问:
Averaging Over Hamming Weights
Question: Let $\Sigma:=\{0,1\}.$ A finite string $,s,$ over $\Sigma$ is any finite sequence composed of $0s$ and $1s.$  The length of $s$ denoted $|s|$ counts the number of symbols comprising $s.$ The Hamming weight of $s;$ which I will denote by $\mathcal{H}(s),$ counts the number of symbols in $s$ that differ from $0.$ What are some mathematical and more importantly physical interpretations of
$$
\frac{1}{|s|}\mathcal{H}(s) \label{a}\tag{1}
$$
?
Note I am not asking for an account of histories or common usages of Hamming weights. I am asking what averaging over Hamming weights informs us.

Edit: See Phicar's answer. It would appear $\ref{a}$ gives a density, not a probability, of the number of $1s$ in $s.$

I would think $\ref{a}$ gives the probability of drawing a $1$ among the symbols ( -bits) of $s.$

答: Notice that $\frac{\mathcal{H}(s)}{|s|}$ gives you a value in between $0$ and $1$ that defines the density of $1's$ in the string. This is not quite a probability(you are evaluating non random points), but lets you compare strings of different sizes. If you do not divide by the length, the word $00100$ would have the same value as $1$ but the later seems more populated of $1's$ than the former.

Title: Averaging Over Hamming Weights


问:
Integer solutions to $z^2=10x^2-2y^2$
I need to show that a $(67, 12, 2)$-design does not exist. I tried to use the Bruck, Ryser and Chowla theorem which in this case states that if a $(67, 12, 2)$-design exists, then the equation
$$z^2=10x^2-2y^2,$$
has a non-trivial integer solution $(x,y,z)\neq \overline{0}$. Checking from Wolfram I found that it does not have any, which would prove my original problem, but I would like to see proof why it doesn't have any.

It is easy to see that $z$ should be even so I tried to plug in $z=:2k$ and split it in some cases but it did not go anywhere.

答: If $5$ divides either $y$ or $z$, then it divides all three of $x$, $y$, and $z$, in which case it can be factored out, leaving a smaller solution, so we may as well assume that $5\not\mid yz$, in which case $y^2=10x^2-2z^2$ implies $y^2\equiv-2z^2$ mod $5$, which can be written as $3\equiv(yz)^2$ mod $5$ (since $-2\equiv3$ and $z^4\equiv1$ mod $5$). But the only nonzero squares mod $5$ are $1$ and $4$.

ContentLicense: CC BY-SA 4.0

Title: Integer solutions to $z^2=10x^2-2y^2$


问:
If $X_n \longmapsto X$ in law and $Y_n \longmapsto Y$ in law then $X_n +Y_n \longmapsto X+Y$ in law?
We say that given a sequence of random variable $(X_n)_{n=1}^{+\infty}$ defined on $(\Omega_n,\mathcal{A}_n,P_n)$ to $\mathbb{R}^d$ converge in law to $X$ random variable if $PX_n \to P_x$ in the sense $\forall \hspace{0.1cm} f \in C_b(\mathbb{R}^d) \hspace{0.1cm} \mathbb{E}[f(X_n)] \underset{n}{\longmapsto} \mathbb{E}[f(X)]$

Is holds that if $X_n \overset{\mathscr{L}}{\longmapsto} X, \hspace{0.1cm} Y_n \overset{\mathscr{L}}{\longmapsto} Y$ then $X_n+Y_n \overset{\mathscr{L}}{\longmapsto} X+Y$ ?

答:
The answer is negative.

Take, for example, $X_n=Z$ and $Y_n=-Z$, where $Z\sim N(0,1)$. Then $X_n\xrightarrow{L}N(0,1)$ and $Y_n\xrightarrow{L}N(0,1)$, but $X_n+Y_n=0$ for each $n\ge1$.


问:
Do Diophantine solution bounds themselves require a secondary solving operation?
With a Diophantine equation like

$$889a + 90b - 90c - 2000d = 0$$

Sympy provides the following solution:

$$a = 10 t_0$$
$$b = t_1$$
$$c = 79121 t_0 + 1001 t_1 + 200 t_2$$
$$d = -3556 t_0 - 45 t_1 - 9 t_2$$

where the only known constraints on all $t$ is that they're integers.

If I have other constraints: $0 \le a, b, c, d < 10$, what is the best way to determine the range of all $t$? I fear that in the general case (not knowing the degree or linearity of the original equation at compile time) this requires setting up six linear programming problems - double the number of degrees of freedom of the Diophantine solution - to minimize and maximize each degree of freedom $t$.

答:
About as pleasant as it is going to get:
your expression in terms of a reduced basis is
$$ (a,b,c,d) = u(0,1,1,0) + v (-20,21,-21,-7) + w(30,18,-19,15)$$
You may check the dot product of the original $(889, 90, -90, -2000)$
with each of these row vectors.

```<code>? newbasis = colbasis * toreduce
%20 =
[0 -20  30]

[1  21  18]

[1 -21 -19]

[0  -7  15]

? nt = mattranspose(newbasis)
%21 =
[  0  1   1  0]

[-20 21 -21 -7]

[ 30 18 -19 15]

? gramnew = nt * newbasis
%22 =
[ 2    0   -1]

[ 0 1331   72]

[-1   72 1810]

? matdet(gramnew)
%23 = 4806521
</code>```

Body:
About as pleasant as it is going to get:
your expression in terms of a reduced basis is
$$ (a,b,c,d) = u(0,1,1,0) + v (-20,21,-21,-7) + w(30,18,-19,15)$$
You may check the dot product of the original $(889, 90, -90, -2000)$
with each of these row vectors.

? newbasis = colbasis * toreduce
%20 =
[0 -20  30]

[1  21  18]

[1 -21 -19]

[0  -7  15]

? nt = mattranspose(newbasis)
%21 =
[  0  1   1  0]

[-20 21 -21 -7]

[ 30 18 -19 15]

? gramnew = nt * newbasis
%22 =
[ 2    0   -1]

[ 0 1331   72]

[-1   72 1810]

? matdet(gramnew)
%23 = 4806521


问:
simple calculation with Christoffel symbols on Poincare half-plane
Equip $H=\{(x,y):y>0, x,y \in \mathbb{R}\}$ with the metric $$ds^2=\frac{dx^2+dy^2}{y^2}.$$
(https://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model). I want to show that the sectional curvature
$$ K(\partial_i,\partial_j) = \frac{\langle R(\partial_i,\partial_j)\partial_j,\partial_i\rangle}{\det(g)}=\frac{R_{ijji}}{\det(g)}$$
is $-1$. I know this is a simple calculation, but for some reason I'm off by a sign and it's driving me nuts. Here $g$ is the metric in matrix form:

$$g=(g_{ij})=
\begin{pmatrix}
\frac{1}{y^2} & 0 \\
0 & \frac{1}{y^2}
\end{pmatrix}.$$

Let $\{ \partial_1=\partial/ \partial x,\partial_2=\partial/ \partial y\}$ be the coordinate basis, and just consider the computation of $K(\partial_1,\partial_2).$ So I just need to compute $R_{1221}=g_{1m}R^m_{\ 221}=g_{11}R^1_{\ 221}$, where

$$R^i_{jkl}=\partial_k \Gamma^i_{jl}-\partial_l \Gamma^i_{jk}+\Gamma^p_{jl}\Gamma^i_{pk}-\Gamma^p_{jk}\Gamma^i_{pl}.$$

I already have the Christoffel symbols, namely

\begin{equation*}
\Gamma^1_{12}=\Gamma^1_{21}=-\frac{1}{y},\\
\Gamma^2_{12}=\Gamma^2_{21}=0,\\
\Gamma^1_{11}=\Gamma^1_{22}=0,\\
\Gamma^2_{11}=\frac{1}{y},\\
\Gamma^2_{22}=-\frac{1}{y}.
\end{equation*}

So
\begin{align*}
R^1_{\ 221} &= \partial_2 \Gamma^1_{21}-\partial_1 \Gamma^1_{22}+\Gamma^p_{21}\Gamma^1_{p2}-\Gamma^p_{22}\Gamma^1_{p1}\\
&= \partial_2 \left( -\frac{1}{y}\right)+\Gamma^1_{21}\Gamma^1_{12}-\Gamma^2_{22}\Gamma^1_{21}\\
&= \frac{1}{y^2}+\left(-\frac{1}{y}\right)^2-\left(-\frac{1}{y}\right)^2\\
&= \frac{1}{y^2}.
\end{align*}

But this gives $R_{1221}=1/y^4$, and hence $K(\partial_1,\partial_2)=1$.

Please tell me what I'm doing wrong here or what formula is wrong.

答:
When using the definition of $R(X,Y)Z$ ending with $-\nabla_{[X,Y]}Z$, the sectional curvature $K(X,Y)$ is defined as

$$ K(X,Y) = \frac{\langle R(X,Y)Y,X\rangle}{\langle X,X \rangle \langle Y,Y \rangle-\langle X,Y \rangle^2}$$

for $X,Y$ linearly independent. We derive the coordinate expression for $R_{ijk}^{\ \ \ \ \ l}$ in terms of the Christoffel symbols. Following Lee's Riemannian Manifolds, we take the components $R_{ijk}^{\ \ \ \ \ l}$ to be defined by $R(\partial_i,\partial_j)\partial_k=R_{ijk}^{\ \ \ \ \ l} \partial_l$, and the Christoffel symbols $\Gamma^k_{\ ij}$ to be defined by $\nabla_{\partial_i}\partial_j=\Gamma^k_{\ ij}\partial_k$. Applying the coordinate dual basis element $dx^l$ to both sides of the defining equation for $R_{ijk}^{\ \ \ \ \ l}$ gives
\begin{align*}
R_{ijk}^{\ \ \ \ \ l} &= dx^l(R(\partial_i,\partial_j)\partial_k) \\
&= dx^l(\nabla_{\partial_i}\nabla_{\partial_j}\partial_k-\nabla_{\partial_j}\nabla_{\partial_i}\partial_k-\nabla_{[\partial_i,\partial_j]}\partial_k)\\
&= dx^l(\nabla_{\partial_i}\nabla_{\partial_j}\partial_k-\nabla_{\partial_j}\nabla_{\partial_i}\partial_k)\\
&= dx^l(\nabla_{\partial_i}(\Gamma^m_{\ jk}\partial_m)-\nabla_{\partial_j}(\Gamma^m_{\ ik}\partial_m))\\
&= dx^l((\partial_i\Gamma^m_{\ jk})\partial_m+\Gamma^m_{\ jk}\nabla_{\partial_i}\partial_m-(\partial_j \Gamma^m_{\ ik})\partial_m-\Gamma^m_{\ ik}\nabla_{\partial_j}\partial_m)\\
&= dx^l((\partial_i\Gamma^m_{\ jk})\partial_m+\Gamma^m_{\ jk}\Gamma^p_{\ im}\partial_p-(\partial_j \Gamma^m_{\ ik})\partial_m-\Gamma^m_{\ ik}\Gamma^p_{\ jm}\partial_p)\\
&= (\partial_i\Gamma^m_{\ jk})dx^l(\partial_m)+\Gamma^m_{\ jk}\Gamma^p_{\ im}dx^l(\partial_p)-(\partial_j \Gamma^m_{\ ik})dx^l(\partial_m)-\Gamma^m_{\ ik}\Gamma^p_{\ jm}dx^l(\partial_p)\\
&= \partial_i\Gamma^l_{\ jk}+\Gamma^m_{\ jk}\Gamma^l_{\ im}-\partial_j \Gamma^l_{\ ik}-\Gamma^m_{\ ik}\Gamma^l_{\ jm}.
\end{align*}

Now for the coordinate basis $\{ \partial_1=\partial/ \partial x,\partial_2=\partial/ \partial y\}$, we can write
$$ K(\partial_i,\partial_j) = \frac{\langle R(\partial_i,\partial_j)\partial_j,\partial_i\rangle}{\det(g)}=\frac{R_{ijji}}{\det(g)},$$
where $\det(g)=1/y^4$. Clearly, $K(\partial_i,\partial_j)=0$ for $i=j$. For $i \neq j$, we have
$$ K(\partial_1,\partial_2)=y^4 R_{1221} \quad \text{and} \quad K(\partial_2,\partial_1)=y^4 R_{2112}.$$
By symmetry properties of the components $R_{ijkl}$, $R_{1221}=R_{2112}$, so in fact $K(\partial_1,\partial_2)=K(\partial_2,\partial_1)$. Since
$$R_{1221}=g_{1m}R^{\ \ \ \ \ m}_{122}=g_{11}R^{\ \ \ \ \ 1}_{122},$$
we use the coordinate expression for the components $R_{ijk}^{\ \ \ \ \ l}$ derived at the outset to compute $R^{\ \ \ \ \ 1}_{122}$ as follows:
\begin{align*}
R^{\ \ \ \ \ 1}_{122}&=\partial_1\Gamma^1_{\ 22}+\Gamma^m_{\ 22}\Gamma^1_{\ 1m}-\partial_2 \Gamma^1_{\ 12}-\Gamma^m_{\ 12}\Gamma^1_{\ 2m}\\
&=\Gamma^2_{\ 22}\Gamma^1_{\ 12}-\partial_2\left(-\frac{1}{y}\right)-\Gamma^1_{\ 12}\Gamma^1_{\ 21}\\
&=\left(-\frac{1}{y}\right)^2-\frac{1}{y^2}-\left(-\frac{1}{y}\right)^2\\
&=-\frac{1}{y^2}.
\end{align*}
Therefore, $R_{1221}=-1/y^4$, and so $K(\partial_1,\partial_2)=K(\partial_2,\partial_1)=-1.$


问:
Is the direct image functor on sheaves of a point inclusion exact?
Given the inclusion of a point $i_x:\{x\}\to X$, is the associated direct image functor $(i_x)_*: Sh(\{x\}) \to Sh(X)$ an exact functor?

I know that in general this is not the case and the direct image functor is only left exact. I also know that if $\{x\}$ is a closed subset in X, then it should be an exact functor. So I was trying to construct a counterexample by looking at the inclusion of an open subset, but to be honest, I just lack the intuition for the subject to come up with something. Any hints?

答:
I gave it a go, and I think I might have solved it, but I only know the basics on sheaves so you should definetely double check my proof.

we have that $ \mathcal{F}_{x} = \mathcal{F}(\{x\}), \ \mathcal{G}_{x} = \mathcal{G}(\{x\}), \ \mathcal{H}_{x} = \mathcal{H}(\{x\}). \ $ Since $\ \ \large0 \rightarrow \mathcal{F} \rightarrow \mathcal{G} \rightarrow \mathcal{H} \rightarrow \large0 \ $ is exact, also $ \ \large0 \rightarrow \mathcal{F}_{x} \rightarrow \mathcal{G}_{x} \rightarrow \mathcal{H}_{x} \rightarrow \large0 \ $ is exact.

In order to prove that $ \ \ \large0 \rightarrow i_{\star}(\mathcal{F}) \rightarrow i_{\star}(\mathcal{G}) \rightarrow i_{\star}(\mathcal{H}) \rightarrow \large0 \ $ is exact, we can just prove that it is exact on the stalks. But this is indeed the case:

Let $y \in X \ \ \text{s.t.  there's an open set} \ U \subset X \ \text{with } x \notin U.$ Then $i_{\star}(\mathcal{F})_{y} = i_{\star}(\mathcal{G})_{y} = i_{\star}(\mathcal{H})_{y} = \large0$

If $ \ \  \forall \ \  U \subset X \text{ such that } y \in U \text{ we also have } x \in U, \ $ then $ \ i_{\star}(\mathcal{F})_{y} = \mathcal{F(\{x\})}, \ i_{\star}(\mathcal{G})_{y} = \mathcal{G(\{x\})}, \ i_{\star}(\mathcal{H})_{y} = \mathcal{H(\{x\})}$.

In both cases the sequence $ \ \ \large0 \rightarrow i_{\star}(\mathcal{F})_{y} \rightarrow i_{\star}(\mathcal{G})_{y} \rightarrow i_{\star}(\mathcal{H})_{y} \rightarrow \large0 \ $ is exact. Therefore $ \ \ \large0 \rightarrow i_{\star}(\mathcal{F}) \rightarrow i_{\star}(\mathcal{G}) \rightarrow i_{\star}(\mathcal{H}) \rightarrow \large0 $ is also exact.


问:
Translation invariance of $\sum_{n\in\mathbb{Z}} \operatorname{sinc}(n+c)$
$\newcommand{\sinc}{\operatorname{sinc}}$

Let $\sinc:\mathbb{R}\to\mathbb{R}$, $\sinc(x) = \frac{\sin (x)}{x}$ with $\sinc(0)=1$. A classical result is that the sum and integral of $\sinc$ are the same, namely
$$
\sum_{n=-\infty}^{\infty} \sinc(n) = \int_{-\infty}^{\infty} \sinc(x)\,dx = \pi
$$The integral is clearly translation invariant: were we to replace $x$ with $x+c$, a quick substitution $y=x+c$ gets us right back where we started. More surprising to me is that the series is also translation invariant: for any $c\in\mathbb{R}$, $\sum_{n\in\mathbb{Z}} \sinc(n) = \sum_{n\in\mathbb{Z}} \sinc(n+c).$ If $c$ is an integer, a similar reindexing trick makes it obvious, but it's unclear to me why this should work for any fixed $c$. I asked Mathematica why this might be true and it returned the following:
$$
\sum_{n\in\mathbb{Z}}\sinc(n+c) = \frac{1}{2} i \left(B_{e^{-i}}(1-c,0)+B_{e^{-i}}(c,0)-B_{e^i}(1-c,0)-B_{e^i}(c,0)\right);
$$here $B_z(a,b)$ is the incomplete beta function, the analytic continuation of $\int _0^z t^{a-1}(1-t)^{b-1}\,dt$.

I'm looking for one of two things: either a clean and sensible explanation for why the sum is translation invariant or, failing that, a derivation of Mathematica's output and a walkthrough of the functional identities to show that it collapses to $\pi$ for $c\in\mathbb{R}$.

答:
With the residue theorem or the Fourier inversion theorem, for $a> 0$ $$\int_{-\infty}^\infty  \frac{\sin(\pi a t)}{\pi t} e^{-2i\pi kt}dt = 1_{k\in [-a/2,a/2]}$$
For $a\in (0,2)$ let $$f(t) = \sum_n \frac{\sin(\pi a (t+n))}{\pi (t+n)} $$
It converges, it is $1$-periodic and it is equal to its Fourier series
$$f(t)=\sum_k c_k e^{2i\pi kt}, \qquad c_k = \int_0^1 f(t)e^{-2i\pi kt}dt=\int_{-\infty}^\infty  \frac{\sin(\pi a t)}{\pi t} e^{-2i\pi kt}dt = 1_{k\in [-a/2,a/2]}$$
ie. $$f(t)=1$$


问:
Why do we say A given B in Bayes theorem?
I enjoyed this blog on visualizing Bayes theorem but I have a question on the wording for .

$p(A\mid B) = p(A,B)/p(B)$

In words I understand this to be "probability of A given B"

if we are "given B" why do we need to divide by p(B) ?   Surely if I am "given" something then its probability is $1$?

Why do we say it this way?

[Update]

I am seeing that p(B) = 1 (because we are "given" it )

thus p(A,B)/p(B) = p(A,B)

I am looking for the right words to articulate that we still need to divide by p(B) and that it is not 1

If we were saying "given we know the probability of B" instead of "given B" I would not be so confused.

答:
I suspect that you may not get a satisfactory answer to your question since you are bothered by a single word that has several  meanings in informal discourse but just one in this mathematical context. That one disagrees with your favorite daily use meaning. But I will try.

The right way to think of "given" here is that "the probability of $A$ given $B$" means "what would the probability of $A$ be if I knew that $B$ had occurred?".

So for example the probability that a coin shows "heads" given that it shows "tails" is $0$. I think that's a reasonable English sentence. It's pretty clearly correct. It also follows from the formula in your question.

Dividing by the probability of $B$ in that formula is the precise way to take into account the hypothesized occurrence of $B$. It tells you what fraction the event ($A$ and $B$) is of the event $B$.


问:
Factorizing polynomials in a field.
With the following field, $F=\mathbb{F}_2[y]/(y^3+y+1)$, I need to factorize $x^3+x+1$ and $x^3+x^2+1$  in the ring $F[x]$. I have done this and found out the roots of the first polynomial is $y,y^2,y^4$. Then the second is just the first polynomial with $x+1$ substituted into the first, so factorization is clear there as well.

However, I achieved this with the help of the following hint,

$\textit{Hint: if $r$ in $F$ is a root of such a polynomial, why is $r^2$ also a root?}$

I don't really know why this is the case, any hints as to why this is the case would be appreciated.

答:
Quite simple: $r$ is a root, by definition, if and only if $\;r^3=r+1$.

This implies that $$(r^2)^3=(r^3)^2=(r+1)^2=r^2+1^2=r^2+1\qquad(\text{we're in characteristic }2).$$

Title: Factorizing polynomials in a field.


问:
I need to find all functions $f:\mathbb R \rightarrow \mathbb R$ which are continuous and satisfy $f(x+y)=f(x)+f(y)$
I need to find all functions $f:\mathbb R \rightarrow \mathbb R$ such that $f(x+y)=f(x)+f(y)$. I know that there are other questions that are asking the same thing, but I'm trying to figure this out by myself as best as possible. Here is how I started out:

Try out some cases:

$x=0:$
$$f(0+y)=f(0)+f(y) \iff f(y)=f(0)+f(y) \iff 0=f(0) $$
The same result is for when $y=0$

$x=-y:$
$$f(-y+y)=f(-y)+f(y) \iff f(0)=f(-y)+f(y) \iff 0=f(-y)+f(y)\iff \quad f(-y)=-f(y)$$
I want to extend the result of setting $x=-y$ to numbers other that $-1$, perhaps all real numbers or all rational numbers. I got a little help from reading other solutions on the next part:

Let $q=1+1+1+...+1$. Then
$$f(qx)=f((1+1+...+1)x)=f(x+x+...+x)=f(x)+f(x)+...+f(x)=qf(x)$$
I understood this part, but I don't understand why this helps me find all the functions that satisfy the requirement that $f(x+y)=f(x)+f(y)$, but here is how I went on:

Thus
$$f(qx)=qf(x)$$ and it should follow that
$$f \bigg (\frac {1}{q} x\bigg)= \frac{1}{q}f(x)$$ where $q\not =0$, then it further follows that
$$f \bigg (\frac {p}{q} x\bigg)= \frac{p}{q}f(x)$$ where $\frac{p}{q}$ is rational, and lastly it further follows that
$$f (ax)= af(x)$$ where $a$ is real. Thus functions of the form $f(ax)$ where $a$ is real satisfies the requirement of  $f(x+y)=f(x)+f(y)$.

I don't know how much of what I did is correct\incorrect, and any help would be greatly appreciated. Also is there any way that I can say that functions of the form $f(ax)$ where $a$ is real are the only functions that satisfy the requirement of  $f(x+y)=f(x)+f(y)$? Or do other solutions exist?

Again, thanks a lot for any help! (Hints would be appreciated, I'll really try to understand the hints!)

答:
If you do not assume continuity of $f$, then it depends on the Axiom of Choice, stated as Zorn's Lemma.  Consider the family $\cal L$ of linearly independent over $\mathbf Z$ subsets of $\mathbf{R}$. Given a chain of such sets,

$$L_1 \subset L_2 \subset \dotsm \subset L_n \subset \dotsm$$

we can prove $L = \bigcup_n L_n \in \cal L$.  Suppose $L$ were not linearly independent over $\mathbf Z$. Then, there would be distinct $x_i \in L$ and nonzero integers $a_i$ such that

$$
a_1 x_1 + \dotsm + a_m x_m = 0
$$

But each $x_i$ must be in $L_n$ for some $n$. Let $N$ be the maximum of all such $n$.  Then, each $x_i \in L_N$, and so $L_N$ is a $\mathbf Z$-dependent set, contrary to the assumption.

Since every chain has an upper bound, Zorn's Lemma states that there is a maximal member, $M$, of $\cal L$.  Every real number $r$ must be a $\mathbf Z$=linear combination of a finite set of elements of $M$; otherwise, $M \cup \{r\}$ would be $\mathbf Z$-independant, contrary to $M$ being maximal. $M$ is a Hamel basis of $\mathbf R$.

And now, any permutation of $M$ extends by linearity to a discontinuous $f$ such that $f(x+y) = f(x) + f(y)$.

ContentLicense: CC BY-SA 3.0


问:
Plane of symmetry and geodesics in surfaces
Let $(M,n)$ be an oriented regular surface of $\mathbb{R}^{3}$ such that the intersection of $M$ with the plane
$\Pi:=\{(x,y,z) \in \mathbb{R}^{3}:z=0\}$ is the image of a curve parametrized by arc length $\alpha$.

I am trying to prove that if $\varphi:M \rightarrow M$ defined by $\varphi (x,y,z)=(x,y,−z)$ is an isometry of M then the curve $\alpha$ must be a geodesic of M.

I would like to prove that the normal vector $n \circ \alpha(t)$ is orthogonal to the vector $e_3 = (0,0,1)$ (for any $t \in \operatorname{dom}(\alpha)$).
I know that the vector $n \circ \alpha(t)$ is an eigenvector of the linear transformation $R: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ given by  $R(x,y,z)=(x,y,−z)$, but I don't know how to prove that the tangent space $T_{\alpha(t)}M$ can't be the vector space $\operatorname{span}\{e_1,e_2\}$.
I also know that the hypothesis $M \cap \Pi$ equal to the image of a curve is important and necessary.

I have tried many things but I'm not making any progress. So I would really appreciate any help.

答:
The reason that the tangent plane $T_{\alpha(t)}$ can not be $span\{e_1, e_2\}$ is that in that case the projection to the $x-y$ plane would be a chart near $\alpha(t)$, i.e. $M$ would be locally a graph of $z=f(x,y)$; but the only graph invariant w.r. to $R$ is the graph of the zero functions $z=0$. However, in such a case the intersection of $M$ with $\Pi$ would not be a curve (a smooth curve can not have an image that fills in any open set of a plane). So this is excluded.

Now, (as you have perhaps already figured out) the derivative of any map $\varphi$ mapping $M_1$ to $M_2$ sends the tangent plane $T_p M_1$ to the tangent plane $T_{\varphi(p)} M_2$. In our case, when $p$ is on the curve $\alpha$ we have $\varphi(p)=p$ and $D\varphi$, being the linearization of $R$, which is already linear and fixes $p$, is $R$. Thus the tangent plane is invariant, and so the normal vector is invariant (up to sign), i.e. an eigenvector with eigenvalue $\pm 1$, and the case of it being $\pm e_3$ is excluded, so it lies in the $x-y$ plane. Now the acceleration vector of $\alpha$ is also in the $x-y$ plane and, since $\alpha$ is unit speed it is orthogonal to the velocity vector of $\alpha$, hence parallel to the normal vector to $M$, making $\alpha$ a geodesic.


问:
Conformal map onto a circle, from an identification space composed of five square regions.
I am looking to derive a conformal map for the problem illustrated in this image. I've read a bit about how to map a square onto a circle, but I'm struggling to extend the concepts for the domain at hand. I don't have a rigorous mathematical background (mech. engineer in computational fluid dynamics), so I would appreciate if someone here could advise me on the route that I should take in order to derive such a conformal mapping.

The end application is to generate a smooth computational mesh that looks like this. I have generated a mesh like this using other means, but the smoothness of the mesh vertices is not sufficient for extremely fine meshes. This results in spurious oscillations in the numerical problem I am trying to solve.

答:
This question is a couple of years old, but so far no analytic solutions have been offered.

If you're willing to consider a computational solution, You might want to try this Boundary First Flattening app.

Boundary First Flattening (BFF) is a free and open source application
for surface parameterization. Unlike other tools for UV mapping, BFF
allows free-form editing of the flattened mesh, providing users direct
control over the shape of the flattened domain—rather than being stuck
with whatever the software provides. The initial flattening is fully
automatic, with distortion mathematically guaranteed to be as low or
lower than any other conformal mapping tool. The tool also provides
some state-of-the art flattening techniques not available in standard
UV mapping software such as cone singularities, which can dramatically
reduce area distortion, and seamless maps, which help eliminate
artifacts by ensuring identical texture resolution across all cuts.
BFF is highly optimized, allowing interactive editing of meshes with
millions of triangles.

The BFF application is based on the paper, “Boundary First Flattening”
by Rohan Sawhney and Keenan Crane.

You can supply it with a mesh model of a cube that is missing one of its faces.  This is a surface with a boundary.
You should be able to use the app to create a mesh whose boundary is a circle corresponding to the boundary of your model and whose interior is a nearly conformal flattening of the remaining five faces of the cube.


问:
Using the converse if Cayley Hamilton theorem
My textbook says

: Let $M$ be a $3 \times 3$ Hermitian matrix which satisfies the matrix equation
$$
M^{2}-5 M+6 I=0
$$
Where $I$ refers to the identity matrix. Which of the following are possible eigenvalues of the matrix $M$
(a) (1,2,3)
(b) (2,2,3)
(c) (2,3,5)
(d) (5,5,6)

Then it proceeds as:

According to Cayley-Hamilton theorem, we can write $\lambda^{2}-5 \lambda+6=0 \Rightarrow \lambda=2,3$ Correct option is (b)

It's clear that the author has used the Cayley Hamilton theorem but in reverse but how can we use the converse Cayley Hamilton theorem? I've read that the converse of Cayley Hamilton theorem doesn't hold in general so what's the author doing here?

I'd be glad if someone pointed out my mistake. Much thanks.

答:
Suppose that $p(M)=0$ for some square matrix $M$ and some polynomial
$$
     p(\lambda)=\lambda^k+a_{k-1}\lambda^{k-1}+\cdots + a_{1}\lambda+a_0.
$$
Then
$$
         p(M)-p(\lambda)I = -p(\lambda)I.
$$
You can rewrite the left side in order to obtain an inverse for $M-\lambda I$ for any $\lambda$ for which $p(\lambda)\ne 0$ as follows:
$$
        (M-\lambda I)q(\lambda,M)=q(\lambda,M)(M-\lambda I)=-p(\lambda)I
$$
Therefore $M-\lambda I$ is invertible if $p(\lambda)\ne 0$. So the only possible eigenvalues of $M$ are the solutions of $p(\lambda)=0$. That does not mean every root of $p(\lambda)$ is an eigenvalue because $q(\lambda,M)=0$ could occur. But it is certainly the case that every eigenvalue of $M$ is a root of $p(\lambda)$.

In your case, $p(M)=0$ where $p(\lambda)=\lambda^2-5\lambda+6$. So, the eigenvalues of $M$ must be roots of $p$, which are $3$ and $2$. That does not mean that both $2$ and $3$ are eigenvalues. But $2$ and $3$ are the only possible eigenvalues. Out of your possible  answers, the only possible legitimate answer is (b) $2,2,3$ because neither $1$, nor $5$, nor $6$ are possible eigenvalues, as they are not roots of the annihilating polynomial $p$.


问:
Probability of frog jumping
I'm trying to understand the solution given to this problem on AOPS site:

A frog makes $3$ jumps, each exactly $1$ meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than $1$ meter from its starting position?

Solution (Complex Numbers) :

We will let the moves be complex numbers $a$, $b$, and $c$, each of magnitude one. The frog starts on the origin. It is relatively easy to show that exactly one element in the set

$$ \{|a+b+c|,|a+b-c|,|a-b+c|,|a-b-c|\} $$

has magnitude less than or equal to $1$. Hence, the probability is $\dfrac{1}{4}$.

..........

I only understand $|a+b+c|$. I don't know where other three terms come from. Also why only one of these has magnitude less than or equal to $1$?

Thank you.

答:
Where these terms come from: the idea is to modify the problem to get an equivalent experiment.

Simply following the word problem, we could choose jumps $a,b,c$ uniformly from the unit circle and then end up at $a+b+c$.

Instead, we could choose $b$ in two steps: first choose an angle $\theta \in [0,\pi)$, then flip a coin to let $b$ be one of $e^{i\theta}$ or $-e^{i\theta}$. This is also uniform on the unit circle. We could do the same thing for $c$.

If, using the second method, we choose the angles for $b$ and $c$, but wait to choose the sign until the end, then we are flipping two coins to choose between $\{a+b+c, a+b+(-c), a+(-b)+c, a+(-b)+(-c)\}$.

Why exactly one of these is in the unit circle: I can't promise this is the best argument, but here's one. We'll show that at most one of these points is in the unit circle, then show that at least one is in the unit circle.

The four points are the vertices of a rhombus with sides parallel to $2b$ and $2c$, which is centered at $a$. Here's a typical picture:

Assume none of the corners are on the boundary of the unit circle, which holds with probability $1$. Then two adjacent corners can't both be in the circle, because they're two units apart. Two opposite corners can't both be in the circle, because the midpoint between them (which is $a$) would also be in the circle. So at most one of these four points is in the unit circle.

On the other hand, from $a$, the four directions $b+c, b-c, -b+c, -b-c$ are perpendicular, so two of them point into the circle; without loss of generality, say that it's $b+c$ and $b-c$. Now consider the following two right triangles:

The right triangle with vertices $a$, $a+b+c$, and $a+b-c$.

The right triangle inscribed in the unit circle with one vertex at $a$ and the two sides being the two chords from $a$ going in the same directions as $b+c$ and $b-c$.

Both right triangles have hypotenuse $2$: the first one because $|(a+b+c)-(a+b-c)| = |2c|=2$, and the second one because its hypotenuse is a diameter of the unit circle. So the first right triangle has a side that's shorter than the corresponding side of the second one. This means one of its vertices lies on the side of the second right triangle, and is in the interior of the circle, QED.

Title: Probability of frog jumping


问:
Showing $\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$
I found the following exercise reading my calculus notes:

If $x,y$ and $z$ are positive real numbers, show that $$\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$$

I've been trying to solve it for a while. However, I have no idea how to approach it. Any help is welcome.

答:
Note that

$(x-1)^2 \geq 0 \implies x^2+1 \geq 2x $

$(y-1)^2 \geq 0 \implies y^2+1 \geq 2y $

$(z-1)^2 \geq 0 \implies z^2+1 \geq 2z $

Multiplying these three inequalities you will get $(x^2+1)(y^2+1)(z^2+1)\geq 8xyz \implies \frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\ge8$


问:
How to find the Laurent expansion for $\frac{\exp\left(\frac{1}{z^{2}}\right)}{z-1}$ about $z=0$?
I want to find the Laurent expansion for  $\frac{\exp\left(\frac{1}{z^{2}}\right)}{z-1}$ about $z=0$,

I've tried to apply this formula $\frac{1}{1-\omega}=\sum_{n=0}^{\infty }\omega^{n}$ and the usual Taylor series of the exponential function, but I don't know how to continue:

$$\begin{align}f(z)&=\frac{1}{z-1}\exp\left(\frac{1}{z^{2}}\right)\\
&=-\frac{1}{1-z}\exp\left(\frac{1}{z^{2}}\right)\\&=-\left (\sum_{n=0}^{\infty }z^{n}  \right )\left ( \sum_{n=0}^{\infty}\frac{1}{n!z^{2n}} \right )\end{align}$$
Thanks in advance.

Ps: I tried applying a Cauchy product, but I think this is not appropriate.

Edit 1: If it is useful at the end of the text, the authors say that the Laurent expansion is:

$\sum_{k=-\infty }^{\infty }a_{k}z^{k}$ with $a_{k}=-e$ if $k\geq 0$ and
$a_{k}=-e+1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{(j-1)!}$if $k=-2$ or $k=-2j+1$ where $j=1,2,...$

Body:
First, we can write two series for $\frac1{z-1}$ in the two regions $|z|<1$ and $|z|>1$ as

$$\frac1{z-1}=\begin{cases}
-\sum_{n=0}^\infty z^n&,|z|<1\\\\
\sum_{n=1}^\infty z^{-n}&,|z|>1\tag1
\end{cases}$$

Second, the Laurent series for $e^{1/z^2}$ for $0<|z|$ is given by

$$e^{1/z^2}=\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}\tag2$$

where $a_n$ the sequence such hat

$$a_n=\begin{cases}
1&,n\,\text{even}\\\\
0&,n\,\text{odd}
\end{cases}$$

Putting $(1)$ and $(2)$ together reveals

$$\frac{e^{1/z^2}}{z-1}=
\begin{cases}
-\sum_{m=0}^\infty z^m \sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}&,0<|z|<1\tag3\\\\
\sum_{m=1}^\infty z^{-m}\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}&,1<|z|
\end{cases}
$$

For $|z|>1$, the Laurent series of $\frac{e^{1/z^2}}{z-1}$ can be written

$$\begin{align}
\frac{e^{1/z^2}}{z-1}&=\sum_{m=1}^\infty z^{-m}\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}\\\\
&=\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,\sum_{m=1}^\infty z^{-(n+m)}\\\\
&\overbrace{=}^{p=n+m}\sum_{n=0}^\infty  \frac{a_n}{(n/2)!}\sum_{p=n+1}^\infty\,z^{-p}\\\\
&=\sum_{p=1}^\infty\left(\sum_{n=0}^{p-1}  \frac{a_n}{(n/2)!}\right)\,z^{-p}
\end{align}$$

For $0<|z|<1$, the Laurent series of $\frac{e^{1/z^2}}{z-1}$ can be written

$$\begin{align}
\frac{e^{1/z^2}}{z-1}&=-\sum_{m=0}^\infty z^{m}\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\,z^{-n}\\\\
&=-\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\sum_{m=0}^\infty z^{m-n}\\\\
&\overbrace{=}^{p=m-n}-\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\sum_{p=-n}^\infty z^{p}\\\\
&=-\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\left(\sum_{p=-n}^{0} z^{p}+\sum_{p=1}^\infty z^{p}\right)\\\\
&=-e \sum_{p=1}^\infty z^{p}-\sum_{n=0}^\infty \frac{a_n}{(n/2)!}\sum_{p=0}^{n} z^{-p}\\\\
&=-e \sum_{p=1}^\infty z^{p}-\sum_{p=0}^{\infty}\left(\sum_{n=p}^\infty \frac{a_n}{(n/2)!} \right)z^{-p}\\\\
&=-e \sum_{p=0}^\infty z^{p}-\sum_{p=1}^{\infty}\left(\sum_{n=p}^\infty \frac{a_n}{(n/2)!} \right)z^{-p}
\end{align}$$


问:
troubles to understand the embedding theorem
I try to understand the proof of the embedding theorem:

"An $m-$dimensional differentiable manifold can be embedded as a closed subset of the Euclidean space $\Bbb R^n$, if $2m<n$."

The proof is given as follow: We know that we can find for a manifold a proper function $f:M^m \to \Bbb R^n$ for an $n>0$ and that we can approximate this by an injective immersion $g:M^m \to \Bbb R^n$, so that $\vert g - f \vert \leq 1$ and $A= \emptyset$. If $K \subset \Bbb R^n$ is compact, then $K \subset K(r)$ for some radius $r$, hence $g^{-1}(K)$ is closed in the compact set $f^{-1}(K\overline{(r+1)})$, hence compact. Therefore $g$ is proper, hence an embedding.

Now I have some questions in this proof:

Why we have that if $K \subset \Bbb R^n$ is compact, then $K \subset K(r)$ for some radius $r$?

Why it follows that $g^{-1}(K)$ is closed in the compact set $f^{-1}(K\overline{(r+1)})$?

Why we can now conclude that a proper is an embedding?

EDIT:

I guess, he claims that $g^{-1}(K) \subset f^{-1}(K\overline{(r+1)})$ to conclude that $g^{-1}(K)$ is compact. But Why is this inclusion right? I guess that it comes from the fact, that $\vert g - f \vert \leq 1$. But I am not sure.

Many thanks for your help!

答:
Compact sets in $\mathbb{R}^n$ are bounded.

Preimage of a closed set by continuous $g$ is closed, and preimage of a compact set  by proper $f$ is compact.

Proper injective immersion is exactly an embedding with closed image.

$g^{-1}(K)\subseteq f^{-1}(K(r+1))$ if and only if $[g(x)\in K\implies f(x)\in K(r+1)]$. But $g(x)\in K\implies g(x)\in K(r)$ i.e. $|g(x)|<r$ and by triangle inequality $|f(x)|\leq |g(x)|+|f(x)-g(x)|<r+1$, so indeed $[g(x)\in K\implies f(x)\in K(r+1)]$. Thus $g^{-1}(K)\subseteq f^{-1}(K(r+1))$ and $g^{-1}(K)\subset f^{-1}(\overline{K(r+1)})$.


问:
Let $I=\langle 240,x^3-x^2\rangle$ be an ideal of $\mathbb Z[x]$, then does $x^3 -x+30\in\sqrt{I}$ hold?
The question can basically be re-written as, does there exist $f,g\in\mathbb Z[x]$ and $n\in\mathbb N$ such that $240f(x) +(x^3-x^2)g(x)=(x^3-x+30)^n$?

I firstly tried to compare constants and the coefficients of $x$ on the left and right hand side but couldn’t reach a contradiction. I think there are no such polynomials or $n$ since it would be difficult to find them explicitly, but not sure if there are some quick tricks or properties of radical ideals that are able to help answer the question

答: Since the lowest power of $30$ that is divisible by $240$ is four, I tried dividing $(x^3-x+30)^4$ by $x^3-x^2$. This gave a remainder of $108000 x^2-108000 x+810000$, which, lo and behold, is divisible by $240$.


问:
How to show that poisson process with linear drift diverges to $\infty$
If $({N_t})_{t\in\mathbb{R}}$ is a poisson point process at rate $\lambda$, I want to show that $N_t-c t$ diverges to infinity as $t\to\infty$ if $c<\lambda$. I can show using the law of large numbers that for $n\in\mathbb{N}$ we have $\frac{N_n}{n}\to\lambda$ as $n\to\infty$. And from there I can conclude $N_n-cn\to\infty$ as $n\to\infty$. But I don't know how to show this for general $t\in\mathbb{R}$ when $t\to\infty$.

答:
Let $n$ denote $\lfloor t\rfloor$, for ease of reading. Use
$$
\frac{N_n}{n+1}\le \frac{N_t}t\le \frac{N_{n+1}}n
$$
to show that $N_t/t\to \lambda$ as $t\to\infty$ through the reals.


问:
Finding $\sin(x+y)$ if $3\sin x+4\cos y=5$ and $4\sin y+3 \cos x=2$

Suppose that $x$ and $y$ satisfy the equations
$$\begin{align}
3\sin x+4\cos y &=5 \\
4\sin y+3 \cos x &=2
\end{align}$$
What is the value of $\sin(x+y)$?

I'm quite rusty at trigonometry, and this practice question confused me so much. Can someone give me a detailed explanation on how to solve this?

答:
We want $ \sin(x+y) = \sin x \cos y + \cos x \sin y$.

Note that the first equation has a linear combination of $ \sin x $ & $\cos y$ ... but we want their product ... so square it !

Is this enough of a hint ?

 $ ( 3 \sin x + 4 \cos y )^2 + ( 4 \sin y + 3 \cos x )^2 = 5^2+2^2 \\ 9(  \sin^2 x +  \cos^2 x ) + 24(\sin x \cos y + \cos x \sin y) +16 ( \sin^2 y +  \cos^2 y ) = 5^2+2^2 \\ \sin(x+y) =\color{red}{\frac{1}{6}}$.


答: Square the equations then add them to get $\sin x\cos y+\cos x\sin y=1/6$.

Title: Finding $\sin(x+y)$ if $3\sin x+4\cos y=5$ and $4\sin y+3 \cos x=2$

Body:
Suppose that $x$ and $y$ satisfy the equations
$$\begin{align}
3\sin x+4\cos y &=5 \\
4\sin y+3 \cos x &=2
\end{align}$$
What is the value of $\sin(x+y)$?

I'm quite rusty at trigonometry, and this practice question confused me so much. Can someone give me a detailed explanation on how to solve this?


答:
Factoring is your friend. The two parts $x^3-x$ and $30$ of your polynomial each look similar in form to one of the two generators of $I$. Factoring the two parts, and factoring the two generators, and comparing, we find that $30\in\sqrt I$ as $30^4=3375\cdot240$. And we get $x^3-x\in\sqrt I$, as
$$
(x^3-x)^2=(x-1)(x+1)^2\cdot(x^3-x^2)
$$
Finally, $\sqrt I$ is closed under addition, so $x^3-x+30$ is also contained in $\sqrt I$.

Title: Let $I=\langle 240,x^3-x^2\rangle$ be an ideal of $\mathbb Z[x]$, then does $x^3 -x+30\in\sqrt{I}$ hold?

Body:
The question can basically be re-written as, does there exist $f,g\in\mathbb Z[x]$ and $n\in\mathbb N$ such that $240f(x) +(x^3-x^2)g(x)=(x^3-x+30)^n$?

I firstly tried to compare constants and the coefficients of $x$ on the left and right hand side but couldn’t reach a contradiction. I think there are no such polynomials or $n$ since it would be difficult to find them explicitly, but not sure if there are some quick tricks or properties of radical ideals that are able to help answer the question


问:
Proving function for Stirling Numbers of the Second Kind
I need to proof the following formula for Stirling Numbers of the Second Kind:

$\sum\limits_{n \geq 0} S(n,k) x^n = \frac{x^k}{(1-x)(1-2x)\cdots(1-kx)}$

It is used widely around formularies but I neither have found any proof nor was I able to figure it out on my own.

Could you please help me finding a solution?

Thank you in advance!

答:
This is set out in the initial part of section 1.6 of Geneatingfunctionology with the result in equation 1.6.5.

In essense, you start with the recurrence

$$S(n,k) = S(n-1,k-1) + kS(n-1,k)$$

with $S(0,0)=1$.  Of the natural generation functions which might express this, the simplest which deals with the factor of $k$ is likely to be of the form

$$B_k(x) = \sum_n S(n,k) x^n$$

which with the recurrence leads to

$$B_k(x) = xB_{k-1}(x) + kxB_k(x) = \frac{x}{1-kx} B_{k-1}(x)$$

for $k \ge 1$; $B_0(x)=1$.  That in turn leads to the desired formula by multiplying successive terms.


答:
Notice that:

$$\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz} = \left(x + \frac{1}{x}\right)\left(y + \frac{1}{y}\right)\left(z + \frac{1}{z}\right).$$

The minimum of the function $f(w) = w + \frac{1}{w}$ is attained for $w=1$. More specifically, $f(1)= 2.$ See addendum section below for further details.

Then:

$$\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz} = \left(x + \frac{1}{x}\right)\left(y + \frac{1}{y}\right)\left(z + \frac{1}{z}\right) \geq 2 \cdot 2 \cdot 2 = 8.$$

Addendum

This addendum uses calculus tools, not just pre-calculus ones!

$w=1$ is the minimum of $f(w) : \mathbb{R}^+ \to \mathbb{R}^+.$
Indeed $f(w)$ is continuous for $w>0$. Moreover, notice that:

$$f'(w) = \frac{w^2-1}{w^2}.$$

Hence, $f'(w) = 0 \Rightarrow w = 1$.

Moreover:

$$f''(w) = \frac{2}{w^3} > 0$$  for all $w > 0.$

Finally, notice that:

$$\lim_{w=0^+} f(w) = +\infty,$$

and

$$\lim_{w=+\infty} f(w) = +\infty.$$
Then, $w=1$ is the unique minimum for $w > 0.$

Title: Showing $\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$

Body:
I found the following exercise reading my calculus notes:

If $x,y$ and $z$ are positive real numbers, show that $$\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$$

I've been trying to solve it for a while. However, I have no idea how to approach it. Any help is welcome.


问:
Finding a real $3\times3$ matrix with eigenvalues $1$, $i$ and $-i$ geometrically
In a problem I'm trying to find a real-valued $3\times3$ matrix that has the eigenvalues $1$, $i$ and $-i$ (which must mean that the corresponding eigenvectors for $i$ and $-i$ must be complex). I already know that one way to approach this is to write out a general $3\times3$ matrix of the form

$$A=
\begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & j
\end{pmatrix}$$

then evaluate the determinant of $A-\lambda I$ and set it equal to $0$ to form a characteristic polynomial for the eigenvalues and then choose $a,b,c,d,e,f,g,h,j$ such that the polynomial becomes $(\lambda-1)(\lambda^2+1)$ to give the required eigenvalues as roots.

However, I'm looking for a geometric way to find such a matrix (which is the approach hinted at in the problem). I thought maybe some kind of complex plane transformation might work, but I wasn't sure how a $3\times3$ matrix would apply in such a situation.

How can I find such a matrix geometrically, without having to do lots of algebra as in the method outlined above?

答:
The $2\times2$ matrix $\pmatrix{0&1\\-1&0}$, which corresponds geometrically to $90^\circ$ rotation

(multiplication by $i$), has eigenvalues $i$ and $-i$.

Put a $\color{blue}1$ on the diagonal, and you're done:   $\pmatrix{\color{blue}1&0&0\\0&0&1\\0&-1&0}$.


问:
Are these equations derivations of the other?
It is stated   P(A|B) = {P(A and B)}{P(B)}

I would like to know if this is a derived from the Bayes' theorem P(A|B) = { P(A)* P(B|A) }/{P(B) }

答:
No, it is the converse.  Bayes' Theorem may be derived from the definition of conditional probability.  $$\mathsf P(A\mid B)=\dfrac{\mathsf P(A\cap B)}{\mathsf P(B)}\qquad\text{and}\qquad\mathsf P(B\mid A)=\dfrac{\mathsf P(A\cap B)}{\mathsf P(A)}\\\text{therefore}\\~\\\mathsf P(A\mid B)=\dfrac{\mathsf P(A)\,\mathsf P(B\mid A)}{\mathsf P(B)}$$

However, the factor of $\mathsf P(A\cap B)$ cannot be deduced from Bayes' Theorem.


问:
How to find the eigenvalues of $x x^T - y y^T$?

If $x$ and $y$ are two linearly independent column $n$-vectors, where $n \geq 2$, find all the eigenvalues of $x x^{T} - y y^{T}$

Could you please tell me whether what I have written is correct or not?

Using the identity

\begin{align*}
\lambda^n\det(\lambda I_{(m)} - AB) = \lambda^m\det(\lambda I_{(n)} - BA)
\end{align*}

for $A \in F^{m \times n}$ and $B \in F^{n \times m}$, we can calculate the characteristic polynomial of $xx^T - yy^T$ by setting $A = (x, y) \in F^{n \times 2}$ and $B = (x^T, -y^T)^T \in F^{2 \times n}$ directly as:

\begin{align*}
\varphi(\lambda) &= \det(\lambda I_{(n)} - (xx^T - yy^T)) =
\lambda^{n - 2}\det\left(\lambda I_{(2)} - \begin{pmatrix} x^T \\ -y^T \end{pmatrix}\begin{pmatrix} x & y \end{pmatrix}\right) \\
&=
\lambda^{n - 2}\begin{vmatrix}
\lambda - x^Tx & -x^Ty \\
y^Tx & \lambda + y^Ty
\end{vmatrix} \\
&= \lambda^{n - 2}[(\lambda - x^Tx)(\lambda + y^Ty) + (x^Ty)^2] \\
&=
\lambda^{n - 2}(\lambda^2 - (x^Tx - y^Ty)\lambda - (x^Txy^Ty - (x^Ty)^2))
\end{align*}

Since $x$ and $y$ are linearly independent, by Cauchy-Schwarz inequality $(x^Tx)(y^Ty) > (x^Ty)^2$ (that is, the equality of C-S inequality cannot hold), whence the determinant $\Delta$ of the quadratic equation $\lambda^2 - (x^Tx - y^Ty)\lambda - (x^Txy^Ty - (x^Ty)^2) = 0$ equals to

\begin{align*}
\Delta = (x^Tx - y^Ty)^2 + 4(\|x\|^2\|y\|^2 - (x^Ty)^2) > 0.
\end{align*}
Hence the two non-zero eigenvalues are two distinct real numbers
\begin{align*}
\lambda_1 = \frac{y^Ty - x^Tx + \sqrt{\Delta}}{2}, \quad
\lambda_2 = \frac{y^Ty - x^Tx - \sqrt{\Delta}}{2}.
\end{align*}

答:
This looks correct, except for the sign error in the final formulas, and fact that you could (and do) have $\lambda=0$. Think about it geometrically: $xx^T$ is the orthogonal projection onto $x$ times $x^Tx$, $yy^T$ is the orthogonal projection onto $y$ times $y^Ty$. So everything orthogonal to $x$ and $y$ is mapped to zero (you can also check this algebraically), giving you an $n-2$ dimensional zero eigenspace. Then you can restrict to the plane spanned by $x$ and $y$, effectively reducing to the 2D problem, which gives the two eigenvalues you have derived: namely, in the basis $v_1=x$ and $v_2=y$ we have $(1,0) \to (x^Tx, -y^Tx) $,  $(0,1) \to (x^Ty, -y^Ty)$ giving the matrix in your second line.

Moreover, a geometric analysis leads to the following: let $u=x+y, v=x-y$. Then $M=xx^T-yy^T=\frac{1}{2} [ vu^T+uv^T ]$ and let

$w_1=v |u| +u|v|$

$w_2=v|u|-u|v|$

I claim these are the eigenvectors. (The geometry of the problem is that an eigenvector is any vector which after reflection through the line of $u$ along direction of $v$ is rotated by 90 degrees; then drawing a picture reveals that it must be a sum of two vectors of equal length directed along the lines of $u$ and $v$, leading to the above formulas.)

We check

$$ 2M w_1=[ vu^T+uv^T ] (v |u| +u|v|)= v [u^Tv |u| +u^Tu |v| ]+u[v^Tv|u|+v^Tu|v| ]= $$
$$v |u| [u^Tv+|u||v|]+u|v|[|v||u|+v^Tu]= [|v||u|+v^Tu] w_1$$

Similarly

$$ 2M w_2=[ vu^T+uv^T ] (v |u| -u|v|)= v [u^Tv |u| -u^Tu |v| ]+u[v^Tv|u|-v^Tu|v| ]$$
$$= v |u| [u^Tv-|u||v|]-u|v|[-|v||u|+v^Tu]= [-|v||u|+v^Tu] w_2$$

Thus the eigenvalues are $\lambda_1=\frac{1}{2} [|v||u|+v^Tu] $ and $\lambda_2=\frac{1}{2}[ -|v||u|+v^Tu]$.

You can check their sum is $\lambda_1+\lambda_2=x^Tx-y^Ty$, matching the trace of your matrix, and their product is

$\frac{1}{4}[(v^Tu)^2-|u|^2|v|^2]= \frac{1}{4}[(|x|^2-|y|^2)^2-(|x|^2+|y|^2)^2- (2x^Ty)^2]=|x|^2|y|^2- (x^Ty)^2$

matching the determinant of your matrix.


问:
why does this loan calculation have no direct equation?

The problem

Find the monthly payment at which a given "simple" loan would be paid off in $n$ months.  I was told there is no direct equation for this.  I would like to understand why.

Inputs

$P$: the loan's starting principal balance

$I$: the loan's starting interest balance

$r$: the monthly interest rate

$n$: the target number of months by which to pay off the loan (i.e. both the principal balance and the interest balance are reduced to zero)

Output

$p$: the fixed monthly payment made on the loan

Rules

Interest is calculated at the beginning of the month, on the remaining principal balance only.  This newly accrued interest is immediately added to the running interest balance, $I$.

The fixed monthly payment is applied at the end of the month, always first towards the remaining interest balance, then towards the remaining principal balance.

Reduced case, for study

Consider a loan that has no starting interest balance.  I found this known equation.  I then solved for $p$.

$$p = \frac{P \cdot r}{1 - 10^{-N \cdot log (1 + r)}}$$

Some test values, everything checks out.

```<code>P=1000, I=0, r=0.01, n=1 --> p = 1010
P=1000, I=0, r=0.01, n=2 --> p = 507.51
P=1000, I=0, r=0.01, n=3 --> p = 340.02
P=1000, I=0, r=0.01, n=4 --> p = 256.28
</code>```

Common case

Now consider a loan with both starting principal and interest balances.

Let $I_i$ be the interest balance at the start of month $i$.  I worked out this equation:

$$I_i = I + (i - 1)(P \cdot r - p)$$

My stubborn intuition is that you should be able join the above two formulas together somehow?

I think of the life of the loan as having three phases.

Phase one - the monthly payment only goes towards interest.  Here, the phase begins with a principal balance of $P$ and an interest balance of $I$.  This phase will last $n_0$ months, which I worked out as being

$$n_0 = \lfloor{\frac{I_1}{p - P \cdot r} + 1}\rfloor$$

Phase two - the crossover or transition month.  This phase is exactly one month long (I'm ignoring the edge case where this month may not even happen).  For this month, part of the monthly payment is used to zero out the interest balance.  The remainder of the payment is used to start whittling down the principal balance.  Here, this phase begins with a principal balance of $P$ and an interest balance of

$$I_{n_o} = I_1 + (n_0 - 1)(P \cdot r - p)$$

Phase three - the monthly payment only goes towards principal.  This is the same as our above reduced case, for which there is a known formula for $p$.  Here, this phase begins with an interest balance of zero, and a principal balance I'm calling $P'$.  This phase will last $n_1$ months.  I worked out these equations:

$$P' = P(1 + r) - p + I_{n_o}$$

$$n_1 = N - n_0 - 1$$

The impasse

Given the above formulas, I tried combining the three phases together to solve for $p$.  But I soon became stuck, and ended up with $p$ inside a floor function.  Is this the specific mathematical reason why there's no direct equation here?  Because it's impossible to isolate a variable that's inside a floor function?

Again, my only goal below is to get $p$ all alone on the left side of the equation.  I defined several constants below, none of which depend on $p$.

$$p = c_3 + c_4 \cdot \frac{\lfloor{\frac{I_1}{p - P \cdot r} + 1}\rfloor}{1 + c_1} - c_5 \cdot \frac{\lfloor{\frac{I_1}{p - P \cdot r} + 1}\rfloor \cdot P}{1 + c_1} + c_5 \cdot \frac{P}{1 + c_1}$$

where

$$c_0 = 1 - 10^{-N \cdot log (1 + r)}$$

$$c_1 = \frac{r}{c_o}$$

$$c_2 = \frac{P \cdot r \cdot (1 + r)}{c_0}$$

$$c_3 = \frac{c_2 + c_1 \cdot I_1 - c_1 \cdot P \cdot r}{1 + c_1}$$

$$c_4 = \frac{c_1 \cdot P \cdot r}{1 + c_1}$$

$$c_5 = \frac{c_1}{1 + c_1}$$

答:
First of all, it is important to understand exactly where the initial formula (based on a zero interest balance) came from.

$$p = \frac{P \cdot r}{1 - 10^{-N \cdot log (1 + r)}}$$

Let $s = (1 + r),~$  and assume that $log$ is referring to logarithms base 10.  Then:

$$10^{-n \cdot log (s)} ~=~ \frac{1}{(s)^n}.$$

where it is assumed that $n$ payments will be made.

Therefore, the formula is asserting that

$$p \left[1 - \frac{1}{(s)^n}\right] ~=~ Pr.$$

You can reasonably assume that since the payment is always greater than the running interest balance, that (for simplicity) the running interest balance can be lumped in with the loan balance.

I highlighted the above sentence to emphasize that in your first scenario, it is true.  However, in your second scenario, where the beginning interest balance is not zero, things will get more complicated.

Anyway, staying within the first scenario, let $P_k$ represent what is happening at the end of the $k$-th month, under the assumption that the payment is always received before the end of the month, and then applied at the end of the month.

$P_1 ~: ~Ps - p.$

$P_2 ~: ~(Ps - p)s - p = Ps^2 - ps - p.$

$P_3 ~: ~(Ps^2 - ps - p)s - p = Ps^3 - ps^2 - ps - p.$

$\cdots$

$P_n ~: ~Ps^n - p\left[s^{(n-1)} + s^{(n-2)} + \cdots s^2 - s - 1\right].$

Note that:

$$\left[s^{(n-1)} + s^{(n-2)} + \cdots s^2 - s - 1\right]  ~=~ \frac{s^n - 1}{s - 1}.$$

This means that $$Ps^n ~=~ p\left[\frac{s^n - 1}{s - 1}\right].$$

Noting that $~r = (s-1),~$ this means that

$$Pr ~=~ p\left[\frac{s^n - 1}{s^n}\right] ~=~ p \left[1 - \frac{1}{(s)^n}\right].$$

Assuming that you understand the source of the formula, a key point must now be addressed.  Consider the following rule, which is (presumably) intended to apply to loans with a beginning interest balance of zero, where the fixed monthly payment is greater than the incurred monthly interest charge:

Interest is calculated at the beginning of the month, on the remaining principal balance only.

Banks are in business to make money.  The bank accepts the above rule because (in effect) the running interest balance is always zero.

This means that to consider what happens when the initial interest balance is not zero, you have to consider two separate cases:

The bank will allow the interest rate to only be applied to the loan balance, regardless of whether the interest balance is itself significant.

The bank will insist that the interest rate also be applied to the
(significant) interest balance.

Although I consider the second case above more realistic, I will go ahead and attack each case separately, starting with the first case.

Note that your work assumes that the second case pertains.  The problem with your work in the second case is that you made things overly complicated.  Anyway, I am going to attack each case from scratch, starting with the first case:

$\underline{\text{Case 1: the interest rate is applied only to the loan balance}}$

First of all - pretend that you have a zero interest balance, and use that to calculate $p$ under the assumption that $n_1$ monthly payments will be made (after the interest balance is reduced to zero).

Let $$n_0 = \left\lfloor \frac{I}{p} \right\rfloor$$

This means that

$$n_0~\leq ~\left(\frac{I}{p}\right) ~< (n_0 + 1).$$

Under this scenario, after the payment is received at the end of  month $n_0$, the interest balance will have been reduced so that it is now $~< p$.  The easiest way to handle this scenario is to include a second payment (in this month only) that is sufficient to bring the interest balance to zero.

This will keep the math simple and imply that your payments are all constant except in month $~n_0,~$ where you are making a second (interest) payment.  This means that except for this single occurrence of a second payment, you will end up making $(n_0 + n_1)$ payments of $p$.

If you are lucky enough to find a mentally retarded bank loan officer, then you can pursue Case 1.

$\underline{\text{Case 2: the interest rate is also applied to the interest balance}}$

I studied your work.  As you discovered, you are chasing your tail as follows:

If you calculate $n_0$ and $n_1$ via the algorithm in Case 1, you will find that at the end of $(n_0 + n_1)$ payments, you haven't paid off the loan because the beginning interest balance was itself accruing interest.

You can't perform the computation of $n_0 = \left\lfloor\frac{I}{p}\right\rfloor$ without knowing what $p$ is.  And you can't determine what $p$ is without knowing what $(n_0 + n_1)$ is.

You need a way out of this mess that the bank will accept.  All the bank cares about is money.

Simply construe $L_1 = L + I,$ where $L$ is the initial (or outstanding) loan balance, $I$ is the current interest balance (if any), and $L_1$ is the new loan balance.  Then calculate the payment $p$ assuming that the loan is $L_1$ and the number of months of the loan, from this point forward, is to be $n$.

From the bank's perspective this is a realistic and acceptable construance because:

Interest will be charged against $L_1$.

Assuming that $p$ is caculated on the basis of $n$ months, $p$ will (as in your initial scenario) be greater than the monthly interest charges.

This is because with interest being charged against $L_1,$ the new interest balance $I_1$ will start at zero.  Therefore, as before, the running interest balance will always be zero.


问:
Help me solve a trigonometric equation
I am doing some work in RF circuit design. I need to solve an equation for my design:

$$\frac 1{\cos(t_1)}+\frac 1{\sin(t_1)} =\frac 1{\cos(t_2)}+\frac 1{\sin(t_2)}$$

(I created a nicely typed image of this equation but setting don't allow me to upload unless I get at least 10 !]

Kindly help me find a general solution of the above equation (i.e. to find relation between $t_2$ and $t_1$). I may solve easier equations (like $\,\cos t=0.5$ !), but seems complex to me.
I observe that at $t_1=t_2=\dfrac {\pi}4$, this equation holds true. But I am unable to proceed further.

Thanks.
Mohammad A Maktoomi.

答:
Put $\sin\bigl(t+{\pi\over4}\bigr)=:u$. Then
$${1\over\cos t}+{1\over\sin t}={\sqrt{2}\sin\bigl(t+{\pi\over4}\bigr)\over{1\over2}\sin(2t)}=-{2\sqrt{2} u\over\cos\bigl(2(t+{\pi\over4})\bigr)}=-2\sqrt{2}{u\over 1-2u^2}\ .$$
Disregarding the cases $u=\pm{1\over\sqrt{2}}$ we therefore have to consider the equation
$${u_1\over 1-2u_1^2}={u_2\over 1-2u_2^2}\ ,$$
or
$$(u_1-u_2)(1+2u_1u_2)=0\ .$$
There is the obvious solution $u_1=u_2$, which leads to (i) $t_1=t_2$ or  (ii) $t_1+t_2={\pi\over2}$.

The solutions to $1+2u_1u_2=0$ can be parametrized by
$$u_1=\pm {1\over\sqrt{2}} e^\tau\ ,\quad u_2=\mp {1\over\sqrt{2}} e^{-\tau}\qquad(-\log\sqrt{2}\leq \tau\leq\log\sqrt{2})\tag{1}$$
(note that $u_1$ and $u_2$ are sines). Each pair $(u_1,u_2)$ produced by $(1)$ will lead to two pairs $(t_1,t_2)$ solving the original equation; see the cases (i) and (ii) above.

Title: Help me solve a trigonometric equation


问:
Operator norm of integral operator of continuous functions
Let $A:C^0[-1,1] \rightarrow \mathbb{R} $ be defined by $$A(f) =\int_{-1}^{1} \gamma (t) f(t) dt $$ for $\gamma \in C^0[-1,1] $.
It's easy to show that $$||A||\leq \int_{-1}^{1}|\gamma (t) | dt$$ but how can the reverse inclusion be shown.

I was thinking something along the lines of constructing a function $f$ based on the roots of $\gamma $, where $\gamma (t) >0 $ let $f(t) =1 $ and where $\gamma (t) <0 $ let $f(t) =-1 $. But to make this $f$ continuous I was thinking of joining - 1 and 1 (near a root of $\gamma $) by a line segment which becomes arbitrarily steep. Its intuitively clear that then we can make $|A(f) |$ arbitrarily close to the value above that I am looking to prove.
Seems a bit difficult to make formal though.

Any help or better ideas? By the way I'm using the supremum norm on the space of continuous functions.

答:
Here is a more formal argument presenting your idea. This is taken from one of my old questions.

Let $\varepsilon > 0$. Since $\gamma$ is continuous, $\gamma^{-1}(\langle 0, +\infty\rangle)$ and $\gamma^{-1}(\langle -\infty,0\rangle)$ are open sets so they can be written as countable (or finite, that case is easier) disjoint unions of intervals:

$$\gamma^{-1}(\langle 0, +\infty\rangle)) = \bigcup_{n=1}^\infty \langle a_n, b_n\rangle, \quad \gamma^{-1}(\langle -\infty,0\rangle) = \bigcup_{n=1}^\infty \langle c_n, d_n\rangle$$

Now define $$S = \bigcup_{n=1}^\infty \left\langle a_n+\min\left\{\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}, \frac{b_n-a_n}{2}\right\}, b_n-\min\left\{\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}, \frac{b_n-a_n}{2}\right\}\right\rangle \\
\cup \bigcup_{n=1}^\infty \left\langle c_n+\min\left\{\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}, \frac{d_n-c_n}{2}\right\}, d_n-\min\left\{\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}, \frac{d_n-c_n}{2}\right\}\right\rangle\\
\cup \gamma^{-1}(\{0\})$$

and define $f_\varepsilon \in C[-1,1]$ to be equal to $\operatorname{sgn} f$ on $S$, and to an affine function on $[-1,1]\setminus S$ connecting $\pm 1$ with $0$ so that $f_\varepsilon$ is continuous.

Now we have:

$$\begin{align}Af_\varepsilon &= \int_{[-1,1]} \gamma(t)f_\varepsilon(t)\,dt\\
&= \int_{S} \underbrace{\gamma(t)f_\varepsilon(t)}_{=|\gamma(t)|}\,dt + \underbrace{\int_{[-1,1]\setminus S} \gamma(t)f_\varepsilon(t)\,dt}_{\ge 0}\\
&\ge \int_{S} |\gamma(t)|\,dt\\
&= \int_{[-1,1]} |\gamma(t)|\,dt - \int_{[-1,1]\setminus S} |\gamma(t)|\,dt\\
&\ge \int_{[-1,1]} |\gamma(t)|\,dt - \|\gamma\|_\infty \lambda([0,1]\setminus S)\\
&\ge \int_{[-1,1]} |\gamma(t)|\,dt - \varepsilon\\
\end{align}$$

since

$$\lambda([-1,1]\setminus S) = \lambda\left(
\bigcup_{n=1}^\infty
\left[a_n,
a_n+\min\left\{\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}, \frac{b_n-a_n}{2}\right\}
\right]
\cup
\left[b_n-\min\left\{\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}, \frac{b_n-a_n}{2}\right\}, b_n
\right]
\right)\\
+\lambda\left(
\bigcup_{n=1}^\infty
\left[a_n,
c_n+\min\left\{\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}, \frac{d_n-c_n}{2}\right\}
\right]
\cup
\left[d_n-\min\left\{\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}, \frac{d_n-c_n}{2}\right\}, d_n
\right]
\right)\\
\le \sum_{n=1}^\infty \lambda\left(\left[a_n, a_n+\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}\right]\right) + \lambda\left(\left[b_n-\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}, b_n\right]\right) + \lambda\left(\left[c_n, c_n+\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}\right]\right) + \lambda\left(\left[d_n- \frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}, d_n\right]\right)\\
=\frac{\varepsilon}{\|\gamma\|_\infty}\sum_{n=1}^\infty\frac{1}{2^{n+1}} = \frac{\varepsilon}{\|\gamma\|_\infty}$$

Thus:

$$\|A\| \ge \frac{\|Af_\varepsilon\|_\infty}{\|f_\varepsilon\|_\infty} \ge |Af_\varepsilon| = Af_\varepsilon = \int_{[-1,1]} |\gamma(t)|\,dt - \varepsilon \xrightarrow{\varepsilon\to 0} \int_{[-1,1]} |\gamma(t)|\,dt.$$


问:
Why do we say A given B in Bayes theorem?
I enjoyed this blog on visualizing Bayes theorem but I have a question on the wording for .

$p(A\mid B) = p(A,B)/p(B)$

In words I understand this to be "probability of A given B"

if we are "given B" why do we need to divide by p(B) ?   Surely if I am "given" something then its probability is $1$?

Why do we say it this way?

[Update]

I am seeing that p(B) = 1 (because we are "given" it )

thus p(A,B)/p(B) = p(A,B)

I am looking for the right words to articulate that we still need to divide by p(B) and that it is not 1

If we were saying "given we know the probability of B" instead of "given B" I would not be so confused.

答:
You can think of this in another way. Think of an Excel sheet with 4 columns as follows:

\begin{array} {|r|r|}\hline Event 1 & P(Event 1) & Event 2 & P(Event 2) \\ \hline A & 0.5 & B & 0.5 \\ \hline F & 0.5 & B & 0.25 \\ \hline A & 0.25 & B & 0.25 \\ \hline F & 0.5 & K & 1 \\ \hline A & 0.25 & J & 1 \\ \hline  \end{array}

Now, the question is what is the probability of A, knowing that B has occurred?

Since we know B has already occurred, we can just filter the Event2 column for 'B' and we get this:

The "Given B" table

\begin{array} {|r|r|}\hline Event 1 & P(Event 1) & Event 2 & P(Event 2) \\ \hline A & 0.5 & B & 0.5 \\ \hline F & 0.5 & B & 0.25 \\ \hline A & 0.25 & B & 0.25 \\    \hline\end{array}

Ok, so we see that A and F occur given that B has occured from the above table. So we want to know only about A and we don't care about F. So let's filter on the Event1 column above for 'A' only.

\begin{array} {|r|r|}\hline Event 1 & P(Event 1) & Event 2 & P(Event 2) \\ \hline A & 0.5 & B & 0.5 \\ \hline A & 0.25 & B & 0.25 \\ \hline   \end{array}

But really, the table above tells you the Probability of A occurring and B occurring. I think the "given B" part can be thought of using this table. Here you know the probability of A "given B". So the probabilities can be represented as:

total probability of A occurring * total probability of B occurring, which is
$$(0.5 + 0.25) \times (0.5 + 0.25)$$

But, the question again is what is the probability of A, knowing that B has occurred?

So we don't really care about B here. And to get the value of A and kick B out of the picture, we divide by the total probability of B

$$\frac{(0.5 + 0.25) \times (0.5 + 0.25)}{(0.5 + 0.25)}$$

$$=0.75$$


问:
Finding a constant which makes a Homogeneous ODE into a Non-Homogeneous ODE
I have a Non-Homogeneous ODE of the form $y'' + 8y' + 17y = 5$  which I have obtained the solution $y(x) = ae^{-4x}[\cos(x) + i\sin(x)] + be^{-4x}[\cos(x)-i\sin(x)] + \frac{5}{17}$ where $a$ and $b$ are complex constants.

I am unsure of how to find a constant, $c$, which satisfies the equation $z(x) = y(x) - c$  where $c$ is a real constant and $z(x)$ is a Homogeneous Differential Equation. I originally thought that $c = \frac{5}{17}$ but that seems too simple.

How should I go about finding $c$?

答:
You are correct that $c=\frac{5}{17}$, which gives the particular solution to be $y_{p}(x)=\frac{5}{17}$.

Since the RHS is a constant, you would try a constant as your particular solution $y_{p}(x)=c$ and substituting gives $c=\frac{5}{17}$. See here.

Note your solution can be simplified further, since $a$ and $b$ are arbitrary constants we can write
$$y(x) = c_{1}e^{-4x}\cos(x)+c_2e^{-4x}\sin(x) + \frac{5}{17}$$

where $c_1=a+b$ and $c_2=i(a-b)$.


问:
Proving $\mathrm{det}\, T = \mathrm{det}\, R\, \mathrm{det}\, S$ given $T: V\to V$ by $T{\bf x}=R{\bf w}_1+S{\bf w}_2$
I've been learning linear algebra and I came across this question that I don't know how to solve. The question begins by letting $W_1$ and $W_2$ be subspaces of a finite-dimensional vector space $V$. Suppose that $V=W_1 \oplus W_2$. Then, for any $x∈V$ we may express $x$ uniquely as $x=w_1+w_2$ where $w_1∈W_1$ and $w_2∈W_2$. Let $R: W_1 \to W_1$ and $S: W_2\to W_2$ be linear transformations and define the linear transformation$T = \mathrm{det}\, R\, \mathrm{det}\, S$ given $T: V\to V$ by $T{\bf x}=R{\bf w}_1+S{\bf w}_2$

The question asks to prove $\mathrm{det}\, T = \mathrm{det}\, R\, \mathrm{det}\, S$

If anyone could give some guidance or help solve this problem, I would really appreciate it

答:
Hint: If $A$ and $B$ are square matrices, then
$$\det\pmatrix{A&0\\0&B}\ =\ \det A\cdot\det B\,.$$


问:
A property on compact Hausdorff space
Let $X$ be a Hausdorff compact space and let $A$ and $B$ be two disjoint closed sets in $X$.

Suppose that for any $z\in X$ we can find a neighborhood $U_z$ such that $U_z\cap A=\emptyset$ or $U_z\cap B=\emptyset$.
Since $\{U_z\mid z\in X\}$ is a cover of $X$, we can find $\{z_1,z_2,\dots,z_n\}\subseteq X$ such that $\bigcup\limits_{k=1}^nU_{z_k}=X$.

A chain from $x$ to $y$ on $X$ is a family of open sets $\{U_k\}_{k=1}^n$ such that $x\in U_1,$ $y\in U_n$ and $U_k\cap U_{k+1}\neq\emptyset$, $\forall k\in\{1,2,\dots,n-1\}$.

I am asking if we can construct such a cover $\{U_{z_k}\}_{k=1}^n$ so that for $a\in A$ and $b\in B$ there is no chain made by elements of $\{U_{z_k}\}_{k=1}^n$ from $a$ to $b$?

P.S. I haven't used the Hausdorff condition yet!

Update:

Let me explain my whole purpose of the question: It is a well known fact that if $Q$ is a quasi-component in a Hausdorff compact space, then it must be connected. I want to prove this using the concept of chains.

There is a result that states: For $x\in X$, the set $Q(x)=\{y\in X|$ for every possible cover of $X$ there is a chain from $x$ to $y$ consisting of elements of the cover $\}$ is a quasi-component of $x$.

I want to prove that if $Q=A\cup B$ where $A$ and $B$ are disjoined closed sets, then there must exists at least one $z\in X$ s.t. every neighborhood of $z$ has nonempty intersection with both $A$ and $B$, and since $A$ and $B$ were closed they cannot be disjoined, which will contradict the assumption that $Q$ is disconnected.

So, I suppose that if $\forall z\in X$ there is a neighborhood s.t. does not intersect both $A$ and $B$, then there must be a cover of $X$ for which you cannot find a chain from an element in $A$ to an element in $B$.

答:
Let $X$ be connected and $A, B$ nonempty. Assume such a cover $(U_{z_k})_{k=1}^n$ exists where no chain between $A$ and $B$. Then define $S$ to be the set of all indices $k$ such that there exists a chain to $A$. Furthermore, we define
$$M=\bigcup_{k\in S} U_{z_k}$$
and
$$ N= \bigcup_{k\in S^c} U_{z_k}.$$
By construction we get that $M, N$ are disjoint and open. As $A,B$ are nonempty, so are $M, N$. However, then $M$ is clopen, nonempty and different from $X$ which contradicts that $X$ is connected.

Title: A property on compact Hausdorff space


问:
Showing $\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$
I found the following exercise reading my calculus notes:

If $x,y$ and $z$ are positive real numbers, show that $$\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$$

I've been trying to solve it for a while. However, I have no idea how to approach it. Any help is welcome.


问:
Jensen's Formula and the number of zeros.
I am doing a question that requires me to find the number of zeros in a disk when $f$ is non-constant, bounded and analytic on $D$. I saw that the Wikipedia page of Jensen's formula has the following information:

Jensen's formula can be used to estimate the number of zeros of analytic function in a circle. Namely, if f is a function analytic in a disk of radius R centered at $z_0$ and if |f| is bounded by M on the boundary of that disk, then the number of zeros of f in a circle of radius r < R centered at the same point $z_0$ does not exceed

Does anyone know how this formula is derived?

答:
Lemma:

For an analytic $f$ with no zeros on $|z|=r$ and whose zeros inside $|z|=r$ are located at $\left\{\alpha_k\right\}_{k=1}^n$ with multiplicities, define
$$
g(z)=f(z)\prod_{k=1}^n\frac{r^2-\bar\alpha_kz}{r(z-\alpha_k)}\tag1
$$
Then $|g(z)|=|f(z)|$ when $|z|=r$ and $g$ has no zeros in $|z|\le r$.

Proof:

The factors of $z-\alpha_k$ remove the zeros inside $|z|=r$, so $g$ has no zeros for $|z|\le r$.

If $|z|=r$, then
$$
\begin{align}
\left|\frac{r^2-\bar\alpha z}{r(z-\alpha)}\right|
&=\left|\frac{\bar z}r\frac{r^2-\bar\alpha z}{r(z-\alpha)}\right|\tag{2a}\\
&=\left|\frac{\bar zr^2-\bar\alpha|z|^2}{r^2(z-\alpha)}\right|\tag{2b}\\
&=\left|\frac{r^2(\bar z-\bar\alpha)}{r^2(z-\alpha)}\right|\tag{2c}\\[6pt]
&=1\tag{2d}
\end{align}
$$
Explanation:

$\text{(2a)}$: $|\bar z|=|z|=r$

$\text{(2b)}$: distribute $\bar z$ over the denominator

$\text{(2c)}$: collect the factors of $|z|^2=r^2$

$\text{(2d)}$: $|\bar z-\bar\alpha|=|z-\alpha|$

Thus, $\prod\limits_{k=1}^n\frac{r^2-\bar\alpha_kz}{r(z-\alpha_k)}$ has absolute value $1$ when $|z|=r$; and therefore, $|g(z)|=|f(z)|$ when $|z|=r$.

$\large\square$

Jensen's Formula

Taking the average over $|z|=r$ gives
$$
\begin{align}
\frac1{2\pi}\int_0^{2\pi}\log\left|f\!\left(re^{i\theta}\right)\right|\,\mathrm{d}\theta
&=\frac1{2\pi}\int_0^{2\pi}\log\left|g\!\left(re^{i\theta}\right)\right|\,\mathrm{d}\theta\tag{3a}\\
&=\log|g(0)|\tag{3b}\\
&=\log|f(0)|+\sum_{k=1}^n\left(\log(r)-\log|\alpha_k|\right)\tag{3c}
\end{align}
$$
Explanation:

$\text{(3a)}$: the Lemma says $|g(z)|=|f(z)|$ when $|z|=r$

$\text{(3b)}$: the Lemma implies that $\log|g(z)|$ is harmonic for $|z|\le r$

$\text{(3c)}$: take the log of the absolute value of $(1)$ at $z=0$

$(3)$ is Jensen's Formula

Counting Zeros

If we know that all the zeros are inside $|z|=r$ and we know the average of $\log|f|$ on $|z|=R$,
$$
\begin{align}
\frac1{2\pi}\int_0^{2\pi}\log\left|f\!\left(Re^{i\theta}\right)\right|\,\mathrm{d}\theta-\log|f(0)|
&=\sum_{k=1}^n\left(\log(R)-\log|\alpha_k|\right)\tag{4a}\\
&\ge n(\log(R)-\log(r))\tag{4b}
\end{align}
$$
So we have that
$$
n\le\frac1{\log(R)-\log(r)}\left(\frac1{2\pi}\int_0^{2\pi}\log\left|f\!\left(Re^{i\theta}\right)\right|\,\mathrm{d}\theta-\log|f(0)|\right)\tag5
$$


问:
Munkres' Definition of Simply Connected
Munkres defines simply connected as follows:

A space $X$ is said to be simply connected if is a path-connected space and if $\pi(X,x_0)$ is the trivial (one-element) group for some $x_0 \in X$, and hence for every $x_0 \in X$.

I am not quite sure what is meant by the trivial group exactly. I know in a path-connected space it is possible to create loops since every two points are connected and if you start from $x_0$ you can get to $x_1$ and similarly you can go back from $x_1$ to $x_0$. What does it mean that it has to be trivial? It seems path connectedness does not necessarily imply simply connectedness. Could you please provide a example that a space is path connected but not simply connected?

答: Trivial group is one which consists of only the identity element. $\mathbb C \setminus \{0\}$ is path connected but not simply connected.

Title: Munkres' Definition of Simply Connected


问:
Successions and series, exercise.
It is true that

if $\;a_n>0\;$ and $\;\lim\limits_{n\to\infty}\dfrac{a_{n + 1}}{ a_n}<1\;,\;$ then $\;\lim\limits_{n\to\infty}a_n=0\;?$

Is that true or false?

答:
Yes, it is true.

Since $\;\lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_{n}}<1\;,\;$ there exist $\;h\in\left]0,1\right[\;$ and $\;\nu\in\mathbb{N}\;$ such that

$\dfrac{a_{n+1}}{a_{n}}<h\quad$ for all $\;n>\nu\;.\quad\color{blue}{(*)}$

Since $\;a_n>0\;$ for all $\;n\in\mathbb{N}\;,\;$ from $(*)$ we get that

$0<a_{n+1}<h^{n-\nu}a_{\nu+1}\;$ for all $\;n>\nu\;.\quad\color{blue}{(**)}$

Moreover, $\;\lim\limits_{n\to\infty}h^{n-\nu}a_{\nu+1}=0\;$ and, by applying the squeeze theorem to the inequalities $(**)$, we get that

$\lim\limits_{n\to\infty}a_{n+1}=0\;,\;$ hence

$\lim\limits_{n\to\infty}a_n=0\;.$

Another way to prove that $\;\lim\limits_{n\to\infty}a_n=0\;.$

From $(*)$ it follows that

$a_{n+1}<a_n\quad$ for all $\;n>\nu\;.$

Consequently, there exists $\;\lim\limits_{n\to\infty}a_n=\inf\limits_{n>\nu}\;\{a_n\}=l\ge0\;.$

If $\;l>0\;,\;$ then $\;\lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_{n}}=\dfrac{l}{l}=1\;,\;$ but it is impossible because, for hypothesis, $\;\lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_{n}}<1\;.$

Consequently, it results that $\;l=0\;$ that is

$\lim\limits_{n\to\infty}a_n=0\;.$

Title: Successions and series, exercise.


问:
Gradient with a change of variables?
Given $w\in\mathbb{R}^d$, $X\in\mathbb{R}^{n\times d}$ and $f(w) = \frac{1}{2}||y-Xw||^2_2$. My other function is $z\in\mathbb{R}^n$ such that $g(z) = \frac{1}{2}||y-z||^2_2$, the gradients are,

\begin{align}
\nabla_w f(w) &= X^T(Xw-y)\\
\nabla_zg(z) &= z-y.
\end{align}
What happens when I evaluate the gradient of $g(z)$ at $z=Xw$?

Solution

My though is that because $z\in\mathbb{R}^n$, the gradient should produce a value in $\mathbb{R}^n$, meaning,
\begin{align}
\left. \nabla_zg(z)\right|_{z=Xw} = Xw-y.
\end{align}

But my confusion comes from, if I evaluate, $\nabla_wg(Xw)$, then I get,
\begin{align*}
\left. \nabla_zg(z)\right|_{z=Xw} = \nabla_wg(Xw) = X^T(Xw-y).
\end{align*}
But now this value, $X^T(Xw-y)\in\mathbb{R}^d$, which makes me think its wrong. But I am not exactly sure how to approache this substitution of variables.

答:
Use $\nabla_zg$ to write the differential of the function
$$dg = (z-y) \cdot dz$$
Now use the relationship $\,z=Xw\,$ to switch the independent variable from $z\to w$
$$\eqalign{
dg &= (Xw-y) \cdot (X\,dw) \\
   &= \big(X^T(Xw-y)\big) \cdot dw \\
}$$
and recover the gradient from that last equation
$$\eqalign{
\nabla_wg  &= X^T(Xw-y) \\
}$$
This is identical to the other gradient calculation using
$$f(w) = g(Xw)$$

Title: Gradient with a change of variables?


问:
How to do the given Line Integral problem in Maple ? Find the work done by the force F along C.
How to do the given Line Integral problem in Maple ?
Find the work done by the force F along C.
F(x, y, z) = (y, −x, −z), C consists of the line segments from (0, 1, 0) to (3, 2, −1), (3, 2, −1) to (−1, 4, 2), (−1, 4, 2) to (2, −1, 4), and (2, −1, 4) to (0, −2, −2).

I did the following by hand and got the answer but I have no idea to do it in maple
W = -X -2X +5X + X + 3Y -4Y +3Y -2Y +Z -3Z -2Z + 6Z
W = 3X + 0 + 2Z
W = 3X + 2Z

答:
Since you mentioned Maple,</p>
```<code>restart;
with(VectorCalculus):

SetCoordinates('cartesian'[x,y,z]):

pts := <0,1,0>, <3,2,-1>, <-1,4,2>,
       <2,-1,4>, <0,-2,-2>:

LineInt(VectorField(<y,-x,-z>), LineSegments(pts), inert):

     Int(3-t,t = 0 .. 1) + Int(-11-9*t,t = 0 .. 1)
   + Int(3-4*t,t = 0 .. 1) + Int(28-36*t,t = 0 .. 1)

value(%);

                  -2

# Perhaps you tried something like this (by hand)?

LineInt(VectorField(<Y,-X,-Z>), LineSegments(pts), inert):

     Int(3*Y-X+Z,t = 0 .. 1) + Int(-4*Y-2*X-3*Z,t = 0 .. 1)
   + Int(3*Y+5*X-2*Z,t = 0 .. 1) + Int(-2*Y+X+6*Z,t = 0 .. 1)

value(%);

                3 X + 2 Z
</code>```

Body:
Since you mentioned Maple,

restart;
with(VectorCalculus):

SetCoordinates('cartesian'[x,y,z]):

pts := <0,1,0>, <3,2,-1>, <-1,4,2>,
       <2,-1,4>, <0,-2,-2>:

LineInt(VectorField(<y,-x,-z>), LineSegments(pts), inert):

     Int(3-t,t = 0 .. 1) + Int(-11-9*t,t = 0 .. 1)
   + Int(3-4*t,t = 0 .. 1) + Int(28-36*t,t = 0 .. 1)

value(%);

                  -2

# Perhaps you tried something like this (by hand)?

LineInt(VectorField(<Y,-X,-Z>), LineSegments(pts), inert):

     Int(3*Y-X+Z,t = 0 .. 1) + Int(-4*Y-2*X-3*Z,t = 0 .. 1)
   + Int(3*Y+5*X-2*Z,t = 0 .. 1) + Int(-2*Y+X+6*Z,t = 0 .. 1)

value(%);

                3 X + 2 Z


问:
Natural language translation
Vocabulary:

$R(x)$ - "x is a river"

$L(x)$ - "x is a lake"

$S(x)$ - "x is a sea"

$F(x, y)$ - "x flows into y"

The domain consists only of waterbodies (rivers, lakes and seas).

"Some lakes flow into other waterbodies, and some do not."

$ \exists x \exists y (x \ne y \wedge L(x) \wedge L(y) \wedge \exists z F(x, z) \wedge \neg \exists z F(y, z))$

Have I captured this correctly?

答:
Yes... though you don’t have to point out that $x \neq y$ ... and in fact I would leave it out.

Given that $\exists z \ F(x,z)$ and $\neg \exists z \ F(y,z)$, it logically follows that $x \neq y$. Put differently, you can remove the $x \neq y$ and it will still be logically equivalent.

And, since the English sentence does not explicitly state that we are dealing with two different lakes, I think it is better to just leave it out.

Title: Natural language translation


问:
Integrating $\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^n \, d\theta$, for $n=2$ and $n=3$
How do you integrate the following functions:
$$\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^2 \, d\theta$$ and $$\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^3 \, d\theta
$$
respectively?

Note: Initially, I tried integrating the function without the power and obtained the result below.
$$
\int \frac{\cos\theta}{1+\sin^2\theta} \, d\theta = \arctan(\sin\theta)+ C
$$
However, from here it is difficult to proceed. Integration by substitution doesn't seem to work.

How should I go on from here? Any pointers would be greatly appreciated.

答:
The latter integral may be done by substituting $\tan{\phi} = \sin{\theta}$; the result is the integral

$$\int d\theta \left ( \frac{\cos{\theta}}{1+\sin^2{\theta}}\right)^3 = \int d\phi\, (\cos^4{\phi}-\cos^2{\phi} \sin^2{\phi})$$

which I trust is an improvement.


答:
An element in the fundamental group is trivial if it is homotopic to the constant loop, that's null homotopic.

Look at a circle or torus ($S^1\times S^1$) for a path connected space that is not simply connected.

For intuition, it's pretty easy to see that any closed curve on a sphere can be shrunk, on the sphere, to a point, but not so on a circle or torus.

Title: Munkres' Definition of Simply Connected

Body:
Munkres defines simply connected as follows:

A space $X$ is said to be simply connected if is a path-connected space and if $\pi(X,x_0)$ is the trivial (one-element) group for some $x_0 \in X$, and hence for every $x_0 \in X$.

I am not quite sure what is meant by the trivial group exactly. I know in a path-connected space it is possible to create loops since every two points are connected and if you start from $x_0$ you can get to $x_1$ and similarly you can go back from $x_1$ to $x_0$. What does it mean that it has to be trivial? It seems path connectedness does not necessarily imply simply connectedness. Could you please provide a example that a space is path connected but not simply connected?


问:
Does a eulerian cycle exist in this context?
I have a picture of a rectangle and a vertex off to the side. I'd like to know if, since a Euler cycle includes all edges of a graph exactly once, if a Euler cycle exists in this graph.

Couldn't the rectangle itself be the Euler cycle?
I know Euler cycles exist in connected graphs only, but this graph isn't connected and it appears the rectangle can be used as the Euler cycle for the graph.

What am I missing here? Why can't this be a Euler cycle? Any help appreciated.

答:
It really depends on what definition you go by. Some definitions require the graph to be connected (in which case your example is not an Euler cycle), some do not require that but just require all edges to be visited, in which case your example is correct.

Often the assumption of connectedness is not explicitly stated, and that can indeed lead to confusion.


问:
Mathematics iteration
Hi I recently was learning the newton raphson method but before I started learning that I came across a method for solving equations that cannot be solved using formulae or other methods called iteration.i don't understand how iteration works to solve for roots of equations for example the equation  x^2-x-1 is set equal to 0 and rearranged in some way to make x the subject my question is how does plugging in a certain value of x in this equation converge to a root ?</p>

答:
By IVT, you are sure that the equation $$F(x)=x^2-x-1=0$$
has a root in the intervall $ \alpha $  $ [1,2] $.

There are many many other ways to write your equation in an equivalent form

$$x=x^2-1$$
or
$$x=\frac{1}{x-1}$$
...

or

$$x=x+2(x^2-x-1)$$

many and many

or in a general form
$$x=g(x)$$

an iterative method is a repetitive computation, starting by a value $ x_0 $,  near the root,  and you find successively $ x_1 $, $x_2 $, ....

using the recursive formula
$$x_{n+1}=g(x_n)$$

for example
$$x_1=g(x_0)$$
$$x_2=g(x_1)$$
and so on.

To be sure that the sequence $ x_1,x_2,x_3,....$ converges to the root $ \alpha $, you must choose the function $ g $ which satisfies $$|g'(x)|\le K<1$$

when $ x $ is close to $\alpha$.

Newton-Raphson corresponds to the case

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

Title: Mathematics iteration


问:
P(sqrt(X) > Y) Using Joint PDF
I'm having difficulty understanding a homework solution from my probability class.  The problem provides the following PDF:

And the problem statement is:

Last but not least, here's the given solution:

I'm struggling to understand 2 things.  First, the logic behind it.  Why would we use the PDF to find P(sqrt(X) > Y)?  From my understanding of the PDF and its uses, it doesn't quite seem to follow.  Secondly, I don't really understand how a constant is derived as the answer.  I use u-sub integration and I still get an answer in terms of x.  Am I just bad at calc 2?  Appreciate any help, sorry I can't embed the images directly.

EDIT: I got 10 points so I've embedded the images.  Thanks whoever gave me the points!

答:
Two random variables, $X$ and $Y$, are jointly continuous if there exists a function $f_{XY}(x,y):\mathbb R^2\rightarrow \mathbb R$ which is non-negative and is called the joint probability density function. It satisfies $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f_{XY}(x,y)dxdy=1$$

and for any set $A$ in $\mathbb R^2$ we have $$P((x,y)\in A)=\int\int_{A}f_{XY}(x,y)dxdy$$

See 1 or 2.

So we have since $x>0$ and $y>0$ here then  $$P(\sqrt{X}>Y)=\int_{0}^{\infty}\int_{0}^{\sqrt{x}}f_{XY}(x,y)dydx$$
$$=\int_{0}^{\infty}\int_{0}^{\sqrt{x}}e^{-x/2}ye^{-y^2}dydx$$
$$=\int_{0}^{\infty}\int_{0}^{\sqrt{x}}e^{-x/2}\frac{d}{dy}\big(-\frac{1}{2}e^{-y^2}\big)dydx$$
$$=\int_{0}^{\infty}e^{-x/2}\big[-\frac{1}{2}e^{-y^2}\big]_{y=0}^{y=\sqrt{x}}dx$$
$$=-\frac{1}{2}\int_{0}^{\infty}e^{-x/2}\big[e^{-x}-1\big]dx$$
$$=-\frac{1}{2}\int_{0}^{\infty}\big[e^{-3x/2}-e^{-x/2}\big]dx$$
$$=-\frac{1}{2}\big[-\frac{2}{3}e^{-3x/2}+2e^{-x/2}\big]_{x=0}^{x\rightarrow\infty}$$
$$=-\frac{1}{2}\big[0-(-\frac{2}{3}+2)\big]=-\frac{1}{2}\cdot-\frac{4}{3}=\frac{2}{3}.$$

Title: P(sqrt(X) > Y) Using Joint PDF


Title: Finding a constant which makes a Homogeneous ODE into a Non-Homogeneous ODE

Body:
I have a Non-Homogeneous ODE of the form $y'' + 8y' + 17y = 5$  which I have obtained the solution $y(x) = ae^{-4x}[\cos(x) + i\sin(x)] + be^{-4x}[\cos(x)-i\sin(x)] + \frac{5}{17}$ where $a$ and $b$ are complex constants.

I am unsure of how to find a constant, $c$, which satisfies the equation $z(x) = y(x) - c$  where $c$ is a real constant and $z(x)$ is a Homogeneous Differential Equation. I originally thought that $c = \frac{5}{17}$ but that seems too simple.

How should I go about finding $c$?

Body:
$$y'' + 8y' + 17y = 5$$
$$y'' + 8y' + 17y - 5=0$$
$$y'' + 8y' + 17(y-\dfrac 5 {17}) = 0$$
Since $(y-\dfrac 5{17})'=y'$:
$$(y-\dfrac 5 {17})'' + 8(y-\dfrac 5 {17})' + 17(y-\dfrac 5 {17}) = 0$$
$$z'' + 8z' + 17z = 0$$
Where $z=y-\dfrac 5 {17}$.


问:
How to find the eigenvalues of $x x^T - y y^T$?

If $x$ and $y$ are two linearly independent column $n$-vectors, where $n \geq 2$, find all the eigenvalues of $x x^{T} - y y^{T}$

Could you please tell me whether what I have written is correct or not?

Using the identity

\begin{align*}
\lambda^n\det(\lambda I_{(m)} - AB) = \lambda^m\det(\lambda I_{(n)} - BA)
\end{align*}

for $A \in F^{m \times n}$ and $B \in F^{n \times m}$, we can calculate the characteristic polynomial of $xx^T - yy^T$ by setting $A = (x, y) \in F^{n \times 2}$ and $B = (x^T, -y^T)^T \in F^{2 \times n}$ directly as:

\begin{align*}
\varphi(\lambda) &= \det(\lambda I_{(n)} - (xx^T - yy^T)) =
\lambda^{n - 2}\det\left(\lambda I_{(2)} - \begin{pmatrix} x^T \\ -y^T \end{pmatrix}\begin{pmatrix} x & y \end{pmatrix}\right) \\
&=
\lambda^{n - 2}\begin{vmatrix}
\lambda - x^Tx & -x^Ty \\
y^Tx & \lambda + y^Ty
\end{vmatrix} \\
&= \lambda^{n - 2}[(\lambda - x^Tx)(\lambda + y^Ty) + (x^Ty)^2] \\
&=
\lambda^{n - 2}(\lambda^2 - (x^Tx - y^Ty)\lambda - (x^Txy^Ty - (x^Ty)^2))
\end{align*}

Since $x$ and $y$ are linearly independent, by Cauchy-Schwarz inequality $(x^Tx)(y^Ty) > (x^Ty)^2$ (that is, the equality of C-S inequality cannot hold), whence the determinant $\Delta$ of the quadratic equation $\lambda^2 - (x^Tx - y^Ty)\lambda - (x^Txy^Ty - (x^Ty)^2) = 0$ equals to

\begin{align*}
\Delta = (x^Tx - y^Ty)^2 + 4(\|x\|^2\|y\|^2 - (x^Ty)^2) > 0.
\end{align*}
Hence the two non-zero eigenvalues are two distinct real numbers
\begin{align*}
\lambda_1 = \frac{y^Ty - x^Tx + \sqrt{\Delta}}{2}, \quad
\lambda_2 = \frac{y^Ty - x^Tx - \sqrt{\Delta}}{2}.
\end{align*}

答:
Every vector simultaneously orthogonal to $x$ and $y$ is an eigenvector of the matrix $xx^T - yy^T$ of eigenvalue 0 (note that the matrix has rank 2, so his kernel must have dimension $n-2$ and this is always an eigenspace). So, the remaining eigenvectors must be linear combination of $x$ and $y$.

Solve
$$ (xx^T - yy^T)(\alpha x + \beta y) = \lambda (\alpha x + \beta y) \qquad \lambda,\alpha,\beta \in \mathbb{R}. $$

You will find two polinomial equation in $\lambda,\alpha,\beta$. It is easy to see that it can't be $\alpha=0$ and $\beta=0$. So, from the two equation you can find a polinomial equation of degree 2 in $\frac{\alpha}{\beta}$. Then you can compute the two eigenvalues and the corrisponding eigenvectors.


问:
Maximizing the probability of choosing a green ball from two boxes
I have this excercise from a book of bayesian Analysis

Consider two boxes A and B each of which contains both red balls and green balls. It is known that, in one of the boxes, $\frac{1}{2}$ of the balls are red and $\frac{1}{2}$ are green and that in the other box, $\frac{1}{4}$ of the balls are red and $\frac{3}{4}$ are green.

Let the box in which $\frac{1}{2}$ are red to be denoted box W and suppose:

P(W=A)=ξ

P(W=B)=1−ξ

Suppose that the statistician may select one ball at random from either box A or box B and that, after observing its color, he must decide whether W=A or W=B

Prove that if $\frac{1}{2}$<ξ<$\frac{2}{3}$, then in order to maximize the probability of making a correct decision, he should select the ball from box B.

Prove also that if $\frac{2}{3}$<ξ<1, then it does not matter from which box the ball is selected.

My attempt to solve the excercise is the next:

Using the bayesian analysis

$\pi(\theta|x)= \frac{P(x | \theta) \cdot \pi (\theta)}{P(x)}$

We have 4 cases:

A) When we select a ball from A

Select 1 Ball from the box A and is Red

P(W=A|Red from A) = $\frac{2ξ}{ξ+1}$

P(W=B|Red from A) = $\frac{1-ξ}{1+ξ}$

Select 1 Ball from the box A and is Green

P(W=A|Green from A) = $\frac{2ξ}{3-ξ}$

P(W=B|Green from A) = $\frac{3(1-ξ)}{3-ξ}$

B) When we select a ball from B

Select 1 Ball from the box B and is Red

P(W=B|Red from B) = $\frac{2-2ξ}{2-ξ}$

P(W=A|Red from B) = $\frac{ξ}{2-ξ}$

Select 1 Ball from the box B and is Green

P(W=B|Green from B) = $\frac{2-2ξ}{2+ξ}$

P(W=A|Green from B) = $\frac{3ξ}{2+ξ}$

I have this but I´m Stuck in the part of optimize the probability

Thanks in andvance.

答:
If we choose to draw from $A$ and get red, we should say $W = A$ iff $P(W = A | \text{red from A}) > P(W = B | \text{red from A})$. If it's so, then $P(\text{red from A} \wedge \text{we are right}) = P(\text{red from A} \wedge W = A)$, otherwise $P(\text{red from A} \wedge \text{we are right}) = P(\text{red from A} \wedge W = B)$. And if we choose to draw ball from $A$, total probability of us guessing right is $P(\text{red from A} \wedge \text{we are right}) + P(\text{green from A} \wedge \text{we are right})$.

From your probabilities, if we draw ball from $A$, we should always (if $\xi \geq \frac{1}{2})$) say $W = A$ if we draw red, and if we draw green we should still say $W = A$ if $\xi > \frac{3}{5}$.

So, if $\xi < \frac{3}{5}$, we get answer correct with probability $\xi \cdot \frac{1}{2} + (1 - \xi) \cdot \frac{3}{4}$. If $\xi > \frac{3}{5}$, we just always say $W = A$, and we will be right with probability $\xi$.

We can find similar probabilities for drawing ball from $B$ and compare probabilities of success depending on $\xi$.


问:
$A=2^3×5^2×7^3$ find number of divisors of A
My attempt :

$A=2^3×5^2×7^3$

Let's determine number of numbers primes with A, and
Smaller than A

$\rho (A) = 2^2 ×4×5×6×7^2 =23520$

23520 is a number of numbers primes with A and smaller than A

$\forall k$ $ \in {23520} $ $gcd(A, k)=1$

So : number of numbers doesn't prime white A is $A-23520=45080$

$\forall k $ $\in {45080} $ $gcd(A, k) ≠1$

let's search   about  The number of numbers that divide A

$A=0[k]$
$\Leftrightarrow$ $68600=0[k]$

In the first we can see  K can take the following values ${1,2,4, 5,7,8,10,20,40,50,70,100}$

After using this theory $(a=b[n] $and $a=b[m] $ and $gcd(m, n) =1$$\Rightarrow$ $a=b[MN] $)

We can see that

$K=(1, 2,4,5,7,8,10,20,40,50,70,100,35,56,28,14,140,280,350,700)$

So igot 20 value of k

But I know, there are 48 values of k

I want to know how can I continue for find 48 value of k

Note : I don't want any
other  solution, I want to develop of my attempt

答:
Let the prime factorization of $n$ be $n=p_1^{a_1}..p_n^{a_n}$.

Any prime factor of $n$, must be one of $p_1,..,p_n$. If $p_i$ is one of them, there are $i+1$ ways ($0,1,..,i$) to choose the exponent. So the number of factors is $(a_1+1)..(a_n+1)$.


问:
Limit of a sequence within a logarithm
Let $(x_n)_{n \in \mathbb{N}}$ be a sequence of real numbers.

(1) Assume that $x_n>-1 \space \space \forall \space n \in \mathbb{N}$ and $x_n \to 0$ as $n \to \infty$, show that $\log(x_n+1) \to 0$ as $n \to \infty$.

(2) Assume that $x_n>0 \space \space \forall \space n \in \mathbb{N}$ and $x_n \to x>0$ as $n \to \infty$, show that $\log(x_n) \to \log(x)$ as $n \to \infty$.

I managed to do the first one but I'm not sure how to go about the second one. A similar method didn't seem to apply.

In this question, $\log(x)=y$ where $y$ is the unique real number with $\exp(y)=x$. Where $\exp(x)$ is the limit of $(1+x/n)^n$ as $n$ tends to infinty.

The method I've used so far involves the inequality $1+x\leq e^x \leq 1/(1-x) \space \space \forall \space x <1$.

答:
If you have proven that $\ln\left(\frac{x}{y}\right)=\ln(x)-\ln(y)$ for $x$,$y>0$, then the proof is quite easy. We'll need the following result:

Lemma: If $a_n$ is any positive sequence satisfying $\lim\limits_{n\to\infty}a_n=1$, then $\lim\limits_{n\to\infty}\ln(a_n)=0$.

This follows from (1). If $\lim_{n\to\infty}a_n=1$, then $\lim_{n\to\infty}(a_n-1)=0$, so by (1) we have that $\lim_{n\to\infty}\ln(a_n-1+1)=0$. This is equivalent to $\lim_{n\to\infty}\ln(a_n)=0$, so we are done.

With this established, let's now consider the sequence $x_n$. It was given that $\lim_{n\to\infty}x_n=x>0$, so $\lim_{n\to\infty}\frac{x_n}{x}=1$. It follows from our lemma that $\lim_{n\to\infty}\ln\left(\frac{x_n}{x}\right)=0$, so

$$\lim_{n\to\infty}\left[\ln(x_n)-\ln(x)\right]=0$$

Thus,

$$\lim_{n\to\infty}\ln(x_n)=\ln(x)$$

Title: Limit of a sequence within a logarithm


问:
Let $I=\langle 240,x^3-x^2\rangle$ be an ideal of $\mathbb Z[x]$, then does $x^3 -x+30\in\sqrt{I}$ hold?
The question can basically be re-written as, does there exist $f,g\in\mathbb Z[x]$ and $n\in\mathbb N$ such that $240f(x) +(x^3-x^2)g(x)=(x^3-x+30)^n$?

I firstly tried to compare constants and the coefficients of $x$ on the left and right hand side but couldn’t reach a contradiction. I think there are no such polynomials or $n$ since it would be difficult to find them explicitly, but not sure if there are some quick tricks or properties of radical ideals that are able to help answer the question

答:
Let $P$ be a prime ideal containing the ideal $I=(240,x^3-x^2)$.

From $240\in P$ it follows that $P$ contains one of $2,3,5$, hence $30\in P$.

From $x^3-x^2\in P$ it follows that $P$ contains one of $x,x-1$, hence $x^3-x\in P$.

From $30\in P$ and $x^3-x\in P$, we get $x^3-x+30\in P$.

Thus $x^3-x+30$ is an element of every prime ideal $P$ containing $I$.

Therefore $x^3-x+30\in\sqrt{I}$.


问:
Primitive function of $x^3 \sin x^2$
I'm trying to find the primitive function of $x^3 \sin x^2$, and I've come to a variable exchange ($t = x^2$) which led me to $\frac{1}{2} \int t \sin t dt$.

According to my text book, the primitive function is $\frac{1}{2}(-t \cos t + \sin t) + C$, but I can't see why. Isn't the derivative of that $\frac{1}{2}(t \cos t \sin t + \sin t \cos t)$?

答: Hint: Make $u(x)=x^2$ and note that  $x^3\sin x^2\mbox{d}x= u(x)\sin u(x)  \frac{1}{2}u'(x) \mbox{d}x=u\sin u\, \mbox{d}u$ and use integration by parts.

Title: Primitive function of $x^3 \sin x^2$


问:
Bounded Geodesics in a Quadrangle of Fixed Length.
The following question is a slightly weaker version of the question presented here: Geodesic Quadrangle in a Hyperbolic Space. However, this bound suffices for the statement mentioned in the original question.

Suppose $[p,q]\cup[q,r]\cup[r,s]\cup[s,p]$ is a geodesic quadrangle in a $\delta$-hyperbolic space $X$, such that $d(p,q)=d(s,r)$. Then for any pair of points $x\in [p,q], y\in [r,s]$ with $d(p,x)=d(s,y)$ we have
$$d(x,y)≤3\max\{d(p,s),d(q,r)\}+10δ.$$

答:
We break the proof into three cases.

Case 1: $d(s,x)\leq\max\{d(s,p),d(r,q)\}+5\delta$.

In this case $x$ and $y$ are close enough to the "pq-side" of the quadrangle that following the path along this side serves as a sufficient bound.

Formally, the path $[x,s]*[s,p]*[p,y]$ has $$l([x,s]*[s,p]*[p,y])\leq l([x,s])+l([s,p])+l([p,y])$$ $$\leq (\max\{d(s,p),d(r,q)\}+5\delta)+d(s,p)+(\max\{d(s,p),d(r,q)\}+5\delta) \leq 3\max\{d(p,s),d(q,r)\}+10δ.$$

Case 2: $d(x,r)\leq\max\{d(s,p),d(r,q)\}+5\delta$.

This is the same as Case 1, but on the other side of the quadrangle.

Case 3: $d(s,x)>\max\{d(s,p),d(r,q)\}+5\delta$ and $d(x,r)>\max\{d(s,p),d(r,q)\}+5\delta$.

First, by the $\delta$-slimness of quadrangles, there exists $w\in[s,p]\cup[p,q]\cup[q,r]$ with $d(x,w)\leq2\delta$. Note that $w\not\in[s,p]$. If it were, we would have $d(s,x)\leq d(s,w)+d(w,x)\leq d(s,p)+2\delta$, contradicting our assumption on $d(s,x)$. A similar, symmetric argument shows that $w\not\in[q,r]$. Therefore, $w\in[p,q]$.

Our goal is to now show that $w$ is close to $y$, as then the geodesics from $x$ to $w$, then to $y$ will be small. We will do this by showing that a construction where $w$ is far away from $y$ is impossible.

Let $w\in[p,y]$.$^{[1]}$ For the sake of contradiction, let us assume that $d(w,y)>d(p,s)+5\delta$. Then $$d(w,p)=d(p,y)-d(y,w)< d(p,y)-(d(p,s)+5\delta).$$
Which is to say,
$$d(w,s)+5\delta\leq d(w,p)+d(p,s)+5\delta<d(p,y).$$
However this quickly shows a contradiction to the definition of $w$:
$$d(w,s)+5\delta<d(p,y)=d(s,x)\leq d(s,w)+d(w,x)\quad\Rightarrow\quad 5\delta<d(x,w).$$

We therefore conclude that $d(w,y)\leq d(s,p)+5\delta$, which gives $d(x,y)\leq d(x,w)+d(w,y)\leq 2\delta+d(s,p)+5\delta$. $\square$

[1] If we instead assume that $w\in[y,q]$, then comparing $d(x,w)$ to $d(r,q)$ produces a similar argument.


问:
Curious short pattern in least common multiple of binomial coefficients
$$f(n) = \text{lcm}\Bigg(\binom n 1, \binom n 2, \dots,\binom n n\Bigg)$$

If we list $f(n) =\; $$\text{A002944}$$(n)$ it  starts of kind of boring, but at $n = 14$ we see a curious pattern in base $10$, only for four numbers (I added leading zeros for clarity):

\begin{array}{rcr}
f(14) &=& 024024\\
f(15) &=& 045045\\
f(16) &=& 720720\\
f(17) &=& 680680
\end{array}

Is this just dumb luck and I'm seeing a pattern in noise? Or is there an explanation why this 'echo' pattern occurs?

答:
This pattern of $xy0xy$ (which occurs often in pascals triangle) comes from the fact that $1001$ is the product of a sequence of small primes that don't cancel out for small values.

$1001=7 \times 11 \times 13$


问:
$A \mbox{ is an infinite set } \Leftrightarrow \forall n \in \omega, \exists B \subset A \mbox{ such that }B \mbox{ is equipotent to }n$
$A \mbox{ is an infinite set }  \Leftrightarrow \forall n \in \omega, \exists B \subset A \mbox{ such that }B \mbox{ is equipotent to }n$

$\Rightarrow]$ Let $G=\{n \in \omega: \exists B \subset A \mbox{ such that } B \mbox{ is equipotent to } n\}$

$0 \in G$ because we can take $0 \subset A$ then the function empty is a bijective function from $0$ to $0$.

The problem that I have is prove $n+1 \in G$ when we suppose $n \in G$. If we have that proposition then by induction we finish.

$\Leftarrow ]$ If $A$ is a finite set then $|A| \in \omega$ then exists $B \subset A$ such that $B$ is equipotent to $|A|$, therefore $|B|=||A||=|A|$, so $A=B$. But I don't see any contradiction.

答:
If $\omega\le|A|$, then by definition there is an injection $f:\omega\to A$. Given $n\in\omega$, let $B_n=\{f(k):k<n\}$; you can easily check that $|B_n|=n$. Thus, you don’t actually need to use induction at all.

For the other direction, if $A$ is finite, then $|A|=n$ for some $n\in\omega$, and there is a bijection $f:A\to n$, where $n=\{0,1,\ldots,n-1\}$. If $A$ had a subset $B$ such that $|B|=n+1$, there would be a bijection $g:B\to n+1$, and $f\circ g^{-1}$ would then be a bijection from $n+1$ to $n$, which is impossible.


问:
Exercise of motonone sequences.
If $ a_n $ is increasing and $ a_n <0 $ for all $ n $, does $ a_n $ converge?

I think so, since $ a_n $ is increasing and bounded by $ 0 $, that is, it is a monotone sequence. And by theorem, all these sequences converge. True?

答:
Welcome to MSE ^_^

Yes. You are correct that all bounded monotone sequences converge. In particular, your monotone sequence is bounded (it is entirely inside $[a_0, 0]$) so it converges.

We can say slightly more, actually. The limit $\lim a_n$ of a monotone sequence is its supremum $\sup a_n$ (Do you see why?). After you've thought about it, you can see this proven at the wikipedia page
for monotone convergence.

I hope this helps ^_^

Title: Exercise of motonone sequences.