问:
Why are upper bounds only defined on subsets of preordered sets?
So I recently was reading through spivaks calculus and just learned the least upper bound property.
I noticed that in the definition the sets in consideration are always subsets of $\mathbb{R}$
This made me curious as to why only subsets of $\mathbb{R}$ were considered and why they have a notion of upper bounds.

Having looked up the definition of what an upper bound of a set is on Wikipedia I came across this rather confusing definition.

an upper bound or majorant
of a subset S of some preordered set (K, ≤) is an element of K which is
greater than or equal to every element of S.

I do not know what a preordered set is but im willing to guess that $\mathbb{R}$ is an example of one .

So basically my question is what is a preordered set and what about them makes them able to have upper and lower bounds?

Thanks in advance.

答:
$\mathbb R$ is a well-ordered set.  For every $x,y \in \mathbb R$ either $x > y$ or $y<x$ or $x=y$  And, for every non-empty subset of $\mathbb R$ that subset inherits the ordering.  So, those sub-sets are well-orderd.

However, just because the subset is well-ordered, doesn't mean it has the least upper bound property.  The set of rational numbers such that $x\in\{\mathbb Q|x<\sqrt 2\}$ does not have a least upper bound in $\mathbb Q.$  i.e. $\sqrt 2$ is the least upper bound, but $\sqrt 2$ is not in $\mathbb Q$ yet there are infinitely many elements in $\mathbb Q$ that are arbitrarily close to $\sqrt 2$

Partially ordered sets and pre-ordered sets are not as restricted as well ordered and totally ordered sets.

There will be $x,y \in S$ such that $x>y$ however it may not be true for every $x,y$

For example, consider a tree.  We might say $x>y$ if $y$ if there is a path down the branch from $x$ to $y.$  But, if $x,y$ lie on different branches we cannot say anything about how $x$ relates to $y.$

Hope this helps.


问:
A monotone sequence of sets is convergent
I'm having trouble understanding the following part of a proof from Leadbetter's measure and probability book:

THEOREM 1.4.2: A monotone increasing (decreasing) sequence {$E_n$} is convergent and lim$E_n=\bigcup_{n=1}^∞ E_n (\bigcap_{n=1}^∞ E_n)$.

Proof:

$$\bar{lim}E_n=\bigcap_{n=1}^∞ (\bigcup_{m=n}^∞ E_m )=\bigcap_{n=1}^∞ (\bigcup_{m=1}^∞ E_m )$$
But $\bigcup_{m=1}^∞ E_m$ does not depend on n and thus, $$\bar{lim}E_n=\bigcup_{m=1}^∞ E_m $$

In particular, I don't understand the logic behind moving from $\bar{lim}E_n=\bigcap_{n=1}^∞ (\bigcup_{m=1}^∞ E_m )$ to $\bar{lim}E_n=\bigcup_{m=1}^∞ E_m $. I understand that $\bigcap_{n=1}^∞ (\bigcup_{m=1}^∞ E_m )$ does not depend on n, but I don't see how removing the intersection wouldn't also change the union of all members of {$E_n$}. Since a monotone increasing sequence of sequences contains more elements per $E_n$ as n increases, wouldn't the union of all subsequences of {$E_n$} also be increasing, whereas the intersection wouldn't?

答:
To see what’s going on, let $A=\bigcup\limits_{m=1}^\infty E_m$; this is simply some fixed set. Then

$$\bigcap_{n=1}^\infty\left(\bigcup_{m=1}^\infty E_m\right)=\bigcap_{n=1}^\infty A\,,$$

and no matter how many copies of the set $A$ you intersect, you’re just going to get $A$. Thus,

$$\bigcap_{n=1}^\infty\left(\bigcup_{m=1}^\infty E_m\right)=A=\bigcup_{m=1}^\infty E_m\,.$$

ContentLicense: CC BY-SA 4.0


问:
Trigonometric function integration: $\int_0^{\pi/2}\frac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$
How to integrate $$\int_0^{\pi/2}\dfrac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$$

What's  the approach to it?

Being a high school student , I don't know things like counter integration.(Atleast not taught in India in high school education ).I just know simple elementary results of definite and indefinite integration. Substitutions and all those works good. :)

答:
In this case, the following trick also works: Dividing both the numerator and the denominator by $\cos^4 x$, we can use the substitution $ t = \tan x$ to obtain

\begin{align*}
\int_{0}^{\frac{\pi}{2}} \frac{dx}{(a^2 \cos^2 x + b^2 \sin^2 x)^2}
&= \int_{0}^{\frac{\pi}{2}} \frac{1 + \tan^2 x}{(a^2 + b^2 \tan^2 x)^2} \sec^2 x \, dx \\
&= \int_{0}^{\infty} \frac{1 + t^2}{(a^2 + b^2 t^2)^2} \, dt \\
&= \frac{1}{a^2}\int_{0}^{\infty} \left( \frac{1}{a^2 + b^2 t^2} + \frac{(a^2 - b^2) t^2}{(a^2 + b^2 t^2)^2} \right) \, dt.
\end{align*}

The first one can be evaluated as follows: Let $bt = a \tan\varphi$. Then

$$ \int_{0}^{\infty} \frac{dt}{a^2 + b^2 t^2} = \frac{1}{ab} \int_{0}^{\frac{\pi}{2}} d\varphi = \frac{\pi}{2ab}. $$

For the second one, we perform the integration by parts:

\begin{align*}
\int_{0}^{\infty} \frac{t^2}{(a^2 + b^2 t^2)^2} \, dt
&= \left[ - \frac{1}{b^2}\frac{1}{a^2 + b^2 t^2} \cdot \frac{t}{2} \right]_{0}^{\infty} + \int_{0}^{\infty} \frac{1}{2b^2}\frac{dt}{a^2 + b^2 t^2} \\
&= \frac{1}{2b^2} \int_{0}^{\infty} \frac{dt}{a^2 + b^2 t^2} \\
&= \frac{\pi}{4ab^3}.
\end{align*}

Putting together, the answer is

$$ \frac{(a^2 + b^2)\pi}{4(ab)^3}. $$

ContentLicense: CC BY-SA 3.0


问:
Lebesgue Integral of coordinate function is integral of individual functions
Lang writes in his book Real and Functional Analysis that it is "obvious" for $f:\Omega\rightarrow\mathbb R^n$ with $f(\omega) = (f_1(\omega), f_2(\omega), \dots, f_n(\omega))$ the integral of $f$ is given by $$\int_\Omega f(\omega)\,\mu(\mathrm d\omega) = \begin{pmatrix}\int_\Omega f_1(\omega)\,\mu(\mathrm d\omega) \\ \int_\Omega f_2(\omega)\,\mu(\mathrm d\omega) \\ \vdots \\ \int_\Omega f_n(\omega)\,\mu(\mathrm d\omega) \end{pmatrix},$$
i.e. the integral of the coordinate function is the integral of the individual functions. I don't see why this result holds or is even obvious?

答: For any linear map $L: \mathbb R^{n} \to \mathbb R$ we have $L(\int fd\mu)=\int (L\circ f) d\mu$. This is an easy consequence of the definition of the integral. Apply this to the coordinate maps $L_i(a_1,a_2,...,a_n)=a_i, i=1,2...,n$.


问:
How to estimate a Lipschitz constant for a transformation? $T(x,y)=(u,v)$
Consider a transformation $T(x,y)=(u,v)$ defined on (the unit square $E\ $): $0\leq x\leq 1$, $0\leq y\leq 1$. Where $$u=2x^2+6xy-4x^3/3-3xy^2, \ \ \ v=x^3-y^2 $$
Show that an estimate for the Lipschitz constant $M$ on $E$ is $M=\sqrt{65}$

By Lipschitz constant $M$ we mean $$ |T(p)-T(q)|\leq M|p-q|,\ \ p,q\in E$$ Since $T$ is of class $C'$ we know that $T$ is Lipschitz, I tried doing $|(u,v)-(u',v')|$ but it doesn't seems to be a good way. The hint says maximize the absolute value of derivatives arriving at the matrix $$\left(\begin{array}{rc} 4 & 6\\
3 & 2 \end{array}\right) $$ But I don't see how to arrive to this matrix nor how this can help. Please any guide.

答:
The Jacobian of $T$ at $(x,y)$ is $$J(T)(x,y) = \begin{bmatrix}4x+6y-4x^2-3y^2 & 6x - 6xy \\ 3x^2 & -2y\end{bmatrix}.$$

Maximizing the absolute value of each entry (over $x,y \in [0,1]$) individually leads to the matrix in the hint.

In general, we have the linear approximation $T(q) \approx T(p) + [J(T)(p)](q-p)$ when $q$ is close to $p$, so the idea is that
$$|T(q)-T(p)| \approx |[J(T)(p)](q-p)|.$$

The right-hand side is bounded by $M |q-p|$ where $M$ is the operator norm (largest singular value) of the Jacobian. The operator norm is bounded by the Frobenius norm (square root of sum of squares of entries, or equivalently, square root of sum of squares of singular values). Using the matrix in the hint, the Frobenius norm is $\sqrt{4^2+6^2+3^2+2^2}=\sqrt{65}$.


问:
Given two sets, finding two non trivial homomorphisms that are not isomorphisms
Is it possible to have two non trivial homomorphisms that are not isomorphisms for given two Groups?

I am specially interested in additive/remainder Group of Integers and multiplicative (arithmetic multiplication as group operator) group of complex roots of unity.

Thank You.

答: Yes. Consider the group $(\mathbb Z_7)^*$ of nonzero elements mod $7$, where the group operation is multiplication. Both multiplication by $2$ and multiplication by $3$ give different nontrivial homomorphisms from the group to itself. </p>


问:
Show that $2^n<2^{\lceil n \log_23\rceil}-3^n<3^n-2^n$
Is there a way to show that (with integer $n>2$) $$2^{\lceil n \log_23\rceil}-3^n<3^n-2^n$$
I tried to figure a way with derivative or by looking how both side are growing, but I have some trouble with the ceiling function.

答:
Start with $$2^{\lceil n \log_23\rceil}-3^n<3^n-2^n$$

Bring things to the right side: $$2^{\lceil n \log_23\rceil}<2\cdot 3^n \cdot (1-2^n / (2\cdot 3^n))$$

Take base-2 logarithm: $${\lceil n \log_23\rceil}<1 + n \log_23 + \log_2 ( (1-2^n / (2\cdot 3^n)))$$

Since $n \log_23$ is irrational for n >= 1: $${\lfloor n \log_23\rfloor + 1}<1 + n \log_23 + \log_2 ( (1-2^n / (2\cdot 3^n)))$$

$$n \log_23 - {\lfloor n \log_23\rfloor}< - \log_2 ( (1-2^n / (2\cdot 3^n)))$$

The right hand side is about $$2^n / (2\cdot 3^n) / \ln 2$$ or $${(2/3)}^n / (2\cdot  \ln 2)$$

The original inequality is true as long as $n \log_23$ is more than ${(2/3)}^n / (2\cdot  \ln 2)$ above the next lower integer. This series shrinks very fast, so it’s most likely true for all n.

Using Euclid’s algorithm as if you were calculating $\gcd(\log_23, 1)$ you can find consecutive n where the distance is as small as possible, and using some software that can perform these calculations in very high precision you should be able to prove this for very large n, say $n < 10^{1,000,000}$. Which doesn’t prove it’s true for all n.


问:
Interpreting a homotopy as a function from $[0,1]$ to a function space?
Let $X,Y$ be topological spaces. A homotopy is a continuous $H:X\times[0,1]\to Y$. Currying, we get $H:[0,1]\to X\to Y$. Is there a way to interpret $H$ as a continuous function of type $H:[0,1]\to (X\to Y)$, where $(X\to Y)$ is the space of continuous function from $X$ to $Y$?

There are a few questions.

Given topological spaces $X,Y$, is there natural space of continuous functions from $X$ to $Y$?

Given a function out of a product space $f:X\times Y\to Z$, is there a natural way to curry?

If the above doesn't work for general spaces, is it possible for $Y=[0,1]$?

答:
The answer is in [1].
$\newcommand{\Top}{\mathbf{Top}} % topological spaces$

Yes, the compact open topology.

No. A topological space $X$ is exponentiable if the functor $-\times X: \Top\to\Top$ has a right adjoint $(-)^X$. [1] shows that $X$ is exponentiable iff it is core compact. If $X$ is Hausdorff, $X$ is core compact iff it is locally compact.

Again, no. What matters is $X$, not $Y$.

[1]: Escardó, Martín; Heckmann, Reinhold, Topologies on spaces of continuous functions., Topol. Proc. 26(2001-2002), No. 2, 545-564 (2002). ZBL1083.54009.


问:
Kernel of Jordan Normal Form
Let $J_{m\times m} $ be the Jordan normal form with real $c$ on the diagonal and 1 on the super diagonal. Show that

$$null (J-cI) \subset null (J-cI)^2 \subset ... \subset null(J-cI)^{m-1} \subset null (J-cI)^m $$

I was wondering if there is a shorter and alternative proofs than the one I managed to come up with.

The proof goes as follows:

Since $J$ is a Jordan normal form,
$\exists x_m : (J-cI)^m x_m = 0$ and $(J-cI)^{m-1} x_m \neq 0$.

By induction on m,

Base case $r=2$ then if $$k \in null (J-cI) \\
(J-cI)k = 0 \\
(J-cI)^2= (J-cI)(J-cI) k = 0 $$
so $k \in null (J-cI)^2 $
But since $\exists x_m : (J-cI)^{m-1} x_m \neq 0 $ then
$x_2 = (J-cI)^{m-2} x_m$ and $x_2 \in null (J-cI)^2$ and $x_2 \notin null (J-cI)$ thus a strict subset.

Suppose true for r=m-1 i.e
$$null (J-cI) \subset null (J-cI)^2 \subset ... \subset null(J-cI)^{m-1}$$

Then for $k \in null (J-cI)^{m-1}$, by similar procedure $k \in null (J-cI)^m$ and $x_m \in null (J-cI)^m, x_m \notin null(J-cI)^{m-1} $ so a strict subset.

So the proposition is true by mathematical induction.

Main question is whether there's a more direct method of proving this?

答:
Let $e_1,e_2,\dots,e_n$ be the basis where the matrix of $J$ becomes a single Jordan block
$$J=\pmatrix{c\\ 1& c \\ &\ddots &\ddots\\ &&1&c}
\text{ so }J-cI=\pmatrix{0\\ 1& 0 \\ &\ddots &\ddots\\ &&1&0}\,.$$
Call $A:=J-cI$, the values it maps to the basis elements can be read from its columns, specifically, $Ae_i=e_{i+1}$ if $i<n$ and $Ae_n=0$.

Rank of $A$ is clearly $n-1$, so explicitly $\ker A={\rm span}(e_n)$.

For $A^k$, observe that $Ae_i=e_{i+k}$ if $i+k\le n$ and $Ae_i=0$ otherwise. By a similar argument as above, we get
$$\ker A^k={\rm span}(e_{n-k+1},\dots,e_n)\,.$$

Title: Kernel of Jordan Normal Form


问:
Why does plotting Collatz sequences in polar coordinates produce a cardioid and nephroid?
I generated the Collatz sequences for the first 2,000 starting integers, and plotted these sequences "on top of each other" in polar coordinates, using a fixed radius and with each element in the sequence as theta (converted to radians, i.e., modulo 360). Successive elements within each sequence are connected by (semi-transparent) lines.

Following these steps produced the image below, where we can clearly see two shapes (a cardioid and nephroid). (The Python code that generates this image is available here. The total number of elements in the 2,000 sequences is 136,100, so the total number of lines plotted in the image below is 136,100 - 2,000 = 134,100.)

Link to generated image (high-quality) or lower quality

My question is: Why do a cardioid and nephroid clearly appear? Is there something notable about the Collatz sequences that produce these shapes, or is this unrelated to the intricacies of the Collatz sequences, and that other unrelated integer sequences also produce similar shapes?

As of now, I haven't seen any substantive references online to nephroids and the Collatz conjecture together.

A post here describes how you can generate a cardioid in polar coordinates by drawing lines between evenly spaced points on a circle, but this doesn't relate to the Collatz sequences, and doesn't construct a nephroid.

Update:

Based on the discussion, the cardioid can be generated based on the envelope of lines drawn between points around a circle, with each line drawn between point line between points n and 2n (mod N), as described here. The nephroid is based on the same concept for points between n and 3n, as described here. The Collatz sequence contains both behaviors: some elements in sequence follow 2n->n (the order of the two doesn't matter if we are connecting lines), and some elements follow n->3n+1 (but the +1 component is negligible for the overall shape), resulting in the cardioid and nephroid.

Update #2:

Per a suggestion by heropup below, I've redrawn the image where for each line connecting terms (N,N+1) in a sequence, the line is red if N+1 > N and blue if N+1 < N. From this and per the update above, we can clearly see that the nephroid is red and the cardioid is blue.

Link to colored generated image (high-quality) or lower quality

(This now probably looks better on a white background but takes a long time to generate).

答:
Update from OP

Based on the discussion, the cardioid can be generated based on the envelope of lines drawn between points around a circle, with each line drawn between point line between points n and 2n (mod N), as described here. The nephroid is based on the same concept for points between n and 3n, as described here. The Collatz sequence contains both behaviors: some elements in sequence follow $2n\to n$ (the order of the two doesn't matter if we are connecting lines), and some elements follow $n\to3n+1$ (but the $+1$ component is negligible for the overall shape), resulting in the cardioid and nephroid.


问:
What does it mean for one function to be big O and/or little, Omega, or big Theta of another function?
Lets say I have two functions f(n) and g(n) and I would have to determine whether f(n) is Big O and/or little o, Omega, or Big Theta of g(n).

How would you determine that using f(n)=sqrt(n) and g(n)=log(n)?

I was thinking that since f(n) grows faster it is Omega of g(n).  Am I understanding that correctly or do I have that backwards?

答:
In your examples(taking $\log=\ln$) as we have $g(n)=\ln n \lt \sqrt{n}=f(n)$, then we can say, that $g(n)=\ln n = O(\sqrt{n}) = O(f(n))$. So, $f$ "grows faster" we can understand as $g \in O(f)$. In this case we can say, that also $\ln n = o(\sqrt{n})$, because $\lim\limits_{n\to\infty}\frac{\ln n}{\sqrt{n}}=0$, so, $f$ "grows strictly faster" we can understand as $g \in o(f)$.

But we cannot generalize this property as shows example $1+\frac{1}{n}\in O(1)$, but $1+\frac{1}{n}\notin o(1)$ .

In general we have $f\in o(g) \Rightarrow f\in O(g)$ i.e. $o(g) \subset O(g)$.


问:
Is the set made up of all points that satisfy $g_{1}(x, y, z)=x^2 + 2y^2=1 $, $\:g_{2}(x, y, z)= xy + xz=2$ compact?
I want to find the least value of a function whose domain is made up of all points that satisfy the following:

$g_1(x, y, z)= x^2 + 2y^2=1$

$g_2(x, y, z)= xy + xz=2$

In my book it is written that "a minimum exists since the domain is a closed set". However, previously there is a theorem that one can only be certain that function obtains a max and min value if the domain is compact, i.e. both closed and bounded. From what I see, the domain is not bounded, how can we then be certain that "a minimum exists"?

Any help would be appreciated.

$\textbf{Edit:}$: I want to find the least distance form the origin, so my function is $f(x, y,z)=^2+^2+^2$

答: Given the additional information, set forth in your comment, that the function you're trying to minimize is $f(x, y, z)=x^2+y^2+z^2$, the solution is easy.  Just note that because $(1, 0, 2)$ is in your solution set, the minimum must fall within the (compact) sphere of radius, say, $953.8$ (or pick your favorite number that's at least as large as $\sqrt 5$).  Since the solution sets of $g_1$ and $g_2$ are closed, their intersection with this compact sphere must be compact and the minimum must occur somewhere in that intersection.


问:
Describe object geometry in 2D space
How would I describe geometry (position + orientation) of an object in 2 dimensions?
I'm familiar with a certain method of describing an object's geometry in 3 dimensions, using a 4x4 matrix with its position Vector, look Vector, right Vector, and top Vector included, but I don't see a need to include anything more than position and look vector in the case of a 2d matrix, correct? If so, how would the orientation of that look? An explanation or link to additional resources on this topic would be greatly appreciated. Thanks!</p>

答:
A 2D transformation matrix $\mathbf{R}$,
$$\mathbf{R} = \left[ \begin{matrix}
u_x & - u_y & t_x \\
u_y &   u_x & t_y \\
  0 &     0 &   1 \\
\end{matrix} \right] \tag{1}\label{G1}$$
describes translation by $\vec{t} = ( t_x, t_y )$, rotation by $\theta$, and scaling by $\lambda \gt 0$,
$$\left\lbrace \, \begin{aligned}
u_x &= \lambda \cos \theta \\
u_y &= \lambda \sin \theta \\
\end{aligned} \right . \quad \iff \quad \left\lbrace \, \begin{aligned}
a &= \sqrt{ u_x^2 + u_y^2 } \\
\theta &= \operatorname{atan2}\left( u_y, u_x \right) \\
\end{aligned} \right . \tag{2}\label{G2}$$
where $\operatorname{atan2}$ is the two-argument form of arcus tangent (covering all four quadrants).

Transforming vector $\vec{p} = (p_x, p_y)$ to vector $\vec{q} = (q_x , q_y)$ is done using
$$\left[ \begin{matrix} q_x \\ q_y \\ 1 \end{matrix} \right] = \left[ \begin{matrix} u_x & -u_y & t_x \\ u_y & u_x & t_y \\ 0 & 0 & 1 \end{matrix} \right] \left[ \begin{matrix} p_x \\ p_y \\ 1 \end{matrix} \right]$$

Thus, for rotation, translation, and scaling, you need four scalars: $\theta$, $\lambda$, $t_x$, and $t_y$; or equivalently (through $\eqref{G2}$) $u_x$, $u_y$, $t_x$, and $t_y$.  You can call $\vec{u} = (u_x, u_y)$ the "right" vector; then the "up" vector is that one rotated 90° counterclockwise, i.e. $\vec{v} = (-u_y, u_x)$.

Given a rotation matrix
$$\mathbf{R} = \left[ \begin{matrix}
r_{11} & r_{12} & r_{13} \\
r_{21} & r_{22} & r_{23} \\
 0     &  0     &  1     \\
\end{matrix} \right]$$
you can recover the four scalars (considering numerical stability) via
$$\left\lbrace \, \begin{aligned}
u_x &= \displaystyle \frac{r_{11} + r_{22} }{2} \\
u_y &= \displaystyle \frac{r_{21} - r_{12} }{2} \\
t_x &= r_{13} \\
t_y &= r_{23} \\
\end{aligned} \right.$$
and optionally apply $\eqref{G2}$ if you want the rotation angle $\theta$ and scale factor $\lambda$.

Title: Describe object geometry in 2D space


答:
I don't know why are you doing $$||A-B|| = ||A|| - ||B||$$ at certain places because it's certainly not true.

Start with the condition number of $A$ and write the RHS of your inequality as:
$$k(A) = \|A\|\|A^{-1}\|\geq\dfrac{\|A\|\|B^{-1} - A^{-1}\|}{\|B^{-1}\|\|B-A\|}$$
can you take it from here?

You would need to use that fact that operator norms are sub-multiplicative:
$$\|XY\|\leq\|X\|\|Y\|$$

Title: Prove that $\frac{1}{K(B)}\frac{\|A-B\|}{\|A\|} \le \frac{\|A^{-1}-B^{-1}\|}{\|B^{-1}\|} \le K(A)\frac{\|B-A\|}{\|A\|}.$

Body:
Let $A$ and $B$ be invertible matrices and $K(C)$ the condition number of an invertible matrix $C$. Prove that $$\frac{1}{K(B)}\frac{\|A-B\|}{\|A\|} \le \frac{\|A^{-1}-B^{-1}\|}{\|B^{-1}\|} \le K(A)\frac{\|B-A\|}{\|A\|}.$$

I tried
\begin{align}
\frac{1}{K(B)}\frac{\|A-B\|}{\|A\|} &\le \frac{\|A^{-1}-B^{-1}\|}{\|B^{-1}\|} \le K(A)\frac{\|B-A\|}{\|A\|}  &&\Leftrightarrow \\[5pt]
\frac{1}{K(B)}\frac{\|A\|-\|B\|}{\|A\|} &\le \frac{\|A^{-1}\|-\|B^{-1}\|}{\|B^{-1}\|} \le K(A)\frac{\|B\|-\|A\|}{\|A\|}  &&\Leftrightarrow \\[5pt]
\frac{\|A-B\|}{\|A\|\|B\|\|B^{-1}\|} &\le \frac{\|A^{-1}\|}{\|B^{-1}\|}\le \frac{\|A\|\|A^{-1}\|(\|B\|-\|A\|)}{\|A\|}  &&\Leftrightarrow \\[5pt]
\frac{\|A-B\|}{\|A\|\|B\|} &\le \|A^{-1}\|\le \frac{\|A^{-1}\|(\|B\|-\|A\|)}{\|A\|}  &&\Leftrightarrow \\[5pt]
\frac{\|A-B\|}{\|B\|} &\le K(A)\le \|A^{-1}\|(\|B\|-\|A\|) &&\Leftrightarrow \\[5pt]
\frac{\|A-B\|}{\|B\|} &\le K(A)\le \|A^{-1}\|\|B\|-K(A)  &&\Leftrightarrow \\[5pt] \frac{\|A-B\|}{\|B\|} &\le 0 \le \|A^{-1}\|\|B\|-2K(A) &&\Leftrightarrow \\[5pt] \|A-B\| &\le 0 \le \|A^{-1}\|\|B\|-2K(A) &&\Leftrightarrow \\[5pt]
\|A-B\| &\le 0 \le \|B- 2A\|
\end{align}

I don't know how to continue. It doesn't look correct.


问:
Inequalities bound
If $b$, $c$, $d$ and $e$ are all real numbers and positive and $b\leq c$ and $d \leq e$. Under what condition is $\frac{b}{d} \leq \frac{c}{e}$ ? I have been trying to think about it but cannot come up with a condition. Thanks.

答:
If we assume all numbers are positive then

$\frac bd \le \frac ce\iff$

$be \le dc$.

If we let $\delta = c-b \ge 0$ and $\gamma = e-d \ge 0$ then this is true if and only if

$b(d+\gamma) \le  d(b+\delta) \iff$

$b\gamma \le d\delta \iff$

$\frac {\gamma}{\delta} = \frac {e-d}{c-b} \le \frac db$.

That's something, I guess.  The inequality holds if the ratio of the differences is less than the ratio of the smaller values....

But ... is it useful.... Or easy to express?

Six of one half a dozen of the other, I guess.

(It's got the poetry of sounding like a formula "The ratio of the difference between the pairs is less or equal to the ratio of the smaller values of the pairs".  That sounds like it's important.  Even if it isn't.)

But we need to take cases where they aren't all positive.

Title: Inequalities bound


问:
Do Lipschitz convex functions have at least one "small" subgradient at each point?
Let $X \subset \mathbb R^d$ be closed and convex and $f: X \to \mathbb R$ be convex and Lipschitz. That means there is $L > 0$ such that $|f(x)-f(y)| \le L \|x-y\|$ for all $x,y \in X$. Here $\|\cdot\|$ is the Euclidean norm. We define the subgradient sets:

$$\partial_f(x) = \{v \in \mathbb R^d: f(y) \ge f(x)+ v \cdot(y-x) \ \forall y \in X\}.$$

It seems common practice (For example Lemma 2.6 of this survey ) to claim  each subgradient of $f$ has norm at most $L$. Unfortunately this is not true $-$ the proof in the survey only works under the assumption that $x$ is an interior point of $X$.

If we allow boundary points here is a   counterexample: Let $X=[0,1]$ and $f(x) = x$ and $L=1$. Then at each interior point $\partial_f(x) = \{1\}$ which is clearly bounded by $L$. However at the boundary we have $\partial_f(0) = (-\infty,1]$ since for any $y \in [0,1]$ and $v \le 1$ we have $y \ge vy$ and so
$$f(y) = y \ge 0 + vy = 0+v(y-x) =f(x)+v(y-x).$$
Likewise $\partial_f(0) = [1,\infty)$.

However I notice there still exists a small subgradient at each endpoint, even if there are many more large ones. Namely $1$ is still a subgradient at the endpoints. I wonder is this fact true in general?

Conjecture: Let $X \subset \mathbb R^d$ be closed and convex and $f: X \to \mathbb R$ be convex and Lipschitz with constant $L$. For each $x \in X$ there is $v \in \partial_f(x)$ with $\|v\| \le L$.

答: Yes.  We may assume that $X$ has nonempty interior (otherwise it's contained in an affine subspace, and we can extend the function by making it constant on the orthogonal complement...).  Now any $x \in X$ is the limit of a sequence $x_n$ in the interior of $X$, and at each $x_n$ there is a subgradient $v_n$ with norm  $\le L$.  The $v_n$ have a limit point, which is a subgradient at $x$ and has norm $\le L$.


答: Consider the homomorphism $g_n:\mathbb{Z}\to\mathbb{Z}$ such that $g_n(1)=n$, where $n>0$. Then all the $g_n$'s are distinct and they are definitely not isomorphisms because $g_n(\mathbb{Z})=n\mathbb{Z}\subset\mathbb{Z}.$</p>

Title: Given two sets, finding two non trivial homomorphisms that are not isomorphisms

Body:
Is it possible to have two non trivial homomorphisms that are not isomorphisms for given two Groups?

I am specially interested in additive/remainder Group of Integers and multiplicative (arithmetic multiplication as group operator) group of complex roots of unity.

Thank You.


问:
How to express these expressions in integral form? $\frac{1}{3n+2}$ or $\frac{1}{(2n+1)^2}$
How to express these expressions in integral form? $\dfrac{1}{3n+2}$ or $\dfrac{1}{(2n+1)^2}$

In another thread, it is shown to me that you can rewrite $\dfrac{1}{3n+2}$ as $\displaystyle\int_{0}^{1}x^{3n+1}dx$. It has been a long time since I evaluate integrals so I don't know how to turn this into that form.

Can you help me, showing steps, of expressing these expressions like $\dfrac{1}{3n+2}$ or $\dfrac{1}{(2n+1)^2}$ in integral forms? I think it is highly useful when you sum series.

答:
Note that $$\int_{0}^{1} x^{b} dx = \frac {1}{b+1}$$

Thus for the expression $$\frac {1}{(2n+1)^2}$$ you need to solve for $b$

That is $$ b+1= (2n+1)^2$$
Which gives you $$b=4n(n+1)$$

The result is then $$ \int _{0}^{1} x^{4n(n+1)}=\frac {1}{(2n+1)^2}$$


问:
4D Left Isoclinic Rotations as double rotations
According to this page, 4D left-isoclinic rotation matrices are double rotations of the same angle (and same sign). But in the same page, in the context of quaternions, it is mentioned that left-isoclinic matrices have the form:
$$R=\begin{bmatrix}a&-b&-c&-d \\ b&a&-d&c\\c&d&a&-b\\d&-c&b&a\end{bmatrix}$$ where $a,b,c,d\in\mathbb{R}$ such that $$a^2+b^2+c^2+d^2=1.$$ My question is, how can we view $R$ as a double rotation? I mean, how can we find the common angle? If two of $b,c,d$ are $0$ then it is easy to see. But what if this is not the case?

答:
Notice the first column! This is the matrix of left-multiplication-by-$(a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k})$. We may write this quaternion in polar form as $ a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}=\exp(\theta\mathbf{u}) $.

Solving for the unknowns $\theta$ and $\mathbf{u}$ with Euler's formula yields

$$ \begin{cases} \cos\theta=a \\ \sin\theta=\sqrt{b^2+c^2+d^2} \end{cases},
\qquad \mathbf{u}=\frac{~~~~b\mathbf{i}+c\mathbf{j}+d\mathbf{k}}{\sqrt{b^2+c^2+d^2}}. $$

The angle $\theta$ is convex and $\mathbf{u}$ is a unit vector. If we extend $\{\mathbf{u}\}$ to an (appropriately oriented) orthonormal basis $\{\mathbf{u},\mathbf{v},\mathbf{w}\}$ for $\mathbb{R}^3$, then $\{1,\mathbf{u},\mathbf{v},\mathbf{w}\}$ is an orthonormal basis for $\mathbb{H}$ with the same orientation and multiplication table as $\{1,\mathbf{i},\mathbf{j},\mathbf{k}\}$.

With respect to this basis, the transformation acts as a rotation by $\theta$ in the oriented $1\mathbf{u}$-plane and its orthogonal complement, the oriented $\mathbf{vw}$-plane. (Exercise.)

There is more than one way to choose a pair of oriented, orthogonal, stable planes. The stable planes for $\exp(\theta\mathbf{u})$ are the same as for those of multiplying-by-$\mathbf{u}$. Multiplying by $\mathbf{u}$ is a complex structure on $\mathbb{H}$ making it a left complex vector space, so the stable planes are precisely the 1D complex subspaces! This forms the complex projective line $\mathbb{CP}^1\simeq S^2$, diffeomorphic to the Riemann sphere by stereographic projection.

(The space of all oriented 2D planes in 4D real space is the real Grassmanian $\mathrm{Gr}_2\mathbb{R}^4\simeq S^2\times S^2$, which can be shown in a couple ways using quaternions too.)


问:
Covariance matrix of a random variable and its affine transformation
Given:

the random variable $\mathbf{X} \sim \mathcal{N}(\mathbf{c, \Gamma})$ where $\mathbf{X} \in \mathbb{R}^L$

its affine transformation $\mathbf{Y} = \mathbf{A} \mathbf{X} + \mathbf{b} + \mathbf{E}$ where , $\mathbf{Y} \in \mathbb{R}^D$, $\mathbf{E} \sim \mathcal{N} (\mathbf{0}, \mathbf{\Sigma})$ and $\mathbf{A} \in \mathbb{R}^{D \times L}$ and $\mathbf{b} \in \mathbb{R}^D$

I want compute the joint distribution of $\mathbf{X}, \mathbf{Y}$.

To do that, I am computing the following:

$$\begin{bmatrix}
\mathbf{X}
\\
\mathbf{Y}
\end{bmatrix} \sim \mathcal{N} \left(
\begin{bmatrix}
\mathbf{c}
\\
\mathbf{A c + b}
\end{bmatrix},
\begin{bmatrix}
\mathbf{\Gamma} &  \mathbf{R_{X,Y}}
\\
\mathbf{R_{Y,X}}  & \mathbf{\Sigma + A \Gamma} \mathbf{A}^T
\end{bmatrix}
\right)$$

Where the I have computed the variance thanks to the properties of the affine transformations.
However I am having troubles on how to compute $\mathbf{R_{X,Y}}$ ( $ = \mathbf{R_{Y,X}}^T $).

Can someone help me? Or at least give some hint on solve using the following formula?

$$\mathbf{R_{X,Y}} = \mathbb{E}[(\mathbf{X} - \mathbb{E}[\mathbf{X}])([\mathbf{Y} - \mathbb{E}[\mathbf{Y}])^T]$$

I will need it to compute later the conditional distribution of $\mathbf{X}$ given $\mathbf{Y}$ using well know formulas (e.g. conditional distribution of gaussian process )

答:
We have
\begin{align*}
\mathbf{R}_{\mathbf{X}, \mathbf{Y}} &= \text{Cov}(\mathbf{X}, \mathbf{Y}) = \text{Cov}(\mathbf{X}, A\mathbf{X} + \mathbf{b} + \mathbf{E}) \\
&=\text{Cov}(\mathbf{X}, A\mathbf{X}) + \text{Cov}(\mathbf{X}, \mathbf{b}) + \text{Cov}(\mathbf{X}, \mathbf{E}) \\
&=\text{Cov}(\mathbf{X}, \mathbf{X})A^\intercal + \mathbf{O} + \text{Cov}(\mathbf{X}, \mathbf{E}) \\
&=\mathbf{\Gamma}A^\intercal + \text{Cov}(\mathbf{X}, \mathbf{E})
\end{align*}
If $\mathbf{X}$ and $\mathbf{E}$ are known to be independent, then $\text{Cov}(\mathbf{X}, \mathbf{E}) = \mathbf{O}$ and we have $\mathbf{R}_{\mathbf{X}, \mathbf{Y}} = \mathbf{\Gamma}A^\intercal$.


问:
Theorem 3.2.14 (Probability: Theory and Examples, Durrett, 5th ed.)
The proof of the theorem is a single line and thus seemingly straightforward, yet I cannot see how to justify it. I tried to rearrange some of the pieces, but that approach failed to improve my understanding. Any help in justifying the statement would be greatly appreciated.

$\textbf{Theorem 3.2.14.}$
If there is a $\varphi\ge0$ so that $\varphi(x)\to\infty$ as $|x|\to\infty$ and
$$
C=\sup_n\int\varphi(x)dF_n(x)<\infty,
$$
then $F_n$ is tight.

$\textbf{Proof.}$
$$
1-F_n(M)+F_n(-M)\le\frac{C}{\inf_{|x|\ge M}\varphi(x)}.
$$

答:
I think Durret's book is good but sometimes you find thigs like these. Anyway He's using $$\text { Theorem 1.6.4. Chebyshev's inequality. Suppose } \varphi: \mathbf{R} \rightarrow \mathbf{R} \text { has }  \varphi \geq 0 \text { , let } A \in \mathcal{R} \text { and let } \; i_{A}=\inf \{\varphi(y): y \in A\}. \;\; \text{ Then } \; i_{A} P(X \in A) \leq E(\varphi(X) ; X \in A) \leq E \varphi(X)$$

So you write $A:=\{|x|\geq M\}$. You have that $$\inf_{|x|\ge M}\varphi(x)\cdot \big(1-F_n(M)+F_n(-M)\big)= i_A P(X_n\in A)\leq E (\varphi(X_n))=\int \varphi(x)F_n(dx)<C$$

So if $\varepsilon>0$ you can found $M$ big enough in such way that $\frac{C}{\inf_{|x|\ge M}\varphi(x)}<\varepsilon$ and you finished.


问:
How to prove that a hexagon is the regular polygon with the most sides that can tile a plane
I need to know the answer to this question to find out why bees use hexagonal cells in hives. I know that a circle takes up the most area using the least perimeter, so bees would try to make shapes as close to a circle as possible to use the most space without wasting too much material on walls. However, bees don't use circles because using circles creates a lot of waste space between cells, so bees use a shape that can tile a plane without overlap. The shape that meets the conditions of having a lot of sides and being regular to look like a circle, and being able to tile a plane  is the hexagon, so bees use this in their hives. However, I want to know whether or not this is the polygon with the most sides that fits this requirement. I try to prove there is no bigger polygon like this.

First, I note that an integer number of interior angles must meet at a site and add up to $360$ degrees to tile a plane. For example, squares can tiles a plane because they all have $90$ degree angles, and $4$ of these make $360$. All polygons have exterior angles adding up to $360$, so a polygon with $n$ sides has $\frac{360}n$ degrees per side. However, interior and exterior angles are supplementary, thus the interior angle of an $n$ side polygon measures $180- \frac{360}n$ degrees. Now, let's test whether this quantity divides $360$ degrees. We get $360 \over 180- \frac{360}n$, which simplifies down to $\frac {2n}{n-2}$. Now, I need to prove that $n=6$ is the biggest number such that this quanitity is an integer, but I am not sure how I would go about  doing that.

答: The simplest way is to begin as you did, and consider how many polygons are meeting at a corner.  It has to be at least 3 to be a corner; if it were 2 it would be an edge, not a vertex.  The minimum is 3. If all the angles are the same size, then the size of each interior angle that joins at the vertex must be 360/3 = 120.  Therefore a hexagon has the most sides.</p>


问:
Convert Rotation Velocity Respect from a Certain Coordinate Sytem into a Rotation Velocity Respect from another Coordinate Sytem
So I have a Case:

I have two 3D Coordinate Sytems:

Real World(Rigid) Coordinate System ~ the Origin of all objects.

Certain Object's Local Coordinate System ~ since the object can move or rotate, this 3D coordinate system changes its position and orientations.

An Object is found at the Origin(0.0,0.0,0.0).

The Variables are:

α, β, γ = Pitch(X-Axis), Roll(Y-Axis), Yaw(Z-Axis) ~ Intrinsic Rotations of the Object's Local Coordinate System, from the Real World Coordinate System.

u, v, w = rotation velocity of the Object Relative to the Real World Coordinate System.

x, y, z = rotation velocity of the Object Relative to the Object's Local Coordinate System.

The First Trial was:

When the two coordinate systems are in phase(α = 0.0, β = 0.0, γ = 0.0)

Then the rotation velocities of the object on both coordinate systems are equal(x = u, y = v, z = w)

The Second Trial was:

When Object's Local Coordinate System was rotated(α = 131.42, β = -36.7, γ = 317.22)

And u = 14.2, v = 5.2, w = -1.32

Then x = ?, y = ?, z = ?

I am trying to get the Rotation Velocity of the Object relative to the its Local Coordinate System. I will appreciate the help of the community!

答:
It's worth remembering at this point that the instantaneous rotational velocity of a rigid three-dimensional object is always a rotational speed around a single axis.

If you have two non-rotating frames of reference, the rotational speed is the same in both frames, so you need merely convert the coordinates of the directional vector of the axis from one frame to another.

Since the roll, pitch, and yaw axes are orthogonal, you can combine and decompose the rotational velocities as if they were vectors; that is, the rotation rates $u,$ $v,$ and $w$ around the orthogonal axes combine into a single rotation of magnitude
$\sqrt{u^2 + v^2 + w^2}$ around the axis parallel to the vector $(u,v,w).$
In fact, you might as well just use the vector $(u,v,w)$ as a representation of the rotational velocity.
Convert this to the other coordinate system and decompose it into its components along that system's orthogonal axes.

If one rotational system is rotating then you have to think about how you're measuring rotation.


答: Well, $\lim_{n \rightarrow \infty} \frac{2n}{n-2} = 2$.  Its derivative, $\frac{-4}{(n-2)^2}$ is negative everywhere, so this expression decreases asymptotically to $2$.  So $3$ is the least integer it can take and you have shown that it does.

Title: How to prove that a hexagon is the regular polygon with the most sides that can tile a plane

Body:
I need to know the answer to this question to find out why bees use hexagonal cells in hives. I know that a circle takes up the most area using the least perimeter, so bees would try to make shapes as close to a circle as possible to use the most space without wasting too much material on walls. However, bees don't use circles because using circles creates a lot of waste space between cells, so bees use a shape that can tile a plane without overlap. The shape that meets the conditions of having a lot of sides and being regular to look like a circle, and being able to tile a plane  is the hexagon, so bees use this in their hives. However, I want to know whether or not this is the polygon with the most sides that fits this requirement. I try to prove there is no bigger polygon like this.

First, I note that an integer number of interior angles must meet at a site and add up to $360$ degrees to tile a plane. For example, squares can tiles a plane because they all have $90$ degree angles, and $4$ of these make $360$. All polygons have exterior angles adding up to $360$, so a polygon with $n$ sides has $\frac{360}n$ degrees per side. However, interior and exterior angles are supplementary, thus the interior angle of an $n$ side polygon measures $180- \frac{360}n$ degrees. Now, let's test whether this quantity divides $360$ degrees. We get $360 \over 180- \frac{360}n$, which simplifies down to $\frac {2n}{n-2}$. Now, I need to prove that $n=6$ is the biggest number such that this quanitity is an integer, but I am not sure how I would go about  doing that.


问:
Let V be an upper triangular matrix and W and lower one, both of order n. How many operations in VW?

Let V be an upper triangular matrix and W a lower triangular
matrix, both square with n lines. Find the number of multiplications
and additions performed to calculate the product VW.

When I multiply the first line of V by the cols of W I have

$$m = n; n-1; n-2; ... ; 1 = n^2-(n-1)!\\
a = n-1; n-1; n-3;...;0 = n^2-n!$$

For the second line

$$m = n-1; n-2; ...; 1 = n^2-(n-1)!\\
a = n-2; n-3; n-4;...;0 = n^2-n!$$

In total:

$$m = (n^2-(n-1)!)n \\
a = (n^2-n!)n$$

Is this correct?

答:
With $V$ you have $v_{ij}=0$ if $i\gt j$.
With $W$ you have $w_{ij}=0$ if $i\lt j$.

If $VW=Z$, then
$$z_{ij}=\sum_{k=1}^n v_{ik}w_{kj}.$$
But whenever $k\lt \min(i,j)$, the summand is $0$. So the sum is really
$$z_{ij}=\sum_{k=\min(i,j)}^n v_{ik}w_{kj}.$$
This is $n+1-\min(i,j)$ products, and $n-\min(i,j)$ sums.

So you want to calculate
$$\sum_{i=1}^n\sum_{j=1}^n (n+1)-\min(i,j).$$
And this in turn means calculating
$$\sum_{i=1}^n\sum_{j=1}^n \min(i,j).$$
This sum is
$$1(n) + 2(n-1) + 3(n-2)+\cdots +n(1).$$
The calculation of the number of sums is going to likewise turn on this sum.


问:
If $|E| = 0$, then there exists $h \in \mathbb{R}$ such that $E+h$ does not contain a rational point.
Here we are trying to show that if a set $E$ has zero (Lebesgue) measure, then there is a point $h \in \mathbb{R}$ such that $E+h$ does not contain a rational point.

I tried to solve this using contradiction. The argument goes as follows: Suppose that for every $h \in \mathbb{R}$ we can find a rational point in $E+h$. From here I am trying to show that if this is the case, then the measure of $E$ would not be zero, thus proving our claim. However, I am not sure how to formulate the argument to get to the end.

I already looked at the previous questions posted on here for this type of problem, but it did not make much sense to me.

Any help would be appreciated!

答: Hint: $r \in E + h$ iff $-h \in E +(- r)$.  The set of rationals is countable. What do you know about the measure of a countable union?


问:
Convergence of coordinates to zero
Consider a normed finite-dimensional vector space $V$ with some norm $|| .||$

Say a sequence of vectors in this vector space $v_m \rightarrow 0$ where $0$ is the zero vector.

Let $\{b_1,b_2,\ldots,b_n\}$ is a basis for $V$.

Therefore $v_m = \sum_{i=1}^{n} c_i ^m b_i$ for some scalars $c_i$

I am trying to prove that $\forall m$ the sequences $\{ c^m_i \}_{i=0}^{\infty}$
converge to the scalar $0$.

This seems like an 'obvious' theorem, but I am having difficulty proving it.
Obviously some triangle inequality has to be used here but I don't know how.
Using both the inequalities $ | ||x||- ||y|| |\leq ||x-y|| \leq ||x|| + ||y||$
on $||\sum_{i=1}^{n} c_i ^m b_i || < \epsilon$ does not give me an upper bound
on each of the $c^i_m$ sequences.

Any hints?

答:
You can do that by first showing that norms are equivalent in a finite dimensional space (see https://math.stackexchange.com/a/345661/10385) and then you have $M\in \Bbb R, \| \cdot\|_\infty \le M\|\cdot\|$ so if for a sequence of vectors $\left(v_n\right)_{n\in \Bbb N}$, $v_n\underset{n\to +\infty}{\longrightarrow}0$ for the norm $\|\cdot\|$, that is $\|v_n\|\underset{n\to +\infty}{\longrightarrow}0$, you also have $\|v_n\|_\infty\underset{n\to +\infty}{\longrightarrow}0$ and since $0 \le |c_i^n| \le \|v_n\|_\infty\underset{n\to +\infty}{\longrightarrow}0$, $|c_i^n|\underset{n\to +\infty}{\longrightarrow}0$ so $c_i^n\underset{n\to +\infty}{\longrightarrow}0$.

And no, it doesn't work in infinite dimensional spaces.

Take $\Bbb R\left[X\right]$ the space of polynomials.

Take $\forall P \in \Bbb R\left[X\right],\left\|P\right\|=\int\limits_0^1 \left|P(t)\right|\space dt$

And $\left(\left(1-X\right)^n\right)_{n\in \Bbb N}\in \Bbb R\left[X\right]^\Bbb N$

$\left(\left\|\left(1-X\right)^n\right\|\right)_{n\in \Bbb N}$ is decreasing and bounded below so it converges to $l$. We also have $0\le l$.

$\forall \varepsilon > 0,\left\|\left(1-X\right)^n\right\|=\int\limits_0^1 \left|\left(1-t\right)^n\right|\space dt=\int\limits_0^\varepsilon \left|\left(1-t\right)^n\right|\space dt+\int\limits_\varepsilon^1 \left|\left(1-t\right)^n\right|\space dt \le \varepsilon +\int\limits_\varepsilon^1 \left|\left(1-t\right)^n\right|\space dt\underset{n\to +\infty}{\longrightarrow}\varepsilon$

So $\forall \varepsilon >0, l \le \varepsilon$ which means $l=0$.

So $\left(1-X\right)^n\underset{n\to +\infty}{\longrightarrow}\Bbb 0$

But $\left(1-X\right)^n = \sum\limits_{k=0}^n {n \choose k} 1^{n-k}\left(-X\right)^k = 1 + \sum\limits_{k=1}^n {n \choose k} \left(-X\right)^k = 1 - X\sum\limits_{k=0}^{n-1} {n \choose k+1}\left(-X\right)^k$

So since we know that $\Bbb R\left[X\right]=\Bbb R_0\left[X\right] \oplus X\Bbb R\left[X\right]$, in the canonical basis $\left(X^n\right)_{n\in \Bbb N}$, the first coordinate will remain 1.

Title: Convergence of coordinates to zero


问:
Solving for an unknown variable in an index without using logarithms
I have come across the following equation in chapter 1 of a book that is testing applications of index laws. Logarithms are not covered until chapter 17. I can't figure out how to solve for $t$ without using logarithms.

$$ 1000 = 20 \times  10^{0.3t} $$

I have tried to find a common base and then solve for $t$.

Motivation: In what year will the population of Koala's exceed 1000?

答:
You cannot solve this "exactly" without logarithms. But maybe they want you to apply an approximation.

You may be expected to "know" that $2 \approx 10^{0.3}$

Applying that we get, $10^3 \approx 10^{0.3} \cdot 10 \cdot 10^{0.3t}$ which gets us to $1.3 + 0.3t \approx 3$ and $t \approx \frac {3-1.3}{0.3} = \frac{17}3$.

The result is within $0.06 \%$ of the "exact" value you get by taking logarithms.


答: $1000 < 20 \times  10^{0.3t} \implies 50 < 10^{0.3t} \implies 50 < (10^{0.3})^t. $ Now, we find the year such that $50 < (10^{0.3})^t$. $10^{0.3}$ is close to cube root of $10$ which is $2.15$, so we have $50 < (2.15)^t$. Now we can easily guess that $t=5$ doesn't solve this, but $t=6$ does, so in year $5$ is when population first exceeds $1000$ because the population reached $1000$ between years $5$ and $6$.

Title: Solving for an unknown variable in an index without using logarithms

Body:
I have come across the following equation in chapter 1 of a book that is testing applications of index laws. Logarithms are not covered until chapter 17. I can't figure out how to solve for $t$ without using logarithms.

$$ 1000 = 20 \times  10^{0.3t} $$

I have tried to find a common base and then solve for $t$.

Motivation: In what year will the population of Koala's exceed 1000?


答:
which simplifies down to $\frac {2n}{n-2}$. Now, I need to prove that n=6 is the biggest number such that this quanitity is an integer

Well, $\frac {2n}{n-2} = \frac {2n -4 + 4}{n-2} = 2 + \frac 4{n-2}$.

That can only be an integer if $n-2|4$.  Or in other words if $n-2 = 1,2$ or $4$ or if $n = 3, 4$ or $6$.

In particular if $n > 6$ then $4{n-2} < 1$ and can't be an integer.

......

Alternatively $\frac {2n}{n-2} > \frac {2n}{n} =2$ so $3$ is the smallest possible integer option.  And $\frac {2n}{n-2} -3$ occurs if $n = 6$.

If $n > 6$ then $\frac {2n}{n-2} < 3$


问:
Hahn-Banach Theorem applied to sphere
I need to know if what I do is right. And if what I do is ok, how to obtain "or $f(x)\geqslant c$, for all $x\in S_r$".

This is the exercise in my list that I'm solving:

Let $x_0\in S_r=\{x\in X;\|x\|=r\}$, where $X$ is a real normed space. Show that there exists a hyperplane $H=\{x\in X;f(x)=c\}$, with $f\in X^\prime$, which contains $x_0$ and keep the sphere $S_r$ in one of his sides, i.e., or $f(x)\leqslant c$, for all $x\in S_r$, or $f(x)\geqslant c$, for all $x\in S_r$.

My partial solution: Since $x_0\in S_r$, we have $\|x_0\|=r$, which implies $x_0\ne 0$. Then, it follows by a corollary of Hahn-Banach Theorem that there exists $f\in X^\prime$ such that $$\|f\|=1\qquad\text{and}\qquad f(x_0)=\|x_0\|.$$ Now, let $$H:=\{x\in X;f(x)=r\}.$$ We have that $x_0\in H$, as required, because $f(x_0)=\|x_0\|=r$. Besides, for all $x\in S_r$, it follows that $$f(x)\leqslant|f(x)|\leqslant\|f\|\|x\|=1\cdot r=r$$ $$\Longrightarrow\quad f(x)\leqslant r,\ \forall x\in S_r.$$ And we are almost done. But, what about the "or $f(x)\geqslant r,\forall x\in S_r$"?

答:
You are done. You needed to satisfy $f(x) \le r$ for all $x \in S_r$, or you needed to satisfy $f(x) \ge r$ for all $x \in S_r$. You don't need to satisfy both at the same time.

If you wanted to obtain the other condition, you could arbitrarily have taken $g = -f$, and $s = -r$. Then
$$x \in S_r \implies f(x) \le r \implies -f(x) \ge -r \implies g(x) \ge s.$$
So, we have a different functional where the inequality is swapped. The actual hyperplane has not been altered, but you can see why $\le$ or $\ge$ are interchangeable here, so long as you stick with the same one for all $x \in S_r$.

Title: Hahn-Banach Theorem applied to sphere


问:
Showing $\sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+r) = \frac{n(n+1)(n+2)\cdots(n+r+1)}{r+2}$
I was solving a question and saw a pattern. Can someone prove it, please?

We know

$$\sum_{k=1}^{n}k = \frac{n(n+1)}{2}$$
$$\sum_{k=1}^{n}k(k+1) = \frac{n(n+1)(n+2)}{3}$$
$$\sum_{k=1}^{n}k(k+1)(k+2) = \frac{n(n+1)(n+2)(n+3)}{4}$$

So we see the pattern... Can anyone give a proof of why:

$$\sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+r) = \frac{n(n+1)(n+2)\cdots(n+r+1)}{r+2}$$

Body:
You have

\begin{align}
k(k+1)(k+2)\cdots(k+r) &= k(k+1)(k+2)\cdots(k+r)\frac{r+2}{r+2}\\
&= k(k+1)(k+2)\cdots(k+r)\frac{(k+r+1)-(k-1)}{r+2}\\
&= \frac{k(k+1)(k+2)\cdots(k+r)(k+r+1)}{r+2}-\frac{(k-1)k(k+1)(k+2)\cdots(k+r)}{r+2}\\
\end{align}

Make the sum, you deduce easily that
$$\sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+r) = \frac{n(n+1)(n+2)\cdots(n+r+1)}{r+2}$$


问:
Having deposited $\$5,000$ at $9\%$ annual interest, for how many years can you withdraw $\$500$? What's a good way to iterate this?
NOTES:
I hope the tags are appropriate, this is a homework problem EDIT (another note): if it makes a difference, this question may involve calculus since it's in my calculus homework... I doubt it would make sense to use calculus in any way here

If you deposit $5,000$ dollars at $9\%$ annual interest, for how many years can you withdraw $500$ dollars (starting after the first year) without running out of money?

I've tried several equations, such as: $f(x)=5000(1.09)^x-500x$ but this never reaches $0$

The most recent thing I've tried is setting up an iteration equation: $x_{n+1}=x_n(1.09)-500$ where $x_0=5000$ which I will iterate until $x_{n+1}\le0$.

Question: Can I easily make Desmos (and the like) repeat iteration equations to infinity? AND Is there a way I can solve this problem without having to iterate (an equation that doesn't use the previous output as the next input)?

Bonus: If there's a way to turn my iteration into a "regular"<--(lack of technical terms) equation, can all iteration equations be made that way as well? Is there a general way to take an iteration equation and turn it into a simple $f(x)$ function?

答:
The problem with your formulation of $f(x)$ is that you're compounding the 9% for $x$ years and then removing $ \$500$ per year as an aggregate after $x$ years, not each year. You want to reformulate the function by compounding the $ \$500 $ every year too:

$$ f(x) = 5000*1.09^x - \sum_{k=1}^{x} {500*1.09^{k-1}} . $$

This is a geometric series, so you can find a closed-form solution to the sum, then iterate.


答:
This answer gives a proof by induction. Here instead I will give a proof using Hockey-stick identity:

$$\sum_{i=r}^n \binom ir = \binom{n+1}{r+1}$$

Note that

$$k(k+1)\cdots(k+r) = \frac {(k+r)!}{(k-1)!} = (r+1)!\binom{k+r}{r+1}$$

Hence your sum

$$=(r+1)!\sum_{k = 1}^{n} \binom {k+r}{r+1} =(r+1)!\sum_{k = {r+1}}^{n+r} \binom {k}{r+1} = (r+1)!\binom{n+r+1}{r+2} \\= \frac {(r+1)!(n+r+1)!}{(n-1)!(r+2)!} = \frac {n(n+1)\cdots(n+r+1)}{r+2}$$

Title: Showing $\sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+r) = \frac{n(n+1)(n+2)\cdots(n+r+1)}{r+2}$

Body:
I was solving a question and saw a pattern. Can someone prove it, please?

We know

$$\sum_{k=1}^{n}k = \frac{n(n+1)}{2}$$
$$\sum_{k=1}^{n}k(k+1) = \frac{n(n+1)(n+2)}{3}$$
$$\sum_{k=1}^{n}k(k+1)(k+2) = \frac{n(n+1)(n+2)(n+3)}{4}$$

So we see the pattern... Can anyone give a proof of why:

$$\sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+r) = \frac{n(n+1)(n+2)\cdots(n+r+1)}{r+2}$$


问:
The number of groups of order 32
There are 51 groups of order ($32=2^5$). My question is how this number was computed.

Graham Higman and Charles Sims gives an estimate for the number of $p$-goups (i.e. groups of order $p^n$ where $p$-is a prime number).

I will be very grateful for any suggestion.

答:
The number of groups is computed by explicitly constructing all groups. This is done (originally by hand, nowadays on the computer -- see the paper by Besche/Eick/Obrien mentioned in Dietrich Burde's reply) by considering the ways groups of the given order can be formed as extensions. BAsically the steps are:

Assume (by induction) groups of smaller order have been classified

For each group $G$, classify the irreducible modules. For each module $M$ such that $|M|\cdot|G|$ has the right order. For prime powers only the trivial module needs to be considered.

Compute the 2-cohomology group $H^2(G,M)$ and the corresponding extensions

Eliminate further isomorphisms (i.e. isomorphisms that do not preserve the extension structure. (This is basically the costly part and the achilles heel of any such classification.)

Groups that have no solvable normal subgroup need to be handled separately (with simple groups as a seed). Essentially one needs to classify subgroups of wreath products $Aut(T)\wr S_m$ for $T$ simple and small $m$ (and subdirect products of these groups). For any plausible order bound this mosly degenerates to groups $T\le G\le Aut(T)$.

Title: The number of groups of order 32


问:
Terminology for graphs whose vertices can be graphs
Let $W_0$ be some fixed, finite set of "atomic vertices". Noting $2^X$ the power set of some set $X$, we consider $\bigl\{ (V, E) \bigm| V \in 2^{W_0}, E \in 2^{\{ (s, t) | s, t \in V \}} \bigr\}$, the set of all graphs whose vertices come from $W_0$. We are interested in the case where such graphs could appear as vertices in a graph (together with "atomic vertices" from $W_0$), and recursively consider such graphs-of-graphs as vertices in "higher-order" graphs

$$\forall n \in \mathbb{N}, \quad W_{n+1} = W_n \cup \bigl\{ (V, E) \bigm| V \in 2^{W_n}, E \in 2^{\{ (s, t) | s, t ∈ V \}} \bigr\}\quad .$$

I'm considering a Computer Science and/or Knowledge Graph and/or Philosophy application involving such "graphs-of-graphs". Before reinventing the wheel, I would like to know what the literature has to say about this, but I don't know where to start looking.

Are you aware of any terminology/conventions/notations for such "graphs"? Are there issues/paradoxes that I should be aware of in the limit $n \to \infty$? Are there works, especially accessible ones, that you would recommend on the topic? Thanks!

Following Mike's Comment/Answer, I decided to add a few explicit examples.

First suppose that $W_0 = \emptyset$, in which case $W_1 = \bigl\{\bigr (\emptyset, \emptyset) \}$ because the only graph you can build with no vertices is the empty graph (with no vertices nor edges). But for $W_2$, we may treat the graph $(\emptyset, \emptyset)$ as a vertex, giving

\begin{align}
W_2 = \Bigl\{ & (\emptyset, \emptyset),\\ & \bigl(\{(\emptyset, \emptyset)\}, \emptyset\bigr),\\ & \bigl(\{(\emptyset, \emptyset)\}, ((\emptyset, \emptyset), (\emptyset, \emptyset))\bigr) \Bigr\} \quad ,
\end{align}

i.e., the set comprised of 1. the empty graph, 2. a graph with no edge and the empty graph as a vertex, and 3. a graph with a single edge joining the empty graph to itself (self-loop). We could then continue with $W_3$, this time with 3 vertices available as building blocks.

In the limit $n \to \infty$, I believe that recursive definitions like the following would be valid

$$g = \Bigl( \bigl\{(\emptyset, \emptyset), g\bigr\}, \bigl\{ \bigl((\emptyset, \emptyset), g\bigr) \bigr\} \Bigr) \quad ,$$

i.e., the graph $g$ has the empty graph and itself as vertices, with a single edge oriented from the empty graph to the graph $g$. But this is the kind of "hand waving" that could eventually lead to issues/paradoxes...

Up to this point, I considered $W_0 = \emptyset$, but what if we allow for some "atomic vertices"? My understanding is that if $W_0$ only contains graphs that would have been valid in the above (e.g., $W_0 = \{(\emptyset, \emptyset)\}$), this will affect "shallow" layers but leave the limit $n \to \infty$ unaltered. However, if we instead add something new (and perhaps not even a graph, say $W_0 = \{42\}$), then this would increase the expressiveness of limit $n \to \infty$.

Please note that I have a very "applied" perspective here: I'm thinking of such graphs as a candidate tool to represent concepts. As an example of where I'm coming from, please consider Figure 19(c) from https://arxiv.org/abs/2003.02320 . Ignoring the edge labels "flight" and "valid from", this knowledge graph could be represented as

\begin{align}
\Bigl(\Bigl\{ & \bigl( \bigl(\{\text{Santiago}, \text{Arica}\},\{\bigl(\text{Santiago}, \text{Arica}\bigr)\}\bigr), 1956 \Bigr\},\\
\Bigl\{& \bigl( \bigl(\{\text{Santiago}, \text{Arica}\},\{\bigl(\text{Santiago}, \text{Arica}\bigr)\}\bigr) , 1956\bigr)\Bigr\}\Bigr) \quad .
\end{align}

This kind of representation is already quite powerful, and I'm wondering how far we can push it.

答:
For whatever reason, my internet browser won't let me get into writing a comment, which is what this is.

As far as this specific example, I am not sure. But, for comprehension, better to both write in in concise formal mathematical language, and then explain a bit less formally with words. Meanwhile, is the graph you wrote out even well-defined? So $S,T \in \cal{P}(S_{n-1})$ are adjacent in $G_n$
iff $|S|=|T|$ and $s_it_i \in E(G_{n-1})$ for each $i=1,2,\ldots, |S|$ and some ordering of $S$ and $T$?

IN GENERAL:Your best bet is to just say explicitly what you mean. Say "Let $G$ be a graph. And then in turn, we replace each vertex $v \in V(G)$ with a graph $H_v$ of the following structure, and then we put an edge between a vertex $u_v \in H_v$
and $u_w \in H_w$ iff $vw$ is an edge in $G$, and ...."

The term commonly used for what you are describing is "graph product", but there are many flavors of such graph products i.e., do you replace each edge $vw$ in $G$ with a complete bipartite graph between $V(H_v)$ and $V(H_w)$ or just a matching between $V(H_v)$ and $V(H_w)$, or even just one edge; are the $H_v$s; $v \in V(G)$ all the same graph, and even, are the $H_v$s vertex-disjoint.


问:
Curve defined by a vector
http://i.stack.imgur.com/tD4Bn.png

I'm studying line integrals with a curve as a vector, but I couldn't understand the 'dr' part.

 First of all: the curve isn't really a curve, it's like some points where a vector points from the origin. So this 'curve' does not really exists, rigth?

 And then, when we take the integral with respect to 'dr', shouldn't we take it do ||dr|| (the magnitude of dr)? I really didn't understand, so I need some better and intuitive explanation...

Thank you so much!

答:
It depends on if you're integrating over a scalar field or a vector field.

Define a path $C:\vec{r}(t)$ for $a\le{t}<b$. If you are integrating over a vector field $\vec{F}(\vec{x})$, the integral is

$$\int_C\vec{F}(\vec{x})\cdot{d\vec{r}}$$

If you are integrating over a scalar field, the integral is

$$\int_C{F(\vec{x})|d\vec{r}|}$$

Title: Curve defined by a vector


问:
Show that $\lim_{n \to \infty} (\int_{0}^{\infty}f_n(x)dx) \not= \int_0^\infty f(x)dx$
There is a sequence of functions, $f_{1},f_{2},f_{3},\ldots$, whose domain is $[0, \infty)$, such that $$f_{n}(x)=\begin{cases}
      \frac{1}{n} & \text{if } x \in [0, n) \\
      \frac{-1}{n}x+1+\frac{1}{n} & \text{if } x \in [n, n+1) \\
      0 & \text{if } x \in [n+1, \infty) \\
   \end{cases}$$

Show that $\lim_{n \to \infty} (\int_{0}^{\infty}f_n(x)dx) \not= \int_0^\infty f(x)dx$. So they should not be the same as the integral of the limit. As to how I can solve it, I tried but is not successful. Actually, I try to convert this is improper integral, to normal integral. I am stuck though.

答: The functions get shorter and shorter as $n$ increases, and each slants down to zero between $n$ and $n+1$. The integrals of the individual functions are all, one plus the area of the small triangle, which gets smaller as $n$ increases.  So the limit of the integrals is $1$.  The integral of the limit, however is different, because the limit of the sequence of functions is the zero function, which integrates to zero.


问:
Show that $\exists \delta > 0, \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\sin(xk^n)|\ge \delta$.

Let $k \ge 2$, $k \in \Bbb N$. Show that $\exists \delta > 0, \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\sin(xk^n)|\ge \delta$.

My intuition tells me that we can pick $\delta=1/2$. I tried to study $u_n = \sin(xk^n)=U_{k^n}(\sin(x))=U_k(u_{n-1})$. I thought of introducing a variant of Chebyshev polynomials $U_n(\sin(x))=\sin(nx)$. But it didn't help much. Does someone have a hint? Thanks. (here $\Bbb N = \{0,1,2,\ldots\}$.)

I needed this result for solving this exercice (which is a oral exam that was posed during the competitive exams of Polytechnique, #1 engineering school in France).

Show that for $f(z)=z^k$ with $k \ge 2$
$$
\exists \delta >0, \forall x,y \in \Bbb U, x \neq y \implies \exists n \in \Bbb N, |f^{(n)}(x)-f^{(n)}(y)| \ge \delta.
$$

Suppose $x \neq 0$ and $y/x=e^{i \theta}$,
\begin{align}
| f^{(n)}(x) - f^{(n)}(y) | &= |x^{k^n} - y^{k^n}|
=
 | 1- e^{i \theta k^n}|\\
&= | e^{-i \theta k^n/2} - e^{i \theta k^n/2}|
= |2i \sin(\theta k^n/2)| = 2|\sin (\theta k^n/2)|
\end{align}
So
$$
|f^{(n)}(x)-f^{(n)}(y)| \ge \delta \iff |\sin (\theta k^n/2)| \ge \delta/2
$$

答:
I only really see one way to solve this problem, though how you write the details can vary considerably. You asked for an "elegant" solution, so here's my arrangement.

Key Lemma: Let $y \in (0, 1)$ and $k \in \mathbb{Z}_{\geq 2}$. Then
$$\{yk^n \pmod{1} : n \in \mathbb{Z}_{\geq 0}\} \cap [1/k^2, 1-1/k^2] \neq \varnothing.$$

Proof: Consider the base $k$ expansion of $y$, so $y = 0{.}y_1 y_2 y_3{\ldots}$ and $yk^n \pmod{1} = 0{.}y_{n+1} y_{n+2}{\ldots}$. If the latter expression is not in $[1/k, 1-1/k]$, we must have $y_{n+1} = 0$ or $y_{n+1} = k-1$. So, we may suppose that $y_{n+1} \in \{0, k-1\}$ for all $n \in \mathbb{Z}_{\geq 0}$. Since $y \neq 0, 1$ by assumption, we have some $N$ where $y_{N+1} \neq y_{N+2}$. Hence $yk^N \pmod{1} = 0{.}y_{N+1} y_{N+2}{\ldots} = 0{.}0(k-1){\ldots}$ or $0{.}(k-1)0{\ldots}$. In the first case, $yk^N \pmod{1} \in [1/k^2, 1/k]$, while in the second case, $yk^N \pmod{1} \in [1-1/k, 1-1/k^2]$. $\Box$

Now set $\delta = \sin(\pi/k^2) > 0$. For all $x \in (0, \pi)$, by the key lemma applied to $y = x/\pi$, we have some $n \in \mathbb{Z}_{\geq 0}$ such that $xk^n \pmod{\pi} \in [\pi/k^2, \pi-\pi/k^2]$. From the unimodality and symmetry of $\sin$, it follows that $|\sin(xk^n)| \geq |\sin(\pi/k^2)| = \delta$.


问:
What happens in the long run?
Suppose we have a round table with 4 boys, each having a bowl with cereals. There is an adult in the same room. When the adult leaves the room, the boys start stealing each other's cereal, but they can $\textbf{only}$ steal from its neighbor as they $\textbf{cannot}$ leave their seat. When the adult comes back, they stop stealing each other's cereal.

${\bf question:}$ If the adult leaves and comes back $N$ times, can we say about the number of cereals in each boy's bowl? Are they eventually all have the same number of cereals?

${\bf thought}:$ We can start by letting $x_i$ be the number of cereal in ith bowl. So, initially we can encode our situation in a vector ${\bf x} = (x_1,x_2,x_3,x_4)$. After the adult leaves the room, now $x_1$ becomes $x_1 + k_1 x_2 + k_2 x_4$ ;
$x_2$ becomes $x_2 + k_3 x_1 + k_4 x_3$; $x_3$ becomes $x_3 + k_5 x_2 + k_6 x_4$ and $x_4$ becomes $x_4 + k_7 x_3 + k_8 x_1$, so after one step we obtain new vector

$$ {\bf y }  = \left( \begin{matrix}  x_1 + k_1 x_2 + k_2 x_4 \\ x_2 + k_3x_1 + k_4 x_3 \\  x_3 + k_5 x_2 + k_6 x_4 \\ x_4 + k_7 x_3 + k_8 x_1 \end{matrix} \right) = \underbrace{ \left( \begin{matrix} 1 & k_1 & 0 & k_2 \\ k_3 & 1 & k_4 & 0 \\ 0 & k_5 & 1 & k_6 \\ k_8 & 0 & k_7 & 1 \end{matrix} \right) }_{A} {\bf x}$$

So, in order to solve this problem, we may have to compute $A^N {\bf x}$ which seems pretty laborious.

Is this a correct approach to the problem?

${\bf Update:}$
Let us assume that at each round, each boy takes half the number of cereals from the neighbor. In this case, then

$$ {\bf y} = \left( \begin{matrix} (x_2+x_4)/2 \\ (x_1+x_3)/2 \\ (x_2+x_4)/2 \\ (x_1+x_3)/2 \end{matrix} \right) $$

答: In the special case, notice that each boy doesn't keep any of his orignal cereal - it gets shared evenly between his two neighbours. So even if boy 1 and boy 3 initially have different amounts, boys 2 and 4 each have $\frac{x_1+x_3}{2}$ cereal after 1 round, and boys 1 and 3 each have $\frac{x_2+x_4}{2}$ cereal after 1 round. After that, the cereal amounts keep alternating - after an even number of rounds, boys 1 and 3 have $\frac{x_1+x_3}{2}$ and boys 2 and 4 have $\frac{x_2+x_4}{2}$, and after an odd number of rounds, boys 1 and 3 have $\frac{x_2+x_4}{2}$ cereal and boys 2 and 4 have $\frac{x_1+x_3}{2}$.

Title: What happens in the long run?


问:
Maximization of the product of two inner products
I have an optimization problem of the form

$\max_\gamma \langle f\circ\gamma,w_f\rangle\langle g\circ\gamma,w_g\rangle$

where $f\circ\gamma$ is the composition of $f$ and $\gamma$ and the inner product is defined as

$\langle f\circ\gamma,w_f\rangle = \int (f\circ\gamma)(t) w_f(t)\,dt$

In other words I am trying to find a $\gamma$ that maximizes the product of the two inner products above. I know one could pursue a gradient approach, which would be slow. Is there a more direct optimization method?

答:
If $f$ and $g$ are differentiable, you could do something like the following.

Define $J(\gamma)$ as your objective function. Then, the variation of $J$ in direction $h$ is given by
\begin{align*}J'(\gamma; h)
&= \int f'(\gamma(t)) \, h(t) \, w_f(t) \, \mathrm d t \, \int g(\gamma(t)) \, w_g(t) \, \mathrm d t + \int f(\gamma(t)) \, w_f(t) \, \mathrm d t \, \int g'(\gamma(t)) \, h(t) \, w_g(t) \, \mathrm d t\\
&= \int \Bigg\{\Big( \int g(\gamma(s)) \, w_g(s) \, \mathrm d s \Big) f'(\gamma(t)) \, w_f(t) + \Big(\int f(\gamma(s)) \, w_f(s) \, \mathrm d s\Big) \, g'(\gamma(t)) \, w_g(t) \, \Bigg\} \, h(t) \mathrm d t\\
\end{align*}
Now, this has to be zero for all $h$. Hence, for all $t$, you get the equation
$$\Big( \int g(\gamma(s)) \, w_g(s) \, \mathrm d s \Big) f'(\gamma(t)) \, w_f(t) + \Big(\int f(\gamma(s)) \, w_f(s) \, \mathrm d s\Big) \, g'(\gamma(t)) \, w_g(t) = 0$$
Now, it depends on the functions how this equation could be solved.

Maybe, this is of some use.


答:
The special case is fairly straightforward with the theory of probability matrices (matrices with nonnegative entries, all of whose columns add up to $1$). You have a transition matrix that is
$$A = \left(\begin{array}{cccc}
0 & \frac{1}{2} & 0 & \frac{1}{2}\\
\frac{1}{2} & 0 & \frac{1}{2} & 0\\
0 & \frac{1}{2} & 0 & \frac{1}{2}\\
\frac{1}{2} & 0 & \frac{1}{2} & 0
\end{array}\right).$$
Because every column adds up to $1$, this is a probability matrix; all eigenvalues $\lambda$ satisfy $|\lambda|\leq 1$, $1$ is definitely an eigenvalue, and the long-term behavior is dominated by the eigenvectors corresponding to eigenvalues of absolute value $1$.

Here, the matrix clearly has rank $2$, so the eigenvalue $\lambda=0$ has algebraic multiplicity $2$: the nullspace has basis $\mathbf{v}_3=(1,0,-1,0)^t$ and $\mathbf{v}_4=(0,1,0,-1)^t$.

The eigenvalue $\lambda = 1$ has eigenvector $(1,1,1,1)^t$, which we will want to normalize (make the entries add up to $1$) to the eigenvector $\mathbf{v}_1=(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4})^t$. And the fourth eigenvalue in this case is $\lambda=-1$, which has corresponding eigenvector $\mathbf{v}_2=(1,-1,1,-1)^t$.

If $\mathbf{x}_0$ is the initial distribution, then we can express $\mathbf{x}_0$ in terms of the basis $\mathbf{v}_1,\ldots,\mathbf{v}_4$. If we have
$$\mathbf{x}_0 = \alpha_1\mathbf{v}_1 + \alpha_2\mathbf{v}_2 + \alpha_3\mathbf{v}_3 + \alpha_4\mathbf{v}_4,$$
then after applying $A$ to $\mathbf{x}_0$ $n$ times you will get
$$A^n\mathbf{x}_0 = \alpha_1\mathbf{v}_1 + (-1)^N\alpha_2\mathbf{v}_2.$$
Now, because the entries of $\mathbf{v}_0$ add up to $1$ and the entries of $\mathbf{v}_2$ add up to zero, we see that $\alpha_1$ is just the total amount of cereal we started with.

The summand $(-1)^N\alpha_2$ makes this a bit tricky (it’s easier when you have an odd number of kids, because then you don’t get an eigenvalue $-1$). A way to simplify the analysis is to replace the basis $\mathbf{v}_1,\mathbf{v}_2$ of the image with
$$\begin{align*}
\mathbf{w}_1 &= \mathbf{v}_1 + \frac{1}{4}\mathbf{v}_2 = \left(\frac{1}{2},0,\frac{1}{2},0\right)^t\\
\mathbf{w}_2 &= \mathbf{v}_1 - \frac{1}{4}\mathbf{v}_2 = \
\left(0,\frac{1}{2},0\frac{1}{2}\right)^t.
\end{align*}$$
Then the action of $A$ is to swap $\mathbf{w}_1$ and $\mathbf{w}_2$.

So if we have an initial cereal distribution of $(x_1,x_2,x_3,x_4)$, then after $N$ steps, you will have:

If $N$ is odd, then each of kids 1 and 3 have the average amount of cereal that kids 2 and 4 had (that is, half the sum of their totals), while kids 2 and 4 will have the average amount that kids 1 and 3 had.

If $N$ is even, then each of kids 1 and 3 has the average amount of cereal that kids 1 and 3 had to start with, and kids 2 and 4 will have half the average amount of cereal that kids 2 and 4 had to start with.

Basically, after the first step, you are just swapping cereal amounts between adjacent kids; you could just shuffly the bowls clockwise at each step and get the right distribution.

(If there were five kids, each taking half from its neighbor, then the limiting process has each kid with one fifth the total amount of cereal served. More generally, if you have $K$ kids, if $K$ is odd then at the limit they all have the same amount, and if $K$ is even, then you divide the cereal into that served to odd kids and that served to even kids, and eventually you will be swapping the averages of each half between the kids)

For the general case you will have something similar: every column will add up to $1$ so you still have a probability matrix, so the long-term behavior is dominated by the eigenvalue $1$ (and possibly $-1$, if it is an eigenvalue). The eigenvectors corresponding to eigenvalues different from $1$ all will have entries that add up to $0$, so the coefficient of the eigenvector corresponding to $1$ (once normalized) is just the sum of the entries of the initial condition. That multiple of the eigenvector of $1$ is the limiting behavior. So all you need to do is (1) determine if $-1$ is an eigenvalue; and (2) find the eigenvectors corresponding to $1$ (and if necessary to $-1$).

Title: What happens in the long run?

Body:
Suppose we have a round table with 4 boys, each having a bowl with cereals. There is an adult in the same room. When the adult leaves the room, the boys start stealing each other's cereal, but they can $\textbf{only}$ steal from its neighbor as they $\textbf{cannot}$ leave their seat. When the adult comes back, they stop stealing each other's cereal.

${\bf question:}$ If the adult leaves and comes back $N$ times, can we say about the number of cereals in each boy's bowl? Are they eventually all have the same number of cereals?

${\bf thought}:$ We can start by letting $x_i$ be the number of cereal in ith bowl. So, initially we can encode our situation in a vector ${\bf x} = (x_1,x_2,x_3,x_4)$. After the adult leaves the room, now $x_1$ becomes $x_1 + k_1 x_2 + k_2 x_4$ ;
$x_2$ becomes $x_2 + k_3 x_1 + k_4 x_3$; $x_3$ becomes $x_3 + k_5 x_2 + k_6 x_4$ and $x_4$ becomes $x_4 + k_7 x_3 + k_8 x_1$, so after one step we obtain new vector

$$ {\bf y }  = \left( \begin{matrix}  x_1 + k_1 x_2 + k_2 x_4 \\ x_2 + k_3x_1 + k_4 x_3 \\  x_3 + k_5 x_2 + k_6 x_4 \\ x_4 + k_7 x_3 + k_8 x_1 \end{matrix} \right) = \underbrace{ \left( \begin{matrix} 1 & k_1 & 0 & k_2 \\ k_3 & 1 & k_4 & 0 \\ 0 & k_5 & 1 & k_6 \\ k_8 & 0 & k_7 & 1 \end{matrix} \right) }_{A} {\bf x}$$

So, in order to solve this problem, we may have to compute $A^N {\bf x}$ which seems pretty laborious.

Is this a correct approach to the problem?

${\bf Update:}$
Let us assume that at each round, each boy takes half the number of cereals from the neighbor. In this case, then

$$ {\bf y} = \left( \begin{matrix} (x_2+x_4)/2 \\ (x_1+x_3)/2 \\ (x_2+x_4)/2 \\ (x_1+x_3)/2 \end{matrix} \right) $$


答:
Let us count the number of subsets of $\{ 0, 1, \dots, n + r\}$ of size $r+2$ in two different ways. One way, of course, is $$\binom{n+r+1}{r+2} = \frac{n(n+1)\cdots (n+r+1)}{(r+2)!}.\tag{1}$$
For the other way, let us first pick the largest element that will be in our set, and call it $j$. Thus, the remaining $r+1$ elements are chosen from $\{0, 1, \dots, j-1\}$. Summing over all values of $j$, we obtain $$\sum_{j=r+1}^{n+r} \binom{j}{r+1}.$$ Writing $k = j - r$, we find that this sum equals $$\sum_{k=1}^n \binom{k+r}{r+1} = \sum_{k=1}^n\frac{k(k+1)\cdots(k+r)}{(r+1)!}.\tag{2}$$
Equating $(1)$ and $(2)$ and cancelling $(r+1)!$ in the denominator, we obtain the result.


问:
Equation with Logarithm in Exponent
How to I solve the following exercise with a logarithm? I've forgotten the "trick" for doing so:

$x^{log_{10} x} =10^4$

答:
The trick: $a^b=e^{b\ln a}$.  Applied to this problem: $$x^{\log_{10}x}=e^{\ln x \log_{10}x}=e^{\ln x \frac{\ln x}{\ln 10}}=e^{(\ln x)^2/(\ln 10)}$$

On the other hand $10^4=e^{4\ln 10}$, so we may take the log of both sides to get $$\frac{(\ln x)^2}{\ln 10}=4\ln 10$$

Cross-multiply, take the square root, exponentiate, and you're done.

Title: Equation with Logarithm in Exponent


问:
proving that $89 \mid 2^{44}-1$
i tried to prove that $2^{44} \equiv 1\pmod{89}$.

I noticed that by Fermat's little theorem $2^{88} \equiv 2^{44}\cdot 2^{44} \equiv 1\pmod{89}$
which means that $2^{44}$ is the inverse of itself $\rightarrow 2^{44} \equiv 1 \pmod{89}$ or $2^{44}\equiv 88 \pmod{89}$.

how can I rule out the second option?

答:
By Second Supplement to Law of Quadratic Reciprocity, $\left(\dfrac2p\right) = 1$ for $p \equiv 1 \pmod 8$.

i.e. there exists some $a \in \mathbb Z$ such that $a^2 \equiv 2 \pmod {89}$.

$a^{88} \equiv 1 \pmod {89}$ follows from Fermat's Little Theorem.

Hence $2^{44} \equiv a^{88} \equiv 1 \pmod {89}$.

Title: proving that $89 \mid 2^{44}-1$


问:
Step-by-step process to find $3f(x)+f(g(x))$ and $4f(g(x))+2f(-3)$ when $f(x)=e^{x^2}-1$ and $g(x) = \sqrt{x^2-1}$
I can't understand my teacher explanation about how to solve this kind of operations with functions (I'm sorry for my english)

He gave us these functions

$$f(x)=e^{x^2}-1 \qquad g(x) = \sqrt{x^2-1}$$

And I need to find those values in these other functions

a) $\quad 3f(x)+f(g(x))$

b) $\quad 4f(g(x))+2f(-3)$

I will not lie, I have no idea how to solve this, I know i need to find the first functions inside the others but not idea how or what steps to follow, can you explain it to me please? I'm begging you, thanks.

Answer to A)

$\quad 3f(x)+f(g(x))$

$\quad 3f(e^{x^2}-1)+f(\sqrt{x^2-1} )$

$\quad 3(e^{x^2}-1)+(e^{x^2}-1) $

$\quad 3e^{x^2}-3+e^{x^2}-1$

答:
You do not really "solve" anything here.

If the value of $x$ is given, you can compute the value of $f(x), g(x)$ etc., right?

Once you know the values of $f(x)$ and $g(x)$, you can easily compute the other expressions involving $f(x)$ and $g(x)$

Let $f(x) = e^{x^2} - 1$ and $g(x) = \sqrt{x^2 - 1}$

For example, if you are to calculate $f(1)$, you need to replace $x$ by $1$ in the expression for $f(x)$

$f(1) = e^{1^2} - 1 = e - 1$ and $g(1) = \sqrt{1^2 - 1} = 0$

Similarly, you replace $x$ by $g(x)$ in the expression for $f(x)$ when you need to evaluate $f(g(x))$

$f(g(x)) = e^{{g(x)}^2} - 1 = e^{x^2 - 1} - 1$

$3f(x) + f(g(x)) = 3(e^{x^2} - 1) + e^{x^2 - 1} - 1$


问:
Convergence of square series
Let $\sum a_n$ be a series that converges absolutely. Prove that $\sum a_n^2$ also converges.

My attempt:

If $\sum |a_n|$ converges, then $\sum an$ converges and $\lim_{n\to\infty} a_n = 0$. There is an $M$ so that for all $n \geq M$

$$|a_n| < 1 \implies |a_n^2| < |a_n|$$

By the direct comparison test, $\sum |a_n^2|$ converges because $\sum |a_n|$ converges. We may conclude that $\sum a_n^2$ converges.

QED.

Is this proof correct? If so, were the absolute values necessary or could this have been done without the absolute values? Any assistance much appreciated.

答:
Your proof is mostly correct. There is one fairly inconsequential error:
$$|a_n| < 1 \implies |a_n^2| < |a_n|.$$
It really should be
$$|a_n| < 1 \implies |a_n^2| \color{red}\le |a_n|,$$
as the strict inequality doesn't account for the possibility that $|a_n| = 0$. But, with or without the strict inequality, the comparison test still applies.

Note that the comparison test requires that the terms of both series be non-negative. That is, the comparison test is a kind of squeeze theorem for series, where a series is squeezed between a convergent series and the $0$ series. Indeed, you can generalise the comparison test so that the lower bound need not be the $0$ series.

So, for this reason, we need absolute values for this argument to work. It turns an absolutely convergent series into a non-negative series. As a specific counterexample where we don't have absolute convergence, recall that
$$\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}$$
converges conditionally by the Leibniz Alternating Series test. Its square is the harmonic series, which famously diverges. So, $\sum a_n$ converging doesn't mean $\sum a_n^2$ converges, when conditional convergence is involved.

Title: Convergence of square series


问:
Showing sequence is Cauchy in L^1 norm
Let

$$f_n(x)=\begin{cases}
\left(x+\frac{1}{2}\right)^n,&\text{if }0\le x<\frac{1}{2}\\\\
1,&\text{if } \frac{1}{2}\le x\le 1\;.
\end{cases}$$

Then clearly $(f_n)$ is sequence in $C[0,1]$. I need to show this is cauchy sequence in $C[0,1]$ with $L^1$ norm.

My attempt  By definition of cauchy sequence, i think we need establish

$||f_n-f_m||<\epsilon$ for sufficiently large $m, n$

i.e. $\int^{1}_0 |f_n(x)-f_m(x)|dx<\epsilon$ for sufficiently large $m, n$

But,

$\int^{1}_0 |f_n(x)-f_m(x)|dx\\
=\int^{1/2}_0 |(x+1/2)^n-(x+1/2)^m|dx$

(Since on $[1/2,1]$ both $f_n$, $f_m$ are equal for all $m, n$ )

In next step i had used traingle inequality and spit the integral into two parts and obtained,

$\int^{1}_0 |f_n(x)-f_m(x)|dx\\
≤\int^{1/2}_0 |(x+1/2)^n| dx+\int^{1/2}_0 |(x+1/2)^m| dx$\
$=\frac{1}{n+1}[1-(1/2)^{n+1}]+\frac{1}{m+1}[1-(1/2)^{m+1}]$

But i am unable to obtain conclusion

Further, i like to know is there is any direct approach to conclude it is Cauchy sequence?

答:
Let $\varepsilon > 0$ be fixed. Note that
\begin{eqnarray*}
\int_{0}^{1/2} \left| (x + 1/2)^n - (x + 1/2)^m \right| \, dx & \leq & \int_{0}^{1/2}  \left| (x + 1/2)^n \right| \, dx + \int_{0}^{1/2}  \left| (x + 1/2)^n \right| \, dx \\
& = & \left[\frac{1}{n+1} - \frac{(1/2)^{n+1}}{n + 1} \right] + \left[ \frac{1}{n+1} - \frac{(1/2)^{m+1}}{m + 1} \right]\\
& < & \frac{1}{n} + \frac{1}{m}.
\end{eqnarray*}
From here, can you determine what constraints you must put on $n$ (resp. $m$) so tha $\dfrac{1}{n} < \dfrac{\varepsilon}{2}$?

Title: Showing sequence is Cauchy in L^1 norm


Title: Step-by-step process to find $3f(x)+f(g(x))$ and $4f(g(x))+2f(-3)$ when $f(x)=e^{x^2}-1$ and $g(x) = \sqrt{x^2-1}$

Body:
I can't understand my teacher explanation about how to solve this kind of operations with functions (I'm sorry for my english)

He gave us these functions

$$f(x)=e^{x^2}-1 \qquad g(x) = \sqrt{x^2-1}$$

And I need to find those values in these other functions

a) $\quad 3f(x)+f(g(x))$

b) $\quad 4f(g(x))+2f(-3)$

I will not lie, I have no idea how to solve this, I know i need to find the first functions inside the others but not idea how or what steps to follow, can you explain it to me please? I'm begging you, thanks.

Answer to A)

$\quad 3f(x)+f(g(x))$

$\quad 3f(e^{x^2}-1)+f(\sqrt{x^2-1} )$

$\quad 3(e^{x^2}-1)+(e^{x^2}-1) $

$\quad 3e^{x^2}-3+e^{x^2}-1$

Body:
Let $f,g:I \to \mathbb{R}$, when $I\subset \mathbb{R}$we define

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

$(f \cdot g)(x)=f(x)g(x)$

$(f \circ g)(x)=f(g(x))$ it is first apply $g$ and then $f$ it iff $g(x)\in \text{Dom $f$}$

In your case $f(x)=e^{x^2}-1$ and $g(x)=\sqrt{x^2-1}$ I will help you with $b)$

$$4f(g(x))+2f(-3)=4f(\sqrt{x^2-1})+2(e^{9}-1)$$
$$ =4(e^{x^2-1}-1)+2(e^{9}-1)$$
$$ =4e^{x^2-1}-4+2e^{9}-2$$
$$ \fbox {$ 4e^{x^2-1}+2e^{9}-6 $}$$
You should be able to do the $a)$
Hint
Notice that $f(a)$ is the value of put the number $a$ in the formula given by $f$

$$3f(x)+f(g(x))=3(e^{x^2}-1)+e^{x^2-1}-1$$
$$=3e^{x^2}-3+e^{x^2-1}-1$$
$$\fbox { $3e^{x^2}+e^{x^2-1}-4$ }$$


Title: proving that $89 \mid 2^{44}-1$

Body:
i tried to prove that $2^{44} \equiv 1\pmod{89}$.

I noticed that by Fermat's little theorem $2^{88} \equiv 2^{44}\cdot 2^{44} \equiv 1\pmod{89}$
which means that $2^{44}$ is the inverse of itself $\rightarrow 2^{44} \equiv 1 \pmod{89}$ or $2^{44}\equiv 88 \pmod{89}$.

how can I rule out the second option?

Body:
You have $2^{11}= 2048 \equiv 1 \pmod {89} $.
Hence,
$$2^{44}=(2^{11})^4 \equiv 1 \pmod {89} $$


答:
Notice that $$\lim_{n\to \infty}f_n(x)=f(x)\equiv 0 $$ so $$\int_0^\infty f(x)dx=0 $$ on the other hand, $$I_n=\int_0^\infty f_n(x)dx =\int_0^n f_ndx+\int_n^{n+1}f_ndx+\int_{n+1}^\infty f_ndx$$
so $$I_n= \int_0^n \frac{1}{n}  dx-\int_n^{n+1} \frac{1}{n}x-1-\frac{1}{n}  dx+\int_{n+1}^\infty  0dx$$ then
$$ I_n = 1+\frac{1}{2n}$$ so we see taking $\lim_{n\to\infty} I_n=1\neq 0$. And the result follows.

Body:
There is a sequence of functions, $f_{1},f_{2},f_{3},\ldots$, whose domain is $[0, \infty)$, such that $$f_{n}(x)=\begin{cases}
      \frac{1}{n} & \text{if } x \in [0, n) \\
      \frac{-1}{n}x+1+\frac{1}{n} & \text{if } x \in [n, n+1) \\
      0 & \text{if } x \in [n+1, \infty) \\
   \end{cases}$$

Show that $\lim_{n \to \infty} (\int_{0}^{\infty}f_n(x)dx) \not= \int_0^\infty f(x)dx$. So they should not be the same as the integral of the limit. As to how I can solve it, I tried but is not successful. Actually, I try to convert this is improper integral, to normal integral. I am stuck though.


问:
Derivation of Burgers' Equation
I'm aware that it is possible to reduce Navier-Stokes to Burgers by neglecting pressure, and that one can derive the inviscid form by considering an ideal gas and concluding that the convective acceleration is zero. My question is whether there is a way to extend the second example to include kinematic viscosity $\nu$, i.e. to derive
$$ \frac{\partial u}{\partial t} + u\cdot \nabla u  = \nu\nabla^2 u $$
in one dimension. I came across an exercise in James Haberman's "Applied Partial Differential Equations" but think I lack the prerequisites to tackle it. Any references or suggestions would be greatly appreciated.

答:
Professor Falkovich seems to provide a general derivation for Berger's equation starting from Navier-Stokes equation at about 19:00 in http://scgp.stonybrook.edu/video_portal/video.php?id=3160
Although it's not clear to me yet what exactly $u$ corresponds to in the end.

Title: Derivation of Burgers' Equation


问:
Integrate $\iint_D 16\cos((4x+2y)^2)dxdy$ with help of contour lines.
$$\iint_D 16\cos((4x+2y)^2)dxdy, \\D: 12\leq4x+2y\leq24\\x\geq0\\y\geq0$$

So one method I've learned to deal with this kind of problem is by integration along the contour lines. So if I substitute as $t=4x+2y$ and define my area function as $A(t)=\frac{t^2}{2}$.

From here I'm trying to use: $$\iint_D f(g(x,y))dxdy=\int_a^bf(t) \ A'(t) dt$$

In my case this yields: $$\iint_D 16\cos((4x+2y)^2)dxdy=\int_{12}^{24}16\cos(t^2)\ t\ dt=8(\sin(24^2)-\sin(12^2))$$

I am not quite sure where my mistake is.

答:
I've never seen this method for evaluating a double integral before, but I think I get the idea. Pretty cool stuff! Here's my take on the problem.

For $a,b\in[12,24]$ with $a<b$ define a set $\Omega(a,b)$ by $$\Omega(a,b)=\{(x,y)\in \mathbb{Q}_1\big|a<4x+2y<b\}$$ The area of $\Omega(a,b)$ is easily seen to be $\frac{b^2-a^2}{16}$. Next, fix $t\in [12,24)$ and let $\Delta t$ be a very small positive number. Note the area of $\Omega(t,t+\Delta t)$ is approximately $\frac{t\Delta t}{8}$. Now since $f(x,y)\approx 16\cos(t^2)$ for any $(x,y)\in \Omega(t,t+\Delta t)$ we see that the volume between of the region of space beneath (or above) $z=f(x,y)$ over our set $\Omega(t,t+ \Delta t)$ is approximately $$16\cos(t^2) \cdot \frac{t \Delta t}{8}=2t\cos(t^2)\Delta t$$ Therefore, $$\iint_{D}f(x,y)dA=\int_{12}^{24}2t\cos(t^2)dt=\sin(25^2)-\sin(12^2)$$ After reading your post I see your issue is your area function, $A(t)$.


问:
Orthogonal polynomials equations
Let ${Q_n(X)}_{n \in N_0}$ sequence of orthogonal polynomials with respect to the weight function $p(x)$ on interval $(a,b)$ . Let $x_i$, $1 \leq i \leq n$ be zeros of $Q_n$.
How to prove that
$$ \sum_{i=0}^{n} \frac {Q_i(x_k)}{\|Q_i\|^{2}} \int_{a}^bp(x)Q_i(x)dx=1 $$ for fixed $k$, $1 \leq k\leq n$.
My proffesor used for Cristoffel-Gaus numerical integration, but I can't figure out why it is true.

答:
If $\{Q_k\}_{0\leq k\leq n}$ is an orthogonal basis of the space of polynomial of degree equal or less than $n$, then $\left\{\frac{Q_k}{\|Q_k\|}\right\}_{0\leq k\leq n}$ is an orthonormal basis of the same space.

Thus any polynomial $P$ of degree $\leq n$, can be decomposed in that basis:
$$P(u)=\sum_{i=0}^n \frac{Q_i(u)}{\|Q_i\|}\int_a^bP(x)\frac{Q_i(x)}{\|Q_i\|}p(x)dx$$
Applying this for $P=1$, and plugging in $u=x_k$ yields the result.

Title: Orthogonal polynomials equations


答:
By Euler's criterion, $2^{44}\equiv\left(\dfrac2{89}\right)\bmod89$, and $\left(\dfrac2{89}\right)=1$ because $89\equiv1\bmod8$,

where $\left(\dfrac2{89}\right)$ is the Legendre symbol.

Title: proving that $89 \mid 2^{44}-1$


答:
$x^{\log_{10} x} = 10^{\log_{10}x\log_{10}x }=10^4$

So $\log_{10}x=\pm 2$ and it follows that both $x=100$ or $x=0.01$ solve the equation.

Title: Equation with Logarithm in Exponent

Body:
How to I solve the following exercise with a logarithm? I've forgotten the "trick" for doing so:

$x^{log_{10} x} =10^4$


问:
show that u(x, t) = $e^{At}$ + B$e^{-3t}$cos(x) + C$e^{Dt}$ cos(2x)

The temperature $u(x,t)$ of a copper wire of length $\pi$ obeys the one-dimensional heat equation
$$
u_t = 3u_{xx}
$$
in the domain $0 < x < \pi$ for all $t > 0$, with the initial condition
$$
u(x,0) = 1 − \cos(2x), quad \text{ for } 0 < x < \pi.
$$
(a) Suppose the temperature $u(x,t)$ also satisfies the boundary conditions
$$
u_x(0,t)=u_x(\pi, t) = 0, \quad \text{for } t \ge 0.
$$
By substitution or otherwise, show that
$$
u(x, t) = e^{At} + Be^{-3t}\cos(x) + Ce^{Dt}\cos(2x)
$$
is a solution to the heat equation satisfying the initial and boundary conditions for appropriate values of the constants $A, B, C$ and $D$.

So far I have identified that I have to use Neumann boundary conditions and have
$$
\begin{split}
A_0 &= \frac{2}{\pi} \int_0^L f(x)dx
     = \frac{2}{\pi} \int_0^\pi (1-\cos(2x))dx
     = 2\\
A_n &= \frac{2}{\pi} \int_0^L f(x)\cos(\frac{nx\pi}{L})dx \\
    &= \frac{2}{\pi} \int_0^\pi (1-\cos(2x))\cos(nx)dx \\
    &= \frac{2}{\pi} \times \frac{4\sin(n\pi)}{n^3-4n} \\
    &= 0
\end{split}
$$
as n must be an integer and for all integers of n, so $\sin(n\pi)=0$.

I know I have gone wrong somewhere and am not sure if it is my whole approach or if I have messed up somewhere can anyone please help.

feel free to edit tags if I have this in the wrong place and sorry this is so poorly written

答:
One way, likely the simplest one, is to substitute directly into the PDE and note that if
$$
u(x, t) = e^{At} + Be^{-3t}\cos(x) + Ce^{Dt}\cos(2x)
$$
then
$$
u_t = Ae^{At} -3Be^{-3t}\cos x + DCe^{Dt}\cos(2x)
$$
and
$$
u_{xx} = -Be^{-3t}\cos x-4Ce^{Dt}\cos(2x)
$$
so you can solve for $A,B,C,D$ which make $u_t=3u_{xx}$ hold.

UPDATE

So $u_t=3u_{xx}$ implies we have
$$
\require{cancel}
Ae^{At} -\cancel{3Be^{-3t}\cos x} + DCe^{Dt}\cos(2x)
= - \cancel{3Be^{-3t}\cos x} - 12Ce^{Dt}\cos(2x)
$$
and we must have $A=0,D=-12$ with any values for $B,C$.


问:
Understanding example 3 on pg. 369 in Dummit & Foote (3rd edition)
Here is the example:

$(3)$ In general,  $$ \mathbb Z_m \otimes_Z \mathbb Z_n \cong \mathbb Z/d\mathbb Z,$$where $d$ is the $g.c.d$ of the integers $m$ and $n.$ To see this, observe first that $$a \otimes b = a \otimes (b.1) = (ab) \otimes 1 = ab (1 \otimes 1),$$
from which it follows that $ \mathbb Z_m \otimes_Z \mathbb Z_n$ is a cyclic group with $1 \otimes 1$ as generator. Since $m (1 \otimes 1) = m \otimes 1 = 0 \otimes 1= 0 $ and similarly $n (1 \otimes 1) = 1 \otimes n = 0,$ we have $d (1 \otimes 1) = 0,$ so the cyclic group has order dividing $d.$ The map \varphi: $ \mathbb Z_m \times \mathbb Z_n \rightarrow \mathbb Z/d\mathbb Z$ defined by $\varphi(a \operatorname{mod}m, b \operatorname{mod}n) = ab \operatorname{mod} d$ is well defined since $d$ divides both $m$ and $n.$ It is clearly $\Bbb Z$-bilinear. The induced map $\Phi: \mathbb Z_m \otimes_Z \mathbb Z_n \rightarrow \mathbb Z/d\mathbb Z$ from corollary 12 maps $1 \otimes 1$ to the element $1 \in \mathbb Z/d\mathbb Z,$ which is an element of order $d.$ In particular $\mathbb Z_m \otimes_Z \mathbb Z_n$ has order at least $d.$ Hence $1 \otimes 1$ is an element of order $d$ and $\Phi$ gives an isomorphism $$ \mathbb Z_m \otimes_Z \mathbb Z_n \cong \mathbb Z/d\mathbb Z.$$

My questions are:

1- How can I prove that $\Phi$ is an isomorphism? what is the definition of it? what is a good candidate for the definition of its inverse?

2- In the fourth line, why we can move the $b$ between the first and second coordinate of the tensor product? I know that this is correct for elements of ring $R$ only i.e. $(rm) \otimes n = m \otimes (rn)$ but I do not know that we can also move the $m,n$'s.

3- Why $d(1 \otimes 1) = 0$?

4-why the cyclic group has order $d$ ? Is not the order of the group is its cardinality? why our group has $d$ elements? I can not see any proof to that.

5- Why $d$ divides both $m$ and $n$ leads to that the map $\varphi$ is well defined?

Answers for those queries will be greatly appreciated.

答:
1.) Dummit and Foote give the definition of $\Phi$ in the same sentence as it first appears. Explicitly, they define $\Phi : (\mathbb Z / m \mathbb Z) \otimes_{\mathbb Z} (\mathbb Z / n \mathbb Z) \to \mathbb Z / d \mathbb Z$ by $\Phi[(1 + m \mathbb Z) \otimes_{\mathbb Z} (1 + n \mathbb Z)] = 1 + d \mathbb Z.$

2.) Each of the quotient rings is an abelian group  (i.e., a $\mathbb Z$-module), and the tensor product is considered with respect to $\mathbb Z,$ so it is $\mathbb Z$-bilinear. Check the definition of $- \otimes_{\mathbb Z} -.$

3.) By definition, $d$ is the greatest common divisor of $m$ and $n,$ hence in particular, there exist integers $k$ and $\ell$ such that $mk + n \ell = d.$ Consequently, we have that $d + m \mathbb Z = n \ell + m \mathbb Z$ so that $$d[(1 + m \mathbb Z) \otimes_{\mathbb Z} (1 + n \mathbb Z)] = (d + m \mathbb Z) \otimes_{\mathbb Z} (1 + n \mathbb Z) = (n \ell + m \mathbb Z) \otimes_{\mathbb Z} (1 + n \mathbb Z).$$ Can you see why this is equal to $0?$

 We have that $$(n \ell + m \mathbb Z) \otimes_{\mathbb Z} (1 + n \mathbb Z) = n[(\ell + m \mathbb Z) \otimes_{\mathbb Z} (1 + n \mathbb Z)] = (\ell + m \mathbb Z) \otimes_{\mathbb Z} (n + n \mathbb Z) = 0.$$

4.) Once you have shown that $(1 + m \mathbb Z) \otimes_{\mathbb Z} (1 + n \mathbb Z)$ generates $(\mathbb Z / m \mathbb Z) \otimes_{\mathbb Z} (\mathbb Z / n \mathbb Z)$ as an abelian group, it suffices to find the order of $(1 + m \mathbb Z) \otimes_{\mathbb Z} (1 + n \mathbb Z).$ (Why?) By (3.) above, it is $d.$

5.) Consider the map $\varphi : (\mathbb Z / m \mathbb Z) \times (\mathbb Z / n \mathbb Z) \to \mathbb Z / d \mathbb Z$ defined by $\varphi(a + m \mathbb Z, b + n \mathbb Z) = ab + d \mathbb Z.$ Given that $a + m \mathbb Z = a' + m \mathbb Z$ and $b + n \mathbb Z = b' + n \mathbb Z,$ we must show that $ab + d \mathbb Z = a' b' + d \mathbb Z.$ By hypothesis, we have that $m \,|\, (a - a')$ and $n \,|\, (b - b')$ so that $a - a' = mk$ and $b - b' = n \ell$ for some integers $k$ and $\ell.$ Using these two identities, can you verify that $ab + d \mathbb Z = a' b' + d \mathbb Z$?

 Considering that $d \,|\, m$ and $d \,|\, n,$ we have that $ab + d \mathbb Z = a'b + d \mathbb Z$ and $a' b + d \mathbb Z = a' b' + d \mathbb Z$ so that $ab + d \mathbb Z = ' b' + d \mathbb Z,$ and $\varphi$ is well-defined.

One other way to see this is to prove the following more general fact.

Fact. If $R$ is a commutative unital ring with ideals $I$ and $J,$ then $$\frac R I \otimes_R \frac R J \cong \frac R {I + J}.$$

To prove this, you might show that for any $R$-module $M,$ we have that $(R / I) \otimes_R M \cong M / IM.$ I believe that you have already proved this in a separate exercise. From this, it follows that $(R / I) \otimes_R (R / J) \cong (R / J) / [I(R / J)].$ It suffices to prove that $I(R / J) = (I + J)/J.$

For our particular case, we have that $R = \mathbb Z,$ $I = m \mathbb Z,$ and $J = n \mathbb Z.$ In a Euclidean domain $R,$ we have that $aR + bR = \gcd(a, b)R.$ Considering that $\mathbb Z$ is a Euclidean domain, it follows that $m \mathbb Z + n \mathbb Z = d \mathbb Z.$ Putting this all together gives the desired result.


问:
showing that $\sigma(\mathbf{XYX}^{−1}) = \sigma(\mathbf{Y})$
I wanted to show the above ($\mathbf{X}$ is nonsingular btw), but I don't think my current proof is complete/quite right, wondering if this is ok:

Let $\mathbf{M} = \mathbf{XYX}^{-1}$. Given that we can represent a matrix $\mathbf{M}$ this way, then we know $\mathbf{M}$ is diagonalizable. Notice that $\mathbf{M}$ and $\mathbf{Y}$ are similar matrices. So, $\mathbf{Y}$ is a diagonal matrix, that shows the eigenvalues of $\mathbf{M}$ on its diagonal.

Taking $\mathbf{Y}$ on its own, we know that one of the properties of a diagonal matrix is that its eigenvalues are on the diagonal. Since $\mathbf{Y}$ is a diagonal matrix, its eigenvalues are on its diagonal.
$$ \therefore \sigma(\mathbf{M}) = \sigma(\mathbf{XYX}^{-1}) = \sigma(\mathbf{Y}) $$

(Note: I realize $\mathbf{Y}$ is not always diagonal, how would I circumvent this?)

答:
I guess $\sigma$ represents the eigenvalues of a matrix.

If that is the case, the eigenvalue $\lambda$ should be an eigen value for both $\mathbf{XYX}^{−1}$ and $\mathbf{Y}$. That means, there are two vectors $u$ and $v$ such that

$$ Yv=\lambda v $$ and
$$ XYX^{-1} u= \lambda u
\\ \Longrightarrow YX^{-1}u = X^{-1}(\lambda u) = \lambda X^{-1} u $$

So if we define $v$ as $X^{-1} u$ we find the associated eigenvectors for $\mathbf{XYX}^{−1}$ and $\mathbf{Y}$ for the eigenvalue $\lambda$. So for each eigenvalue $\lambda$ (and associated eigen vector $u$) of $\mathbf{XYX}^{−1}$, $\mathbf{Y}$ has the same eigen value $\lambda$ (and associated eigen vector $X^{-1}u$)


Title: Equation with Logarithm in Exponent

Body:
To start, take $\log_{10}$ of both sides to get:

$$\log_{10}(x)\log_{10}(x) = \log_{10}(10)^{4} = 4$$

Then just solve $$(\log_{10}(x))^{2} = 4$$

$$\implies \log_{10}x = \pm 2$$

Exponentiating both sides, we get:

$$x = 10^{2} = 100$$ OR

$$x = 10^{-2} = \frac{1}{100}$$


问:
Why does this Infinite Series have contradictory convergences?
In my Calculus II homework, I encountered the following exercise:

If the $n$th partial sum of a series $\sum_{n=1}^\infty a_n$ is
$$s_n = \frac {n-1}{n+1}$$
find  $a_n$ and $\sum_{n=1}^\infty a_n$.

I solved the exercise this way: ( I took $S_n=(n-1)/(n+1)$ to be equation (1))

Equations (2) and (3) answer the exercise's questions. However, when I tried to corroborate my answer in equation (3) by using the found $a_n$ in equation (2), I encountered an inconsistency that I haven't yet been able to harmonize. This is what I tried to do:

Equations (3) and (4) should have yielded the same answer but, this disparity I have been so far unable to harmonize. Your kind comments on what to do will be greatly appreciated.

答:
Your derivation of $a_n$ using $S_n - S_{n-1}$ is only valid for $n > 1$, i.e. $S_0$ is not defined, or rather, it is defined to be zero, but if you substitute zero in $S_n = \dfrac {n-1}{n+1}$ you get $-1$, which is the cause of the disparity.

We hence need to define $a_1$ separately: In fact $a_1 = S_1 = 0$. Hence:

$$S = \sum_{n=1}^\infty a_n = \sum_{n=\color{red}2}^\infty\frac 2{n(n+1)} = 1$$


问:
${\lim}_{a \to \infty} {\int}_{a}^{a+1} \frac{x}{x+\ln(x)} |\sin(x\pi)|dx$
I would like to ask the following questions about "The problem", below from the following perspectives;

(1) Does this limiting value exist?

(2) What is this limiting value, if exist?

The problem;

$${\lim}_{a \to \infty} {\int}_{a}^{a+1} \frac{x}{x+\ln(x)} |\sin(x\pi)|dx$$

The above "The problem" is my modified version of a problem (Henceforth referred to as the "Original Problem") introduced in a Youtube video (In Japanese). The "Original Problem" is as follows .

Original Problem

$${\lim}_{a \to \infty} {\int}_{a}^{a+1} \frac{x}{x+\ln(x)} dx$$

The "Original Problem" is an entrance exam problem of the University of Tokyo, Japan.
The answer to the "Original Problem" is;
$${\lim}_{a \to \infty} {\int}_{a}^{a+1} \frac{x}{x+\ln(x)} dx=1 .$$

The essence of the solution to the "Original Problem" is that the
integrand is constant in the limit;　At the a→∞ limit, the problem is just integrating y=1 over an interval of length 1.Note that,
$$ {\lim}_{a \to \infty} \frac{x}{x+\ln(x)} =1 .$$

The problem is also probably that in the a→∞ limit, the problem is just integrating |y=sin(xπ)| over an interval of length 1.　However, it seems that evaluation methods such as the Squeeze theorem, which were available in the original, cannot be used.

答:
For large values of $a$, your integral behaves like $\int_a^{a+1}|\sin(\pi x)| dx$.

Indeed: For $a\geq 1$, $$
\left |\int_{a}^{a+1} |\sin(x\pi)|dx-\int_{a}^{a+1} \frac{x}{x+\ln x} |\sin(x\pi)|dx\right|= \int_{a}^{a+1} \frac{\ln x}{x+\ln x} |\sin(x\pi)|dx
\leq \frac{\ln (a+1)}{a+\ln a}$$
and the right-hand side goes to zero as $a\rightarrow +\infty$.
Since  $$\int_a^{a+1}|\sin(\pi x)dx |=\frac 2 \pi$$ the answer is: The value exists and it's $\frac 2 \pi$.