问:
For matrices (why) is it true that $\left\Vert P^{-1}MP\right\Vert \le\left\Vert P\right\Vert ^{-1}\left\Vert M\right\Vert \left\Vert P\right\Vert ?$
I'm assuming the answer is embarrassingly obvious, but I'm not seeing it.  In this context the linear mapping and matrix norms are treated as effectively synonymous.  The discussion pertains to real numbers.

Is it the case in general that, given a non-singular matrix
$M$ and a conforming diagonal matrix $P$ with all diagonal components
positive, we have

$$
\left\Vert P^{-1}MP\right\Vert \le\left\Vert P\right\Vert ^{-1}\left\Vert M\right\Vert \left\Vert P\right\Vert ?
$$

Notice that I have written $\left\Vert P\right\Vert ^{-1},$ and not
$\left\Vert P^{-1}\right\Vert .$

This is what I have so far: Denote by $p_{i}$ the diagonal components
of $P.$ Evidently we have

$$
\left\Vert P\right\Vert =max\left(p_{i}\right),
$$
and
$$
\left\Vert P^{-1}\right\Vert =\frac{1}{min\left(p_{i}\right)}.
$$

Context:
From
https://www.scribd.com/read/282634061/Advanced-Calculus-of-Several-Variables

In the proof of Theorem IV 5.4, near the end we encounter the following:

Write

the origin-centered n-cube of radius $1$ as $\mathcal{C}_{1}\subset\mathbb{R}^{n};$

the volume function as $\mathit{v}:\mathbb{R}^{n}\to\mathbb{R};$

the identity mapping as $I;$

and the linear mapping and matrix norms as $\left\Vert \dots\right\Vert .$

Let there be

a non-negligible interval $\mathcal{Q}\subset\mathbb{R}^{n}$ centered on $\mathbf{a}\in\mathbb{R}^{n};$

a mapping $T:\mathbb{R}^{n}\to\mathbb{R}^{n}$ which is $\mathscr{C}^{1}$-invertible
on a neighborhood containing $\mathcal{Q};$

a linear transformation
$\rho:\mathbb{R}^{n}\to\mathbb{R}^{n}$ such that $\rho\left(\mathcal{C}_{1}\right)$
is an origin-centered interval congruent with $\mathcal{Q};$

a translation
$\tau_{\mathbf{a}}\circ\rho\left(\mathcal{C}_{1}\right)=\mathcal{Q};$

and a composite mapping $S=T\circ\tau_{\mathbf{a}}\circ\rho.$

In addition, it is given that

$$
\mathbf{y}=\tau_{\mathbf{a}}\left(\mathbf{x}\right)\in\mathcal{Q}\implies\left\Vert dT_{\mathbf{a}}^{-1}\circ dT_{\mathbf{y}}-I\right\Vert \le\epsilon\in\left(0,1\right).
$$

So, by the chain rule and the linearity of $\rho$ we have

$$
dS_{\mathbf{0}}^{-1}\circ dS_{\mathbf{x}}=\left(dT_{\mathbf{a}}\circ\rho\right)^{-1}\circ\left(dT_{\mathbf{y}}\circ\rho\right)=\rho^{-1}\circ dT_{\mathbf{a}}^{-1}\circ dT_{\mathbf{y}}\circ\rho.
$$

Therefore

$$
\left\Vert dS_{\mathbf{0}}^{-1}\circ dS_{\mathbf{x}}-I\right\Vert \le\left\Vert \rho\right\Vert ^{-1}\left\Vert dT_{\mathbf{a}}^{-1}\circ dT_{\mathbf{y}}-I\right\Vert \left\Vert \rho\right\Vert .
$$

It's the last part that I'm not following.

The part I do (believe I) understand is

$$
dS_{\mathbf{0}}^{-1}\circ dS_{\mathbf{x}}-I=\rho^{-1}\circ dT_{\mathbf{a}}^{-1}\circ dT_{\mathbf{y}}\circ\rho-I
$$

$$
=\rho^{-1}\circ\left(dT_{\mathbf{a}}^{-1}\circ dT_{\mathbf{y}}-I\right)\circ\rho.
$$

The norms I am using:

Just for fun, this is a partial depiction of what is going on in the proof.

答:
As far as I can see, the statement
$$
\left\Vert P^{-1}MP\right\Vert \le\left\Vert P\right\Vert ^{-1}\left\Vert M\right\Vert \left\Vert P\right\Vert
$$

actually is correct, but it is misleading in the current context.  It's correct because
$$
\left\Vert P^{-1}MP\right\Vert \le\left\Vert P^{-1}\left\Vert M\right\Vert  P\right\Vert=\left\Vert M\right\Vert,
$$
which my have been what Edwards actually intended.


问:
Finding the area of a triangle from subareas
A triangle is being divided by two lines. The newly created smaller triangles have the area shown in the picture. What is the area of the whole triangle?

My progress so far:
The two triangles $EJB$ and $BJC$ share the line $BJ$. Additionally, $EJ$ and $JC$ share the same height. Given this and the fact that the area of $EJB$ and $BJC$ are the same, $EJ$ and $JC$ must be of the same length. Therefore, $EJB$ and $BJC$ must be congruent. Am I right so far, and if so, where do I go from here?

答:
You are correct that $EJ$ and $JC$ have the same length. However, this doesn't imply that $EJB$ and $BJC$ are congruent triangles.

Hint: For the next step, connect points $E$ and $K$ with segment $EK$, and deduce that triangles $EJK$ and $KJC$ have the same area. This leaves one triangle whose area you need to determine. Say the area of this remaining triangle is $y$. Look at the two sets of triangles whose bases are $AE$ and $EB$. Now explain why the following equation is true:
$$
\frac y{1+3}=\frac{y+1 +1}{3 + 3}
$$

ContentLicense: CC BY-SA 4.0


问:
zero map of hermitian forms
$n \times n$ matrix $A$ with complex entries is called Hermitian if $A^{*}=A,$ where $A^{*}=\bar{A}^{T}$

$$
H(\Bbb{C}):=\left\{A \in M_{2}(\Bbb{C}) \mid A^{*}=A\right\}
$$
$H(\Bbb{C})$ consists of 2 by 2 the matrices $$
A=\left(\begin{array}{ll}
a & b \\
\bar{b} & d
\end{array}\right)
$$
where $a, d \in \Bbb{R}$ and $b \in \Bbb{C}$. If $A \in H(\Bbb{C})$ then the associated binary Hermitian form is the semi quadratic map
$$
Q: \mathbb{C} \times \mathbb{C} \rightarrow \Bbb{R}
$$
defined by
$$
Q(X, Z)=(X, Z)\left(\begin{array}{c}
a & b \\
\bar{b} & d
\end{array}\right)(X, Z)^{*}=a|X|^2+b X Z+\bar{b} \bar{X} Z+d|Z|^2
$$

Let  $H^{+}(\mathbb{C})$ denote the set of positive definite binary quadratic hermitian forms and
$\Delta(Q):=\operatorname{det}(Q)=a d-|b|^{2}$

Definition: Let $\mathcal{H}_3=\Bbb{C}\times (0,\infty)$ be the upper half space. The map $\xi: H^{+}(\mathbb{C}) \rightarrow \mathcal{H}_{3}$ defined by
$$
\xi\begin{pmatrix}a & b \\
\overline{b} & d \end{pmatrix}=\frac{b}{a}+\frac{\sqrt{\Delta(Q)}}{a} \cdot j
$$
is called the "zero map" for binary quadratic Hermitian forms.

My question: Why is it called zero map? I expect that $Q(X,1)=a|X|^2+bX+\overline{bX}+d$ admits $\frac{b}{a}+\frac{\sqrt{\Delta(Q)}}{a} j$ as a zero since the zero of an positive definite binary quadratic form $f(x,y)=ax^2+bxy+cy^2$ is defined as $\frac{-b+\sqrt{\Delta}}{2a}$, which is really a zero of $f(x,1)$.   But when I set $X=\frac{b}{a}+\frac{\sqrt{\Delta(Q)}}{a} j$, I get  $Q(X,1)=\frac{a^2+b^2+\overline{b^2}+d^2}{d}$ using the algebraic properties of quaternions. Do I have wrong expectation?

答:
No, it isn't a quaternionic zero of  $H(X,1)=0$.  For instance (and I think you can reduce to this case with a linear change of variable) consider $H(X,Z)=|X|^2+|Z|^2$.   Clearly $|X|^2=-1$ has no solutions in $\mathbb{H}$.

More important is the equivariance of the $PSL(2,\mathbb{C})$-action on forms/zero sets.  Take a look at Elstrodt, Grunewald, Mennicke, Groups Acting on Hyperbolic Space for more on the subject.

Title: zero map of hermitian forms


问:
homotopy of a real projective plane
By Seifert-Van Kampen theorem, $\pi_1(P) = \mathbb Z/2$. If $a$ is a generator of the group, this means that $aa = 1$. In terms of homotopy, this implies that $c_{x_0} \cong w*w$, where $w \in a$ is a loop. How can I visually construct homotopy between these two paths? i.e. make a square-like diagram, where the $x$-axis represents the space $P$ and $y$-axis represents time interval.

答:
Let $q : D^2 \to P$ be the quotient map which identifies antipodal points of $S^1$.  We can regard $P$ as the adjunction space $D^2 \cup_\phi S^1$, where $\phi : S^1 \to S^1, \phi(z) = z^2$.

Let $u : [0,1] \to D^2, u(t) = e^{\pi it}$ and $v : [0,1] \to D^2, v(t) = -e^{\pi it}$. Then $q \circ u = q \circ v$. Let us denote this map by $w : [0,1] \to P$. It is a loop based at $q(1)$ representing the generator of $\pi_1(P,q(1))$. The path $\gamma = u * v$ is a loop in $D^2$ (it runs once counterclockwise around $S^1 \subset D^2$) based at $1 \in S^1$ and
$$H: [0,1] \times  [0,1] \to D^2, H(t,s) = (1-s)\gamma(t)+ s$$
is a path homotopy to the constant path based at $1$. Consider the homotopy $H' = q \circ H : [0,1] \times  [0,1] \to P$. This is a homotopy of paths in $P$. We have $H'(t,0) = q \circ (u * v) = (q \circ u) * (q \circ u) = w * w$ and $H'(t,1) = q(1)$.

This proves your claim.

Title: homotopy of a real projective plane


问:
Modes of free vibration in a fixed-fixed beam
I am trying to find the mode shapes of vibration on a fixed-fixed beam.
this can be characterised as a PDE:
$$ EI \frac{\partial^4 v(x,t)}{\partial x^4} + \rho A \frac{\partial^2 v(x,t)}{\partial t^2} = 0 $$
with the boundary conditions
$$v(0,t) = 0  \hspace{2mm} \hspace{2mm} v_x(0,t) = 0 $$
$$v(l,t) = 0 \hspace{2mm}  \hspace{2mm} v_{x}(l,t) = 0 $$
We know that $E$, $I$, $\rho$ and $A$ are constant along the length

I have used the separation of variables method and then substituted in my boundary conditions and reached two different solutions:
$$\text{Re}\left[e^{ItB^2} (\cosh[Bx] - \cos[Bx]) + \frac{\sinh[Bl] + \sin[Bl]}{\cosh[Bl] -\cos[Bl]} (\sin[Bx] - \sinh[Bx])\right]$$
Plot of above showing fixed at l
$$\text{Re}\left[e^{ItB^2} \left((\sinh[Bx] - \sin[Bx]) - \frac{\sinh[Bl] - \sin[Bl]}{\cosh[Bl]
+ \cos[Bl]} (\cosh[Bx] - \cos[Bx])\right) \right]$$
Plot of above showing no slope at l

where $B$ is a constant that combines the natural frequency and the other constants

Each of these solutions satisfies a different three of the boundary conditions but I cant work out how to satisfy them all at the same time

答:
If you use the separation of variables, such that:
$$v(x,t)=V(x)\exp(i\omega t),$$
it is only necessary to write the space equation
$$\frac{d^4}{dx^4}V(x)+k^4V(x)=0$$
where $k=\left(\omega^2 \frac{\rho A}{EI}\right)^{1/4}$.
the solution can be written as
$$V(x)=A\cos(kx)+B\sin(kx)+C\cosh(kx)+D\sinh(kx)$$
and applying your boundary conditions:
$$V(0)=0, V'(0)=0, V(l)=0, V'(l)=0$$
The four equations can be written in matrix format:
$$\begin{pmatrix}1 & 0 & 1 & 0\\
\cos{\left( kl\right) } & \sin{\left( kl\right) } & \cosh{\left( kl\right) } & \operatorname{sinh}\left( kl\right) \\
0 & 1 & 0 & 1\\
-\sin{\left( kl\right) } & \cos{\left( kl\right) } & \operatorname{sinh}\left( kl\right)  & \cosh{\left( kl\right) }\end{pmatrix}\begin{pmatrix} A\\B\\C\\D\end{pmatrix}=0$$
For nontrival solution, the determinant of the matrix in this system of equations reduces to
$$\cos(kl)\cosh(kl)-1=0$$
which needs to be solved numerically. For numerical computation of the solution is better to rewrite this equation as
$$\cos(kl)=\frac{1}{\cosh(kl)}$$
so you can also plot a graph of the functions on the both sides of equation and check where they are equal. The first root, for instance, occurs when $kl\approx 4.73004$. It is possible to check that for large $kl$, the roots approximate to $\cos(kl)=0$.

Now, to calculate the mode shapes, it is necessary to use three of the four equations and write the mode shape as a function of one of the constants ($A,B,C$ or $D$).

If we choose to write the constants in terms of $A$, it is found that

$$B=-\frac{\cosh{\left( kl\right) }-\cos{\left( L k\right) }}{\operatorname{sinh}\left( kl\right) -\sin{\left( kl\right) }}A $$
$$C=-A$$
$$D=\frac{\cosh{\left( kl\right) }-\cos{\left( kl\right) }}{\operatorname{sinh}\left(  kl\right) -\sin{\left( kl\right) }}A $$
The expression for $B,C$ and $D$ are than replaced in the solution $V(x)$

$$V(x)=\left[\cos(kx)-\cosh(kx) + \frac{\cosh(kl)-\cos(kl)}{\sinh(kl)-\sin(kl)}\left(\sinh(kx)-\sin(kx)\right)\right]A$$

since $V(x)$ is a mode shape, it can be scaled by making $A=1$, to give:

$$\Psi(x)=\cos(kx)-\cosh(kx) + \frac{\cosh(kl)-\cos(kl)}{\sinh(kl)-\sin(kl)}\left(\sinh(kx)-\sin(kx)\right)$$

which is not unique.

The first mode shape is shown:


问:
Central Limit Theorem implies Law of Large Numbers?
Let $X_i$ be iid random variables and let $\overline{X}_n=(X_1+\cdots+X_n)/n$. If $EX_i=\mu$ and $\operatorname{Var}X_i = \sigma^2$ then the central limit theorem says that with some conditions we have convergence in some sense

$$\frac{\overline{X}_n -\mu}{\sigma/\sqrt{n}}\to N(0,1)$$

On the other hand the law of large numbers says that if $X_i$ are iid random variables with $EX_i=\mu$, then

$$\lim_{n\to\infty}P(|\overline{X}_n-\mu|<\epsilon)=1$$

Can the central limit theorem be used to prove a form of the law of large number? (obviously weakened since it would have the assumption of existence of $\operatorname{Var} X_i$). Since we expect $\frac{\overline{X}_n -\mu}{\sigma/\sqrt{n}}$ to look like $N(0,1)$ for large $n$, shouldn't we expect $\overline{X}_n -\mu$ to look like $N(0,\sigma^2/n)$. So then the variance of $\overline{X}_n -\mu$ being $\sigma^2/n$ means that the variance is becoming very small and $\overline{X}_n -\mu$ is focusing around $0$ and so something like the law of large numbers should hold.

答:
This argument works, but in a sense it's overkill.  You have a finite variance $\sigma^2$ for each observation, so $\operatorname{var}\left(\overline{X}_n\right)=\sigma^2/n$.  Chebyshev's inequality tells you that
$$
\Pr\left(\left|\overline{X}_n - \mu\right|>\varepsilon\right) \le \frac{\sigma^2}{\varepsilon^2 n} \to 0\text{ as }n\to\infty.
$$
And Chebyshev's inequality follows quickly from Markov's inequality, which is quite easy to prove.

But the proof of the central limit theorem takes a lot more work than that.

ContentLicense: CC BY-SA 3.0


问:
Show that there exists a sequence with subsequences converging to every rational number.

The total question is:

(a) Give an example of a sequence with subsequences converging to $1, 2$, and $3$.

(b) Give an example of a sequence with subsequences converging to every integer.

(c) Show that there exists a sequence with subsequences converging to every rational
number.

My attempt

For (a) I got $(1,2,3,1,2,3,1,2,3,\dots)$

For (b), I got $(0,1,-1,2,-2,3,-3,4,-4,\dots)$

Both of these I think are right, but I have no idea how to start on (c), much less an example for (c). If someone could explain what I should be doing, I'd be so grateful. Thank you.

答:
For (c), if $n\geq 2$ has the prime factorisation
$$
n=2^\alpha3^\beta 5^\gamma \cdot ...
$$
with $\alpha, \beta, \gamma,....\geq 0$, define
$$
a_n=(-1)^\alpha \frac{\beta}{\gamma+1}
$$
Show that $a_n$ takes each rational number infinitely many times.


问:
isolated point of a real plane curve
$\newcommand\C{\mathcal C}$Let $\C:f(x,y)=0$ be a real plane algebraic curve, we will also assume that $\C$ has real points and in particular that $P=(0,0)$ is a point of $\C$. Let $f_d(x,y)$ be the homogeneous part of the lowest degree of $f(x,y)$ such that $P$ is an isolated (real) point of $\C_d : f_d(x,y)=0$. Can I conclude that $P$ is an isolated point of $\C$?

If the answer above is yes, then can I make a similar statement for real analytic curves: so $f(x,y)$ is not anymore a bivariate polynomial but an analytic function and $f_d(x,y)$ is the lowest order non-trivial term in the Taylor series expansion of $f$ at $P$ (it is a homogeneous polynomial, so $\C_d$ remains algebraic).

答:
The answer to the first question is yes. A point $p$ of a real plane curve $C$ is isolated in $C(\Bbb R)$ iff all the tangent directions at $p$ are non-real (the tangent directions are the fiber of the strict transform of $C$ over $p$ in the blowup $Bl_p\Bbb A^2$). These tangent directions are determined by the lowest term of the Taylor expansion of $f$ at $p$, so going from $f_d$ to $f$ doesn't change the tangent directions, and thus $p$ is isolated as a point of $V(f_d)(\Bbb R)$ iff it's isolated as a point of $V(f)(\Bbb R)$.

The extension to the real-analytic case should proceed in the same fashion, except you'll need to extend $f$ to a complex analytic function in some open neighborhood of $\Bbb R^2\subset \Bbb C^2$ and keep track of the real points.

Title: isolated point of a real plane curve


问:
Mapping from linear functionals to $\mathbb R^n$ is surjective
Let $f$ be a linear functional from $\mathbb R ^n$ to $\mathbb R$. Then, it can be shown that $f$  has the form $f (x) = \sum_{i=1} ^n \alpha_i x(i)$, where $x(i)$'s are the coordinates of $x$. Now consider $\alpha = [\alpha_1, \cdots, \alpha_n] $ and the mapping $A$, which maps every $f$ to some $\alpha$. I need to show that $A$ is linear, injective, and surjective (hence an isomorphism).

The proofs to linearity and injectivity are straightforward. But, I feel uncertain how to approach subjectivity. Here is my idea. I have to show that each $\alpha$ can be written in the form of $A (f)$ for some $f$.

Let $b = [b_1, \cdots, b_n]$ be a basis for $\mathbb R^n.$ Then $f(x) = \sum _{i=1}^n f(b_i) x(i)$. That is, each $\alpha_i = f(b_i)$. Thus, $\alpha = [\alpha_1, \cdots, \alpha_n] = [f(b_1), \cdots, f(b_n)].$ Not sure how to continue form here onward. Is this the proper approach at all?

Or, instead should I start by parsing $f$ into independent $f_i$s, each equal to the corresponding $\alpha_i$? (How can I do that in proper formalism?)

答: Given any $\alpha $ in $\mathbb R^{n}$ simply define $f$ by $f(x)=\sum \alpha_ix_i$. Verify that this is a lnear map. Now $\alpha $ is the image of this $f$.


问:
From second-order differential equation and constraints, prove $y(x)=0$
Prove :
Let $P : R → R$ be a continuous function such that $P(x) > 0$ for all $x ∈ R$. Let $y$ be a twice
differentiable function on R satisfying $y''(x) + P(x)y'(x) − y(x) = 0$ for all $x ∈ R$. Suppose
that there exist two real numbers $a, b$ $(a < b)$ such that $y(a) = y(b) = 0$. Then $y(x) = 0$ for all $x ∈ [a, b]$.

What I've gotten so far is, $p(x) = y(x)-y''(x)/y'(x)>0$, so, $|y(x)| > |y''(x)|$
or, $y''(x)$ is negative...

答: Assume there exists some point $x \in (a,b)$ such that $y(x) > 0\,$ ($y(x) < 0$). This implies there is a maximum (minimum) at some $c \in (a,b)$ where $y'(c) = 0$ and $y''(c) < 0\,$ ($y''(c) > 0$). In either case substituting into $y''(c) + P(c) y'(c) - y(c) = 0$ leads to a contradiction.


问:
$L^1$ convergence in a finite measure space
I encountered this problem while studying a past comprehensive exam.

Let $(X,\mathcal{M}, \mu)$ be a finite measure space. Given measurable $f: X \rightarrow \mathbb{C}$, let $E_f(\lambda) = \mu(\{x: |f(x)| > \lambda\})$ be its distribution function.
Suppose that $\{f_n\}$ is a sequence of measurable functions converging to $f$ $\mu$-a.e. such that there exists a lebesgue measurable function $E(\lambda)$ such that $E: (0,\infty) \rightarrow [0,\infty)$ satisfying $E_{f_n}(\lambda) < E(\lambda)$ and $\int_{(0,\infty)}E(\lambda)d\lambda < \infty$
. Prove that $f_n$ converges to $f$ in $L^1$.

I suspect that Egoroff's theorem needs to be used, and that I need to somehow control the integral of $|f-f_n|$ on the exceptional set using $E(\lambda)$ but I can't get much further than that. I'm not even sure for example why $f\in L^1$ even though the conditions on $E_{f_n}$  force $f_n$ to be.

答:
Since $f_n$ converges to $f$ $\mu$-almost everywhere, it suffices to show that $f, f_n\in L^1$ and that $lim_{n \rightarrow \infty}{||f_n||_1} = ||f||_1$.
Now $lim_{n \rightarrow \infty}{||f||_1} = lim_{n \rightarrow \infty}{\int_{(0,\infty)}{E_{f_n}(\lambda)~d\lambda}} = \int_{(0,\infty)}{lim_{n\rightarrow \infty}{E_{f_n}(\lambda)}~d\lambda}$ by dominated convergence.

By Egoroff's theorem, $f_n$ converges to $f$ in measure, therefore $E_{f_n}$ converges to $E_{f}$ at every point of continuity. Since $E_{f_n}$ is right continuous, it has at most countably many discontinuities. therefore $lim_{n\rightarrow \infty}{E_{f_n}(\lambda)} = E_f(\lambda)$ almost everywhere.

So $\int_{(0,\infty)}{lim_{n\rightarrow \infty}{E_{f_n}(\lambda)}~d\lambda} = \int_{(0, \infty)}{E_{f}(\lambda)~d\lambda} = ||f||_1$. Since everything is finite, this concludes the proof.


问:
Poisson (joint) pdf
Lad $X_1,...,X_n$ be i.i.d. with the distribution $\mathcal{Poiss}(\lambda)$, for $\lambda>0$.

Write the joint pdf $(X_1,...,X_n)$ and argue that for $$P((X_1,..,X_n) \in \left \{ 0,1,2,..., \right \}^n)=1$$

for a fixed $(x_1,...,x_n) \in \left \{ 0,1,2,..., \right \}^n$

Concerns

I am unsure on what I am even asked to show. I get the Poisson distribution definition and the support of it. But I am not sure on what is asked to show to even begin to show it is indeed equal to $1$. I can probably do the calculations and manipulation when I get it in another form.

答:
The probability mass function of a vector of random variables will equal the joint probability mass function of the random variables.

The joint probability mass function of independent random variables shall be the product of the probability mass functions of those random variables.

The probability mass functions of identically distributed random variables are identical, and these are a Poisson distributed random variables with parameter $\lambda$.

So....

$$\begin{align}\mathsf P(\langle X_1,\ldots,X_n\rangle{=}\langle x_1,\ldots, x_n\rangle) ~&=~ \mathsf P\left(\bigcap\limits_{i=1}^n( X_i{=}x_i)\right)\\&=~ \prod_{i=1}^n \mathsf P(X_i{=}x_i)\\[0ex]&~~\vdots\end{align}$$

Title: Poisson (joint) pdf


问:
RREF of matrix and corresponding row space
I recently did an exercise where I was asked to find the reduced row echelon form of a matrix so that I would then be able to find the row and column rank, etc. When attempting the reduced row echelon form for the first time, I noticed that early on in elimination, I obtained a zero row.

$$A = \begin{pmatrix}
    1 & 2 & 1 & 2 & 2 \\
    1 & 2 & 2 & 3 & 4 \\
    1 & 2 & 3 & 4 & 6 \\
    1 & 2 & 4 & 6 & 8 \\
    \end{pmatrix}$$

Row operations: $$-R1+R2, R3+R4 \rightarrow R3, \\-5R2+R3, -3R2+R4, \\-R2+R1, -R3+R1, \\-R3+R2, -R3+R4.$$

$$\text{RREF}(A) = \begin{pmatrix}
    1 & 2 & 0 & 0 & 0 \\
    0 & 0 & 1 & 0 & 2 \\
    0 & 0 & 0 & 0 & 1 \\
    0 & 0 & 0 & 0 & 0 \\
    \end{pmatrix}$$

$$\text{Basis row}A=\begin{pmatrix}
1 & 2 & 1 & 2 & 2 \\
1 & 2 & 2 & 3 & 4 \\
1 & 2 & 3 & 4 & 6 \\
\end{pmatrix}$$

For some reason, I didn't want to permute the matrix, so I started from scratch and added the fourth row to the third row so that I wouldn't need to do any permutations. While I ended up getting the correct $\text{RREF}$ of the matrix, because of my method, I ended up eliminating the wrong row (row $4$ instead of row $3$) and got the wrong row space as a result.

While I am certain that my row operations were correct, I am confused why my method failed in letting me get the correct basis vectors for the row space, as I was utilizing the definition that all nonzero rows are row vectors. I really appreciate any and all help!

答:
Careful, it's the nonzero rows of $\text{RREF}(A)$ that can be used as a basis for the row space of $A$, not the corresponding rows of $A$.  Consider for example $A = \begin{pmatrix} 1 & 1 \\ 2 & 2 \\ 1 & 2 \end{pmatrix} \to \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix} $ where the row space is all of $\mathbb{R}^2$ and one possible basis is $\bigl\{ (1,0) , (0,1) \bigr\}$ and another one is $\bigl\{ (1,1) , (1,2) \bigr\}$ (that is, rows 1 and 3 of $A$), but rows 1 and 2 are multiples hence linearly dependent hence not a basis.

More generally, row reduction preserves the span of the rows but does not preserve (absence of) relations between the rows.  For columns, it's the other way around.


问:
Fractional ideal of a Dedekind domain A is a projective module
It can be known that any fractional ideal $\alpha$ of A can be generated by two elements, and is that enough to construct a reverse of the quotient map from $A^2$ to $\alpha$? And by the way, I wonder whether Dedekind domain is a UFD, and does g.c.d of two elements makes sense.

答: A Dedekind domain is not necessarily a U.F.D. If it is, it actually is a P.I.D. For fractional ideals, a theorem asserts that, over a Dedekind domain, any finitely generated torsion-free module is projective.</p>


问:
Evaluating $\int_0^1 (1-x^2)^n dx$
In an exercise I'm asked the following:

a) Find a formula for $\int (1-x^2)^n dx$, for any $n \in \mathbb N$.

b) Prove that, for all $n \in \mathbb N$: $$\int_0^1(1-x^2)^n dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}$$

I used the binomial theorem in $a$ and got:

$$\int (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) (-1)^k \ \frac{x^{2k + 1}}{2k+1} \ \ \ + \ \ C$$

and so in part (b) i got:

$$\int_0^1 (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) \ \frac{(-1)^k}{2k+1}$$

I have no clue on how to arrive at the expression that I'm supposed to arrive. How can I solve this?

答:
Here is an approach based on your attempt using a series

$$F\left( n \right)=\int\limits_{0}^{1}{{{\left( 1-{{x}^{2}} \right)}^{n}}dx}=\sum\limits_{k=0}^{n}{\left( \begin{matrix}
   n  \\
   k  \\
\end{matrix} \right){{\left( -1 \right)}^{k}}}\int\limits_{0}^{1}{{{x}^{2k}}dx}=\sum\limits_{k=0}^{n}{\left( \begin{matrix}
   n  \\
   k  \\
\end{matrix} \right)\frac{{{\left( -1 \right)}^{k}}}{2k+1}}$$
Now note
$$\left( \begin{matrix}
   n  \\
   k  \\
\end{matrix} \right)=\frac{1}{2\pi i}\int\limits_{\left| z \right|=r}^{{}}{\frac{{{\left( 1+z \right)}^{n}}}{{{z}^{k+1}}}dz}$$
And so
$$F\left( n \right)=\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}}{2k+1}\frac{1}{2\pi i}\int\limits_{\left| z \right|=r}^{{}}{\frac{{{\left( 1+z \right)}^{n}}}{{{z}^{k+1}}}dz}}=\frac{1}{2\pi i}\int\limits_{\left| z \right|>1}^{{}}{\frac{{{\left( 1+z \right)}^{n}}\arctan \left( {{z}^{-1/2}} \right)}{\sqrt{z}}dz}$$

However
$$\arctan \left( z \right)=\sum\limits_{k=0}^{\infty }{\frac{{{2}^{2k}}{{\left( k! \right)}^{2}}}{\left( 2k+1 \right)!}\frac{{{z}^{2k+1}}}{{{\left( 1+{{z}^{2}} \right)}^{k+1}}}}$$
which yields
$$F\left( n \right)=\frac{1}{2\pi i}\int\limits_{\left| z \right|>1}^{{}}{\sum\limits_{k=0}^{\infty }{\frac{{{2}^{2k}}{{\left( k! \right)}^{2}}}{\left( 2k+1 \right)!}\frac{1}{{{\left( z+1 \right)}^{k-n+1}}}}dz}$$
and the residue therefore provides the result
$$F\left( n \right)=\frac{{{2}^{2n}}{{\left( n! \right)}^{2}}}{\left( 2n+1 \right)!}$$

Title: Evaluating $\int_0^1 (1-x^2)^n dx$


问:
Show the points $u,v,w$ are not collinear
Consider triples of points $u,v,w \in R^2$, which we may consider as single points $(u,v,w) \in R^6$. Show that for almost every $(u,v,w) \in R^6$, the points $u,v,w$ are not collinear.

I think I should use Sard's Theorem, simply because that is the only "almost every" statement in differential topology I've read so far. But I have no idea how to relate this to regular value etc, and to solve this problem.

Another Theorem related to this problem is Fubini Theorem (for measure zero):

Let $A$ be a closed subset of $R^n$ such that $A \cap V_c$ has measure zero in $V_c$ for all $c \in R^k$. Then $A$ has measure zero in $R^n$.

Thank you very much for your help!

答:
$u,v,$ and $w$ are collinear if and only if there is some $\lambda\in\mathbb{R}$ with $w=v+\lambda(v-u)$.  We can thus define a smooth function

$$\begin{array}{rcl}f:\mathbb{R}^5&\longrightarrow&\mathbb{R}^6\\(u,v,\lambda)&\longmapsto&(u,v,v+\lambda(v-u))\end{array}$$

By the equivalence mentioned in the first sentence, the image of $f$ is exactly the points $(u,v,w)$ in $\mathbb{R}^6$ with $u,v,$ and $w$ collinear.  Now, because $5<6$, every point in $\mathbb{R}^5$ is a critical point, so that the entire image of $f$ has measure $0$, by Sard's theorem.


问:
Prove we cannot make a homeomorphism from [0, 1] to ℝ by contradiction.
So start off contradiction proofs by assuming the opposite.
So we assume f is a homeomorphism from ℝ to [0, 1].

Since f is a surjection, there exists some  ∈ ℝ with $f() = 0$.
Let $x_1 =  – 1$ and $x_2 =  + 1$.

Since f is an injection, and since $x_1≠ x_2 ≠ $,we know that $f(x_1) ≠ f(x_2) ≠ 0$.

Let $M = min \{ f(x_1), f(x_2) \}$.

I know I have to use the intermediate value theorem at some point but I'm stuck at the next step.

答:
Suppose we have such a homeomorphism.  By Heine-Borel, the compact subsets of  $\mathbb R $ are those that are closed and bounded.  Thus $[0,1] $ is compact and  $\mathbb R $ isn't  (unbounded).  Contradiction.

One can use connectedness  (IVT) to prove that there's no continuous injection from  $\mathbb R $ to $I $.  Or going in the other direction, one can just remove an endpoint to get that there's no continuous surjection.

One can also argue from compactness that there's no continuous surjection from  $I $ to $\mathbb R $.


问:
Decomposition of a sum of cubes
I have to decompose the following polynomial:
$$(x+2y)^3+(x-2y)^3$$
Since I have the sum of two cubes, I have thought:
$$(x+2y)^3+(x-2y)^3=(x+2y+x-2y)[(x+2y)^2+(x-2y)^2-(x+2y)(x-2y)]=2x\color{red}{[(x+2y)^2+(x-2y)^2-(x+2y)(x-2y)]}$$
Now in my book the result is $2x(x^2+12y^2)$, but I can't understand how rewriting the red term as $x^2+12y^2$, without trivially executing  the squares. Can you give me a hint?

答:
Your function is odd in $x$ and even in $y$, to wit

$[(-x)+2y]^3+[(-x)-2y]^3=-[(x-2y)^3+(x+2y)^3]$

$[x+2(-y)]^3+[x-2(-y)]^3=[(x-2y)^3+(x+2y)^3]$

So after factoring out $x$ your remaining quadratic factor must be even in both $x$ and $y$.  Therefore your red expression can only contain three nonzero terms $ax^2+by^2+c$.  Putting $x=0$ then forces $b=12,c=0$ and $y=0$ forces $a=1$.

Title: Decomposition of a sum of cubes


问:
On the Frattini Subgroup
For a prime $p$, let $H=\{x\in \mathbb{C}\colon x^{p^n}=1 \mbox{ for some } n\geq 1\}$ be the Prufer $p$-group, $C_2=\langle y\colon y^2=1\rangle$, and $G=H\oplus C_2$. Then $H$ is the unique maximal subgroup of $G$, hence it is Frattini subgroup of $G$. Let $S=\{e^{2\pi i/p^n} \colon n\geq 1\}$. Then $S$ generates the subgroup $H\leq G$, and $S\cup \{y\}$ generate whole $G$. Now, as $H$ is Frattini subgroup of $G$, it is the set of non-generators of $G$. But as $S\cup \{y\}$ generate $G$, we must have $\{y\}$ generates $G$, I reached to a wrong conclusion. Can one clarify mistake in the arguments, if any?</p>

答: Your mistake is in your interpretation of "non-generators". A non-generator is an element that can be removed from any generating set and still leave a generating set. It is true that $S\cup\{y\}$ generates $G$, but given any $z\in S$, we may put $S':=S\setminus\{z\},$ and $S'\cup\{y\}$ will still generate $G$. We cannot remove all elements of $S$ from $S\cup\{y\}$ and still have a generating set for $G$.

Title: On the Frattini Subgroup


问:
How do I find bases for $M_{2 \times 2}$ vector space
I'm asked to find two different bases for $M_{2 \times 2}$.I have not worked with vectorspaces that consists of matrices before. How many matrices are there in the basis for this type of vectorspace? And how do I find two different bases for this type of vector room? There is unfortunately no information about this in my book.

答:
Consider the set of $2\times 2$ matrices
\begin{equation*} B=\left\{\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\right\} \end{equation*}
and note that this set is linearly independent, spans the set of $2\times 2$ matrices and, continuing the idea from the comment, has four members. Now you just have to find another set of four vectors (matrices) that is distinct from this one and accomplishes the same thing.


问:
Probability of Parking Spot Being Empty
A parking spot is unoccupied 1/3 of the time...
But, you find it empty for nine consecutive days in a row.

Find the probability that it will be empty on the tenth day.

Read more: http://www.businessinsider.com/8-mind-bending-interview-questions-that-google-asks-its-engineers-2012-7?op=1#ixzz2zVa8Qb8w

The answer is 2/3 but I can't figure out how that is possible. It should be 1/3 instead.

答:
The answer is 2/3

The problem is defined as follows: if the probability of being occupied for one day is 1/3, then P(C) = 1/3, thus the probability of not being occupied is P(~C) = 2/3, let us make it easier by naming if P(F) = P(~C) {probability of being free}

The probability of being not occupied (free) for n consequence days is P(Fn) = (2/3)^n

i.e. Probability of being free for 2 consequence days = 4/9, for 10 consequence days = (2/3)^10, and for 9 days = (2/3)^9

Now the problem asks about the probability of being free for the 10th day GIVEN that it is free for 9 consequence days. that is:

P(F_10 | F_9) = P(F_10)/P(F_9) = (2/3)^10 / (2/3)^9 = 2/3

Please let me know if you find a mistake in my answer.

Title: Probability of Parking Spot Being Empty


问:
What is the relationship between the language accepted by DFA $M_1$ and the language accepted by DFA $M$?
Let $M = (Q, Σ, δ, q_0, F)$ be a DFA. Let $M_1 = (Q, Σ, δ, q_0, F_1)$ be an AFD identical to $M$ except for the set of final states, where $F_1$ is defined as the set of states $q ∈ Q$ for which $\hat{δ} (q, z) ∈ F$ for some $z$. What is the relationship between the language accepted by $M_1$ and the language accepted by $M$? Justify your answer.

(Hint: Use the fact that $\hat{δ} (q, xy) = \hat{δ} (\hat{δ}(q, x), y))$

Hi I'm answering this question, and by drawing some DFA, I think the realtionship between the language accepted by $M_1$ and the language accepted by $M$, is that the language acceptes by $M_1$ contains all the prefix from all the strings in $M$.

But I can't seem a way to justify using the hint.

Can someone help?

答:
You are correct.

Consider some arbitrary $s$, and let $q = \hat{\delta}(Q, s)$.

We see that $s \in L(M_1)$ iff $q \in F_1$ iff $\exists z (\hat{\delta}(q, z) \in F)$. Now, let's note that for every $z$, we have $\hat{\delta}(q, z) = \hat{\delta}(\hat{\delta}(Q, s), z) = \hat{\delta}(Q, sz)$, using the hint.

Then we have $\exists z (\hat{\delta}(q, z) \in F)$ iff $\exists z (\hat{\delta}(Q, sz) \in F)$ iff $\exists z (sz \in L(M))$.

Then $s \in L(M_1)$ iff there exists $z$ such that $sz \in L(M)$.

Then $L(M_1)$ is indeed the set of all prefixes of elements of $L(M)$.


问:
To use Etale methods in proofs or not
My question concerns a statement made by Fong and James in "Geometry of the Simple groups..."(1998); the statement says that they offer a proof "independent of etale methods."  Am I to presume that "etale methods" are not as desirable as other methods in writing proofs?   (I'm new to etale and hence may be oversensitive to what seems to be a judgement statement.)  If etale methods are not desirable, obviously I would like to know why not.  What is the down-side of etale methods?</p>

答:
There are two ways a proof can be simple. It can be easy to remember, because it fits into a pretty story. (For example, the proof that subgroups of free groups are free using algebraic topology and the fact that coverings of graphs are graphs.)

Alternatively, it can have few pre-requisites and be easily accessible to anyone with patience and the ability to understand the statement of the theorem. (For example, the "elementary" proofs of the prime number theorem avoiding complex analysis.)

(These don't have to be in conflict, but they often are.)

In the theory of representations of finite groups of Lie type, there is amazing power to using the Deligne-Lusztig theory. At the same time, someone who is mainly interested in finite groups may prefer to digest a longer proof using only elementary techniques rather than spend a few years absorbing the foundations of algebraic geometry.

Of course, this is all subjective. Certainly, though, more proofs are better. This last principle gives a final reason the authors might want to explicitly point out that their proof is "étale-free": it counts as evidence that their proof is genuinely new.

Title: To use Etale methods in proofs or not


问:
Exercise on the logic of propositions, and equivalences.
The domain of the variable consists of the empty set. Determine if $$ \forall x (x ^ 2 + 1 = 9)$$ is true or false.

I believe that if it is true, because when speaking of for all $ x $ in emptiness, then any property is satisfied (by emptiness).

答:
As noted in the comments, this is correct, through the vacuous argument you claimed.

(Mostly just making this community wiki so the question does have an answer and doesn't end up cluttering the unanswered queue down the road.)


问:
expected value of joint density function

Let $X$ and $Y$ be jointly continuous random variables with joint density:$$f_{\small X,Y}(x,y) = \begin{cases} \frac{1}{3}(x+y)&,& {\text{for} \space x\in(0,1) \space \text{and} \space y\in(0,2)} \\[1ex] 0&,& \text{otherwise}. \end{cases}$$

Compute $\mathbb{E}[\sin(\pi XY)]$.

If we define $Z=g_{\small X,Y}(x,y)=\sin(\pi XY)$, then
$$
\mathbb{E}[Z] = \iint_{\mathbb{R}^2}\, g_{\small X,Y}(x,y)\,f_{\small X,Y}(x,y)\,dx\,dy
$$
I am unsure of how to compute this due to the $\sin$ function. Any advice would be greatly appreciated.

答:
To get you started:

$$\begin{align}\mathsf E[\sin(XY\pi)]&=\iint_{\Bbb R^2}\sin(xy\pi)\,f_{\small X,Y}(x,y)\,\mathrm d\langle x,y\rangle \\[1ex]&=\frac 13\iint_{(0,1)\times(0,2)} (x+y)\sin(xy\pi)~\mathrm d\langle x, y\rangle\\[1ex] & = \frac13\left(\int_0^1x \int_0^2\sin(xy\pi)\,\mathrm d y\,\mathrm d x+\int_0^2y \int_0^1\sin(xy\pi)\,\mathrm d x\,\mathrm d y\right)\\&~~\vdots\end{align}$$


问:
how to find the first positive root of a cubic
I have a cubic function:
$$ax^3+bx^2+cx+d$$
What would be the fastest algorithm to find the first positive root (if there is one)?
Ideally the algorithm wouldn't have operations that are too expensive for a computer like square root(although that seems inevitable), cos, etc.

Any suggestion, link to resource is appreciated.
Thanks in advance.

答:
The Jenkins-Traub algorithm for polynomials with real coefficients, is a sound, well tested black box polynomial root finder, that finds the roots of smallest modulus first:

https://dl.acm.org/doi/10.1145/355637.355643

The reference implementation, 493.f, is in FORTRAN and is not easy to read, as it was written to preserve precision and accuracy while being fast.

There is one error in 493.f, on line 209, the initial value for V going into the second stage fixed shift iterations is written as

```<code> V = BND
</code>```

when it should be

```<code> V = BND*BND
</code>```

This error matters for finding roots of very small modulus first, so it's worth correcting for your case.

If you do decide to pick through 493.f, along with the main algorithm, you'll find Horner's rule for polynomial evaluation, Newton-Rhapson iteration for quick estimate of roots for root bounds, Cauchy's bound for the roots of a polynomial, a method to compute the quadratic formula while preserving as much precision and accuracy as possible, and a plethora of numerical techniques related to preserving precision and computing machine error.  Quite a lot of thought went into the implementation.

But if precision doesn't matter to you, then Jenkins-Traub may be overkill.

Body:
The Jenkins-Traub algorithm for polynomials with real coefficients, is a sound, well tested black box polynomial root finder, that finds the roots of smallest modulus first:

https://dl.acm.org/doi/10.1145/355637.355643

The reference implementation, 493.f, is in FORTRAN and is not easy to read, as it was written to preserve precision and accuracy while being fast.

There is one error in 493.f, on line 209, the initial value for V going into the second stage fixed shift iterations is written as

 V = BND

when it should be

 V = BND*BND

This error matters for finding roots of very small modulus first, so it's worth correcting for your case.

If you do decide to pick through 493.f, along with the main algorithm, you'll find Horner's rule for polynomial evaluation, Newton-Rhapson iteration for quick estimate of roots for root bounds, Cauchy's bound for the roots of a polynomial, a method to compute the quadratic formula while preserving as much precision and accuracy as possible, and a plethora of numerical techniques related to preserving precision and computing machine error.  Quite a lot of thought went into the implementation.

But if precision doesn't matter to you, then Jenkins-Traub may be overkill.


问:
Find marginal distribution if conditional is multinomially distributed
I would like to find the marginal distribution of $X=\left(X_1,\dots,X_k\right)$ if
$X|N=n$ is multinomially distributed with parameters $n$ and $p_1, \dots_, p_k$ and $N$ has Poisson distribution with parameter $\lambda$. I was thinking of using the law of total probability but I didn't see how to proceed with it. Could somebody show me how to do this?

I don't know if this is correct, but here's my attempt:

Since $y_1 + \dots y_m = n$ we get $y_m =n - \sum_{j=1}^{m-1}y_j$ and so

$$
P\left( Y_1 = y_1, \dots , Y_m = y_m \right) = \sum_{n=0}^{\infty}
P\left(Y_1 = y_1, \dots, Y_m = y_m | N=n \right)P\left(N=n \right)
=
\sum_{n=0}^{\infty}
\frac{n!}{y_1!\dots y_{m-1}!(n - \sum_{j=1}^{m-1}y_j)!}
\prod_{j=1}^{m-1}p_j^{y_j} \cdot p_m^{n - \sum_{j=1}^{m-1}y_j} \frac{e^{-\lambda}\lambda^n}{n!}
$$

$n$ has to start from $n= y_1+\dots y_{m-1}$ because of the factorial so this becomes (after making the change of variable $j = n -y_1 - \dots -y_{m-1}$
$$
e^{-\lambda} \cdot \frac{\prod_{j=1}^{m-1}(p_j\lambda)^{y_j}}{y_1!\dots y_{m-1}!}
\sum_{j=0}^{\infty}\frac{(\lambda p_m)^j}{j!} =
e^{-\lambda(1-p_m)} \cdot \frac{\prod_{j=1}^{m-1}(p_j\lambda)^{y_j}}{y_1!\dots y_{m-1}!}
=
\prod_{j=1}^{m-1}\frac{(p_j\lambda)^{y_j}e^{-\lambda p_j}}{y_j!}.
$$

What confuses me is that (if my calculations/reasoning is correct) I don't get the marginal distribution of $Y_m$.

答:
Each $X_i$ has Poisson distribution with parameter $\lambda p_i$ and the $X_i$ are independent. This is the Poisson splitting property, that extends Poisson thinning. See e.g. Splitting Poisson process formal proof for a detailed discussion of the case $k=2$. The general case follows by iteration, or see https://probabilityandstats.wordpress.com/2011/07/17/splitting-a-poisson-distribution/ and
https://kristianeschenburg.github.io/2018/11/poisson-multinomial

Directly: Given non-negative integers $x_1,\ldots, x_k$ that sum to $n$, we have

$$P(X_1=x_1,\ldots, X_k=x_k)=\frac{e^{-\lambda} \lambda^n}{n!} \cdot \frac{n!}{\prod_{i=1}^k x_i!}  \prod_{i=1}^k p_i^{x_i} =\prod_{i=1}^k \frac{e^{-\lambda p_i} (\lambda p_i)^{x_i}}{x_i!} \,,$$
and the RHS is the pdf of independent Poisson variables with parameters $\lambda p_i$ evaluated at $x_1,\ldots x_k$.


问:
Mice and poisonous wine brainteaser
The question is:

You have 1000 bottles of wine for a birthday party. 20 hours before the party, the winery indicate 1 bottle of wine is filled with poison, without telling you which bottle. You have 10 lab mice to test this on. The poison is so strong that it'll take only 18 hours to kill the mice who drinks it. Is there a way that you can find the poisoned bottle using the 10 mice before the party.

Now, the part I am confused on is not the solution to this problem, but actually a solution to a slightly different problem (essentially where you have unlimited time to perform the testing) that the author of the question included in answering this question.

The alluded to solution states:

If the mice can be tested sequentially to eliminate half of the bottles each time, the problem becomes a simple binary search problem. Ten mice can identify the poisonous bottle in up to 1024 bottles of wine.

I am confused about the first sentence. How can the mice be tested sequentially to eliminate half of the bottles?

答:
That solution assumes plenty of time for testing.  Sequentially would mean to test a mouse with the first half of the bottles, wait to see if it lives, and then test the next mouse with either half of the first half or half of the second half, wait again, etc, but that takes far too much time for the problem you stated.

The solution for your problem requires simultaneous testing on all ten mice.  Label the bottles in binary.  Feed the first mouse from all of the bottles with a 1 in the 1's place.  Feed the second mouse from all of the bottles with a 1 in the 2's place, and so on.  The survival or not of each mouse gives a digit of the poisoned bottle's number.

Title: Mice and poisonous wine brainteaser


Title: Evaluating $\int_0^1 (1-x^2)^n dx$

Body:
In an exercise I'm asked the following:

a) Find a formula for $\int (1-x^2)^n dx$, for any $n \in \mathbb N$.

b) Prove that, for all $n \in \mathbb N$: $$\int_0^1(1-x^2)^n dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}$$

I used the binomial theorem in $a$ and got:

$$\int (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) (-1)^k \ \frac{x^{2k + 1}}{2k+1} \ \ \ + \ \ C$$

and so in part (b) i got:

$$\int_0^1 (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) \ \frac{(-1)^k}{2k+1}$$

I have no clue on how to arrive at the expression that I'm supposed to arrive. How can I solve this?

Body:
Integrate by parts

$$I_n= \int_0^1 (1-x^2)^n dx=\int_0^1 \frac{(1-x^2)^n}{x^{2n}}d(\frac{x^{2n+1}}{2n+1})=\frac{2n}{2n+1}I_{n-1},\>\>\>I_0=1
$$
which leads to $I_n= \frac{2^{2n}(n!)^2}{(2n + 1)!}$.


问:
Prove $\displaystyle \dfrac{n}{n + 1} > M$ whenever $n > N$ for some integer $N$
Q: If $M$ is a positive number less than 1. Prove the terms in $\displaystyle \left\{\dfrac{n}{n + 1}\right\}_{n = 1}^{\infty}$ exceed $M$ for sufficiently large $n$; that is, prove $\displaystyle \dfrac{n}{n + 1} > M$ whenever $n > N$ for some integer $N$.

Not exactly sure how to prove this but here is my attempt so far.

By the definition of the limit, $\displaystyle \lim_{n \to \infty} \dfrac{n}{n + 1} > M$, we have $\forall M\in(0, 1)\exists N\in\mathbb{N}$ such that if
\begin{equation*}
n > N \hspace{1cm} \rightarrow \hspace{1cm} a_n > M
\end{equation*}

If $a_n = \dfrac{n}{n + 1}$, then we want to show that it exceeds $M$ as $n$ gets larger.

This is the part where I am kind of lost. I am not exactly sure how to proceed from here. Some tips or advice would be useful. Thanks

答: Hint: $\frac{n}{n+1} = 1-\frac{1}{n+1}$. Fix an arbitrary $M < 1$ and write $M = 1 - \epsilon$ for some $\epsilon > 0$. Using archimedian property...


Title: Prove $\displaystyle \dfrac{n}{n + 1} > M$ whenever $n > N$ for some integer $N$

Body:
Q: If $M$ is a positive number less than 1. Prove the terms in $\displaystyle \left\{\dfrac{n}{n + 1}\right\}_{n = 1}^{\infty}$ exceed $M$ for sufficiently large $n$; that is, prove $\displaystyle \dfrac{n}{n + 1} > M$ whenever $n > N$ for some integer $N$.

Not exactly sure how to prove this but here is my attempt so far.

By the definition of the limit, $\displaystyle \lim_{n \to \infty} \dfrac{n}{n + 1} > M$, we have $\forall M\in(0, 1)\exists N\in\mathbb{N}$ such that if
\begin{equation*}
n > N \hspace{1cm} \rightarrow \hspace{1cm} a_n > M
\end{equation*}

If $a_n = \dfrac{n}{n + 1}$, then we want to show that it exceeds $M$ as $n$ gets larger.

This is the part where I am kind of lost. I am not exactly sure how to proceed from here. Some tips or advice would be useful. Thanks

Body:
Both $M$ and $1-M$ are positive

if $$n > \frac{M}{1-M},$$  we see
$$  n+1 > \frac{1}{1-M}  $$
$$ \frac{1}{n+1}  < 1-M  $$
$$ -\frac{1}{n+1}  > M-1  $$
$$ 1-\frac{1}{n+1}  > M  $$
$$ \frac{n}{n+1}  > M  $$


问:
Why we take closed interval in defining curve in complex plane?
A curve in the complex plane is defined as a continuous function from a closed interval of the real line to the complex plane:$ z(t) : [a, b] → \mathbb{C}$, $a\leq t\leq b$.

Doubt: Why we choose closed interval to define curve in the complex plane? Can't we take an open interval to define curve?

Thanks for help

答:
You buy a car to move from point $A$ to point $B$, and you define a curve for the same reason, to move from point $A=z(a)$ to point $B=z(b)$.

One could of course define a curve using an open interval, and then someone may ask why didn't we use a closed interval instead. (There is one answer to that at the very end.) One could also define a curve on a half-open interval, and (now that I wrote it) that actually turns out to be useful to provide an alternative characterization (or definition) of simply connected regions in the plane, due to Donald J. Newman. It is included and used in the textbook

Complex Analysis by Joseph Bak and Donald J. Newman (in section 8).

Springer, New York, 2010. xii+328 pp. ISBN: 978-1-4419-7287-3 (3rd edition).

They adopt a non-traditional definition of simply connected regions. Such definition allows for the simplification of some proofs, e.g. of the following results:

8.4 Theorem. Suppose $f$ is analytic in a simply connected region $D$ and $\Gamma$
is a simple closed polygonal path contained in $D$. Then $\int_\Gamma f = 0.$

8.5 Theorem.
If $f$ is analytic in a simply connected region $D$, there exists a “primitive” $F$, analytic
in $D$ and such that $F'= f .$

8.6 General Closed Curve Theorem.
Suppose that $f$ is analytic in a simply connected region $D$ and that $C$ is a smooth
closed curve contained in $D$. Then $\int_C f = 0.$

The statements of the above results are standard, but in the approach taken in the above textbook, the proofs rely on the following definition. (As usual, a region in the plane is a connected open set, and we define a simply connected region below, where $\tilde D$ denotes the complement of $D$ in the plane. Also, $d(a,B)$ denotes the distance from a point $a$ to a set $B$.)

8.1 Definition.
A region $D$ is simply connected if its complement is “connected within $\varepsilon$ to $\infty$.”

That is, if for any $z_0\in \tilde D$ and $\varepsilon > 0,$
there is a continuous curve $\gamma(t), 0\le t < \infty$
such that:

(a) $d(\gamma(t),\tilde D)<\varepsilon$ for all $t\ge 0,$

(b) $\gamma(0) = z_0,$

(c) $\lim_{t\to\infty} \gamma(t) = \infty$.

A curve $\gamma$, satisfying (b) and (c), is said to “connect $z_0$ to $\infty$.”

Note that the curve $\gamma$ is defined on the half-open interval $[0,\infty)$. The late Professor Newman adopted the above definition to provide for more elementary proofs (and perhaps also his intuition was that this definition of simply connected regions is equivalent to the usual ones).
For the proof of the equivalence of Newman's definition of a simply connected region with the more widely accepted usual definitions, one direction is easy. For the other direction and a complete proof one could see:

Joseph Bak and Strashimir G. Popvassilev,

The evolution of Cauchy's closed curve theorem and Newman's simple proof.

Amer. Math. Monthly 124 (2017), no. 3, 217–231.

https://doi.org/10.4169/amer.math.monthly.124.3.217

https://www.jstor.org/stable/10.4169/amer.math.monthly.124.3.217

Finally (just a comment that gives a good possible answer to the original question), a curve $z : [a, b] \to \mathbb{C}$ could be restricted to the smaller domain
$z : (a, b) \to \mathbb{C}$. Conversely, if $z(t)$ were only defined on $(a,b)$ and
$a<r<s<b$ then of course $z : [r,s] \to \mathbb{C}$ is defined. So one cannot really distinguish between the two cases easily, they are closely related, but it seems that if
$z : [a, b] \to \mathbb{C}$ then we get something more, namely endpoints $z(a)$ and $z(b)$.
This also allows us to compose paths $z : [a, b] \to \mathbb{C}$ and
$w : [p,q] \to \mathbb{C}$ provided $z(b)=w(p)$, and that brings a lot of extra structure
(the fundamental group) to study and use ( I guess I could have started with that...)


问:
Why is the empty set finite?
On page 25 of Principles of Mathematical Analysis (ed. 3) by Rudin, there is the definition (excluding the irrelevant parts for this question):

Definition 2.4: For any positive integer $n$, let $J_n$ be the set whose elements are the integers $1,2,...,n$; let $J$ be the set consisting of all positive integers. For any set $A$, we say:

(a) $A$ is finite if $A \thicksim J_n$ for some $n$ (the empty set is also considered to be finite).

I am unsure why the empty set is considered to be finite. Given the defintion, for the empty set to be finite, then $\emptyset \thicksim J_n$ for some $n$.

The issue is for equivalence, there has to exist a one-to-one mapping of $\emptyset$ onto $J_n$. $\emptyset$ has no elements to correspond and since the definition requires $n \in J$, $J_n$ will always have at least one element.

Thus, my question: Why is the empty set considered finite?

答:
What they are saying here is that the empty set is considered to be finite by convention, or for a different reason. If they were being more precise, they would have said that a set is finite if it is empty or is in bijection with some $J_n$.

Suppose in general that for non-negative integers $n$ we defined $J_n$ to be the set of positive integers no greater than $n$. This certainly matches the definition given for $J_n$ when $n\ge 1$, but what about when $n=0$? Well, there aren't any positive integers no greater than $0$, so $J_0=\emptyset$! Certainly the empty set is in bijection with itself. In that way, we may extend the definition of finite, without relying on a "just roll with it" in the case of the empty set.

Title: Why is the empty set finite?


问:
How do you talk about representations of a subquiver of $A_n$?
Let $Q$ be a linear Dynkin quiver of type $A_n$, which looks like

$$1 \longrightarrow 2 \longrightarrow 3 \longrightarrow \cdots \longrightarrow (n-1) \longrightarrow n$$

Then we know by Gabriel's theorem that the indecomposable representations of $Q$ are given by $M^{r,s}=(M^{r,s}_i,M^{r,s}_a)_{i \in Q_0, a \in Q_1}$ where $M^{r,s}_i=K$ for $r \leq i \leq s$ and $M^{r,s}_i=0$ otherwise.

Now $Q$ has two subquivers of type $A_{n-1}$ with indecomposable representation denoted by $N^{r,s}=(N^{r,s}_i,N^{r,s}_a)$ for $1 \leq r \leq s \leq n-1$. Is it true that for one subquiver it holds $N^{r,s} \cong M^{r,s}$ for all $1 \leq r \leq s \leq n-1$ and for the other $N^{r,s} \cong M^{r,s}$ for all $2 \leq r \leq s \leq n$? Intuitively I would say yes, it makes sense. But I still have some doubts on whether it is possible to make this statement.

答: So yeah, it's not quite right to say they're isomorphic since $M^{r,s}$ and $N^{r,s}$ are representations of different quivers. You're morally right though. To make it more precise you've gotta say something like, looking at the "left" case, you've got a morphism of quivers $\phi\colon A_{n-1} \to A_n$ that maps the nodes of $A_{n-1}$ to the nodes labelled $1, \dotsc, n-1$ of $A_n$. This morphism of quivers $\phi$ induces a functor on the representation categories $\Phi \colon \mathrm{Rep}A_n \to \mathrm{Rep}A_{n-1}$ where $\Phi(M^{r,s})_i = M^{r,s}_{\phi(i)}$ and $\Phi(M^{r,s})_a = M^{r,s}_{\phi(a)}$. Essentially you're just forgetting about that last node and arrow of $A_n$ to get a representation of $A_{n-1}$. The big idea is that since $\phi$ is injective, then $\Phi$ is surjective, and $N^{r,s}$ is isomorphic to $\Phi(M^{r,s})$ for every $r$ and $s$ such that $1 \leq r\leq s < n$.


问:
How to solve this probability problem using Binomial coefficient?
I am looking for a closed form solution with the help of manipulation of Binomial coefficients.

Here is the problem :
We are given three integers $a,b,c$ .

Constraints : $b>0 , b,a <998244353$

$c<=10^x$
, where $x=18$

Consider an array of size $c$ .

It only consists of only zeroes and ones .

Calculate the probability that the array contains an odd number of $1$'s .

Probability of any array element to be zero , is given by $a/b$ .

Example :
$
c=2
$

$
a=1
$
$b=3$

$(p1=a/b)$

Only the following arrays satisfy the condition ,

{0,1} , (probability1) : (1/3)*(2/3) = (2/9)

{1,0) , (probability2) : (2/3)*(1/3) = (2/9)

Final answer : 2/9 + 2/9 = 4/9

答:
If $p(c)$ is the probability of being odd then $p(0)=0$ and $p(1)=\frac {b-a}b$ and $p(c)=\frac{a}{b}p(c-1) + \frac{b-a}{b}(1-p(c-1))$

But as lulu commented, if $0 < a < b$ and there is a limit then it will be $\frac12$ so it might be easier to work with $q(c)=p(c)-\frac12$ so you start with $q(0)=-\frac12$ and $q(1)=\frac{b-2a}{2b}$ and have $q(c)= \frac {2a-b}{b} q(c-1)$ and so as closed forms $q(c)= -\frac12 \left(2\frac{a}{b}-1\right)^c $  and thus  $$p(c)=\frac12 -\frac12 \left(2\frac{a}{b}-1\right)^c$$ and so long as $0 < a < b$ you get $q(c) \to 0$ and $p(c) \to \frac12$ as $c$ increases.


问:
Non-oriented Rubiks Cube Group
Consider the (non-oriented, so we do not care about the center of each side) rubiks cube group, $G$, which is a subgroup of $S_{48}$ and is generated by the letters: $\{F,B,U,D,L,R\}$, i.e. I mean that $G = \langle F,B,U,D,R,L\rangle$ (subject to the canonical relations for the non-oriented $3\times 3$ rubiks cube). I am trying to just manually verify if the superflip is in the center (just by verifying if it commutes with each of the generators) using https://rubikscu.be/#cubesolver , but this is exceptionally cumbersome and time consuming. Let $h\in G$ denote the superflip. I have seen other websites, MSE pages, and pdfs describe the center as simply only containing $1_G$ and $h$, but these pages very rarely distinguish if they mean the oriented cube (the subgroup of $S_{54}$) or non-oriented, I am not really sure if it matters either (but furthermore, I am really sure how one would rigorously prove this in any case, the arguments I can find are mostly fairly ambiguous).

答:
As aschepler mentioned in the comments, it's irrelevant how the superflip is realized in terms of the generators. Since the Rubik's Cube group is formally just a subgroup of $S_{48}$, an element like $h$ is literally just a permutation.

But first, let's choose something other than $\{1,\ldots,48\}$ to serve as the set whose permutations we consider.
Let's label the faces as Front, Bottom, Up, Down, Right, Left, and denote the 48 faces

for corners: like “FUR” to denote the Front sticker (hence the first letter) of the front-up-right cubelet

for edges: like “FU” to denote the Front sticker of the front-up cubelet.

To avoid ambiguity, let's further deviate from the standard notation and add commas so that a cycle looks like $(FU,UF)$.

For any term representation of the superflip, we will be able to calculate
\begin{align*}
h = &(FU, UF)(FR,RF)(FD,DF)(FL,LF)\\
    &(UR,RU)(UL,LU)(DR,RD)(DL,LD)\\
    & (BU,UB)(BR,RB)(BD,DB)(BL,LB).
\end{align*}
Note in particular that $h$ is of order two.

To prove that it commutes with every cycle, let's refresh our memories and see how the generators are defined:
\begin{align*}
R &:=(\text{corner cycles})(RU, RB, RD, RF)(UR,BR,DR,FR)\\
U &:=(\text{corner cycles})(UR,UF,UL,UB)(RU,FU,LU,BU)\\
F &:=(\text{corner cycles})(FU,FR,FD,FL)(UF,RF,DF,LF)\\
L &:=(\text{corner cycles})(LU,LF,LD,LB)(UL,FL,DL,BL)\\
D &:=(\text{corner cycles})(DF,DR,DB,DL)(FD,RD,BD,LD)\\
B &:=(\text{corner cycles})(BU,BL,BD,BR)(UB,LB,DB,RB).
\end{align*}

It is now easy to verify that each of the commutators $R^{-1}h^{-1}Rh$ etc. vanish, or, equivalently (and perhaps easier to verify if you know what conjugation does in terms of permutations :-) ), that $h^R := R^{-1}hR = h$ etc.  Notice in particular that since $h$ leaves corner faces invariant, the respective corner cycles in the generators don't contribute to the commutator, explaining why I left them out above.

As you correctly stated, if $h$ commutates with all generators, it has to lie in the center.

Title: Non-oriented Rubiks Cube Group


答:
The parenthetical remark is just saying that formally the definition of finite set does not apply to the empty set, but the empty set is taken to be finite by convention.

If this bothers you, note it is possible to define a set as infinite precisely when there exists a proper subset of it with a bijection to the set. This captures in one go what it means to be infinite.

Title: Why is the empty set finite?

Body:
On page 25 of Principles of Mathematical Analysis (ed. 3) by Rudin, there is the definition (excluding the irrelevant parts for this question):

Definition 2.4: For any positive integer $n$, let $J_n$ be the set whose elements are the integers $1,2,...,n$; let $J$ be the set consisting of all positive integers. For any set $A$, we say:

(a) $A$ is finite if $A \thicksim J_n$ for some $n$ (the empty set is also considered to be finite).

I am unsure why the empty set is considered to be finite. Given the defintion, for the empty set to be finite, then $\emptyset \thicksim J_n$ for some $n$.

The issue is for equivalence, there has to exist a one-to-one mapping of $\emptyset$ onto $J_n$. $\emptyset$ has no elements to correspond and since the definition requires $n \in J$, $J_n$ will always have at least one element.

Thus, my question: Why is the empty set considered finite?


问:
Degenerate conic composed of two lines
In reading the text Multi-View Geometry for Computer vision, after the notion of a conic is introduced with its corresponding conic coefficient matrix, an example is given on page 32 stating

The conic $C= lm^T + ml^T$ is composed of two lines $l$ and $m$. Points on $l$ satisfy $l^T x = 0$, and are on the conic since $x^TCx = (x^T l)(m^Tx) + (x^Tm)(l^Tx) = 0$. Similarly, points satisfying $m^Tx = 0$ also satisfy $x^TCx=0$.

I'm not sure what this notation means. I'm guessing that $C$ is the sum of two outer products, since otherwise it would be a sum of scalars, and then $x^TCx$ would be a scalar multiple of the squared norm of $x$.

If it is indeed a sum of outer products, then is it generally true that for an outer product $l \bigotimes m^T$ that $x^T l \bigotimes m^T x = (x^T l)(m^Tx)$ ? (where the latter expression is a product of dot products)

Title: Degenerate conic composed of two lines

Body:
Just as a sanity check...

$C := lm^T + ml^T = \begin{pmatrix}
2l_1m_1 & l_1m_2+l_2m_1 & l_1m_3+l_3m_1 \\
l_1m_2+l_2m_1 & 2l_2m_2 & l_2m_3+l_3m_2 \\
l_1m_3+l_3m_1 & l_2m_3+l_3m_2 & 2l_3m_3
\end{pmatrix}.$

Then $C x = \begin{pmatrix}
x_1 (2l_1m_1) + x_2 (l_1m_2+l_2m_1) +x_3(l_1m_3+l_3m_1) \\
x_1(l_1m_2+l_2m_1) + x_2(2l_2m_2) +x_3(l_2m_3+l_3m_2) \\
x_1(l_1m_3+l_3m_1) + x_2(l_2m_3+l_3m_2) +x_3(2l_3m_3)
\end{pmatrix}$.

And $x^T C x = $
$x_1(x_1 (2l_1m_1) + x_2 (l_1m_2+l_2m_1) +x_3(l_1m_3+l_3m_1)) + x_2 (x_1(l_1m_2+l_2m_1) + x_2(2l_2m_2) +x_3(l_2m_3+l_3m_2)) + x_3 (x_1(l_1m_3+l_3m_1) + x_2(l_2m_3+l_3m_2) +x_3(2l_3m_3))$

Whereas $(x^Tl)(m^Tx) + (x^Tm)(l^Tx) = $
$(x_1l_1+x_2l_2+x_3l_3)(m_1x_1+m_2x_2+m_3x_3)$
$ + (x_1m_1 + x_2m_2 + x_3m_3)(l_1x_1 + l_2x_2 + l_3x_3)$

And matching these term by term I see they are equal.


问:
Sequence of Gamma r.v.s converges in probability to 1
Let $\{X_n\}$ be a squence of Gamma distributed random variables with pdf

$$
 f(x;\alpha,\beta) =
  \begin{cases}
      \hfill \dfrac{x^{\alpha - 1}e^{-x/\beta}}{\beta^{\alpha}\Gamma(\alpha)}    \hfill & \text{ for $x>0$} \\
      \hfill 0 \hfill & \text{elsewhere} \\
  \end{cases}
$$
In our case, we have $\alpha = n$ and $\beta = 1/n$. Show that
$$X_n \overset{\mathbb{P}}{\to} 1$$

My try: We have to show that for every $\varepsilon>0$:
$$\mathbb{P}(|X_n - 1|<\varepsilon)=\mathbb{P}(-1+\varepsilon < X_n < 1-\varepsilon)=1$$
Thus:
\begin{align*}
\mathbb{P}(-1+\varepsilon < X_n < 1-\varepsilon) &= \int_{-1+\varepsilon}^{1-\varepsilon} f(x;n,1/n)\mathrm{d}x\\
&= \int_{0}^{1-\varepsilon}\dfrac{x^{n - 1}e^{-nx}}{\dfrac{1}{n^n}\Gamma(n)}\mathrm{d}x \\
&=\dfrac{n^n}{\Gamma(n)}\int_{0}^{1-\varepsilon} x^{n - 1}e^{-nx}\mathrm{d}x
\end{align*}
Now make the substitution $x \mapsto nx$, this yields
\begin{align*}
\mathbb{P}(-1+\varepsilon < X_n < 1-\varepsilon) &= \dfrac{n^n}{\Gamma(n)}\int_{0}^{n(1-\varepsilon)} \dfrac{1}{n}\left(\dfrac{x}{n}\right)^{n - 1}e^{-x}\mathrm{d}x \\
&= \dfrac{1}{\Gamma(n)}\int_{0}^{n(1-\varepsilon)} x^{n - 1}e^{-x}\mathrm{d}x
\end{align*}
Now, taking the limit as $n \to \infty$ we yield

\begin{align*}
\mathbb{P}(-1+\varepsilon < X_n < 1-\varepsilon) &=\lim_{n \to \infty}\dfrac{1}{\Gamma(n)} \int_0^{n(1-\varepsilon)} x^{n - 1}e^{-x}\mathrm{d}x \\
&= \lim_{n \to \infty}\dfrac{\Gamma(n)}{\Gamma(n)} \\
&=1
\end{align*}
Since $\varepsilon>0$ was given arbitrarily, we conclude that
$$X_n \overset{\mathbb{P}}{\to} 1$$
And I think I solved my question...

For the sake of future reference or if someone has the same (kind of) problem, I will post it anyways, even tough I am pretty sure I solved it.

答: Some years later, hehe. I consider that approach is kind of complicated. I'd use the fact that $E[X_n] = 1$, so we can try to calculate $P(|X_n - \mu| \geqslant \epsilon)$, use Chebyshev's theorem and then use the squeeze theorem to conclude that $X_n \overset{P}\to  \mu = 1$.


问:
Understanding a Basis
I'm having a very difficult time understanding a basis.  I understand that it's basically just coordinates that span a subspace, but can someone help me to understand how a set can span a subspace and be a linearly independent set?

If a set spans a subspace doesn't this mean that any vector can be represented by a linear combination of the vectors in the subspace?  If that is the case, isn't this a direct contradiction to the set being linearly independent?

What am I missing here?

答:
I am currently reading through Paul R. Halmos' "Finite dimensional linear algebra". The way a basis is defined in his book is as follows:

Given a finite dimensional vector space $V$, a (not "the") basis of $V$, say $S$, is a linearly independent set of vectors, such that every vector in $V$ is equal to some linear combination of the vectors in $S$.

Now, to answer your question, let us assume that a set of vectors, again say $S$, spans a vector space $V$. i.e, we assume that every vector in $V$ is equal to some linear combination of the vectors in $S$. Linear independence states that, given a linear combination of the vectors in $S$, if this combination is equal to the zero vector, then all the coefficients in the linear combination must be equal to zero.

Your statement is the following: $S$ is linearly independent $\implies$ that $S$ does not span $V$. Taking the contrapositive of this statement, we have that $S$ spans $V$ $\implies$ $S$ is linearly dependent. Now, we can simply provide a counter example. Say then that $V = \mathbb{R}^2$. Then, $V$ forms a vector space, and if we let $S = \{(1, 0), (0, 1)\}$, we have a linearly independent basis that spans $V$. $\square$

Note that some intricacies have been omitted (it's 1am!), such as proving that $S$ does actually span $V$, however I believe my answer to be sufficient as it stands. Do let me know if anything anything is missing. Clearly this idea extends to your question of a subspace if we set $V$ to be equal to the subspace you are speaking of (any will do).

Title: Understanding a Basis


答: That definition is a (rare) example of Rudin doing things inefficiently. He could have defined $J_n$ for each non-negative integer $n$ to be the set of non-negative integers less than $n$, so that $J_0=\varnothing$, $J_1=\{0\}$, $J_2=\{0,1\}$, etc. Then he could have defined a set $A$ to be finite if and only if $A\sim J_n$ for some $n\in\Bbb N$ (where $\Bbb N$ includes $0$). This is essentially the usual set-theoretic definition stripped of some set-theoretic detail that would be out of place here.</p>

Title: Why is the empty set finite?


问:
Probability that each number obtained by throwing a die is no smaller than the preceding number

A fair die is thrown 4 times. Find the probability that the each number obtained is no smaller than the preceding number.

If all numbers obtained are same, number of such outcomes $$=\dbinom{6}{1}=6$$

If 3 numbers obtained are same, number of ways is equal  to the number of ways of choosing any two numbers $=\dbinom{6}{2}$. But these two number can be arranged as $x,y,y,y$ or $x,x,x,y$ where $x<y$. So $$2\times\dbinom{6}{2}$$

If 2 numbers are same, the cases are $a_1<a_2<a_3=a_4$, $a_1<a_2=a_3<a_4$, $a_1=a_2<a_3<a_4$, $a_1=a_2<a_3=a_4$. Number of ways $$=3\times\dbinom{6}{3}+\dbinom{6}{2}$$

if all 4 numbers are different, $$\dbinom{6}{4}$$

Total number of possibilities $=6^4$.

Even though I got the correct answer, is there any shorter method?

答:
Let us count the number of ways in which this is possible.

Suppose between each dice throw, the number that appears on the die increases by a certain amount. Note that since we are only counting the number of favourable outcomes, this is a valid assumption. There are $3$ places between the dice throws where this increase can occur. But the initial throw can also produce a number greater than one, and the last can also produce a number less than six. So, we add $2$ more places for increase: before the first throw and after the last throw. We get the following diagram:

(Start at $1$)

Increase $\#1$

Die roll $\#1$

Increase $\#2$

Die roll $\#2$

Increase $\#3$

Die roll $\#3$

Increase $\#4$

Die roll $\#4$

Increase $\#5$

(End at $6$)

There are $5$ spots for increase, and $5$ units of increase from $1$ to $6$.
So, the answer is equal to the no. of ways of distributing $5$ similar objects into $5$ distinct boxes.
For this problem, the formula is $\binom{5+5-1}{5-1} = \binom94 = 126$
So no. of favourable outcomes is $126$.

Total possible outcomes are $6^4 = 1296$

$$\frac{126}{1296}=\frac7{72}$$

Credit: Keyur Joshi (on Quora)


问:
Proving that given a triangle $ABC$ the line $AB$ is parallel to $CK$
Supposing a triangle $ABC$, where $|AC|\neq |BC|$, denote the incentre $I$ and the points of tangency between the inscribed circle and $BC$,$CA$,$AB$ to be $D,E,F$ respectively. $M$ is the midpoint of $AB$. $K$ is the intersection of the perpendicular to $CM$ passing through $I$, and line $DE$. The task is to prove that $CK||AB$.

I have realized that $DE$ is a side of the Gergonne triangle of $ABC$, but I haven't been able to connect it with angle bisectors or anything pertaining to the incentre, nor with medians. I have also tried constructing the circumscribed circle and looking at the perpendicular bisectors but without much success.

I'd really appreciate your help.

答:
It's well known that $DE, IF, CM$ are concurrent. Call this point $X$.

Then $CX\perp IK$, $CI\perp XK$, meaning $I$ is the orthocenter of $CXK$ and the result follows.

Another interesting thing is that $\triangle CXK$ is self-polar wrt the incircle.

Edit:

The following proof that $DE,IF,CM$ are concurrent is not mine. It's taken from Evan Chen's EGMO.

Let $X$ be the intersection of $DE$ and $FI$ and let $A',B'$ be points on $CA$ and $CB$ such that $A'B'\parallel AB$ and $A'B'$ passes through $X$.

Now $E$ is the foot of $I$ onto $CA'$, $X$ is the foot of $I$ onto $A'B'$ ($B'A'\parallel BA$ and $IF\perp AB$, so $IF\perp AB$), and $D$ is the foot of $I$ onto $CB'$. Then $EXD$ is in fact the Simson line of $I$ wrt $\triangle CA'B'$, which means that $I$ is on the circumcircle of $\triangle CA'B'$.

Since $CI$ is an angle bisector of $\angle A'CB'$, $I$ is the arc midpoint of arc $B'A'$, and so $XI$ is the perpendicular bisector of $B'A'$. This means $X$ is the midpoint of $B'A'$, and so taking a homothety at $A$ which takes $A'B'\mapsto AB$ shows that the $C$-median passes through $X$ as well.

Once seeing $I$ is the orthocenter of $\triangle CXK$, $CK\perp IX$ but we already know $IX\perp AB$ so $CK\parallel AB$.


问:
Why is square root by long division found so?
We were taught the long division method of finding square root in junior classes. The  logic behind the method used to be unclear, it remains so even now! However, we learnt and practiced the algorithm well. To me it appears that the identity $$(a+b)^2=a^2+2ab+b^2$$ lies behind the method. However this seems to be only the minimal idea. So extensions and generalizations are welcome here. See the picture below as I can't type this stuff. Some one else may edit it properly.

答:
For $3$ or $4$ digit numbers you were using $$(10a+b)^2=100a^2+(20a+b)b$$

For $5$ or $6$ digit numbers you would have used the messier $$(100a+10b+c)^2=10000a^2+(2000a+ 100b)b+(200a+ 20 b+c)c$$

So you would separate the number to be square-rooted into $100$s, in your example $27,04$

and it looked like

```<code>       ?  ?
     ------
     |27,04

</code>```

The largest square less than or equal to $27$ is $25=5^2$ so you write

```<code>       5  ?
     ------
   5 |27,04
     |25
     ---
     | 2,04
</code>```

You then double the $5$ you have written at the top to give $10$ and then add an extra digit $X$ so $10X \times X$ is a large as possible but does not exceed $204$: $102 \times 2= 204$ works exactly

```<code>       5  2
     ------
   5 |27,04
     |25
     ---
 102 | 2,04
     | 2,04
     ------
          0
</code>```

So $\sqrt{2704} =52$

If the result had not been exact, you could have continued the same way, bringing down two more digits (possibly $00$).

My school thought this a waste of time and instead taught us to use logarithm tables.

Body:
For $3$ or $4$ digit numbers you were using $$(10a+b)^2=100a^2+(20a+b)b$$

For $5$ or $6$ digit numbers you would have used the messier $$(100a+10b+c)^2=10000a^2+(2000a+ 100b)b+(200a+ 20 b+c)c$$

So you would separate the number to be square-rooted into $100$s, in your example $27,04$

and it looked like

       ?  ?
     ------
     |27,04

The largest square less than or equal to $27$ is $25=5^2$ so you write

       5  ?
     ------
   5 |27,04
     |25
     ---
     | 2,04

You then double the $5$ you have written at the top to give $10$ and then add an extra digit $X$ so $10X \times X$ is a large as possible but does not exceed $204$: $102 \times 2= 204$ works exactly

       5  2
     ------
   5 |27,04
     |25
     ---
 102 | 2,04
     | 2,04
     ------
          0

So $\sqrt{2704} =52$

If the result had not been exact, you could have continued the same way, bringing down two more digits (possibly $00$).

My school thought this a waste of time and instead taught us to use logarithm tables.


问:
Smallest prime of the form 41033333333...?
What is the smallest prime of the form $410333333333...$ ?
 It should have more than $10,000$ digits.

[added from answer posted 2013 May 26 at 20:52 by Peter]

I thought it would be clear, but it seems not to be.
 Of course, after $410$, only $3$'s should follow.
 Otherwise, it would be very easy to find primes.
 I checked the numbers up to about $10,000$ digits,
 but of course, I would be glad if someone checks
 this, too.

I do not understand the question, WHY this number
 is interesting for me. Mersenne primes are not so
 much more interesting, but recently a prize of
 $100,000\$ $was payed for a community finding a
 $17$-million-digit mersenne prime. I would have
 better ideas what to do with all this money ...

答:
I guess $(1231\times 10^{6\times 6233}-1)/3$, that is with 37,398 threes. (PFGW calls it a probable prime to base 2,3,5,7.)

Let $a(n)=(1231\times 10^n-1)/3$. Then if a prime $p$ divides $a(n)$, then
$$
10^n \equiv 1231^{-1} \pmod{p} \\
10^{n+k\cdot\mathrm{ord}_p10} \equiv 1231^{-1} \pmod{p} \\
p \mid a(n+k\cdot \mathrm{ord}_p10)
$$
where $\mathrm{ord}_p10$ is the smallest exponent $i$ for which $10^i\equiv 1\pmod{p}$.
So for example
$$
11 \mid a(2k+1) \\
41 \mid a(5k) \\
35 \mid a(3k+2) \\
47 \mid a(46k+10)
$$
and so forth.

If $n_2$ satisfies one or more of these for a prime $p$ with $k>1$, then there must be a smaller $n_1$ with $k=0$ with $p \mid \gcd(a(n_2),a(n_1))$. Since GCD can be computed quickly, for $n>10368$ and divisible by 6 I identified candidates where $\gcd(a(n),a(i))=1$ for several choices of $i<n$. This eliminated about 95% of cases, I made a list of the rest and tested about 1000 before finding one that reported as a probable prime.


问:
Simple method to prove $a^3+b^3=1$ has no integer solutions if $ab\neq 0$
Any simple method to prove $a^3+b^3=1$ has no integer solutions if $ab\neq 0$ that does not involve Fermat's last theorem?

Body:
$$1 = a^3+b^3 = (a+b)(a^2-ab+b^2) = (a+b)((a+b)^2-3ab)$$
implies $a+b=\pm 1$ and so $(a+b)^2 - 3ab = 1 - 3ab \not\in \{-1,1\}$ if $ab \ne 0$


问:
How to find out how big a ball is?
Ok, This is probably a really simple question but. I need to know how I can find out how big a ball is. For example, a tennis ball is 2 1/2 inches big, but how do you find that?

Though, for reference, the explanation and answer to this question needs to be as simple as it can possibly get. I have a learning disability that heavily affects my mathematics capabilities and severe Dyscalculia. This explanation here probably makes me sound pretentious, but a lot of people don't understand or they throw too many numbers at me and get frustrated when I haven't said anything before hand. I'm sorry if it does, but I'm covering all my bases lol.

Any help on how to figure this question out is very much appreciated!

答: Make a cylinder of paper that is large enough to enclose the ball.  Put the ball on a table and close the cylinder around the ball.  Mark where the paper's (verticle) edge meets the other side of the ball while the cylinder is perpendicular to the table. This allows you to measure the circumference.  Then the division by $\pi$ ($3.1418$) gives the diameter, which is the width of the ball.  The old approximation for $\pi$ is $\frac{22}7$, so you can mulitply by $7$ and divide by $22$ to get the number you are looking for.

Title: How to find out how big a ball is?


问:
Understanding finding the Laurent series
I am having trouble understanding how to find a Laurent series.
$$\sum\limits_{n=-\infty}^{\infty} a_{n}(z-c)^{n}$$
where $a_{n}=\frac{1}{2 \pi i} \int_{C} \frac{f(w)}{(w-c)^{n+1}}dw$, and $C$ is any circle with center $c$.

I am currently trying to solve the following problem:

Find Laurent series, about the given point $z=-2$, of $$\frac{z}{(z+1)(z+2)}$$
Hint: let $z+2=u$

I am trying to find help starting this problem and understanding Laurent series.

Body:
Following the hint yields
$$\frac{u-2}{(u-1)u}=\frac{2}{u}+\frac{1}{1-u}=2u^{-1}+\sum_{n=0}^\infty u^n.$$
Now substitute $u=z+2$ to obtain
$$\frac{z}{(z+1)(z+2)}=2(z+2)^{-1}+\sum_{n=0}^\infty (z+2)^n.$$


问:
If $a \neq b$, under what condition does $(I-C)a = (I-C)b$?
Suppose $a, b \in \mathbb{R}^n$, and $a \neq b$ (that is, $a_j \neq b_j$ for at least one $j$, where $j = 1, \ldots, n)$.

Let $I$ denote a $n \times n$ identity matrix, and let $C \in \mathbb{R}^{n \times n}$.

I want to know whether it's possible for the following to hold:

$$(I-C)a = (I-C)b$$

Here's my attempt:

\begin{align*}
(I-C)a &= (I-C)b\\
\Leftrightarrow a - Ca &= b - Cb\\
\Leftrightarrow a - b &= Ca - Cb\\
\Leftrightarrow a-b &= C(a-b)
\end{align*}

Can I go further or is this it? So far, based on the above, I have $(I-C)a = (I-C)b$ iff $a-b = C(a-b)$.

答:
Your condition is equivalent to saying that one of the eigenvalues of $C$ is $1$, and that $a-b$ is in the eigenspace. If we start with a matrix $C$ that has $1$ as an eigenvalue, we can just pick any $v$ in this eigenspace and write $v=a-b$. For example, an easy matrix that has $1$ as an eigenvalue is:

$$
C = \begin{bmatrix}
1 & 0 \\ 0 & 2
\end{bmatrix}.
$$

A vector in the eigenspace of $1$ is $(1,0)$, so you could just take $a = (0,2)$ and $b=(1,2)$. You can then check that $(I-C)a = (I-C)b$.

Alternatively, you can start with your vectors $a,b$ and construct a matrix $C$. Since $a-b \neq 0$, we can add vectors to get a basis $\{a-b, v_2, ..., v_n\}$ for $\mathbb{R}^n$. We can complete to get another basis $\{a-b, w_2, ..., w_n\}$. Then define a linear map $T : \mathbb{R}^n \rightarrow \mathbb{R}^n$ by setting $T(a-b) = a-b$ and $T(v_i) = w_i$ for $i \geq 2$. Let $C$ be the expression of $T$ in the standard basis, which is given by

$$
C = [a-b, w_2, ..., w_n] \cdot [a-b, v_2, ..., v_n]^{-1}.
$$

Then $C$ satisfies $C(a-b) = a-b$ since this is true of $T$; hence $(I-C)a = (I-C)b$. For example, take $a=(2,3), b=(2,1)$. Then $a-b = (0,2)$, and we can complete this to the bases $\{(0,2), (1,1)\}$ and $\{(0,2), (3,-1)\}$. Then:

$$
C =
\begin{bmatrix}
0 & 3 \\ 2 & -1
\end{bmatrix}
\begin{bmatrix}
0 & 1 \\ 2 & 1
\end{bmatrix}^{-1}
=
\begin{bmatrix}
3 & 0 \\ -2 & 1
\end{bmatrix}
$$

You can check that $C(a-b) = a-b$.


问:
MATHCOUNTS NATIONAL 2014 #12
I don't know how to approach this. The correct answer is $2013$.

What is the value of

$$\frac{2013^3-2\cdot 2013^2\cdot 2014+3\cdot 2013\cdot 2014^2-2014^3+1}{2013\cdot 2014}\,?$$

I thought about using Binomial Theorem in some way as the format looks similar to the results you get, but couldn't follow through with it.

答:
Your idea of using the binomial theorem is a good one, but there’s a little more work to be done. Most of the numerator looks a lot like the result of expanding $(2013-2014)^3$, so you should start by writing out that expansion. Then you can easily see that the numerator is

$$(2013-2014)^3+2013^2\cdot 2014+1\,;$$

Can you finish it from there?

Title: MATHCOUNTS NATIONAL 2014 #12


Title: MATHCOUNTS NATIONAL 2014 #12

Body:
I don't know how to approach this. The correct answer is $2013$.

What is the value of

$$\frac{2013^3-2\cdot 2013^2\cdot 2014+3\cdot 2013\cdot 2014^2-2014^3+1}{2013\cdot 2014}\,?$$

I thought about using Binomial Theorem in some way as the format looks similar to the results you get, but couldn't follow through with it.

Body:
Consider the expression $$\frac{a^3 - 2a^2b + 3ab^2 - b^3 + 1}{ab}$$ where $a = 2013$ and $b = a+1 = 2014$

Put $b = a+1$ and simplify!

$$\frac{a^3 - 2a^2b + 3ab^2 - b^3 + 1}{ab} = \frac{a^2(a+1)}{a(a+1)} = a = 2013$$


问:
Get covariance from log-density function
Problem

Given a following log-density function
$$ \ln p(y| a, b) = a \cdot g(y) + b \cdot h(y) + k(a,b)$$
where $g(y), h(y), k(a,b)$ are difined function and $a,b$ are parameters.

Find $\Bbb Cov( g(Y), h(Y))$ using $k(a,b)$.

My try

I believe that it got something to do with the score function and Fisher information. We basically can calculate the  Fisher information using two methods:

$\Bbb Var[(\ln p(y| a, b))'']$

$\Bbb -E[(\ln p(y| a, b))']$

These are derivatives.
My idea is to use both of these definitions and to make them equal. Since we have two parameters, I assume that it's a vector case.
Fisher information using (1):
$$\begin{bmatrix}
\Bbb Var( g(Y))\\
\Bbb Var( h(Y))
\end{bmatrix}$$
And Fisher infomration using (2)
$$\begin{bmatrix}
k_a''\\
k_b''
\end{bmatrix}$$
Even though I've found each of the variances, I have them separately and don't know how to get to the covariance.

Body:
Here is how I think $\Bbb{Cov}(g(Y),h(Y))$ is calculated (and $\Bbb{Var}(g(Y)), \Bbb{Var}(h(Y))$)

$\Bbb{Var}(g(Y))\\
=\Bbb{Var}(g(Y))+2\cdot0+0\\
=\Bbb{Var}(g(Y))+2\cdot\Bbb{Cov}(g(Y),k'_a)+\Bbb{Var}(k'_a)\\
=\Bbb{Var}(g(Y)+k'_a)\\
=\Bbb{Var}(\partial_a \ln(p(x|a,b))\\
=\Bbb{E}([\partial_a \ln(p(x|a,b)-\Bbb{E}(\partial_a \ln(p(x|a,b))]^2)\\
=\Bbb{E}([\partial_a \ln(p(x|a,b)]^2)-[\Bbb{E}(\partial_a \ln(p(x|a,b)))]^2\\
=\Bbb{E}([\partial_a \ln(p(x|a,b)]^2)-0^2\\
=-\Bbb{E}(\partial^2_a \ln(p(x|a,b))\\
=-\Bbb{E}(k''_a)\\
=-k''_a\\
$

$\Bbb{Var}(h(Y))\\
=\Bbb{Var}(h(Y))+2\cdot0+0\\
=\Bbb{Var}(h(Y))+2\cdot\Bbb{Cov}(h(Y),k'_b)+\Bbb{Var}(k'_b)\\
=\Bbb{Var}(h(Y)+k'_b)\\
=\Bbb{Var}(\partial_b \ln(p(x|a,b))\\
=\Bbb{E}([\partial_b \ln(p(x|a,b)-\Bbb{E}(\partial_b \ln(p(x|a,b))]^2)\\
=\Bbb{E}([\partial_b \ln(p(x|a,b)]^2)-[\Bbb{E}(\partial_b \ln(p(x|a,b)))]^2\\
=\Bbb{E}([\partial_b \ln(p(x|a,b)]^2)-0^2\\
=-\Bbb{E}(\partial^2_b \ln(p(x|a,b))\\
=-\Bbb{E}(k''_b)\\
=-k''_b\\
$

$\Bbb{Cov}(g(Y),h(Y))\\
=\Bbb{Cov}(g(Y),h(Y))+0+0+0\\
=\Bbb{Cov}(g(Y),h(Y))+\Bbb{Cov}(k'_a,h(Y))+\Bbb{Cov}(g(Y),k'_b)+\Bbb{Cov}(k'_a,k'_b)\\
=\Bbb{Cov}(g(Y)+k'_a,h(Y)+k'_b)\\
=\Bbb{Cov}(\partial_a \ln(p(x|a,b),\partial_b \ln(p(x|a,b))\\
=\Bbb{E}([\partial_a \ln(p(x|a,b)-\Bbb{E}(\partial_a \ln(p(x|a,b))][\partial_b \ln(p(x|a,b)-\Bbb{E}(\partial_b \ln(p(x|a,b))])\\
=\Bbb{E}([\partial_a \ln(p(x|a,b)][\partial_b \ln(p(x|a,b)])-\Bbb{E}([\partial_a \ln(p(x|a,b)])\Bbb{E}([\partial_b \ln(p(x|a,b)])\\
=\Bbb{E}([\partial_a \ln(p(x|a,b)][\partial_b \ln(p(x|a,b)])-0\cdot0\\
=-\Bbb{E}(\partial_a\partial_b \ln(p(x|a,b)])\\
=-\Bbb{E}(k''_{ab})\\
=-k''_{ab}
$


问:
Does a bounded function need to be defined everywhere inside an interval in order to be Riemann integrable?
If a bounded function $f$ is not defined on a finite set $A$ of numbers inside an interval $[a,b]$, is it Riemann integrable over that interval $[a,b]$?

I have seen that some people jump over this question by defining the function in an arbitrary way, $0$ for example, on the set of numbers where it is not defined. But this seems wrong to me, since then you are modifying the domain, and therefore integrating a different function, not the original one.

My confusion increases, because it seems to me that the $f$ might be Darboux integrable, without the need of defining the function in the set $A$. But I have read that Riemann and Darboux are equivalent.

Any help would be welcome. Thanks in advance.

答:
First of all, it seems like you might be thinking that all bounded functions are Riemann integrable, so I want to disavow you of that notion. Take $f(x) = 1$ for $x$ rational and $0$ for $x$ irrational. This is bounded on all compact intervals but is not Riemann integrable on any.

Strictly speaking, the definition of a Riemann integral is on a bounded function $f: [a, b] \longrightarrow \mathbb R$, so $f$ must be defined on all of $[a, b]$ no ifs ands or buts. If we had some $f$ defined on $[a, b] - A$ then it is technically incorrect to write $\int_a^b f(x) dx$. In fact, we cannot even ask if $f$ is Riemann integrable on $[a, b]$ as it is not a real function defined on $[a, b]$.

However, we are not always so strict in math. We can choose to abuse the notation of the Riemann integral and write $\int_a^b f(x) dx$ even when $f$ is not defined everywhere on $[a, b]$, but we must justify this abuse. Indeed, let $g_1, g_2: [a, b] \longrightarrow \mathbb R$ such that $g_1$ is Riemann integrable and $g_2$ differs from $g_1$ at only finitely many points. Then $g_2$ is Riemann integrable and $\int_a^b g_1(x) dx = \int_a^b g_2(x) dx$. To see why, we can work with the difference $h = g_1 - g_2$ and prove that $h$ is Riemann integrable with integral $0$. Roughly speaking, as $h$ is nonzero at only finitely many points, shrinking our partition around these points allows our Riemann sums to tend to $0$.

What this fact about $g_1, g_2$ says is that the Riemann integral does not care about the values of your function on any finite set. This is why we feel justified in talking about the Riemann integrability of a partially defined function $f: [a, b] - A \longrightarrow \mathbb R$. Whatever extension to the whole of $[a, b]$ we choose makes no difference - if $g_1, g_2: [a, b] \longrightarrow \mathbb R$ both extend $f$ then $g_1$ is Riemann integrable iff $g_2$ is, in which case their integrals coincide. We can therefore comfortably say that $f$ is Riemann integrable, when we mean that an extension is, and we can comfortably compute its integral by computing the integral of any extension.

To put this a bit more formally, let $R([a, b])$ be the set of all Riemann integrable functions on $[a, b]$. This is actually a real vector space and we have a linear map $\int_a^b: R([a, b]) \longrightarrow \mathbb R$ sending $f \mapsto \int_a^b f(x) dx$. Now, we define an equivalence relation on $R([a, b])$ via $f \sim g$ if $f(x) = g(x)$ at all but finitely many points. The quotient $R([a, b])/{\sim}$ is a real vector space, and the integral descends to a well defined map $R([a, b])/{\sim} \longrightarrow \mathbb R$ sending $[f] \mapsto \int_a^b f(x) dx$. This is just formalism for my above statement that the Riemann integral doesn't care about the values on a finite set. Additionally, take a partially defined function $f: [a, b] - A \longrightarrow \mathbb R$ where $A$ is finite. Suppose that $f$ can be extended to a Riemann integrable function on $[a, b]$. Then there is no canonical choice of $g \in R([a, b])$ to assign to $f$. However, there is a canonical equivalence class to assign to $f$ - that is, $[g]$ where $g$ is any extension of $f$. In fact, we can feel justified referring to this equivalence class as just $[f]$. Point being, this object $[f]$ is still specific enough to talk about Riemann integrability as much as we want. As an aside, this equivalence relation idea is extremely fruitful when dealing with the Lebesgue integral, where we replace "finite" with "measure 0."


问:
A cosine function has maximum value of 14 and a minmum value of 4, a period of 7, and a phase shift of 12.
A cosine function has a maximum value of 14 and a minimum value of 4, a period of 7, and a phase shift of 12. Write an equation representing this cosine function...

Could someone tell me if I'am write and if I'am wrong explain why and a solution my answer $y\,\,\, = \,\,\,14\cos 2\pi {{\left( {x\,\, - \,\,12} \right)} \over 7}\,\,\, + \,\,\,2$

答: If the max value is $14$ and the min value is $4$, what is the midline value? What is the amplitude? Answering these questions will let you fix your errors--specifically, the $+2$ at the end and the $14$. The rest looks good.</p>


问:
Solving $yy''+(y')^2- \frac{2yy'}x + \frac{y^3}{x^3}=0$
$$yy''+(y')^2- \frac{2yy'}x + \frac{y^3}{x^3}=0$$

I have a problem with this differential equation. I would really appreciate any help.

答:
Let $u=y^2$, then
$$u'=2yy'$$
$$\frac{1}{2}u''=yy''+(y')^2$$

So, the equation reduces to
$$\frac{1}{2}u''-\frac{1}{x}u'=-\frac{u^{3\over 2}}{x^3}$$

This is an Emden-Fowler equation, for which there is literature on exact solutions (under some specific conditions) and approximation schemes.


问:
What is the difference between square of sum and sum of square?
What is difference between square of sum $(\sum_{i=1}^{n}x_i)^2$ and sum of square $\sum_{i=1}^{n}x_i^2$?

I think square of sum is bigger than sum of square but i can not find a relation between them.

I mean:

$$\left(\sum_{i=1}^{n}x_i\right)^2=\sum_{i=1}^{n}x_i^2+?$$

答:
Additionally, while the derivation is lengthy to show, the difference when summing over the naturals can be explicitly expressed as a polynomial, quite surprisingly

$(\sum_{i=1}^{n}i)^2-\sum_{i=1}^{n}(i^2)=\frac{n}{12}(n^2-1)(3n+2), n\in\mathbb{N}$


答:
Let us do it with a trinomial
$$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac$$ see the fig.

Title: Why is square root by long division found so?

Body: We were taught the long division method of finding square root in junior classes. The  logic behind the method used to be unclear, it remains so even now! However, we learnt and practiced the algorithm well. To me it appears that the identity $$(a+b)^2=a^2+2ab+b^2$$ lies behind the method. However this seems to be only the minimal idea. So extensions and generalizations are welcome here. See the picture below as I can't type this stuff. Some one else may edit it properly.


问:
Prove a function is onto given some conditions
Suppose there is a $f:A\to B$ onto and $g: A\to B$ that is 1-1. Show that $g:A\to B$ is onto, or give a counterexample.

I think this seems true in the finite case, but when we deal with infinite sets it is no longer the case. Anyone know how to prove it for finite case? For infinite case, does $f:\mathbb{N} \to \mathbb{N}, n \to n$ and $g: \mathbb{N} \to \mathbb{N}, n \to n+1$ suffice as a counterexample?

答:
If $A$ and $B$ are finite you can make a counting argument. Since $f$ is a surjective function from $A$ to $B$, for each $b\in B$ there is an element $a\in A$ such that $f(a)=b$, but since $A$ is not necessarily injective, $f^{-1}(b)$ (the preimage of $b$ under $f$, or the set of $a\in A$ such that $f(a)=b$) could be a set with several elements. Then $\lvert B\rvert \leq \lvert A \rvert$.

Now since $g$ is an injective function from $A$ to $B$, for each $b\in g(A)$ there is a unique $a \in A$ such that $g(a)=b$ (note $g(A)$ is not necessarily equal to $B$ yet, that's what we want to show). This shows $\lvert A\rvert \leq \lvert B \rvert$

Combining the above results notice that $\lvert A\rvert = \lvert B \rvert$

Since $g$ is injective $\lvert g(A) \rvert =\lvert A\rvert =\lvert B \rvert$. Since $B$ is finite and $g(A)\subset B$ with $\lvert g(A) \rvert =\lvert B \rvert$, it must be the case that  $g(A)=B$. This shows that $g$ is surjective as we wanted.


问:
Why does a Reuleaux triangle appear when drilling a hole with a knife
As the season commands, I was blowing out eggs for the kindergarten. In order to have large holes at both ends of the egg I first poke a small hole with a pin and then enlarge the said hole with a pointy knife (by rotating said knife in the hole in the hope of creating a round hole). To my surprise, enlarging the hole in such a fashion ended up giving me (in 7 out of 8 tries) a shape which rather looks like a Reuleaux triangle than a circle.

Now it's obvious that the resulting shape should have constant width. But my [first and vague] question is: are there reasons not to get a circle? (and is the phenomenon confirmed?)

In some sense a "Reuleaux triangle"-like shape is more probable since it has a much smaller symmetry group (so if the medium is not uniform, it might be a more "optimal" solution [to some optimisation problem]). So my [hopefully more precise] question is:

Question: What is a good model for this "drilling process" which could explain a "Reuleaux triangle"-like shape?

Just for details (but I doubt it matters) the knife was only sharp one side and both sides of the knife touch the edge of the hole at all times while the drilling occurs.

答:
As an experiment, I took a sharp knife and stuck it through a piece of paper.
Then I rotated then knife.

At first, the sharp edge of the knife stayed stuck in place while the dull edge carved a circular sector out of the piece of paper,
or more precisely cut along a circular arc and pushed a sector-shaped flap of paper aside.
When the knife had rotated about $60$ degrees, the sharp edge broke free and started to scrape along the previously intact edge of the original cut,
although it curled the paper away slightly rather than cutting it.

After approximately another $60$ degrees, the sharp edge of the knife hit the end of the original cut and the dull edge broke free again and pushed past the sector-shaped flap.

I tried the experiment twice more. The second attempt resulted in something more like the Kakeya set mentioned here, but the third time was much like the first.

The holes I got the first and last time were not Reuleaux triangles. One side was an almost perfect circular arc, another was nearly straight, and the third was hard to identify as any shape because it still had a flap of paper attached.
This shows that the experiment with paper was not a perfect model of what happens with an eggshell.
But my hypothesis is that you get a roughly triangular shape because of the way the shell breaks under the force of the knife, that is, it tends to keep breaking away where it has already broken until the angle between the knife and shell at the other end of the cut has widened so much that the knife edge is able to break free.
At that point you have a roughly triangular shape, and all that remains is for the knife to continue taking off bits of shell, mostly from the middle of each edge of the triangle, until it can turn $360$ degrees.

Paper obviously is not a great model of the shell of an egg.
It seems it would be a particularly poor model of the shell itself,
but I was working on the assumption that the shell is still attached to the membrane underneath.

I still would not expect the paper to necessarily reproduce the properties
of the combined membrane and shell,
although I expect it would model the shell better than a sheet of peanut brittle or a sheet of glass would.
And I did not want to sacrifice any actual eggs at that time.

The value of the experiment is not so much that it confirms the final shape, but rather the idea of closer observation of what happens as the knife turns.

I would not be surprised if different material properties produced a hole more like a higher-order Reuleaux polygon.

I suspect that in the case where I got the more Kakeya-set-like figure I had inadvertently applied a force to the blade (rather than just torque) causing the sharp end to slice further into the paper. Unfortunately I was not so observant that time.

If it turns out that the behavior of a knife in eggshell is similar to its behavior in paper, then the question is why the one edge continues to act as a fulcrum until the knife has made such a large angle with the original edge of the hole.
I think this might be an engineering or materials science question,
difficult to guess by mathematics alone.

Update: Results for heavy-duty corrugated cardboard from a large box of vegetables.

The cardboard has a distinct bias. The knife is difficult to insert with the blade perpendicular to the corrugation.
Rather than the material peeling away toward the side opposite from which the knife was inserted, the cardboard simply crushes in place.
Both edges of the knife move at first, then one edge breaks free and sweeps across the triangular bit of cardboard that was obstructing it.
The resulting hole is roughly triangular.

The knife is much easier to insert parallel to the corrugation but harder to turn. Again both edges push through the material at first.
It seems I have to rotate the knife more than $90$ degrees before one edge breaks free.
When it does, the hole ends up roughly square, loosely capturing the knife blade.


问:
Evaluating $\int_0^1 (1-x^2)^n dx$
In an exercise I'm asked the following:

a) Find a formula for $\int (1-x^2)^n dx$, for any $n \in \mathbb N$.

b) Prove that, for all $n \in \mathbb N$: $$\int_0^1(1-x^2)^n dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}$$

I used the binomial theorem in $a$ and got:

$$\int (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) (-1)^k \ \frac{x^{2k + 1}}{2k+1} \ \ \ + \ \ C$$

and so in part (b) i got:

$$\int_0^1 (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) \ \frac{(-1)^k}{2k+1}$$

I have no clue on how to arrive at the expression that I'm supposed to arrive. How can I solve this?

答:
Beta Integral
$$
\begin{align}
\int_0^1\left(1-t^2\right)^n\mathrm{d}t
&=\frac12\int_0^1\left(1-t\right)^nt^{-1/2}\mathrm{d}t\tag{1a}\\
&=\frac{\Gamma(n+1)\Gamma(1/2)}{2\Gamma(n+3/2)}\tag{1b}\\
&=2^{2n}\frac{\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+2)}\tag{1c}\\
&=\frac{4^n}{2n+1}\frac1{\binom{2n}{n}}\tag{1d}
\end{align}
$$
Explanation:

$\text{(1a)}$: substitute $t\mapsto t^{1/2}$

$\text{(1b)}$: Beta Integral

$\text{(1c)}$: Gauss' Multiplication Formula

$\text{(1d)}$: convert to Binomial Coefficients via Factorials

Integration by Parts
$$
\begin{align}
\int_0^1\left(1-t^2\right)^n\mathrm{d}t
&=\int_0^12nt^2\left(1-t^2\right)^{n-1}\,\mathrm{d}t\tag{2a}\\
&=\frac{2n}{2n+1}\int_0^1\left(1-t^2\right)^{n-1}\mathrm{d}t\tag{2b}\\
&=\prod_{k=1}^n\frac{\color{#C00}{2k}}{\color{#090}{2k+1}}\tag{2c}\\
&=\color{#090}{\frac1{2n+1}\frac{2^nn!}{(2n)!}}\color{#C00}{2^nn!}\tag{2d}\\[3pt]
&=\frac{4^n}{2n+1}\frac1{\binom{2n}{n}}\tag{2e}
\end{align}
$$
Explanation:

$\text{(2a)}$: integrate by parts

$\text{(2b)}$: add $\frac{2n}{2n+1}$ of the left side of $\text{(2a)}$ to $\frac1{2n+1}$ of the right side

$\text{(2c)}$: induction

$\text{(2d)}$: evaluate the products

$\text{(2e)}$: simplify

Title: Evaluating $\int_0^1 (1-x^2)^n dx$


答:
Well $a^3 + b^3 = (a+b)(a^2 -ab + b^2)=1$

The only factors of $1$ are either $1\cdot 1$ or $-1 \cdot (-1)$.

So we must have $a+b = a^2 - ab + b^2 =\pm 1$.

And that just won't work unless $ab=0$.

It would follow that $b= \pm 1 -a$

And therefore  $a^2 - a(\pm 1 - a) + (\pm 1 - a) = \pm 1$ so

$a^2+a^2 \mp a -a \pm 1 = \pm 1$

SO either $2a^2 -2a+1=1$ or $2a^2 -1 = -1$ so

$2a(a-1) = 0;b= 1-a$ or $2a^2 = 0;b = -1-a$.

So we have either $a=0$ and $b=1$.  Of $a =1$ and $b =0$.  Or $a=0$ and $b =  -1$.

In all three cases we must have $ab = 0$.

Title: Simple method to prove $a^3+b^3=1$ has no integer solutions if $ab\neq 0$

Body: Any simple method to prove $a^3+b^3=1$ has no integer solutions if $ab\neq 0$ that does not involve Fermat's last theorem?