(i \pi k x), \hspace{5mm} \hat{u}(k,t) = \frac{1}{2} \int_{-1}^{1}u(x,t) \exp(-i \pi k x)dx,$$ where $i$ is the imaginary unit. Due to the orthogonality and completeness of the Fourier basis functions, we have $$u_t(x,t) = \sum_{k \in \mathbb{Z}} \hat{u}_t(k,t) \exp(i \pi k x) = \sum_{k \in \mathbb{Z}} \exp(-i \pi k x(1 + \pi^2) t + i \pi^3 k^3t) \hat{u}(k,t) \exp(i \pi k x),$$ and $$ u(x,t) = \sum_{k \in \mathbb{Z}} \hat{u}(k,0) \exp(-i \pi k(1 + \pi^2)t + i \pi^3 k^3 t) \exp(i \pi k x) $$ Prove that for any time $ t > 0$, $$\lVert u(x,t) \rVert := \sqrt{\frac{1}{2} \int_{-1}^1 |u(x,t)|^2 dx} = \sqrt{\frac{1}{2} \int_{-1}^1 |u(x,0)|^2 dx} $$

My attempt: I begin by trying to show that $|u(x,t)|^2 = |u(x,0)|^2$ by using that $$|u(x,t)|^2 = u(x,t)^2 = \bigg ( \sum_{k \in \mathbb{Z}} \hat{u}(k,0) \exp(-i \pi k(1 + \pi^2)t + i \pi^3 k^3 t) \exp(i \pi k x) \bigg )^2$$

Then inserting for $\hat{u}(k,0)$ gives $$u(x,t)^2 = \bigg ( \sum_{k \in \mathbb{Z}} \frac{1}{2} \int_{-1}^1 u(x,0) \exp(-i \pi k(1 + \pi^2)t + i \pi^3 k^3 t) \exp(i \pi k x) \exp(-i \pi k x) \bigg ) ^2$$

$$= \bigg ( \sum_{k \in \mathbb{Z}} \frac{1}{2} \int_{-1}^1 u(x,0) \exp(-i \pi k(1 + \pi^2)t + i \pi^3 k^3 t) \bigg )^2$$

But here Im stuck and can't really progress further. Any help would be appreciated.

答:
We will make use of the complex variables identity that $|f(x)|^2=f(x)\overline{f(x)}$. If you don't want to cite Parseval's Theorem/identity, we have the following calculations.

First note that:

\begin{equation}
\begin{split}
&\int_{-1}^1|u(x,t)|^2dx \\
&=\int_{-1}^1 u(x,t)\overline{u(x,t)}dx
\\&=\int_{-1}^1\Big{(}\sum_{k\in \mathbb{Z}}\hat{u}(k,t)\exp(i\pi kx)\Big{)}\overline{\Big{(}\sum_{m\in \mathbb{Z}}\hat{u}(m,t)\exp(i\pi mx)\Big{)}}dx\\
&=\int_{-1}^1\Big{(}\sum_{k\in \mathbb{Z}}\hat{u}(k,t)\exp(i\pi kx)\Big{)}\Big{(}\sum_{m\in \mathbb{Z}}\overline{\hat{u}(m,t)\exp(i\pi mx)}\Big{)}dx\\
&=\int_{-1}^1\Big{(}\sum_{k\in \mathbb{Z}}\hat{u}(k,t)\exp(i\pi kx)\Big{)}\Big{(}\sum_{m\in \mathbb{Z}}\overline{\hat{u}(m,t)}\exp(-i\pi mx)\Big{)}dx\\
&=\int_{-1}^1\Big{(}\sum_{k=m, k\in \mathbb{Z}}\hat{u}(k,t)\overline{\hat{u}(k,t)}\exp(i\pi k x)\exp(-i\pi m x)+\sum_{k\neq m;\\ m,k\in \mathbb{Z}}\hat{u}(k,t)\overline{\hat{u}(k,t)}\exp(i\pi k x)\exp(-i\pi m x)\Big{)}dx\\
&\text{By orthogonality, the summation on the right hand side vanishes}\Rightarrow\\
&=\int_{-1}^1\sum_{k=m, k\in \mathbb{Z}}\hat{u}(k,t)\overline{\hat{u}(k,t)}\exp(i\pi k x)\exp(-i\pi m x)dx\,\,\text{Since $k=m$, this simplifies}\\
&=\int_{-1}^1\sum_{k=m, k\in \mathbb{Z}}|\hat{u}(k,t)|^2dx=2\sum_{k=m, k\in \mathbb{Z}}|\hat{u}(k,t)|^2
\end{split}
\end{equation}

Conversely from the formula you provided, we have that:
\begin{equation}
\begin{split}
&\int_{-1}^1|u(x,t)|^2dx\\
&=\int_{-1}^1 u(x,t)\overline{u(x,t)}dx\\
&=\int_{-1}^1\Big{(}\sum_{k\in \mathbb{Z}}\hat{u}(k,0)\exp(-i\pi k(1+\pi^2)t+i\pi^3k^3t)\exp(i\pi k x)\Big{)}\overline{\Big{(}\sum_{m\in \mathbb{Z}}\hat{u}(m,0)\exp(-i\pi m(1+\pi^2)t+i\pi^3m^3t)\exp(i\pi m x)\Big{)}}dx\\
& \text{Following the same steps as the above by using orthogonality, we arrive at:}\\
&=\int_{-1}^1\sum_{k=m, k\in \mathbb{Z}}|\hat{u}(k,0)|^2dx=2\sum_{k=m, k\in \mathbb{Z}}|\hat{u}(k,0)|^2
\end{split}
\end{equation}
Therefore, since the first equation holds for all t, we get that $\int_{-1}^1|u(x,t)|^2dx=\int_{-1}^1|u(x,0)|^2dx$ as desired.

Title: Proof Involving Fourier Basis Functions


问:
First passage time in Brownian motion
Let $X_t,\, t\geqslant 0,$ be a Brownian motion and consider the stopping times $T_a := \inf \{t \mid X_t = a\}$. Find the probability $\mathbb P\{T_{2}< T_{-1} < T_{3}\}$, for instance.

So we have two events $\{T_2 < T_{-1}\}$ and $\{T_{-1}< T_3\}$. Separately, the probabilities are clear. My intuition says that we can multiply the probabilites for the initial problem, but I'm not really satisfied with this intuitive mumbo-jumbo. What if there are some funny cases, when it doesn't work..

Yet, I have no idea how to formally explain this.

答:
They are definitely not independent.

You didn't specify the starting value for the Brownian motion, but let's say it's 0.  On the event $\{T_3 < T_{-1}\}$, where the Brownian motion reaches 3 before -1, it must have passed through the value 2 even earlier, by continuity.  So we have $\{T_2 < T_{-1}\} \subset \{T_3 < T_{-1}\}$.  As such, $\{T_2 < T_{-1}\}$ cannot be independent of $\{T_3 < T_{-1}\}$, unless they both had probability 0 or 1 which is not the case.  So $\{T_2 < T_{-1}\}$ is also not independent of $\{T_3 < T_{-1}\}^c = \{T_{-1} < T_3\}$.

(This same argument works if the starting point is anything less than 2.  If it's greater or equal to 2, then $P(T_2 < T_{-1}) = 1$ and the events are trivially independent.)

You do end up multiplying probabilities, but not because of independence per se.  The strong Markov property yields a certain conditional independence statement, if you like, but you have to be very careful to frame it the right way.

Title: First passage time in Brownian motion


问:
$f(t)^2<1+2\int_{0}^{t}f(s)\:ds$ Then prove that $f(t)<1+t$
Given that $f:[0,1]\to [0, \infty)$ be a continuous function such that
$$f(t)^2<1+2\int_{0}^{t}f(s)\:ds,\:\forall t \in [0,1] \to (1)$$
Then prove that $$f(t)<1+t ,\forall t \in [0,1]$$

I actually got the solution in this way:
Differentiating $(1)$ both sides with respect to $t$ and using Leibnitz rule we have
$$2f(t)f'(t)<2f(t) \to (2)$$
$\implies$
$$2f(t)\left(f'(t)-1\right)<0$$
Now since the range of $f(t)$ cannot be negative,we get
$$f'(t)-1<0$$
$\implies$
$f(t)-t$ is a strictly decreasing function in $t \in [0,1]$
So we get
$$f(t)-t<f(0)-0 \to (3)$$
Now in $(1)$ plugging in $t=0$ we get
$$f(0)^2<1 \implies f(0)<1$$
So using $f(0)<1$ in $(3)$ we get
$$f(t)-t<1 \implies f(t)<1+t, \forall t \in [0,1]$$
Now my query is in getting $(2)$. After differentiation how can we be sure that inequality is retained. Am i missing any concept there?

答:
Let $g(t)=\int_{0}^{t}f(s)ds$ and  $h(t)=\sqrt{1+2g(t)}-t-1,$ h is a continuously differentiable function.

I observe that
$$h'(t)=\frac{g'(t)}{\sqrt{1+2g(t)}}-1<0.$$
So, $$h(t)\leq h(0)=0\Rightarrow 1+2g(t)\leq (1+t)^2$$ and as a consequence
$$f(t)<1+t.$$


问:
Relationship between probability and odds
While I'm reading Weird Maths: At the Edge of Infinity and Beyond by David Darling and Agnijo Banerjee, I found this part that I can't understand.

Of course,  it’s extremely unlikely that you’ll win the jackpot twice.
But when considering the probability that someone will,  you need to
multiply the odds by the number of people  who play the lottery, which
greatly reduces those odds, as  well as the number of ways they can
win the lottery twice  (approximately half the square of the number of
times  they play the lottery). After all of this, the odds look much
more reasonable that someone, somewhere will scoop the  jackpot twice.

I know the odds is p/(1-p), but still don't understand why the authors did this multiplication.

If you need more text, you can find here.

答:
I don't think working with odds instead of probabilities improves understanding in this problem, so here is an answer using probabilities.

Suppose we have $n$ people playing the lottery, with each having a probability $p$ of winning, independently of all the other players.  Then the total number of winners has a Binomial distribution with parameters $n$ and $p$, so the probability that there will be exactly two winners is
$$\binom{n}{2} p^2 (1-p)^{n-2}$$
Now
$$\binom{n}{2} = \frac{n(n-1)}{2}$$
so for large $n$,
$$\binom{n}{2} \approx \frac{n^2}{2}$$
Also, for large $n$ and small $p$, $(1-p)^{n-2} \approx 1$.

Combining these approximations, we see the probability of having exactly two winners is approximately
$$\frac{n^2 p^2}{2}$$
which seems to be what the quoted text is getting at.


问:
Diophantine equation $2^x-3^y=2021$
$$2^x-3^y=2021$$
where $x,y$ are non-negative integers. I only found $2^{11}-3^3=2021$.

答:
According to Pillai, the equation $2^x-3^y=d$ has at most one solution except for $d=-1,5,13$. Since we have a solution for $d=2021$, it is the only solution. For a proof by elementary methods see the reference below (the corollary to Theorem $4$).

Reference: R. Scott, On the Equations $p^x-b^y=c$ and $a^x+b^y=c^z$, J. Number Theory 44 (2) (1993), 153–165.

Title: Diophantine equation $2^x-3^y=2021$


问:
Suppose 7 people are sitting at a round table. If three people are selecting at random, find the probability that at least two of then are together.
First we find the probability that all three sit together. The string method can be used for this. So
$$P(3)=\frac{4! 3!}{6!} =\frac 15$$

Now for finding probability that two people are together and one of them of is separate. I selected any two people out of the $3$ selected people and string them. Then number of ways to do that should be
$\binom{3}{2} 2!$

Now if this string is at any one position, then only $4$ positions are left to seat the third person. So number of ways should be $\binom{4}{1} 4!$

But multiplying all this gives a value more than 6!, which shouldn’t happen. What am I doing wrong?

I also though about using complementary method, but I didn’t know how to apply it for the circular permutation. So can I get a hint on that too?

Thanks

答:
WLOG, one of the $3$ to be chosen (or choosing seats) can be $A$ at the "head" of the table, and we can choose the remaining two in $\binom62  = 15$ unrestricted ways

If we linearize the table, it, it would look like $\require{cancel}\mathbf A \,\xcancel{:}\, \circ\, \uparrow \circ\, \uparrow\circ\, \uparrow\circ\,\xcancel{:}\,\mathbf A$

with crosses giving barred positions, and uparrows giving allowed  "apart" positions  for the other $2$ to be chosen as $\binom32 = 3$

Thus P(at least two chosen are adjacent) $= 1- \frac3{15} = \frac 45$


问:
the limit of Poisson CDF with diverging rate parameter
The CDF of a Poisson distribution with rate parameter $\lambda$ is
$$
P(n;\lambda)=\sum_{k=0}^n \frac{\lambda^ke^{-\lambda}}{k!}.
$$
As $n$ goes to infinity, the CDF would certainly approach 1.

Now, consider the case when the rate parameter is $\xi n$ with $\xi\in(0,1)$ being a given constant.
According to the last answer in Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$, the following limit should be 1:
$$
\lim_{n\to\infty}P(n;\xi n)=\lim_{n\to\infty} \sum_{k=0}^n\frac{(n\xi)^ke^{-n\xi}}{k!}=1.
$$
Is it possible to check the convergence rate of the above limit?
More specifically, I wonder whether the convergence rate is faster than $n$, i.e., whether $n[1-P(n;\xi n)]=O(1)$?

I checked using software that the convergence rate is faster than $n$, but I don't know how to show it rigorously. Can anyone provide some hints and insights? Thanks!

Body: $$\sum_{k=0}^n\frac{n^k\,e^{-n}}{k!}=\frac{\Gamma (n+1,n)}{\Gamma (n+1)} \to \frac 12$$


问:
How to find mode when modal class is first or last class?
We know that formula of finding mode of grouped data is

Mode = $l+\frac{(f_1-f_0)}{(2f_1-f_0-f_2)}\cdot h$

Where, $f_0$ is frequency of the class preceding the modal class and $f_2$ is frequency of the class succeeding the modal class. But how to calculate mode when there is no class preceding or succeeding the modal class.

答:
we can take their value as 0. The frequency of the succeeding model class is taken as 0 if model class is the last observation.

You can also check it from the equation as-

$l =$ lower limit of the modal class,

$h =$ size of the class interval (assuming all class sizes to be equal),

$f_1 =$ frequency of the modal class,

$f_0 =$ frequency of the class preceding the modal class,

$f_2 =$ frequency of the class succeeding the modal class.

Even if $f_2$ is $0$, the mode can be easily found by using the above expression.


问:
Is this solution correct? (Exercise 1.1.19 in Hatcher's AT)
Exercise 19 on the page 39 in Hatcher's AT says:

Exercise 19. Show that if $X$ is path-connected $1$-dimensional CW complex with basepoint $x_0$ a $0$-cell, then every loop in $X$ is homotopic to a loop consisting of a finite sequence of edges traversed monotonically. [See the proof of Lemma 1.15.]

The proof of Lemma 1.15 says:

Proof: Given a loop $f:I\to X$ at the basepoint $x_0$, we claim that there is a partition $0=s_0<s_1<\cdots <s_m=1$ of $I$ such that each subinterval $[s_{i-1},\,s_i]$ is mapped by $f$ to a single $A_\alpha$. (......the rest omitted)

(In the lemma, the $A_\alpha$'s are assumed to form an open cover of $X$.)

And here is my solution to Exercise 19:

It is enough to solve the problem with an additional assumption that $X$ is a finite CW complex, since the compact set $f(I)$ is contained in a finite subcomplex of $X$.

Let $x$ be a $0$-cell in $X$. The union $N_x$ of $\{x\}$ and all the $1$-cells attached to $x$ is open in $X$. To be sure about this, add a $0$-cell in the middle of each edge of $X$. (Doing this is suggested by JasonDeVito.) This makes no change in the situation. All such $N_x$ forms an open cover of $f(I)$.

Like in the Proof of Lemma 1.15, we can find a partition $0=s_0<\cdots <s_n=1$ of $I$ such that each image $f([s_{i-1},\,s_i])$ of subinterval is contained in an $N_x$. Let $f_i:I\to X$ be the linear reparametrization of $f$ restricted to the subinterval $[s_{i-1},\,s_i]$. Then $f$ is path-homotopic to $f_1 \cdot ... \cdot f_n$.

There is a unique way to connect two points $f(s_{i-1})$ and $f(s_i)$ by a monotonic path $g_i$ in $N_x$, up to reparametriation. This is due to the topology of $N_x$, which is homeomorphic to the finite number of half-closed intervals whose closed endpoints identified. Again by this topology on $N_x$, $f_i$ and $g_i$ are path-homotopic to each other since they have same endpoints. Therefore $f$ is path-homotopic to $g_1 \cdot ... \cdot g_n$. The end.

Is my solution correct? I think it is, but want to be sure.

答:
Your proof has a nice idea, but it suffers a little from the vague meaning of traversing something monotonically.

Note that in your proof traversing monotonically can also mean constant (consider the case that $f(s_{i-1}) = f(s_i)$). By the way, Hatcher's statement has the same problem: Consider the constant loop at $x_0$ which traverses $0$ edges. However, this can be argued away by saying that every loop in $X$ is homotopic to a finite product of paths traversing a single edge monotonically and defining the empty product to be the constant loop.

Here are some suggestions for clarification.

In Hatcher's statement an edge is closed $1$-cell $\bar e_\alpha$, i.e. the image of a characteristic map $\phi_\alpha : [0,1] \to X$. I think we should regard a path traversing $\bar e_\alpha$ monotonically as either the path $\phi_\alpha$ or the path $-\phi_\alpha$ travelled in opposite direction. Or, more generally, as any path obtained from $\phi_\alpha$ or $-\phi_\alpha$ via reparameterization.

You do not need to introduce additional $0$-cells in the $CW$-structure of $X$. Define your open cover $\mathcal U$ for applying Lemma 1.15 (by the way, this only occurs in the edition from 2001) as follows: It contains all open $1$-cells $e_\alpha$ (which are open in $X$ because $\dim X = 1$). Pick a point $y_\alpha \in e_\alpha$ and for each $0$-cell $x$ let $N_x$ be the union of all $C_x(\bar e_\alpha \setminus \{y_\alpha\})$ with $x \in \bar e_\alpha$ and $C_x(\bar e_\alpha \setminus \{y_\alpha\})$ being the component of $\bar e_\alpha \setminus \{y_\alpha\}$ containing $x$. Add these $N_x$ to $\mathcal U$.

Each $\bar e_\alpha \setminus \{y_\alpha\}$ is either homeomorphic to a half-open interval $[0,1)$ with $x$ corresonding to $0$  or to an open interval $(-1,1)$ with $x$ corresonding to $0$. Hence $N_x$ is homeomorphic to the one-point union $V_x$ of half-open intervals $[0,1)$ glued together at the point $0$. In particular $N_x$ is contractible. A monotonic path in $N_x$ connecting two points $a, b \in N_x$ corresponds under $h :  N_x \stackrel{\approx}{\rightarrow} V_x$ to a path in $V_x$ connecting $h(a)$ and $h(b)$ as follows: If $h(a),h(b)$ belong to the same copy of $[0,1)$, then take the linear path in $[0,1)$ (which is constant if $h(a) = h(b)$) and if $h(a),h(b)$ belong to distinct copies of $[0,1)$, then take the linear path in the first copy from $h(a)$ to $0$ followed by the linear path in the second copy from $0$ to $h(b)$. A monotonic path in $e_\alpha \approx (0,1)$ is defined similarly (it is even simpler than above).

Now proceed as in your proof: For an adequate partition of $[0,1]$ each $f_i = f \mid_{[s_{i-1},s_i]}$ maps into a member $U$ of $\mathcal U$ (which is contractible), hence it is homotopic in $U$ to the monotonic path $g_i$ on $[s_{i-1},s_i]$ having the same endpoints as $f_i$. Hence $f$ is homotopic to the product $g$ of the $g_i$. But this does not suffice to prove Hatcher's statement, and that is a big gap in your approach. Note that some $g_i$ may be constant and some $g_i$ may travel back in an $e_\alpha$ from $g_{i-1}(s_{i-1})$ in the direction of $g_{i-1}(s_{i-2})$.

To treat such cases, let us first observe that constant $g_i$ are irrelevant, they can be omitted from the product without changing the homotopy class of $g$. So let us assume that no $g_i$ is constant. Next observe that any $0$-simplex $x$ can be met at most once by $g$ (this can only happen if $f_i$ maps into $N_x$, and then the homotopic $g_i$ goes at most one trough $x$). So w.lo.g. we may assume that if $x \in \text{im}(g)$, then $x = g(s_i)$ for some $i$ (if necessary, add partition points). Let $g^{-1}(X^0) = \{s_{i_0},\dots,s_{i_r}\}$ with $0= i_0 < i_1 < \dots < i_r = n$. Then $g((s_{i_{l-1}},s_{i_l}))$ is a connected subset of $X \setminus X^0$ and must therefore be contained in some open $1$-cell $e_\alpha$. The correponding edge (= closed $1$-cell)  $\bar e_\alpha$ is a circle if $g(s_{i_{l-1}}) = g(s_{i_l}) = x$ and a closed interval if $x = g(s_{i_{l-1}}) \ne g(s_{i_l}) = x'$. The $g_k$ with $s_{i_{l-1}} < k < s_{i_l}$ may travel forwards and backwards, i.e. $g_l = g \mid_{[s_{i_{l-1}},s_{i_l}]}$ need not be traversing  $\bar e_\alpha$ monotonically. But if  $\bar e_\alpha$ is a closed interval, then $g_l$ will be homotopic to the path travelling monotonically from $x$ to $x'$. Similarly,  if  $\bar e_\alpha$ is a circle,  then $g_l$ will be homotopic to either the constant loop or one of $\phi_\alpha$, $-\phi_\alpha$ (note that it cannot have a winding number $\ne \pm 1$  because that would require that $g_l^{-1}(x)$ contains more points than $s_{i_{l-1}},s_{i_l}$).


问:
proof: $l_i: a_i\times (x-x_i)=0$ (i=1, 2) intersects at exactly one point
given: $a_i, x_i\in R^3$, $a_1\times a_2 \neq 0$ and $det(a_1, a_2, x_2-x_1)$=0

prove: $l_i: a_i\times (x-x_i)=0$ (i=1, 2) intersects at exactly one point

答: $a_i\times (x-x_i)=0$ is a line through $x_i$ in the direction of $a_i$. The condition $det (a_1,a_2,x_2-x_1)=0$ means that $a_1,a_2,x_2-x_1$ lie on a plane, therefore both lines lie on that plane. Since $a_1\times a_2\neq 0$, the lines are not parallel so they intersect at exactly  one point.


问:
Why proofs which provide algorithms for construction are valid?
10 days ago I asked question about validity of a proof (Is this a valid proof of the fact that a subspace of an $n$-dimensional vector space has a basis?), but as I later realised I didn't explain well what I really wanted to know and this resulted in the answers that didn′t quite answered my real question, so I will try to rephrase it. So in that previous post there was a proof of the fact that any subspace of finite-dimensional vector space has a finite basis. In it I basically constructed this basis by adding vector if the current set of vectors is not maximal linear independent set. Then I said that this "algorithm" will terminate because of the fact that the vector space is finite dimensional. But my concern about this proof is "Why algorithm for constructing/finding an object is a valid proof of its existence ?". Such question may sound strange, but my concern is the fact that you can′t say by reading a proof how many steps execution of such algorithm will take, because it depends on what vector space and what subspace you take. So basically we don't know how many steps the proof has and it seems kind of strange to me. And this concern is not just about proof of this fact about subspaces, but about any proof which proves existence of an object with an algorithm which has number of steps which depends on something. For example, proof of the fact that any tree has a vertice of degree 1, where you take any vertice and then go along the edges of the tree until you find the vertice of degree 1 (which you will, because tree has only finite amount of vertices and tree is acyclic).

So why are proofs, which provide algorithms for constructing an object, valid and can we formalise (maybe using set theory) them in such a way that number of steps of a proof will be specific ?

答:
Any sort of "iterative process" is really just an informal way of describing a proof by induction.  For instance, in this case, you could formulate your argument as a proof by induction on $j$ that for each $j\in\mathbb{N}$ either $M$ has a basis or there exists a linearly independent subset of $M$ of size $j$.  The induction step of this argument is just exactly what you do in step $j$ of your argument: either your linearly independent subset of size $j$ (which you have by the induction hypothesis) is already a basis, or else you can enlarge it to a linearly independent subset of size $j+1$.  The idea that the algorithm must eventually stop can then be described as the statement that when you take $j=n+1$, it is not possible to have a linearly independent subset of size $j$, so there must exist a basis for $M$.

Or, you could more directly mimic your proof by defining your basis recursively (though this of course is ultimately again a proof by induction, since that's how you prove that recursive definitions make sense, and the most naive formulation of the recursive definition would also use the axiom of choice).  Namely, you recursively define a sequence of sets $(A_j)_{j\in\mathbb{N}}$, where your "step $j$" just tells you how to define $A_{j+1}$ from $A_j$.  Note that in the case that $A_j$ is already a basis, you can simply define $A_{j+1}=A_j$, so that you don't have to worry about not knowing how long the recursion will be.


问:
Square Root Time Complexity
I'm deeply sorry for this, I don't know if it's the appropriate place to post this but I just can't figure out what's the time complexity of this algorithm:

```<code>int fun(int n)
{
    int j = 10;
    while( j < n)
        j+= sqrt(j);
    return j;
}
</code>```

I know that at the $k^{th}$ iteration we have $f(k)=f(k-1)+\sqrt{f(k-1)}$, with $f(0)=10$, I tried solving this function, but I couldn't find any way to do so.

答: Are you looking for the time complexity or the value of the function?  You add at least $3$ every cycle.  The number of loops is then less than or equal to $\frac 13(n-10)+1$ so the time complexity is $O(n)$

Title: Square Root Time Complexity


答:
The recursion $$p_n=1.7p_{n-1}-.7p_{n-2}$$ can be solved in closed form.  The characteristic polynomial has roots $.7, 1$  so the solution is $$p_n=A\times .7^n+B$$  for suitable $A,B$.

Now, $p_1$ is the (unspecified) initial probability.  the information we are given implies that $$p_2=.7p_1+.1\implies A\times .7^2+B=A\times .7^2+.7B+.1\implies B=\frac 13$$

Given that form, regardless of the values of $A$, we can easily check that $$.7p_{n-1}+.1=.7\times A\times .7^{n-1}+.7B+.1=A\times .7^n+B=p_n$$

whence the first recursion implies the second.

Should add, if we are just interested in the limiting value, $L$, of $p_n$, we can read it off from the second recursion. We get $$L=.7L+.1\implies L=\frac 13$$  or you can read it off from the closed form, noting that $.7^n\to 0$ for large $n$.

Title: Probability of stopping smoking

Body:
thank you for taking the time to read my question.

Suppose a smoker decides to stop smoking. We want to model using a recursive relation, the probability of him smoking on the $n$-th day, given $p_1$, probability of him smoking the first day.

If he doesn't smoke on the $n$-th day, the next day the probability of him not smoking is $90\%$

If he smokes on the $n$-th day, probability of him not smoking the next day is $20\%$

Questions:

What is the correct way to model this "game", and is it correct to assume it's a Markov chain with 2 states?

What I tried that netted different answers: (don't know which one is the correct one)

I got $
T=\left[\begin{array}{ll}
0.9 & 0.1 \\
0.2 & 0.8
\end{array}\right]
$ as a transition matrix, $Q$ is the characteristic polynomial.

$$Q(T) = T² - 1.7T +0.7I_2$$

therefore if we multiply by $P_n-2: P_n = 1.7P_{n-1} - 0.7P_{n-2}$

Total probability theorem:
$$p_{n+1} = 0.8p_n + 0.1(1-p_n)\implies p_{n+1}=.7p_n+.1$$


问:
Systems of quadratic forms
I've been working on a problem in field theory, specifically showing that $\mathbb{Q}(\alpha)$ has no intermediate fields when where $\alpha$ is a root of $x^4-nx-1$ whenever $n \not \in \{-4,0,4\}$. The catch is we're NOT allowed to use Galois theory which would make the problem pretty trivial (the resolvent cubic is irreducible in this case so the Galois group of the splitting field is either $A_4$ or $S_4$ and in either case any subgroup of index $4$ is maximal so that there can be no intermediate fields).

My initial thought was to look at $\mathbb{Q}[x]/(x^4-nx-1)$ and show that there can be no quadratic subfield by trying to show that there is no solution to
$$(ax^3+bx^2+cx+d)^2 \equiv k \mod x^4-nx-1$$
for some $a,b,c,d,k\in\mathbb{Q}$. After squaring, and reducing, this is equivalent to an integer solution (since we can clear denominators from $a,b,c,d$ before-hand) of the system of quadratic forms
\begin{align*}
2ad + 2bc+a^2n &= 0\\
c^2+2bd + a^2 + 2abn &= 0\\
2ab + 2cd +nb^2 + 2nac &= 0
\end{align*}
which have matrix forms
$$\begin{bmatrix}
n & 0 &0 &1\\
0&0&1&0\\
0&1&0&0\\
1&0&0&0
\end{bmatrix} \quad \quad \begin{bmatrix}
1 & n &0 &0\\
n&0&0&1\\
0&0&1&0\\
0&1&0&0
\end{bmatrix}\quad \quad \begin{bmatrix}
0 & 1 &n &0\\
1&n&0&0\\
n&0&0&1\\
0&0&1&0
\end{bmatrix}.$$

Is there any simple criterion to show that this system has no simultaneous solutions? Any thoughts on how I can proceed?

答:
Too long for a comment. I would proceed differently, since intersection of quadrics are nasty...

If $E=\mathbb{Q}(\alpha)$ has a quadratic subfield $F=\mathbb{Q}(\sqrt{d}),$ where is a squarefree non zero integer, let $\mu=X^2-uX+v$ be the minimal polynomial of $\alpha$ over $F$. Denoting by $*$ the non trivial automorphism of $F$, you have $\mu\mid X^4-nX+1$, and applying $*$, $\mu^*\mid X^4-nX+1$.

Since $X^4-nX+1$ is irreducible, $\mu$ and $\mu^*$ are distinct irreducible polynomials of $F[X]$ (if $\mu=\mu^*,$ then $\mu\in\mathbb{Q}[X]$, which is not possible). Hence $\mu \mu^*\mid X^4-nX+1$, and comparing degrees and leading coefficients shows that $X^4-nX+1=\mu \mu^*=X^4-(u+u^*)X^3+(v+v^* +uu^*)X^2-(u^*v+uv^*)X+vv^*$.

Hence $u^*+u=0, v+v^*+uu^*=0, u^*v+uv^*=n,vv^*=1.$

Thus $u=a\sqrt{d}$ by the first equality. Write $v=b+c\sqrt{d}$. Then $2b-da^2=0$, $-2acd=n$ and $b^2-dc^2=1$.

Hence $d^2 a^4-4dc^2=4$ and $d^3a^6-n^2=4da^2$.

Thus the polynomial $d^3 X^3-4dX-n^2$ has an integer root $x$

It follows that $X^3-4X-n^2$ has an integer root (replace $X$ by $dX$).

Hence we are reduced to prove that $X^3-4X-n^2$ has no integer roots, for any $n\neq 0,\pm 4$. This should be easier than studying the intersection of three quadrics.

For example, if $n$ is coprime to $3$, reducing modulo $3$ shows a contradiction.

Title: Systems of quadratic forms


问:
A question on transfinite induction
The following were 2 problems given to us on transfinite induction, The transfinite induction I saw on books and books etc was using ordinals but the definition we were given is different and I still couldn't find a proper example on how to use it to a question

1) Let $\preceq$ be a well ordering on $[0,1]$ and for $t \in [0,1]$ let $x_t =\sup(B_t \cup \{t\})$ (sup is taken with respect to the usual ordering on real numbers)

where $B_t=\{x\in W : x \prec t\}$

Show that there exists $a \in [0,1]$ such that $x_t =1$ for each $t \succeq a$

2) Let $(W,\preceq)$ be a well ordered set and $t \in W$. if $f \colon W \to W$ is an order preserving bijection show that $f$ is the identity function on $W$.

Definition of the transfinite induction given:
Let  $(W,\preceq)$ be a well ordered set and $A \subseteq W$ and $B_t \subseteq A \implies t \in A$ Then $A=W$
(where $B_t=\{x\in W \mid x \prec t\}$)

Any help would be appreciated

答:
Ordinals are well-ordered sets, and every well-ordered set is isomorphic to a unique ordinal. So the idea of using general well-orders and ordinals is the same.

Let me give you some hint about the second question (because I can understand it without further clarifications).

We define $A=\{t\in W\mid f(t)=t\}$, and we want to show that $A=W$.

Suppose $t\in W$ such that $\{t'\in W\mid t'<t\}\subseteq A$. Then $f(t')=t'$ for all $t'<t$. Let $x=f^{-1}(t)$, if $x<t$ then by the induction assumption $x\in A$, and therefore $f(x)=x<t$. So we have to have $t\leq x$. Because $f$ is order-preserving we have that $t\leq x\rightarrow f(t)\leq f(x)=t$.

Because $f$ is a bijection it is impossible to have $f(t)<t$ (as those already have preimages) and therefore $f(t)=t$.

Title: A question on transfinite induction


问:
A question on resolutions of coherent sheaves.
If $A$ is a commutative unital ring and $E$ is a finite rank projective $A$-module there is a surjective $A$-linear map $\phi: A^n \rightarrow E$, with kernel $F:=ker(\phi)$ and $F\oplus E \cong A^n$ an isomorphism of $A$-modules. We get a short exact sequence of projective finite rank $A$-modules

$$ 0 \rightarrow F \rightarrow A^n \rightarrow E \rightarrow 0 .$$

There is a canonical projection map (an $A$-linear map)

$$ \psi: A^n \rightarrow F$$

with $ker(\psi)=E$. Using a "splicing-technique" we get an infinite exact sequence of free $A$-modules

$$ \cdots \rightarrow A^n \rightarrow A^n \rightarrow A^n \rightarrow E \rightarrow 0.$$

Hence $E$ has an infinite free resolution by projective (and free) $A$-modules of finite rank.

If $k$ is a field and $X \subseteq \mathbb{P}^n_k$ is a regular projective scheme there is for any coherent sheaf $F$ an infinite resolution by finite direct sums of invertible sheaves:

Hartshorne Corr. II.5.18 says: If $X \subseteq \mathbb{P}^n_k$ is projective, it follows any coherent sheaf $F$ is a quotient

$$ f_1:\oplus_{i(1)=1}^{N(1)} \mathcal{O}(n(1)_{i(1)}) \rightarrow F \rightarrow 0.$$

It follows $X:=Proj(k[x_1,..,x_n]/I)$ with $B:=k[x_i]/I$ noetherian. Hence if $F$ is a coherent module it follow any submodule $E \subseteq F$ is coherent. It follows $\oplus \mathcal{O}(n(1)_i)$ is coherent hence $ker(f_1)$ is a coherent module. Hence there is a surjection

$$ f_2: \oplus_{i(2)}^{N(2)} \mathcal{O}(n(2)_{i(2)} \rightarrow ker(f_1)$$

and continuing this process you get a long exact sequence of $\mathcal{O}_X$-modules

$$S1.\text{   }\cdots \rightarrow  \oplus_{i(2)}^{N(2)} \mathcal{O}(n(2)_{i(2)}) \rightarrow  \oplus_{i(1)=1}^{N(1)} \mathcal{O}(n(1)_{i(1)}) \rightarrow F \rightarrow 0.$$

Example: If $X$ is projective space and $F$ a finite rank locally free sheaf it follows the sequence S1 terminates - it is finite. A similar property holds for the grassmannian. As a consequence it follows the Grothendieck group of projective space (and the grassmannian) is generated by classes of invertible sheaves.

General question: Can you give an explicit example of a projective regular scheme $X \subseteq \mathbb{P}^n_k$ and a finite rank locally trivial (and non-trivial) $\mathcal{O}_X$-module $F$ with a resolution S1 that is infinite? Please give an explicit reference to the litterature.

Special case: If $A \subseteq \mathbb{P}^n_K$ is an abelian variety with $K$ a number field - give an explicit example of a finite rank non-trivial vector bundle $F$ on $A$ with a resolution as in S1 that is infinite.

By "non-trivial" I mean: $F$ is not a direct sum of invertible sheaves.

Note: If $k:=\overline{K}$ is the algebraic closure of $K$ and if $A\subseteq \mathbb{P}^n_k$ is an abelian variety, any homogeneous finite rank vector bundle $E$ is on the form

$$ E\cong \oplus_{L\in Pic^0(A)} L\otimes U_L$$

where $U_L$ is an iterated extension of trivial bundles. Hence in the grothendieck group $K_0(A)$ we get

$$ [L\otimes U_L]=[L]*[U_L]$$

and $[U_L]=rk(U_L)[\mathcal{O}_A]$ hence

$$[L\otimes U_L]=[L]rk(U_L)[\mathcal{O}_A]=rk(U_L)[L]$$

and

$$[E]= \oplus_{L\in Pic^0(A)} rk(U_L)[L].$$

Hence for a homogeneous finite rank vector-bundle $E$ it follows the sequence S1 is finite. Hence in this case it follows the class $[E]\in K_0(A)$ lives in the sub group generated by classes of invertible sheaves. I'm looking for explicit examples where this does not hold.

答: Take a smooth 4-dimensional quadric $Q \subset \mathbb{P}^5$ and a plane $\Pi \subset Q$. Then the structure sheaf $\mathcal{O}_\Pi$ of the plane can't have a finite resolution by direct sums of line bundles because the cohomology class $$[\Pi] \in H^4(Q,\mathbb{Z})$$ does not lie in the subring of $H^\bullet(Q,\mathbb{Q})$ generated by divisors.


问:
A Genetic Introduction to Algebraic Geometry
I am a big fan of "genetic introductions" in mathematics, i.e. where the ideas are introduced in the order they were developed along with why they were introduced, as opposed to the "definition-theorem-proof" kind of introductions on the line of Bourbaki.
For Galois Theory, Algebraic Number Theory and Analytic Number Theory, there are Edwards' wonderful books Galois Theory(Springer), Fermat's Last Theorem(Springer) and Riemann's Zeta Function(Dover). I am a big fan of these books, and this style of exposition.

Now I am looking for a similar exposition in Algebriac Geometry. But all the books I know of in that subject area more or less follows the Bourbaki track. Anyone knows about one?

P.S. if you know of this kind of exposition in other streams of mathematics, please let me to know.

答: Fulton's "Algebraic curves" is a good introductory book that requires some knowledge in commutative algebra.</p>


问:
Quadratic expression in $x$ with roots $\frac gh$ and $-\frac hg$
What quadratic expression in $x$ has roots $$\frac{g}{h}\qquad\text{and}\qquad-\frac{h}{g}?$$ I know that this can be factored as $$\left ( x-\frac{g}{h} \right )\left ( x+\frac{h}{g} \right )=0$$ But I'm really confused with the choices. The choices are as follows:

$(a)\hspace{5 mm}g^2 h^2 x^2 -\frac{g^2}{h^2}$

$(b)\hspace{5 mm}hgx^2+(g^2-h^2)x-hg$

$(c)\hspace{5 mm}hgx^2+(h^2-g^2)x+hg$

$(d)\hspace{5 mm}hgx^2+(h^2-g^2)x-hg$

Is the answer (d)?

答:
$\displaystyle\begin{eqnarray}{\bf Hint}\!: &&0 &\, =& \ \ \left (x-\frac{g}{h} \right )\ \ \left ( x+\frac{h}{g} \right ) \\
\stackrel{\large\times\,\color{#0a0}h\color{#c00}g}\Rightarrow && 0 &\,=& \color{#0a0}h\left ( x-\frac{g}{h} \right ) \color{#c00}g\left ( x+\frac{h}{g} \right )\\
            &&   &\,=&\ \  (h\,x -g)\ \ \ \ (g\,x + h)
\end{eqnarray}$</p>


问:
Countable representation of ${X>0}$ as unions of $1/k^2$
How would I go about proving that the set $\{X>0\}$ can be written in the form firstly, $\bigcup_{\epsilon >0} \{X > \epsilon\}$ and then $\bigcup_{k \geq 1} \{X >1/k^2\}$. It seems so obvious that I don't know how to prove it rigorously.

Many thanks.

答:
$$
\{X>0\}
\iff\{X > \epsilon \}\,\exists\,\epsilon > 0$$

and then this can be writed as $$
\bigcup_{\epsilon>0}\{X>\epsilon\}
$$
The union of $\epsilon$ can be understood as exists $\epsilon$ corresponding to the symbol $\exists$ and the Intersection corresponding to the symbol $\forall$

By the way, $\{X \ge 0\} \iff \{X > 0 -\epsilon\} \text { for any } \epsilon >0$
so , which can be write as following
$$
\bigcap_{\varepsilon>0}\{X>-\epsilon\}
$$


问:
Moment Generating Functions for beginners
Problem: Let  $X$  be a random variable with MGF  $M_X(2)=\frac{1}{3}e^{-t}+\frac{1}{6}+\frac{1}{2}e^{2t}$. What is  $P(X≤1)$?

I tried taking $P(X=0)$ and $P(X=1)$ and plugging those numbers into the equation for $t$ but that didn't work.

I tried:
$$M_X(2)=\frac{1}{3}e^{-(0)}+\frac{1}{6}+\frac{1}{2}e^{2(0)}=0.3333+0.16666+0.5=0.999999 (wrong)$$
$$M_X(2)=\frac{1}{3}e^{-(1)}+\frac{1}{6}+\frac{1}{2}e^{2(1)}=0.1226+0.16666+3.6945=4.1944 (wrong)$$.

I might be confusing what $t$ is with what $X$ is.  Any suggestions would be helpful.

答:
If... the MGF is the following

$$M_X(t)=\frac{1}{3}e^{-t}+\frac{1}{6}+\frac{1}{2}e^{2t}$$

This is the MGF of hte following discrete rv

$$ X =
\begin{cases}
\frac{2}{6},  & \text{if $x=-1$ } \\
\frac{1}{6},  & \text{if $x=0$ } \\
\frac{3}{6},  & \text{if $x=2$ }
\end{cases}$$

Thus

$$\mathbb{P}[X\leq 1]=\frac{3}{6}$$