问:
Show that a matrix has full column rank
Consider a matrix $A$ of size $T\times 2K$. Consider the collection of numbers $k_1,k_2,...,k_T$, where each $k_t\in \{1,...,K\}$.

Each $t$-th row of the matrix $A$ is structured as follows:

its $k_t$-th element is equal to $1$

its $(k_t+K)$-th element is equal to some scalar $\delta_t>0$

all other elements are zero.

Further, $A$ does not contain zero columns.

I believe that the following holds:

$A$ has full column rank $2K$ if and only if there exists $t\in \{1,...,T\}$ and $\tau\in \{1,...,T\}$ with $t\neq \tau$ such that
$$
k_t=k_{\tau} \quad \delta_t\neq \delta_{\tau}
$$

Question: I wrote a proof for this claim (below), which looks to me a bit "naive". Is there anything more formal and, perhaps, simpler that you can suggest?

My proof:

$A$ has full column rank $2K$ if and only if the following system admits as unique solution $x=0_{2K\times 1}$:
$$
(*)\hspace{1cm} x_{k_t} + \delta_{t} x_{k_t+K}=0 \quad \text{ for each $t\in \{1,...,T\}$}.
$$

First, I prove sufficiency.
If there exist $t,\tau$ such that $k_t=k_{\tau}\equiv k$ with $\delta_{t}\neq\delta_{\tau}$, then system $(*)$ contains the equations
$$
\begin{aligned}
& x_{k}+\delta_t x_{k+K}=0,\\
& x_{k}+\delta_{\tau} x_{k+K}=0,\\
\end{aligned}
$$
from which it follows that $x_{k}=x_{k+K}=0$. If the above is true for every $k\in \{1,...,K\}$, then $x_{k}=x_{k+K}=0$ for each $k\in \{1,...,K\}$. Hence, the unique solution of system $(*)$ is $x=0_{2K\times 1}$.

Now, I prove necessity.
Suppose there exist only a $t$ such that $k_t\equiv k$. Then, system $(*)$ contains only one equation involving  any of $x_k,x_{k+K}$, which is
$$
x_{k}+\delta_t x_{k+K}=0,
$$
which is satisfied for any values of $x_k, x_{k+K}$ such that $x_k=-\delta_t x_{k+K}$. Therefore, system $(*)$ has not a unique solution.

答:
Full column rank is equivalent to the column vectors being linearly independent.

After reordering the rows if necessary (which does not change the rank) your matrix can be written in this form (where the $*$'s are positive numbers):
$$\begin{pmatrix}
1 & 0 & 0 & \cdots & 0 & * & 0 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & & \vdots  & \vdots & \vdots & \vdots & & \vdots \\
1 & 0 & 0 & \cdots & 0 & * & 0 & 0 & \cdots & 0 \\
0 & 1 & 0 & \cdots & 0 & 0 & * & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & & \vdots  & \vdots & \vdots & \vdots & & \vdots \\
0 & 1 & 0 & \cdots & 0 & 0 & * & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 & 0 & 0 & * & \cdots & 0 \\
\vdots & \vdots & \vdots & & \vdots  & \vdots & \vdots & \vdots & & \vdots \\
0 & 0 & 1 & \cdots & 0 & 0 & 0 & * & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots  & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1 & 0 & 0 & 0 & \cdots & * \\
\vdots & \vdots & \vdots & & \vdots  & \vdots & \vdots & \vdots & & \vdots \\
0 & 0 & 0 & \cdots & 1 & 0 & 0 & 0 & \cdots & * \\
\end{pmatrix}$$

or, leaving out the zeros,
$$\begin{pmatrix}
1 &  &  &  &  & * &  &  &  &  \\
\vdots& & & & & \vdots & & & & \\
1 &  &  &  &  & * &  &  &  &  \\
 & 1 &  &  &  &  & * &  &  &  \\
& \vdots & & & & & \vdots & & & \\
 & 1 &  &  &  &  & * &  &  &  \\
 &  & 1 &  &  &  &  & * &  &  \\
& & \vdots & & & & & \vdots & & \\
 &  & 1 &  &  &  &  & * &  &  \\
& & & \ddots& & & & & \ddots &  \\
 &  &  &  & 1 &  &  &  &  & * \\
& & & & \vdots  & & & & & \vdots \\
 &  &  &  & 1 &  &  &  &  & * \\
\end{pmatrix}$$

Now it's clear that the column vectors can't be linearly independent if all the $*$'s for some "1"-position are the same (because the $k+K$'th column vector would then be a multiple of the $k$'th one, for some $k$).
On the other hand having not all $*$'s equal in any column (and having at least two rows at every "1"-position) is also clearly sufficient for having linear independence.

Title: Show that a matrix has full column rank


问:
There is no substitution rule?
Lets say there is a formal language $L$ with a bunch of axioms (example from mendelson).

(A1) $A \to (B \to A)$

(A2) $(A \to (B \to C)) \to ((A \to B ) \to (A \to C))$

(A3) $(\neg A \to \neg B) \to ((\neg A \to B ) \to A)$

The only rule of inference is modus ponens.

But then I fall into a confusion.I can't do any proof with this unless i introduce the substitution rule (a inference).And substitution rule is indeed used in all of the proof of theorems of that language.But the substitution rule is not mentioned as a inference rule in the list of inferences(there is only modus ponens).Why?

答:
See Mendelson's textbook, page 3:

let us apply the name statement form to an expression built up from the statement letters $A, B, C$, and so on by appropriate applications of the propositional connectives.

All statement letters (capital italic letters) and such letters with numerical subscripts [footnote: For example, $A_1, A_2, B_1, C_2, \ldots$] are statement forms.

If $\mathscr B$ and $\mathscr C$ are statement forms, then so are $(\lnot \mathscr B), (\mathscr B \land \mathscr C)$ [and so on].

Axioms (A1), (A2) and (A3) [page 28] are expressed with statement forms: $\mathscr B → (\mathscr C → \mathscr B)$ abnd not with statement letters: $B → (C → B)$.

This means that they are schemata: i.e. rules for producing instances writing formulas (made with statement letters) in place of statement forms.

Thus, substitution rule is embedded into the use of schemata.

ContentLicense: CC BY-SA 4.0

Title: There is no substitution rule?


问:
The maximal ideal space of $A$ is contained in the unit ball of $A^\ast$
On page 281 of Rudin's book Functional Analysis, let $\Delta$ be the maximal ideal space of a commutative Banach algebra $A$, $K$ be the norm-closed unit ball of $A^*$, then $\Delta\subset K$ (by Theorem 10.7). Can someone give me more detail?</p>

答: This is true after identification of closed maximal ideals with non-zero characters on $A$. More preciesly each maxiamal closed ideal in $A$ is a kernel of uniqely defined character. One can show that its norm is $1$. For details see this answer.</p>

ContentLicense: CC BY-SA 3.0


问:
Finding a function as a solution to a system of integral equations
In my research, I reached a point where I need to find a nonnegative function, that satisfies the following:

For any $\phi\in[0,\pi]$:

$[\int^\pi_\phi f(\theta)d\theta][ \int^{\pi-2\phi}_{-\phi} f(\theta)d\theta]+[\int^0_{\pi-\phi} f(\theta)d\theta ][\int^{2\phi}_{\pi+\phi} f(\theta)d\theta]-[\int^\phi_0 f(\theta)d\theta]^2-[\int^{\pi-\phi}_\pi f(\theta)d\theta]^2 = \pi^2\cos^2\frac{\phi}{2}$

I would be very glad if somebody can help me, and thank you in advance for your help!

答:
Too long for a comment.

If  $\int_0^{\pi}f(\theta)d\theta = \frac{\pi}{2}$ then the integral equations reduce to

$$
\cases{
\frac{\pi}{2}\int_0^{\pi}f(t-\phi)dt = \frac{\pi^2}{2}\cos^2\left(\frac{\phi}{2}\right)\\
\frac{\pi}{2}+\int_0^{\pi}f(t-\phi)dt = \pi
}
$$

then $\int_0^{\pi}f(t-\phi)dt = \frac{\pi}{2}$. It seems that $f(\cdot)=C_0$. Contradictory.


问:
Difference between a pair of vectors spanning a space, and a pair of vectors being the basis vectors of a space.
Question: $\bar v + \bar w$ and $\bar v - \bar w$ are combinations of $\bar v$ and $\bar w$. The two pairs of vectors span the same space. When are they a basis for the same space?

Answer: They are a basis when $\bar v$ and $\bar w$ are independents.

When the pair of vectors span the same space, doesn't it immediately tell you that they are a basis for that space?

Are these two statements different in some way? I feel that I'm missing something.

答:
The vectors $(1,0)$ and $(-1,0)$ span $\{(x,0)\mid x\in\Bbb R\}$. However, they are not a basis of that space, since they are linearly dependent.

A basis is a linearly independent set that spans the space.


问:
existence of a non continuous and unbounded function
Let $(X , d)$ be a non-discrete metric space and $(Y , d)$ be an unbounded metric space. Then is it true that there always exists an unbounded and non-continuous function from $(X , d)$ to $(Y , d)$. If not then can we find any counter example.

Please note that if $(X , d)$ is a discrete metric space then any function from $(X , d)$ to $(Y , d)$ is a continuous function no matter what is structure of the metric space $(Y , d)$. I can able to prove that there always exist a bounded and non-continuous function from $(X , d)$ to $(Y , d)$, provided $(Y ,d)$ has at least two elements. Any help please. Thanks in advance.

答:
Non-discrete means that there is $x_0\in X$ that is not isolated. And so there is a sequence $(x_n)\subseteq X$ convergent to $x_0$, such that $x_n\neq x_0$. We can also assume that $(x_n)$ is injective, i.e. $x_n\neq x_m$ when $n\neq m$. Because otherwise we can take an injective subsequence of $(x_n)$, which exists by taking element-by-element for smaller and smaller neighbourhoods around $x_0$.

Now let $(y_n)\subseteq Y$ be an unbounded sequence (note that being unbounded guarantees that such sequence exists, finite spaces are always bounded). With that we have all we need. Simply define $f:X\to Y$ by $f(x_n)=y_n$ and define $f(x)$ arbitrarily for any $x\neq x_n$. This function is well defined, unbounded and discontinuous (because unbounded sequences are not convergent). Even more: the discontinuity at $x_0$ is not removable.


问:
Two dimensional continuous random variable and independecy
Q: Let $(X,Y)$ be a two dimensional random variable with the following density function:

$f_{_{X,Y}}(x,y)=\begin{cases}8xy&,  0\le x\le y \le 1 \\ 0&,  o.w.\end{cases}$

Are $X,Y$ independent?

My attempts: in the lecture notes I have, it says that two random variables $X,Y$ are independent $\iff \forall x,y \in \mathbb R: f_{_{X,Y}}(x,y) = f_{_X}(x)\cdot f_{_Y}(y)$.

I found $f_{_X}, f_{_Y}$:

$
  f_{_X}(x)=\int_\limits{-\infty}^\infty f_{_{X,Y}}(x,y) \,dy=
    \begin{cases}
      \int_\limits x^1 8xy\,dy&,  0\le x \le 1 \\
      0&, o.w.
    \end{cases}
  =
  \begin{cases}
    4x-4x^3 &,  0\le x \le 1 \\
    0&, o.w.
  \end{cases}
$

$
  f_{_Y}(y)=\int_\limits{-\infty}^\infty f_{_{X,Y}}(x,y) \,dx=
    \begin{cases}
      \int_\limits 0^y 8xy \,dx &,  0\le y \le 1 \\
      0&, o.w.
    \end{cases}
  =
  \begin{cases}
    4y^3 &,  0\le y \le 1 \\
    0&, o.w.
  \end{cases}
$

this is the part I'm not sure about:
$
  f_{_X}(x)\cdot f_{_Y}(y) =
  \begin{cases}
    (4x-4x^3)\cdot 4y^3 &  ,0 \le x, y \le 1 \\
    0 &, o.w.
  \end{cases}
$

so $f_{_{X,Y}}(x,y) \neq f_{_X}(x)\cdot f_{_Y}(y)$.

It's doesn't seem to be the right way to show the inequality. Also, what if the  equality was true? it's possible that the same function has two ways to be written.

So my question is what should I do to prove or disprove that $ \forall x,y \in \mathbb R: f_{_{X,Y}}(x,y) = f_{_X}(x)\cdot f_{_Y}(y)$, not only in this case but in the more general case.

Please keep in mind that I'm an undergraduate taking my first probability course, so please try to keep it simple.

Thanks

答:
Note that the joint support

$$0\leq x \leq y\leq1$$

is a triangle.

A necessary condition for independence is that the support is Rectangular...thus $X$ and $Y$ cannot be independent


问:
How to prove $\sum\limits_{n=1}^{\infty}\frac{\sin n}n=\frac{\pi-1}{2}$
One of my classmates challenged me to solve $\displaystyle\sum\limits_{n=1}^{\infty}\frac{\sin n}n=\;?$

With a simple c program I found that $\displaystyle\sum\limits_{n=1}^{1048576}\frac{\sin n}n\approx1.070796$.
Later I found $\displaystyle1.070796\approx\frac{\pi-1}{2}$.
My classmate told me I guessed right, but he ask me to prove it, and he gave me a hint that $\displaystyle e^{i\theta} = \cos\theta + i \sin\theta$, though I can't see the relationship between the question and the hint.

So how to prove $\displaystyle\sum\limits_{n=1}^{\infty}\frac{\sin n}n=\frac{\pi-1}{2}$?

答:
Make use of the following fact: For $\vert z \vert \leq 1$ and $z \neq 1$, we have
$$\log(1-z) = - \sum_{k=1}^{\infty} \dfrac{z^k}k$$
Take $z=e^i$ and look at the imaginary part.

We hence have
$$\log(1-e^i) = - \sum_{k=1}^{\infty} \dfrac{e^{ik}}k \implies \text{Imag}(\log(1-e^i)) = \text{Imag}\left(- \sum_{k=1}^{\infty} \dfrac{e^{ik}}k \right)$$
which gives us
$$\sum_{k=1}^{\infty} \dfrac{\sin(k)}k = - \text{Imag}(\log(1-e^i))$$
$$\log(1-e^i) = \log \left(2\sin^2(1/2) - 2i \sin(1/2) \cos(1/2)\right) = \log(2\sin(1/2)e^{-i \pi/2}e^{i/2})$$
Hence,
$$\text{Imag}(\log(1-e^i)) = \dfrac{1-\pi}2$$ which gives us
$$\sum_{k=1}^{\infty} \dfrac{\sin(k)}k = \dfrac{\pi-1}2$$


问:
Finishing the solution to this functional equation: $ f\left(x^2+y\right)=f\bigl(f(x)-y\bigr) + 4f(x)y $
I am trying to solve the below functional equation for $f$ (where $f: \mathbb{R} \to \mathbb{R}$):
$$ f\left(x^2+y\right)=f\bigl(f(x)-y\bigr) + 4f(x)y \text. $$
(Source: https://web.evanchen.cc/handouts/FuncEq-Intro/FuncEq-Intro.pdf: "Iran TST 1996")

Here's what I've got so far:

Substituting $y=0$ we have $f\left(x^2\right) = f\bigl(f(x)\bigr)$.

Substituting $y=f(x)$ we have
$$f\left(x^2+f(x)\right) = f(0) + 4f(x)^2 \text.$$
$$f(0) = 0 \text. $$

Letting $x=0$ we can show the function is even.

Finally letting $y = -x^2$ we can simplify to yield
$$f(x) \left(x^2 - f(x)\right) = 0 \text. $$

And this is where I'm stuck. Given that it's an Olympiad I'm fairly sure that actually the only solutions are $f(x) = x^2$ and $f$ is identically zero and that there will not be any pathological solutions that have discontinuities - though I have not been able to show this.

My ideas that I've tried:

Showing $f$ is injective when not identically zero (on $\mathbb{R}^+$ that is), this doesn't seem to go anywhere useful.

Showing that $f$ is surjective (again onto $\mathbb{R}^+$) when not identically zero, as if that is the case we always must choose $x^2$ instead of zero other we miss out a positive number. However I haven't been able to find any way to show this either.

Use inequalities to show that if $f$ is not identically zero then the inequality $f(x) \ge 0$ that we have from our current information (either $f$ is zero or the square of a real) is strict when $x \ne 0$, but again I couldn't do anything with this either.

答:
Here's another way to avoid the point-wise trap:
Let $P(x,y)$ denote the given assertion, suppose there exist $a,b\ne 0$ such that $f(a)=0, f(b)=b^2$, then $$P(a,-x) : f(a^2+x)=f(-x)=f(x)$$ so $a^2$ is a period of the function $f$. However, we also have $$P(b,a^2) : f(b^2+a^2)=f(b^2-a^2)+4a^2b^2=f(b^2+a^2)+4a^2b^2 \implies 4a^2b^2=0$$ which means either $a=0$ or $b=0$ which is impossible. So, this means that $f(x)\equiv x^2$ and $f\equiv 0$.


答:
The maximal ideal space is in bijection with the set of characters on $A$; if $M$ is a maximal ideal, then the correspondence is given by $M \leftrightarrow \phi$, where the character $\phi: A \rightarrow \mathbb{C}$ is a nontrivial algebra homomorphism whose kernel is precisely $M$.

Because $\phi$ is multiplicative and nonzero, then one can prove that the operator norm of $\phi$ when regarded as a linear operator is at most $1$, so hence each $\phi \in K$. By passing through the above bijection, then it can be regarded that $\Delta \subset K$.

Title: The maximal ideal space of $A$ is contained in the unit ball of $A^\ast$

Body: On page 281 of Rudin's book Functional Analysis, let $\Delta$ be the maximal ideal space of a commutative Banach algebra $A$, $K$ be the norm-closed unit ball of $A^*$, then $\Delta\subset K$ (by Theorem 10.7). Can someone give me more detail?</p>


问:
Where are the formulas for frequency $\omega=\sqrt{\lambda_1 \lambda_2}$, period $T=2\pi/(\sqrt{\lambda_1 \lambda_2)}$
Where are the formulas for frequency $\omega=\sqrt{\lambda_1 \lambda_2}$, period $T=2\pi/(\sqrt{\lambda_1 \lambda_2)}$ got from?

For e.g. Lotka-Volterra system and associated Jacobian and eigenvalues/-vectors?

A Jacobian for a predator-prey model is:

$J=\begin{bmatrix}
\alpha-\beta y& -\beta x\\
\delta y & \delta x - \gamma
\end{bmatrix}$

At $(\gamma/\delta, \alpha/\beta)$ this is:

$J=\begin{bmatrix}
0 & -\frac{\beta \gamma}{\delta}\\
\frac{\alpha \delta}{\beta} & 0
\end{bmatrix}$

For which the eigenvalues are

$\lambda_1=i\sqrt{\alpha \gamma}, \lambda_2=-i\sqrt{\alpha \gamma}$

Then it's said that:

As the eigenvalues are both purely imaginary and conjugate to each
others, this fixed point is elliptic, so the solutions are periodic,
oscillating on a small ellipse around the fixed point,

therefore the frequency and period are the above.

https://en.wikipedia.org/wiki/Lotka%E2%80%93Volterra_equations#Second_fixed_point_(oscillations)

答:
A matrix $$\pmatrix{0&-u\\v&0},~~~u,v>0,$$ has eigenvalues $λ_{1,2}=±iω$ where $ω=\sqrt{uv}$. As a conjugate pair on the imaginary axis or as consequence of the Viete formulas for the characteristic equation, you get also back $λ_1λ_2=ω^2=uv$, but that should only be a secondary insight.

The linearized system
$$
\dot x = -uy\\
\dot y = vx
$$
can be transformed into the scalar second order equation $\ddot x=-uvx=-ω^2x$ which is a harmonic oscillator with frequency $ω$ and thus period $2\pi/ω$.


问:
Subdifferential of locally Lipschitz function locally bounded
Let $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ be locally Lipschitz near $x \in \mathbb{R}^{n}$ with constant $L>0$, i.e. there exist $\varepsilon > 0$ and $L>0$ such that

\begin{equation}
|f(y) - f(z)| \leq L \| y-z\| \qquad \forall y,z \in B_{\varepsilon}(x) \, .
\end{equation}

The generalized Clarke subdifferential $\partial f(x)$ is defined by

\begin{equation}
\partial f(x):=\left\{g \in \mathbb{R}^{n} \mid f^{\circ}(x ; d) \geq g^{\top} d \quad \text {for all } d \in \mathbb{R}^{n}\right\}
\end{equation}

where $f^{\circ}(x ; d)$ denotes the generalized directional derivative which is defined by

\begin{equation}
f^{\circ}(x ; d):=\limsup _{y \rightarrow x, \, \, t \downarrow 0} \frac{f(y+t d)-f(y)}{t}\, .
\end{equation}

It is known that $\partial f(x)$ is a nonempty, convex, compact set such that $\partial f(x) \subset B_{L}(0)$ . According to Clarke and many other authors the Lipschitz property and the fact that $\partial f(x) \subset B_{L}(0)$ imply that for all $x_{i} \in B_{\varepsilon}(x)$

\begin{equation}
\left\|g_{i}\right\| \leq L \text { for all } g_{i} \in \partial f\left(x_{i}\right) \,.
\end{equation}

This is always used without further explanation and therefore I assume that this is beyond trivial but I just dont see it. So I am sorry in advance and very thankful if someone could help me with this.

答: Assuming $B_\epsilon(x)$ is the open ball, for all $y\in B_\epsilon(x)$, there exist $0<\delta\leq\epsilon$ such that $B_\delta(y)\subset B_\epsilon(x)$. Hence $$\lvert f(z_1)-f(z_2)\rvert \leq L\lVert z_1-z_2\rVert, \qquad \forall z_1,z_2\in B_\delta(y).$$ Hence $f$ is locally Lipschitz near $y$ and $\partial f(y)\subset B_L(0)$ by the property you mentioned. Does that seem right?


问:
Show that function constant outside of closed ball is globally Lipschitz
This is a follow-up to my previous question.

Let $I \subseteq \mathbb{R}$, $D \subseteq I \times \mathbb{R}^n$ open, $(t_0,y_0) \in D$. Assume that $f: D \to \mathbb{R}^n$ is locally Lipschitz w.r.t. to $y$ on $D$.

Define $Q:=[t_0,t_0+h] \times \bar{B}_c(y_0) \subset U(t_0,y_0)$ where $U(t_0,y_0)$ is the neighbourhood of $(t_0,y_0)$ from the local Lipschitz condition.

I am trying to prove that the function

$\bar{f}(t,y)=f(t,\Psi(y))$ where $\Psi(y)=y_0+\frac{c}{max\{c,\lVert y-y_0\rVert\}}(y-y_0)$ is globally Lipschitz w.r.t. $y$ on $[t_0,t_0+h] \times \mathbb{R}^n$.

I have been able to prove this for the cases

i) $y_1,y_2 \in \bar{B}_c(y_0)$:

$\lVert \bar{f}(t,y_1)-\bar{f}(t,y_2)\rVert =\lVert f(t,y_1)-f(t,y_2)\rVert \leq K_{t_0,y_0} \lVert y_1-y_2 \rVert$ where $K_{t_0,y_0}$ is the local Lipschitz constant on $Q$.

ii) $y_1,y_2 \notin \bar{B}_c(y_0)$:

$\lVert \bar{f}(t,y_1)-\bar{f}(t,y_2)\rVert =\lVert f(t,y_0+\frac{c}{\lVert y_1-y_0 \rVert}(y_1-y_0))-f(t,y_0+\frac{c}{\lVert y_2-y_0 \rVert}(y_2-y_0))\rVert \leq c \lVert \frac{y_1-y_0}{\lVert y_1-y_0 \rVert} - \frac{y_2-y_0}{\lVert y_2-y_0 \rVert} \rVert \leq \frac{2c}{\lVert y_2-y_0\rVert} \lVert y_1-y_2\rVert < 2 \lVert y_1-y_2\rVert$

Note that I have used the inequality $\left\|\frac{a}{\|a\|}-\frac{b}{\|b\|}\right\|
\le\left\|\frac{a}{\|a\|}-\frac{a}{\|b\|}\right\|+\left\|\frac{a}{\|b\|}-\frac{b}{\|b\|}\right\|
=\|a\|\,\frac{|\|b\|-\|a\||}{\|b\|\,\|a\|}+\frac{\|a-b\|}{\|b\|}
\le\frac{2}{\|b\|}\|a-b\|$.

Now I am stuck on case

iii) $y_1 \in \bar{B}_c(y_0)$, $y_2 \notin \bar{B}_c(y_0)$

$\lVert \bar{f}(t,y_1)-\bar{f}(t,y_2)\rVert =\lVert f(t,y_1)-f(t,y_0+\frac{c}{\lVert y_2-y_0 \rVert}(y_2-y_0))\rVert \leq \lVert y_1 - (y_0+\frac{c}{\lVert y_2-y_0\rVert}(y_2-y_0)) \rVert$

Intuitively, the function $\bar{f}$ should be globally Lipschitz since it extrapolates $f$ constantly outside of $\bar{B}_c(y_0)$, but I cannot seem to prove case iii) gives some upper bound in terms of a constant and $\lVert y_1-y_2 \rVert$ as in the cases i) and ii).

Can anyone please help me out here?

Thanks a lot!

答:
To simplify the notation, let $y_0 = 0$.
All in all, it is enough to prove that
$$
|y_1| \leq c, \
|y_2| > c
\quad\Longrightarrow\quad
|y_1 - z| \leq |y_1-y_2|,
\qquad z := c\, \frac{y_2}{|y_2|}\,.
$$
It holds that
$$
|y_1 - y_2|^2 = |(y_1 - z) + (z - y_2)|^2
= |y_1 - z|^2 + 2 (y_1 - z) \cdot (z - y_2) + |z-y_2|^2\,.
$$
On the other hand, $(y_1 - z) \cdot (z - y_2) \geq 0$ (if you think the two vectors applied in $z$, then both are pointing inside the ball), so that the claim follows.


问:
Kernel of Quadratic Field Norm
Let $D > 0$ be squarefree and consider the field $\mathbb{Q}(\sqrt{D})$. Then the field norm $N : \mathbb{Q}(\sqrt{D}) \to \mathbb{Q}^\times$ is given by $N(a + b\sqrt{D}) = a^2 - D b^2$. Let $K$ be the kernel of this map, consisting of elements of norm $+1$. I'm looking for an example where $K$ is not isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \bigoplus_{i = 0}^\infty \mathbb{Z}$ as abelian groups.

The motivation is the Pell conic group $C$ defined on the curve $x^2 - Dy^2 = 1$. If $D$ is a square of a rational number, then $x \mapsto (\frac{x+x^{-1}}{2},\frac{x-x^{-1}}{2\sqrt{D}})$ gives an isomorphism $\mathbb{Q}^{\times} \cong C$, but otherwise we have the isomorphism $K \cong C$ where $K$ is the kernel of the norm map. I am wondering whether the group structure of $C$ alone can detect the irrationality of a square root of a natural number. Below I've included my current thoughts on this question.

Let's consider $D = 2$. Then the group of units of $\mathbb{Z}[\sqrt{2}]$ is generated by powers of $\pm(1 + \sqrt{2})$, so it is isomorphic $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}$. Since $\mathbb{Z}[\sqrt{2}]$ is a UFD, we have that any element of $\mathbb{Q}(\sqrt{2})^\times$ can be written as a unique product of primes and units of $\mathbb{Z}[\sqrt{2}]$, so $\mathbb{Q}(\sqrt{2})^\times \cong (\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}) \times \bigoplus_{p \text{ prime}} \mathbb{Z}$. Now consider the map $h : \mathbb{Q}(\sqrt{2})^\times \to K$ given by $h(a) = \frac{a^*}{a}$ where $a^*$ denotes the conjugate. My understanding is that Hilbert's Theorem 90 implies this map is surjective. Its kernel is $\mathbb{Q}^\times$, so $K$ should be something like $K \cong (\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}) \times \bigoplus_{p \text{ prime, } \notin \mathbb{Q}} \mathbb{Z}$. This paper uses a similar map to obtain the analogous result for $\mathbb{Q}(i)$. It seems like similar reasoning should apply to any UFD.

Now consider a non-UFD, say $D = 10$. Here we have $2 \cdot 5 = \sqrt{10} \cdot \sqrt{10}$ which at first seems like it might change the structure of $\mathbb{Q}(\sqrt{10})^\times$ compared to $\mathbb{Q}(\sqrt{2})^\times$. However, I think you can throw out $5$ as a generator because you can get it from the other irreducibles as as $(\sqrt{10})^2/2$. It seems like $\mathbb{Q}(\sqrt{10})^\times$ has a chance of being isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \bigoplus_{i = 0}^\infty \mathbb{Z}$ despite not being a UFD. Then applying $h$ should then just kill the factors corresponding to primes in $\mathbb{Q}$, so you might get the same thing for $K$ as with $D = 2$.

答:
In what follows, $L$ is any number field.

We have the following theorem.

Thm 1. Let $L$ be a number field, and let $N:L^\times\to \mathbb{Q}^\times$ be the norm map.

Then $\ker(N)\simeq \mathcal{O}_L^\times\cap \ker(N) \times \mathbb{Z}^{(E)}$, for some countable set $E$.

Dirichlet's theorem says that $\mathcal{O}_L^\times $  is finitely generated, thus so is $\mathcal{O}_L^\times\cap \ker(N) $. Note that the torsion part of this last group is the intersection of the torsion group of $\mathcal{O}_L^\times$ , which is the (cyclic) group $\mu_L$ of roots of $1$ contained in $L$,  with $\ker(N)$. Thus:

Corollary. $\ker(N)\simeq  \mu_L\cap \ker(N)\times \mathbb{Z}^{(E')},$ where $E$ is countable and $\mu_L$ is the ( cyclic )group of roots of unity contained in $L$.

Thus, if you stick to subfields of $\mathbb{R}$ (such as real quadratic fields), you won't get any counterexample. You need $L$ to contain at least a root of unity of order $n\geq 3$.

Example. For all $n\geq 3$, you can take $L=\mathbb{Q}(\zeta_n)$. Then norms are always positive (because complex conjugation is a non trivial automorphism), and $\mu_L=\langle \mu_n\rangle\simeq \mathbb{Z}/n\mathbb{Z}$ .

Let's now move to the proof of the theorem.

For $z\in L^\times,$ I will denote by $(z)$ the fractional ideal $z\mathcal{O}_L$. Abusing notation, I will also denote by $N(I)$ the norm of any fractional ideal $I$.

If $z$ has norm $1$, so is $(z)$, since $N((z))=\vert N(z)\vert$.

Let $\Gamma$ be the set of principal fractional ideals with norm $1$, and let $f:z\in \ker(N)\mapsto (z)\in\Gamma$.

Then $\Gamma$ is an abelian group, and $f$ is a group morphism.

Lemma. $f$ has  kernel $\mathcal{O}_L^\times\cap\ker(N)$.

Proof. Assume that $z\in\ker(N)$ satisfies $(z)=z\mathcal{O}_L=\mathcal{O}_L$. Then $z\in\mathcal{O}_L$, and since $\mathcal{O}_L$ is a domain, $z$ must be a unit.
Hence  $z\in \mathcal{O}_L^\times \cap\ker(N)$. Conversely, elements of  $\mathcal{O}_L^\times \cap\ker(N)$ lie in $\ker(f)$.

Thm 2. The group $\Gamma$ of principal fractional ideals with norm $1$ is  isomorphic to a subgroup of $ \mathbb{Z}^{(\mathcal{P})}$, where $\mathcal{P}$ is the set of nonzero prime ideals of $\mathcal{O}_L$.

Assume thm 2 is proved, and let us prove thm $1$.

By the lemma, we have an exact sequence of abelian groups  $1\to \mathcal{O}_L^\times\cap\ker(N)\to \ker(N)\to Im(f)\to 1$.

By a theorem of Dedekind, if $S$ is any set, any subgroup of $\mathbb{Z}^{(S)}$ is free, and has a basis indexed by a subset of $S$. By Thm2 and this result, $\Gamma$ is then free, and isomorphic to $\mathbb{Z}^{(X)}$ for some subset $X$ of $\mathcal{P}$. Now, $Im(f)$ is a subgroup of $\Gamma$, so we can apply the same result again to get that $Im(f)\simeq  \mathbb{Z}^{(E)}$, for some countable set $E$.

Now by the lemma,  we have an exact sequence of abelian groups $1\to \mathcal{O}_L^\times\cap \ker(N)\to \ker(N)\to Im(f)\to 1$. Since $Im(f)$ is free, this sequence splits, and we are done.

We now have to prove Thm 2 above.

If $I$ is a  nonzero fractional ideal, write $I=\displaystyle  \prod_{\mathfrak{p}}\mathfrak{p}^{r_{\mathfrak{p}}}$, where the $ r_{\mathfrak{p}}'s$ are integers, which are all zero except for a finite number of them. Note that these integers are unique.

Now we have  $N(\mathfrak{p})=p^{f_{\mathfrak{p}}}$, where $p$ is the unique prime number dividing $\mathfrak{p}$ , so $N(I)=1$ if and only if $\sum_{p\mid\mathfrak{p}}f_{\mathfrak{p}}r_{\mathfrak{p}}=0$ for all prime $p$.

Now, we have to understand on which conditions on the $ r_{\mathfrak{p}}'s$ such an $I$ is principal.

Fix a group isomorphism $\mathbb{Z}/d_1\mathbb{Z}\times \cdots\times\mathbb{Z}/d_k\mathbb{Z}\simeq \mathcal{C}\ell(L)$ (the class group of
$L$
), and let $C_1,\ldots,C_k$ be the images of $(1,0,\ldots,0), (0,1,0,\ldots, 0),\ldots,(0,\ldots,0,1)$ in the class group. Then any element of the class group may be written as $m_1\cdot C_1+\cdots+m_k C_k, m_1,\ldots,m_k\in\mathbb{Z}$, and such a linear combination is zero if and only if $d_i\mid m_i$ for  $i=1,\ldots,k$.

Now for each $\mathfrak{p}$, write $[\mathfrak{p}]=\displaystyle\sum_{i=1}^k a_{i,\mathfrak{p}}\cdot C_i$.
Then $[I]=\displaystyle\sum_{i=1}^k (\sum_{\mathfrak{p}} a_{i,\mathfrak{p}}r_\mathfrak{p})\cdot C_i$ ,and $I$ is principal if and only if $[I]=0$ if and only if $d_i\mid (\sum_{\mathfrak{p}} a_{i,\mathfrak{p}}r_\mathfrak{p})$ for $i=1,\ldots,k$.

Hence, denoting by $\mathcal{P}$ the set of nonzero prime ideals of $\mathcal{O}_L$, we see that $\Gamma$ is canonically isomorphic to the abelian group $$\{(r_\mathfrak{p})_\mathfrak{p}\in\mathbb{Z}^{(\mathcal{P})}\mid \sum_{p\mid\mathfrak{p}}f_{\mathfrak{p}}r_{\mathfrak{p}}=0 \mbox{ for all prime }
 p \mbox{ and } d_i\mid (\sum_{\mathfrak{p}} a_{i,\mathfrak{p}}r_\mathfrak{p})\mbox{ for }i=1,\ldots,k\},$$

which is a subgroup of $\mathbb{Z}^{(\mathcal{P})}$.

Title: Kernel of Quadratic Field Norm


问:
Orthogonal complement of a subspace of constant functions

Let $X$ be a finite set and $F(X)$ the space of all functions $f:X \to \mathbb{R}.$ If $\langle \cdot, \cdot \rangle \colon F(X)\times F(X)\to \mathbb{R}, \langle f,g\rangle = \sum_{x\in X} f(x)g(x)$ defines an inner product and $V\subset F(X)$ is the subspace of all constant functions of $F(X)$ show that $V^\bot = \{ f\in F(X) \colon \sum_{x\in X} f(x) = 0 \}.$

I'm slightly confused about how to approach the problem. It seems that what  I would like to achieve is to show that for every $v \in V^\bot$ and every $f \in V$ I have that $\langle v,g\rangle=0$? Any hints would be appreciated. Another approach would probably be that I would indeed show that all vectors of $V^\bot$ would be orthogonal to the vectors of $V$ and then show that $V^\bot \subset V$ and $V \subset V^\bot$?

答:
Hints:

Let $\bf 1$ denote the constant $1$ function. Show that $V={\rm span}({\bf 1})$.

So, $V^\perp=\{{\bf 1}\}^\perp$, i.e. $f\perp V\iff f\perp{\bf 1}$.

Calculate $\langle f,{\bf 1}\rangle$ for generic $f\in F(X)$.

Conclude.


问:
Absolutely continuous then $\nu(A)>0 \Rightarrow \mu(A)>0$
Let $(X,\mathcal{M})$ be a measurable space and $\nu,\mu:\mathcal{M} \to \mathbb{R}$ be additive set functions. I would like to prove or disprove that if $\nu$ is absolutely continuous in terms of $\mu$, then for any $A\in\mathcal{M}$, $\nu(A)>0 \Rightarrow \mu(A)>0$ holds.

What I did:

So firstly, I tried to write down definitions.

Def 1: Let $\Phi: \mathcal{M} \to \mathbb{R}$ be an additive set function. we define

$\Phi^+(E) := \sup_{A\subset E}\Phi(A)$

$\Phi^-(E) := (-\Phi)^+(E)$

$|\Phi|(E) := \Phi^+(E)+\Phi^{-}(E)$

Def 2: Let $\nu,\mu:\mathcal{M} \to \mathbb{R}$ be additive set functions. $\nu$ is said to be absolutely continuous in terms of $\mu$ if $|\mu|(E)=0 \Rightarrow |\nu|(E)=0$ holds for all $E\in\mathcal{M}$.

Note that those definitions came from different sources, so something may be inconsistent. Feel free to correct them if something is wrong.

After I wrote down the definitions, I tried to prove it by contradiction: that is, I first assumed there exists $A\in \mathcal{M}$ such that $\nu(A)>0$ but $\mu(A)\leq 0$. However, I was unsuccessful in this and lost now.

答:
Let $\nu$ be any additive set function that satisfies $\nu(A) > 0$ for some $A \in \mathcal M$. Let $\mu = -\nu$. Then $\nu^+ = \mu^-$ and $\nu^- = \mu^+$, which means that
$$|\nu| = \nu^+ + \nu^- = \mu^- + \mu^+ = |\mu|.$$
This shows that $\nu$ is absolutely continuous with respect to $\mu$ (and vice versa). However, if $A$ is any set for which $\nu(A) > 0$, then by definition $\mu(A) < 0$, so this is a counterexample to the claim.


问:
Speed - way problem
Car Yellow which is of $180km/h$ and Car Purple is of $80km/h$ start going on the same way from the point A. When the more speedy car(Car Yellow) comes to point B and turn immediately(nonstop) back and come to point A, Car Purple reaches point C. What is ${AB} / {BC}$ ?

I don't understand what's wrong with me. If I say $|AB| = x $ and $|BC| = y $, and assume Car Yellow completes $2x$ way at $t$ time, at the same $t$ time I expect Car Purple should be at point C.

I find out then, $x = 90t$ and $x + y = 80t$. I don't understand.

答:
$x=90t \qquad x+y=80t \to y=-10t \qquad \frac{\lvert AB \rvert}{\lvert BC \rvert}=9$

Point C is actually between A and B because Yellow is more than twice as fast as Purple, that's why $y$ is negative.

Title: Speed - way problem


问:
Closed form solution to a linear program
For a given set of nonnegative weights $a_1, \ldots, a_K$, can we find the closed-form solution to the following optimization problem or the lowest computational complexity solution?

$\begin{equation}
\underset{  \{p_k > 0 \} }{\min} \quad \sum_{k=1}^K p_k a_k \quad s.t. \sum_{k=1}^K p_k = P_{max}
\end{equation}
$

Thank you very much!

答: Set $p_k=P_\max$ for the index $k$ for which $a_k$ is minimal, set all other coefficients to zero.


问:
Prove that $QBYZ$ is cyclic
Circles $\omega_1$ and $\omega_2$intersect each other at points $A$ and $B$. Point $C$ lies on
the tangent line from $A$ to $\omega_1$ such that $\angle ABC = 90^{°}$. Arbitrary line
passes through $C$ and cuts $\omega_2$ at points $P$ and $Q$. Lines $AP$ and $AQ$ cut
$\omega_1$ for the second time at points $X$ and $Z$ respectively. Let $Y$ be the foot
of altitude from $A$ to $PQ$.

Prove that $QBYZ$ is cyclic.

I proved that $AYBC$ and $PYBX$ are cyclic but I have no idea how to prove that $QBYZ$ is cyclic.

答: Since quadrilaterals $AZBX$ and $PYBC$ are cyclic, we have $$\angle QYB=\angle PXB=\angle AXB=\angle BZQ$$ which implies that quadrilateral $QBYZ$ is cyclic.

Title: Prove that $QBYZ$ is cyclic


答:
You could also use $t = \frac{s}{v}$ where distance s and velocity v, $|AB| = x$ and $|BC| = y$.
For the yellow car it is $v_1 = 180 \frac{km}{h}$, therefore $$ t = \frac{2x}{v_1}$$ and for the purple car $v_2 = 80 \frac{km}{h}$ $$ t = \frac{x+y}{v_2}$$
Then you see how to eliminate t: $$ t = \frac{2x}{v_1} = \frac{2|AB|}{v_1}= \frac{x+y}{v_2} = \frac{|AB| + |BC|}{v_2}$$

The solution tells you that C must be between A and B.

Title: Speed - way problem

Body:
Car Yellow which is of $180km/h$ and Car Purple is of $80km/h$ start going on the same way from the point A. When the more speedy car(Car Yellow) comes to point B and turn immediately(nonstop) back and come to point A, Car Purple reaches point C. What is ${AB} / {BC}$ ?

I don't understand what's wrong with me. If I say $|AB| = x $ and $|BC| = y $, and assume Car Yellow completes $2x$ way at $t$ time, at the same $t$ time I expect Car Purple should be at point C.

I find out then, $x = 90t$ and $x + y = 80t$. I don't understand.


问:
How do you respond to "I was always bad at math"?
Here in the U.S., it is my experience that over 75% of adults I meet socially will volunteer that phrase or a variation upon learning that I am a mathematician.  I find this frustrating, since almost nobody would  brag about being bad at history or English or really any other subject.   Normally I change the subject, or perhaps say they must have had a bad classroom experience, but that feels unsatisfying and turns the conversation in an ugly direction.

Surely some of the bright minds here on Math.SE have found a good response. Note that I'm not looking to  alienate a new acquaintance, so let's keep it positive.

答:
I usually say something about math being just another language, and that it takes a lot of exposure, practice, and work to become fluent. A very different sort of language, but analogous nonetheless. This usually gets people interested, because it's not a way that they are used to thinking about math.  I don't know how widely this resonates with other math types, but it feels this way to me: whether or not it ought to be called a "language" I don't know, but in terms of how our brains learn it, I think the analogy holds.

Part of why this helps I think is to move sideways from the usual directions, like math is adding up big lists of numbers, or math is beyond the grasp of normal minds, and relating it to something most people have experience with and a whole different set of conceptions about.


问:
Having trouble understanding difference between polyhedron and polytope
Hi i´m reading a pdf about linear programming and i´m having trouble understanding the difference between a polyhedron and polytope between those two definitions

A polyhedron P ⊆ $R^{n}$
is the set of all points x ∈ $R^{n}$
that satisfy a finite set of linear inequalities.
Mathematically,
P = {x ∈ $R^{n}$: Ax ≤ b}
for some matrix A ∈ R
m×n and a vector b ∈ $R^{m}$.

And the definition of polytope is the next one

A polyhedron is called a polytope if it is bounded, i.e., can be enclosed in a ball of
finite radius.

So my problem isn´t the definition of polyhedron already bounded , and also are all polyhedron in these context convex?

And for example

{x ∈ $R^{n}$: Ax ≤ b} if i guide myself from the given definitions i can only assure this is a polyhedron but not a polytipe since it isn't technically bounded but is convex and {x|$x\in\mathbb{R}$} isn't any of those two.

Any insight would be helpful

答:
Consider a set of linear inequalities whose solution set extends to infinity, forming what is called a polyhedral bowl.

This bowl is a polyhedron because it is the solution set. But it is not a polytope because it cannot be enclosed in a sphere of finite radius, i.e. the equalities do not fully bound it.


问:
Solve $x^3-1=2y^2$ integer solution
find the equation  integer solution
$$x^3-1=2y^2$$

I have read this proof:proof, have simple methods? Because this problem is a Olympiad problem. So maybe have other methods to solve it
since $x>0.y>0$ so
$$(x-1)(x^2+x+1)=2y^2$$ since
$(x-1,x^2+x+1)=1$ or$3$.so

(1):if $x-1=2u^2,x^2+x+1=v^2,y=uv,u>0,v>0$,then wehave
$$(2v)^2=(2x+1)^2+3\Longrightarrow (2v-2x-1)(2v+2x+1)=3$$
then
$$\begin{cases}
2v+2x+1=3\\
2v-2x-1=1
\end{cases},~~\begin{cases}
2v+2x+1=1\\
2v-2x-1=3
\end{cases}$$
this two case has no solution $x>0$

(2): $$x-1=6u^2,x^2+x+1=3v^2,y=3uv,u,v>0$$

then $$(2v)^2-3(4u^2+1)^2=1
$$ it's pell equation, maybe this following I can't work. Thanks for your help!

答:
As suggested in the proof you link, it is easier to pass to $\Bbb{Z}[\sqrt{-2}]$ right from the start:

If $x$ and $y$ are integers satisfying $x^3-1=2y^2$, then
$$x^3=1+2y^2=(1+y\sqrt{-2})(1-y\sqrt{-2}),$$
where the two factors on the right hand side are coprime because their sum is $2$ and their product is odd. Then because $\Bbb{Z}[\sqrt{-2}]$ is a unique factorization domain with trivial unit group it follows that
$$1+y\sqrt{-2}=(a+b\sqrt{-2})^3,$$
for some integers $a$ and $b$. Expanding the right hand side and comparing coefficients then shows that
$$1=a^3-6ab^2=a(a^2-6b^2),$$
which clearly has no integral solutions apart from $(a,b)=(1,0)$, corresponding to $(x,y)=(1,0)$.

Title: Solve $x^3-1=2y^2$ integer solution


问:
What is the general pattern for the number of inputs for which this algorithm performs t comparisons?
I am trying to perform an average case analysis over this decision tree, which contains N items where N is a multiple of 4. I am struggling to make a generalization about the number of inputs which perform comparisons.

A sketch of the Decision Tree

Here is a table of the number of comparisons performed, the maximum number of comparisons K can be given by K = 2 + n/4.

Number of comparisons performed for N

答:
The number of decisions for some input is the length of the path from that input node to the root.
If we go from $N$ to $N+4$, we push the original tree down to the right, i.e., increase each path length by $1$, and add four new nodes with path lengths $1$, $2$, $3$, $3$. Thus the number $a_{i,j}$ in your table in row $i$ of column $j$ is simply $a_{i-1,j-1}+a_{i,1}$.

The sum of all numbers of decisions accordingly behaves as follows:
$$ s(N+4)=9+s(N)+N$$
which easily leads to
$$ s(N)=\frac18N^2+\frac 74 N.$$
Divide by $N$ to obtain the average as
$$\frac{s(N)}N= \frac 18N+\frac 74.$$


问:
How to compute the closure of a set of binary strings in term of the "AND" and "OR" operators?
Given a set of $n$ binary strings of length $m$, how many binary strings at least should we add to the set to make sure that the binary "AND" and "OR" operators are closed on the set? The output of the problem is the number of binary strings needed to add.

For example, given the set {001, 010, 110}, we should at least add {000, 011, 111} to the set to make both "AND" and "OR" operators closed. So the output is 3.

Is there any good algorithm or idea related to the problem? Thanks a lot.

答:
I believe you are trying to compute the "sublattice of the boolean lattice of length m" generated by your n binary strings. I don't know a fast way to compute sublattices.  In general, it is exponential in n.

Here is an example:  For any positive integer n, and any positive integer m ≥ n, start with the binary strings of length m which are zero in position i for all i > n, and which are nonzero in all but one other place.  There are exactly n such strings.  How many strings do you have to add?  2n−n, since every string with exactly one nonzero entry (in the first n positions) is an AND of n−1 of the initial strings.  This output is exponential length in n.

If you included "not" or otherwise included "xor", you would suddenly upgrade to finding "ideals in the boolean algebra" and this becomes linear algebra, and things like gaussian elimination should make the problem quadratic in n.

Of course the output itself could still be exponential in n, so you have to settle for a different, but very useful, kind of output: a basis.

As pointed out by Ross Millikan, the critical problem for a practical solution is a better data-type for the output.  Just listing all the words is going to be exponential in general, but sometimes one could produce ridiculously fast membership test functions (and product them ridiculously quickly) — in my example, the membership function simply checks that the positions from n+1 to m are all 0.

However, no data-type can work magic.  Depending on how many possible answers there are, the data-type will have to be at least a certain size. Queyranne–Tardella (2008) attempt to answer the question of how many sublattices are there and in fact study exactly the original problem.  They provide a data-type for the output and prove it is memory optimal up to a constant factor:

Queyranne, Maurice; Tardella, Fabio.
  "Sublattices of product spaces: hulls, representations and counting."
  Discrete Math. 308 (2008), no. 9, 1508–1523.
  MR2392592
  DOI:10.1016/j.disc.2007.03.080


问:
In the solution of $x + e^x = 2$, how does $ln(W(e^2))$ become $2 - W(e^2)$?
$x + e^x = 2$

Take $e$ of both sides

Apply $W(x)$ function then $\ln$ of both sides.

This gives me an answer of $\ln(W(e^2))$, but when I put the equation into Wolfram, it gives $2-W(e^2)$ as answer.

How does this happen?

Body:
$W(z)e^{W(z)}=z$, therefore $$e^{2-W(e^2)}=\frac{e^2}{e^{W(e^2)}}=\frac{e^2}{e^2/W(e^2)}=W(e^2)$$

Which incidentally proves that, if $W(e^2)$ exists, then $W(e^2)>0$ and $2-W(e^2)=\ln W(e^2)$.


问:
Can I get some example coordinates of an equilateral triangle, but only integers?
I'm trying to implement Bresenham's Circle drawing algorithm and I intend to draw a circumscribed and inscribed circle (circumcircle and incircle) in a triangle. I need some simple ways to find coordinates of an equilateral triangle to test my code.

Any help will be appreciated. Thanks.

答: It is possible, the coordinates (a,0,0), (0,a,0) and (0,0,a) (where a is an integer) form an equilateral triangle.</p>


答:
$x=\ln( W(e^2))$ is correct solution, then you need just take

$$2-x=e^{\ln (W(e^2))}=W(e^2)$$

$$\implies 2-x=W(e^2)$$

$$\implies x=2-W(e^2)$$

So, both are equal.

Title: In the solution of $x + e^x = 2$, how does $ln(W(e^2))$ become $2 - W(e^2)$?

Body:
$x + e^x = 2$

Take $e$ of both sides

Apply $W(x)$ function then $\ln$ of both sides.

This gives me an answer of $\ln(W(e^2))$, but when I put the equation into Wolfram, it gives $2-W(e^2)$ as answer.

How does this happen?


问:
Finding $\frac{DK}{DI}$
In triangle $ABC,$ where $AB = 8, AC = 7,$ and $BC = 10,$ $I$ is the incenter. If $AI$ intersects $BC$ at $K$ and the circumcircle of $\triangle ABC$ at $D,$ find $\frac{DK}{DI}.$

I first drew a diagram, but I was unsure where to go from here.

答:
Observe that,  $ID=DB=DC$.

Now, notice that,  $\triangle AKB\sim \triangle ACD$ and hence $\frac{AD}{CD}=\frac{AB}{KB}$. Putting in $ID=CD$ will give the value of $ID$ and thereafter $DK$

Title: Finding $\frac{DK}{DI}$


问:
Two dimensional continuous random variable and independecy
Q: Let $(X,Y)$ be a two dimensional random variable with the following density function:

$f_{_{X,Y}}(x,y)=\begin{cases}8xy&,  0\le x\le y \le 1 \\ 0&,  o.w.\end{cases}$

Are $X,Y$ independent?

My attempts: in the lecture notes I have, it says that two random variables $X,Y$ are independent $\iff \forall x,y \in \mathbb R: f_{_{X,Y}}(x,y) = f_{_X}(x)\cdot f_{_Y}(y)$.

I found $f_{_X}, f_{_Y}$:

$
  f_{_X}(x)=\int_\limits{-\infty}^\infty f_{_{X,Y}}(x,y) \,dy=
    \begin{cases}
      \int_\limits x^1 8xy\,dy&,  0\le x \le 1 \\
      0&, o.w.
    \end{cases}
  =
  \begin{cases}
    4x-4x^3 &,  0\le x \le 1 \\
    0&, o.w.
  \end{cases}
$

$
  f_{_Y}(y)=\int_\limits{-\infty}^\infty f_{_{X,Y}}(x,y) \,dx=
    \begin{cases}
      \int_\limits 0^y 8xy \,dx &,  0\le y \le 1 \\
      0&, o.w.
    \end{cases}
  =
  \begin{cases}
    4y^3 &,  0\le y \le 1 \\
    0&, o.w.
  \end{cases}
$

this is the part I'm not sure about:
$
  f_{_X}(x)\cdot f_{_Y}(y) =
  \begin{cases}
    (4x-4x^3)\cdot 4y^3 &  ,0 \le x, y \le 1 \\
    0 &, o.w.
  \end{cases}
$

so $f_{_{X,Y}}(x,y) \neq f_{_X}(x)\cdot f_{_Y}(y)$.

It's doesn't seem to be the right way to show the inequality. Also, what if the  equality was true? it's possible that the same function has two ways to be written.

So my question is what should I do to prove or disprove that $ \forall x,y \in \mathbb R: f_{_{X,Y}}(x,y) = f_{_X}(x)\cdot f_{_Y}(y)$, not only in this case but in the more general case.

Please keep in mind that I'm an undergraduate taking my first probability course, so please try to keep it simple.

Thanks

答: Your approach is right.  $f_{X,Y}(x,y)=f_X(x)f_Y(y)$ a.e.  is necessary and sufficient for independence of $X$ and $Y$ provided the joint dendisty $f_{X,Y}$ exists.


问:
Does Euclid's demonstration that there are only five Platonic solids need to assume convexity?
At the end of the Elements Book XIII Euclid gives a demonstration that there are only five regular solids, i.e. 'no other figure, besides the said five figures, can be constructed by equilateral and equiangular figures equal to one another' e.g. here.  The demonstration given of this is essentially that any vertex comprised of a number of equal polygons can only be formed by 3, 4 or 5 triangles or 3 squares or 3 pentagons, the reason being that any more of any of these leads to a total angle at the vertex of 360° or more.

It's been pointed out that Euclid also needs the condition of convexity.  This was probably assumed by Euclid, not least because much of the previous discussion in Book XIII was about inscribing solids in spheres.  The essence of my question is that if we don't assume this then the argument appears potentially to fail, i.e. we can easily have more than 5 triangles meeting at a vertex if it can be nonconvex, e.g. as at one of the inner vertices of a solid similar to the small stellated dodecahedron

(from Wikipedia) but composed of 60 equilateral triangular faces rather than 60 isosceles triangle (not thinking of it as twelve intersecting pentagrams either).

The standard way to then show that there are only five Platonic solids is using Euler's formula $V-E+F=2$ applied to a polyhedron comprising $n$ $m$-gons, say, and showing that the only cases that work with the formula are those corresponding to the Platonic solids.

I note that without convexity then there are some other possible solids, e.g. the icosahedron with one vertex pushed in such as

(taken from Richeson's book 'Euler's Gem').  In a sense this is just a varied icosahedron, however, and loses the regularity that we probably want from any Platonic solid so I don't think anyone would call it a 'sixth' Platonic solid.

As above with the small stellated dodecahedron, I'm not thinking of cases where faces 'intersect' such as the Kepler-Poinsot polyhedra, thinking of these as better described by a set of smaller polygons, and also note that in such solids we don't have the same number of smaller polygons meeting at each vertex so they aren't candidates for a sixth Platonic solid.  I'm also not thinking of cases where there are holes in the solid, i.e. genus not equal to 2, or where a cube might be stuck into the middle of a face etc.

To phrase one definite question then: is there a quick, knock down argument as to why a set of nonconvex vertices formed from more than 6 equilateral triangles, or more than 4 squares or pentagons, as described above, could not form a regular solid, meaning a solid with the same number of the same regular polygons meeting at each vertex?

Of course, it doesn't 'feel' as though this would be possible, and we can justify this with Euler's formula, but Euler's formula was not known to Euclid / requires algebraic manipulation not known at that time - I'm wondering if there is a justification of convexity rather than just assuming it.

Any ideas very welcome (or requests for clarification of what I mean, of course).

答:
If you are demanding that faces do not intersect, then the idea of angular deficit offers your proof.

Count the polygonal angles around each vertex of some convex polyhedron and see how much they fall short of 360 deg. For example the cube has 3 x 90 deg corners so is 90 deg short.

Then count the vertices and multiply that by the corner's deficit. For the cube this is 8 x 90 deg = 720 deg or 2 full turns. This is the polyhedron's angular deficit.

It is the same value for all convex polyhedra. In fact it is the same for all polyhedra which have a simple sphere-like surface, such as your dimpled icosahedron (we say they are topological spheres), regardless of any valleys, dimples, etc. and is directly related to the Euler sum V - E + F = 2.

Any skew regular vertex has an angular sum of 360 deg or more, i.e. an angular excess of 0 or more. You cannot obtain an angular deficit using only skew regular vertices. Hence you cannot obtain the necessary Euler value for your solid.


答:
The $W$ function is defined so that $u=W(u)e^{W(u)}$ (for $u\ge0$). Letting $u=e^2$ and taking logs, we have

$$2=\ln(e^2)=\ln(W(e^2)e^{W(e^2)})=\ln(W(e^2))+W(e^2)$$

from which $\ln(W(e^2))=2-W(e^2)$ follows.


问:
Good calculus exercises/problems?
I can't enroll in a university this year, so I'm studying calculus at home, but the only exercises about calculus that I find are the easy ones. Do you know a great page where I can find not only calculus exercises, but problems as well?

I want to find about:

hard limits,

derivatives and integrals,

and some problems

答:
Try this book

B. P. Demidovich: Problems in mathematical analysis

There are lots of exercises that are hard and beautiful.

Title: Good calculus exercises/problems?


问:
An explicit construction of the "Veronese embedding".
I am reading the text "An invitation to Algebraic geometry" by Karen Smith et. al. In the text there is a proposition on Veronese mappings:

Prop: The Veronese mapping $v_d$ defines an isomorphism of $\mathbb{P}^n$ onto its image.

In the proof, it is explicitly shown how $\mathbb{P}^n \rightarrow v_d(\mathbb{P}^n) \rightarrow \mathbb{P}^n$ is an identity map, but the other part of the proof is left for the reader. I am struggling to prove that $v_d(\mathbb{P}^n) \rightarrow \mathbb{P}^n \rightarrow v_d(\mathbb{P}^n)$ is an identity map too. Could someone please help me with this?

I am a beginner at Algebraic Geometry, so a detailed explanation would be appreciated. Thanks in advance!

答:
Here is an "elementary" answer. Start with the projective line:

$$v:=v_2: \mathbb{P}^1:=C \rightarrow \mathbb{P}^2:=S$$

defined by

$$v(u:v):=(u^2: uv: v^2).$$

Let $S$ have coordinates $x,y,z$ and prove that $v(C) \cong Z(xz-y^2)$ is the zero set of the polynomial $F:=xz-y^2$. Let $D(x) \subseteq S$ denote the set of "points" in $S$ with non-zero $x$-coordinate (similar for $D(y),D(z)$). You may define maps

$$v_x: D(x) \rightarrow C$$

by

$$v_x(a:b:c):=(a:b).$$

Since $(a:b:c)\in D(x)$ it follows $a \neq 0$ and hence $(a:b)\in C$ is well defined.

$$v_y: D(y) \rightarrow C$$

by

$$v_y(a:b:c):=(a:b) \text{ or }(b:c)$$

Since $b^2=ac$ we have $(a,c \neq 0)$ and it follows

$$(a:b)\cong (ac:bc)  \cong (b^2:bc) \cong (b:c)$$

hence $v_y$ is a well defined point in $C$.

$$v_z: D(z) \rightarrow C$$

by

$$v_z(a:b:c):=(b:c)$$

Since $(a:b:c)\in D(z)$ it follows $c \neq 0$ and hence $(b:c)\in C$ is a point.

Then you may verify that $v_x,v_y,v_z$ are well defined maps and local inverses to the map $v_2$, hence $C \cong v(C):=S$ is an isomorphism of algebraic varieties.

Hence this exercise gives you an opportunity to work with local coordinates and to understand what it means for a morphism of algebraic varieties to have an inverse morphism. This method generalize to arbitrary Veronese embeddings and you should do this as an exercise.

Question: "In the proof, it is explicitly shown how $P^n→v_d(P^n)→P^n$ is an identity map, but the other part of the proof is left for the reader. I am struggling to prove that $v_d(P^n)→P^n→v_d(P^n)$ is an identity map too. Could someone please help me with this?"

Answer: In the above example I give an open cover $D(x),D(y),D(z)$ of the variety $v(C)$ and I construct morphisms

$v_x,v_y,v_z$ from $D(x) \rightarrow C$ (resp. $D(y),D(z)$) glueing to a global morphism

$$u: v(C) \rightarrow C$$

with the property that $v \circ u= u \circ v = id$ is the identity.
You should verify that the morphisms I define agree on overlaps. It follows $v: C \rightarrow v(C)$ is an isomorphism of varieties.

Note. If $T:=k[x,y,z]/(xz-y^2)$ and $v(C):=Proj(T)$, we have above constructed an isomorphism of varieties

$$ Proj(k[x_0,x_1]):=C \cong^v v(C):=Proj(T),$$

and we may ask for an "inverse ring isomorphism"

$$ f:T \rightarrow k[x_0,x_1]$$

inducing the map $v$ - which is an isomorphism of varieties. You will find there is a natural map $f$ defined by $f(x):=x_0^2, f(y):=x_0x_1, f(z):=x_1^2$ inducing an isomorphism

$$f: k[x,y,z]/(xz-y^2) \cong k[x_0^2,x_0x_1,x_1^2] \subseteq k[x_0,x_1].$$

Hence the "natural" map $f: T \rightarrow k[x_0,x_1]$ is not an isomorphism of $k$-algebras. The correspondence between maps of varieties and maps of coordinate rings is more complicated when studying projective varieties than in the case of affine varieties. In the case of affine varieties $V,W$ with coordinate rings $A(V), A(W)$, there is a 1-1 correspondence between maps $\phi: V \rightarrow W$ and maps of coordinate rings $\phi^*: A(W) \rightarrow A(V)$.  Hence there is an isomorphism $V \cong W$ of algebraic varieties (over $k$) iff there is an isomorphism $A(W) \cong A(V)$ of $k$-algebras. This result does not hold for projective varieties.

I'm not familiar with Smith's book, but you also find this explained in Hartshornes book, Chapter II.5, where a relation between maps of graded rings and the Proj construction is outlined.

Note. If you check Exercise II.2.14 in HH you will find a construction of a morphism of schemes

$$f:U \rightarrow Proj(S)$$

where $U \subseteq Proj(T)$ and $\phi: S \rightarrow T$ is a graded morphism of graded rings. In the case of the projective line $C$ there is for every $d\geq 2$ a "natural" sub ring

$$ i:T:=k[x_0^{d-j}x_1^j:j\in\{0,1,..,d\}] \subseteq k[x_0,x_1]$$

and I believe that in the case of $i$ you get an isomorphism

$$\phi_d: C \rightarrow \mathbb{P}^d$$

onto a curve $\phi_d(C)$ - the $d$-uple embedding of $C$. The construction of the Segre embedding and $d$-uple embedding is explained in the exercices in HH.

Note: There is an alternative approach using invertible sheaves. Let $V^*:=k\{y_0,y_1,y_2\}$ and let $S:=\mathbb{P}(V^*)$ and let $\pi: C \rightarrow Spec(k)$ be the structure map. There is a surjective map of sheaves

$$\phi_v: \pi^*(V^*) \rightarrow \mathcal{O}_C(2) \rightarrow 0$$
defined by

$$\phi_v(y_0):=x_0^2, \phi_v(y_1):=x_0x_1, \phi_v(y_2):=x_1^2.$$

By the universal property of $\mathbb{P}(V^*)$ you get a map $v: C \rightarrow \mathbb{P}(V^*)$ and it follows from HH.II.7.2 that $v$ is a closed immersion.

At $D(y_0)$ you get the map

$$v_0: k[\frac{y_1}{y_0}, \frac{y_2}{y_0}] \rightarrow k[\frac{x_1}{x_0}]$$

defined by

$$v_0(\frac{y_1}{y_0}):=\frac{x_1}{x_0}, v_0(\frac{y_2}{y_0}):=(\frac{x_1}{x_0})^2$$

and it follows $v_0$ is surjective.


问:
Taylor approximation is not optimal
My professor gave a lecture on an orthogonal polynomial based approximation and its advantage over the Taylor series expansion. And his statement was ``in weighted $L_2$ space, Taylor series expansion is not optimal in inner product sense, whereas basis is optimal in the orthogonal polynomial approximation."  Roughly I know that Taylor series approximation has some limitations such as function should be analytic or infinitely differentiable. But how Taylor series is not optimal in inner product sense?

Any suggestions towards finding the reason would greatly be appreciated.

答:
Let us call the weighted $L^2$ space $W$, assume all the polynomials belong to $W$ and let $\Pi_n$ denote the subspace of polynomials of degree $n$ or less.

The essential point to note is that any $p \in \Pi_n$ is the closest approximation for $f \in W$ (in the sense of the $W$ norm) if and only if $f-p$ is orthogonal $\Pi_n$.  If $p_0, p_1, \cdots, p_n, \cdots$ are the $W$ orthogonal polynomials each with unit norm, then the $n$th order approximation
\begin{align*}
f_n = \sum_{k=0}^n \langle f, p_k \rangle  p_k
\end{align*}
is optimal because $p_1, \cdots p_n$ spans $\Pi_n$ and $f-f_n$ is easily verified to be in $\Pi_n^\perp$.  Nor is it hard to see that such an expansion is unique.  In particular the $n$th degree Taylor series cannot be any better.  Note it might be no worse either, an obvious case being when $f$ is itself a polynomial of degree less than or equal to $n$.

The key result is:  if $H$ is an inner product space, $x \in H$ and $V$ a subspace of $H$, then $x \in V^\perp$ if and only if
$$ \lVert v-x \lVert \geqslant \lVert x \rVert$$ for every $v \in V$.  To prove it, first, assume $x \in V^\perp$, then for any $v \in V$,
\begin{align*}
\lVert v-x\rVert^2 &= \langle v-x, v-x\rangle \\
&= \lVert v \rVert^2 + \lVert x \rVert^2 \\
&\geqslant \lVert x \rVert^2
\end{align*}
Conversely if $\lVert v -x \rVert \geqslant \lVert x \rVert$ for all $v \in V$ then for any $u \in V$ and $\alpha \in \mathbb C$, we also have $\alpha u \in V$ and
\begin{align*}
\lVert x \rVert ^2 &\leqslant \lVert x - \alpha u \rVert ^2 \\
&= \lVert x \rVert x ^2 - \alpha  \langle u, x \rangle - \overline{\alpha \langle u, x \rangle} +|\alpha|^2 \lVert u \rVert^2 \\
&= \lVert x \rVert^2 - 2 \Re \Big( \alpha \langle u, x\rangle \Big) + |\alpha|^2 \lVert u \rVert ^2 \tag{1}\label{BPA-1}
\end{align*}
Now choose $\theta$ so that $e^{i\theta}\langle u, x \rangle $ is real and positive and for any $r > 0$ let $\alpha = re^{i\theta}$.  Then cancel $\lVert x \rVert^2$ on each side of inequality \eqref{BPA-1}, then divide by $r > 0$, to get
\begin{align*}
2 \lvert \langle u, x \rangle \rvert \leqslant r \lVert u \rVert^2.
\end{align*}
Now $r$ is arbitrary, so we must have $\langle u, x \rangle = 0$ and $u \in V^\perp$.

Title: Taylor approximation is not optimal


问:
Check if a given mapping is linear or not?

$f:IR_n[X]\rightarrow IR_3[X]$ given by $f(p(X))=p(0)X^2+Dp(0)X^3$

Can you tell me what is the method to check mapping like this.

答:
Since derivation is a linear operation, you have
\begin{eqnarray*}
f((p+q)(X))&=& (p+q)(0) X^2 +D(p+q)(0) X^3\\
&=& (p(0)+q(0)) X^2 +(Dp(0)+Dq(0)) X^3\\
&=& p(0)X^2+q(0) X^2 +Dp(0)X^3 +Dq(0)X^3\\ &=& f(p(X))+ f(q(X))\\
\end{eqnarray*}
for all $p,q \in \mathbb R_n[X]$,  and
\begin{eqnarray*}
f((\alpha \cdot p)(X))&=& (\alpha \cdot p)(0) X^2 +D(\alpha \cdot p)(0) X^3\\
&=& \alpha \cdot p(0) X^2 +\alpha \cdot Dp(0) X^3\\
&=& \alpha \cdot f(p(X)) \\
\end{eqnarray*}
for all $p \in \mathbb R_n[X]$ and $\alpha \in \mathbb R$.


问:
How to rearrange 9 people into sets of 3 so no two people remain together
I study maths as a hobby. I am considering this question on probability.

In how many ways can a group of 9 people attending a conference be split into 3 sets of 3 people for discussion? Later the same people form new sets of 3. In how many ways can this be done if no two people are to remain together?

I have solved the first part. The total ways of selecting people is:

$\binom {9}{3}\ . \binom {6}{3} .\binom {3}{3} = 84.20.1 = 1680$

and since the order of the groups doesn't matter we divide by 3!

$\frac{1680}{3!} = 280$

But I am stuck on the second part. I thought maybe I should first consider the number of selections in which 2 people do remain together and subtract that from 280. This would mean I would treat 2 people as one entity and make selections of 2. So I would get:

$\binom {9}{2}.\binom {7}{2}.\binom {5}{2}$ or perhaps

$\binom {9}{2}.\binom {6}{2}.\binom {3}{2}$ or perhaps

But this is getting me nowhere.

The answer is 36.

答:
Each person in each original group must be placed in a separate new group.

Take one of the groups.  Put each person in that group in separate groups.  It does not matter how since the groups are not labeled.  Say we call the new groups A, B, and C, where A is the group containing the oldest person, B is the group containing the next oldest person, and C is the group containing the youngest person from the group we split apart.

Take another of the original groups.  Place one person in group A, one person in group B, and one person in group C.  We can do this is in $3!$ ways.

Take the last of the original groups.  Again, place one person in group A, one person in group B, and one person in group C.  We can do this in $3!$ ways.

Hence, there are $3!3! = 36$ ways to rearrange the nine people who were originally placed in three groups of three people so that no two people remain together.


问:
Interpret stochastic differential equation as deterministic one
let us assume the following ODE

$$ \frac{dA}{dt} +gA = X  $$

where $X \sim N(0,1)$ and $g$ is a factor. $X$ changes its value in every timestep $dt$ which means that $X$ is a stochastic process.

if we would simply solve this ODE deterministically, then the integrating factor would be found quickly:

$$A = \frac{\int e^{gt}Xdt+C}{e^{gt}}$$

where one knows that

$$\int e^{gt}Xdt = \int e^{gt}dB_t $$ with standard Brownian Motion $B_t$

So:$$A = \frac{\int e^{gt}dB_t+C}{e^{gt}}$$

When i compute the mean and variance of $A$ i get

$$\mathbb{E}(A_t) = Ce^{-gt}$$
$$\mathbb{V}(A_t) = \frac{1}{gt}$$

but when i interprete this ODE as an Ornstein Uhlenbeck process i get the same mean, but different variance...

where is the problem?

答: Your $C$ is not a constant, though it is independent of $t$; it is random variable. So you cannot find the variance by the first method.


问:
Simple example of computing wreath products of groups
I am trying to understand wreath products but cannot find an example of one being computed in depth.

I know that $D_{4}$ is isomorphic to $\mathbb{Z}_{2} \wr \mathbb{Z}_{2}$, but I cannot seem to understand why this would be true from definitions I read in textbooks.

This wreath product is just the semidirect product of the Klein-4 group and $\mathbb{Z}_{2}$. How exactly would this semidirect product be computed, and how can I see that this is isomorphic to $D_{4}$?

答:
The wreath product is defined in terms of a semi-direct product.  There is the restricted and the unrestricted wreath product, depending as one takes a direct sum or product in the definition.  These agree for finite products.

You need a group action.  In this case we get $\Bbb Z_2$ acting on $\Bbb Z_2^2$ by shifts.  Then it is the semi-direct product $\Bbb Z_2^2\rtimes\Bbb Z_2$.  It turns out that this is indeed the usual $D_4\cong\Bbb Z_4\rtimes\Bbb Z_2$.  (Notice that the actions in these two semi-direct products are different.)

$\Bbb Z_2\wr\Bbb Z_2$ is the smallest non-trivial wreath product.  It also equals the hyperoctahedral group $S_2\wr S_2$.

To see the isomorphism, you could work through the notation; or you can convince yourself that this is not the quaternions, the only other nonabelian group of order $8$.


问:
What is a proof?
I am just a high school student, and I haven't seen much in mathematics (calculus and abstract algebra).

Mathematics is a system of axioms which you choose yourself for a set of undefined entities, such that those entities satisfy certain basic rules you laid down in the first place on your own.

Now using these laid-down rules and a set of other rules for a subject called logic which was established similarly, you define certain quantities and name them using the undefined entities and then go on to prove certain statements called theorems.

Now what is a proof exactly? Suppose in an exam, I am asked to prove Pythagoras' theorem. Then I prove it using only one certain system of axioms and logic. It isn't proved in all the axiom-systems in which it could possibly hold true, and what stops me from making another set of axioms that have Pythagoras' theorem as an axiom, and then just state in my system/exam "this is an axiom, hence can't be proven".

EDIT : How is the term "wrong" defined in mathematics then? You can say that proving Fermat's Last Theorem using the number theory axioms was a difficult task but then it can be taken as an axiom in another set of axioms.

Is mathematics as rigorous and as thought-through as it is believed and expected to be? It seems to me that there many loopholes in problems as well as the subject in-itself, but there is a false backbone of rigour that seems true until you start questioning the very fundamentals.

答:
A rigorous mathematical argument which unequivocally demonstrates the truth of a given proposition. A mathematical statement that has been proven is called a theorem.

In mathematics, a proof is a demonstration that if some fundamental statements (axioms) are assumed to be true, then some mathematical statement is necessarily true. Proofs are obtained from deductive reasoning, rather than from inductive or empirical arguments; a proof must demonstrate that a statement is always true (occasionally by listing all possible cases and showing that it holds in each), rather than enumerate many confirmatory cases. An unproven proposition that is believed to be true is known as a conjecture.
Proofs employ logic but usually include some amount of natural language which usually admits some ambiguity.

In fact, the vast majority of proofs in written mathematics can be considered as applications of rigorous informal logic. Purely formal proofs, written in symbolic language instead of natural language, are considered in proof theory. The distinction between formal and informal proofs has led to much examination of current and historical mathematical practice, quasi-empiricism in mathematics, and so-called folk mathematics (in both senses of that term).

http://en.wikipedia.org/wiki/Mathematical_proof and go to methods of proof

Title: What is a proof?


问:
If $X$ and $Y$ are two independent random variables, are $X^{2}$ and $Y$ independent too?
Let $X \sim \mathrm{Unif }[0,1]$ and $Y\sim \mathrm{Gamma}(2,r)$, with $r>0$, be two independent random variables.

I have to find $\mathrm{Cov}(U,V)$ where $U = X$ and $V = XY$.

From previous points, I have calculated that $$f_{U,V}(u,v) = \frac{r^{2}ve^{-r\frac{v}{u}}}{u^{2}} $$ with $0 < u < 1 $ and $ v > 0 $,

and$$f_{V}(v) = re^{-rv}$$ with $v > 0$, hence, $V \sim \textit{Exp}(r)$.

Thus,
\begin{align*}
\mathrm{Cov}(U,V) &= \mathbb{E}(U\cdot V) - \mathbb{E}U \cdot\mathbb{E}V \\
&= \mathbb{E}(X\cdot XY) - \mathbb{E}X\cdot\mathbb{E}(XY) \\
&\overset{indep.}{=} \mathbb{E}(X^{2}\cdot Y) - (\mathbb{E}X)^{2}\cdot\mathbb{E}Y
\end{align*}

If $X^{2}$ and $Y$ are independent, then $$ = \mathbb{E}Y\cdot \textit{Var}(X)$$ which would be pretty convenient. But also how should I prove this?

If they're not independent, then I'm wondering if there is an easy way to evaluate it, without having to find the density function of $UV$ which I predict would end up having the same "problem". I know that there is a way to approximate $\lvert\mathbb{E}(UV)\rvert$ (Cauchy-Schwarz inequality), that is,
$$\lvert\mathbb{E}(UV)\rvert \leq \sqrt{\mathbb{E}(U^{2})\cdot\mathbb{E}(V^{2})}$$ but I'm not sure this is the way to evaluate this.

(I've searched around for an answer, but they all involve Measure Theory which I have no knowledge of).

Any suggestion/help would be appreciated.

答: For any continuous functions $f$ and $g$ (and, in fact, for any Borel measurable functions $f$ and $g$) the variables $f(X)$ and $g(Y)$ are independent. So it is certainly true that $X^{2}$ and $Y$ are independent.


答:
As a simple explanation why: The Taylor polynomial gives you the best possible approximation at one single point, and the further away you are from this point, the worse the approximation. The Taylor polynomial doesn’t even try to keep the error low when you are away from that point.

Polynomials that find an approximation minimising some norm tend to keep the error down over a whole range of values. They have to, since the errors over the whole range contribute to the error norm that is minimised.

Interpolating polynomials can be trouble if you don’t watch out at which points you interpolate. A method that is mostly numerical is minimising the maximum error over an interval; Chebyshev has a nice theorem for that. If you do numerical mathematics and want the highest possible numerical precision, you minimise the sum of polynomial error and rounding error. And you take into account that your polynomial coefficients should be floating point numbers.

Title: Taylor approximation is not optimal

Body:
My professor gave a lecture on an orthogonal polynomial based approximation and its advantage over the Taylor series expansion. And his statement was ``in weighted $L_2$ space, Taylor series expansion is not optimal in inner product sense, whereas basis is optimal in the orthogonal polynomial approximation."  Roughly I know that Taylor series approximation has some limitations such as function should be analytic or infinitely differentiable. But how Taylor series is not optimal in inner product sense?

Any suggestions towards finding the reason would greatly be appreciated.


问:
An algebraic problem I failed to solve
Pls help me with this question. This is a question which came in 2006 Olympiad. Here more than 1 possible options can be correct. I tried like that I assumed that $x = y = z = a =b = c$ but failed. More than one options are correct but I was not able to eliminate any.

Eliminate $x,y,z$ from the equation
$$
\frac{x^2(y+z)}{a^3}=\frac{y^2(z+x)}{b^3}=\frac{z^2(x+y)}{c^3}=\frac{xyz}{abc}=1
$$

Here are the options:

$a^3+b^3+c^3=abc$

$a^3+b^3+c^3+abc=0$

$a^3+b^3+c^3+abc=2abc$

$a^3+b^3+c^3+2abc=abc$

答:
Option $2$ and $4$ are possible. An elimination shows that we have a solution given by
$$
z=\frac{- cx(ab + c^2)}{b^3+c^3},\quad
y=\frac{- bx(ac + b^2)}{b^3+c^3},
$$
and some real number $x$, and
$$
a^3+b^3+c^3+abc=0.
$$

Title: An algebraic problem I failed to solve


问:
Confused with Bounded and Dominated convergence theorem
I am confused with  bounded and dominated convergence theorem.

I have a random variable $Z$  which is $\mathscr{G}$-measurable  and bounded almost sure, ie, there is a positive number $M$ such that $|Z| \leq M $ a.s.Then I have to show that for the conditional expectation $$Y = \mathbb{E}[X|\mathscr{G}]$$ i have :
$$\mathbb{E}[YZ] = \mathbb{E}[XZ]\quad\text{(1)}\quad$$

and more general if $Z$ is $\mathscr{G}$-measurable and $$\mathbb{E}[|ZX|]< \infty$$ then the equality to be proven is still true.

My answer according to David William (Probability with Martingales) page 90 is that :

If $Z$ is the indicator of a set in $\mathscr{G}$, then (1) is true by definition of the conditional expectation $Y$. Linearity then shows that (1) holds for $Z \in \mathscr{G}$ . Next, Monotone Convergence Theorem shows that (1) is true for $Z \in \mathscr{G}$  with the understanding that both sides might be infinite.

or

We know that for integrable r.v.'s $X$, the defining equation for its conditional expectation $Y=E[X|\mathscr{G}]$ given some sub-$\sigma$-field $\mathscr{G}$ is
$$E[X\cdot1_A]=E[Y\cdot1_A],\ \forall A\in\mathscr{G}\quad(1)$$

Changing to:
$$E[XZ]=E[YZ],\ \forall\text{ nonnegative, bounded,} \mathscr{G}\text{-measurable r.v. }Z\quad(2)$$

Now $(2)\implies(1)$ is obvious. For the other direction. My proof is as follows:

Since $Z$ is $\mathscr{G}$-measurable, nonnegative and bounded, we can find an increasing sequence of simple functions $z_n\geq0$ such that $z_n\nearrow Z$ a.s. Moreover, $z_n$ has the form
$$z_n=\sum_{j=1}^{nk}a_j\cdot1_{A_{kj}},\text{ where}\ A_{kj}\in\mathscr{G}$$
Clearly, $E[X z_k]=E[Y z_k]$ by $(1)$. Let $k\to\infty$ and by dominated convergence we have $E[XZ]=E[YZ]$.

But how this can be proven with dominated convergence theorem?
Any help?

答:
Just as you did for a bounded $Z$, you can define a sequence $(z_n)$ of simple functions such that $z_n \nearrow Z$ a.s. You only need $Z$ to be $\mathscr{G}$-measurable for such a sequence to exist, the main difference being that if $Z$ is unbounded then the convergence is not uniform (see e.g. this question with the linked notes for more details).

We now have the following inequality : $\forall n\ge 0$, $|z_nY| = |z_n||Y| \le |Z||Y|$. And we can see that $|ZY|$ is integrable, indeed :
$$\begin{align}
|Y| &\le \mathbb{E}[|X|\,\,|\mathscr{G}] \quad\text{by Jensen's inequality} \\
\implies |YZ| &\le |Z|\mathbb{E}[|X|\,\,|\mathscr{G}] \\
\implies \mathbb{E}[|YZ|] &\le \mathbb{E}\left[|Z|\mathbb{E}[|X|\,\,|\mathscr{G}]\right] \\
&= \mathbb{E}\left[\mathbb{E}[|XZ|\,\,|\mathscr{G}]\right] \quad\text{by $\mathscr{G} $- measurability of $|Z|$}\\
&= \mathbb{E}[|XZ|] < \infty
\end{align}$$

So we have shown that the sequence $(z_nY)$ is dominated by an integrable function.

On top of that, it also holds that $\mathbb{E}[z_nY] = \mathbb{E}[z_nX] \, \, \forall n\ge 0$, by definition and linearity of conditional expectation.

Therefore, by dominated convergence theorem we can conclude that
$$\lim_{n\to \infty} \mathbb{E}[z_nY] = \mathbb{E}[ZY] =\lim_{n\to \infty} \mathbb{E}[z_nX] = \mathbb{E}[ZX] $$


问:
Formula for series $\frac{\sqrt{a}}{b}+\frac{\sqrt{a+\sqrt{a}}}{b}+\cdots+\frac{\sqrt{a+\sqrt{a+\sqrt{\cdots+\sqrt{a}}}}}{b}$
All variables are positive integers.

For:

$$a_1\qquad\frac{\sqrt{x}}{y}$$
$$a_2\qquad\frac{\sqrt{x\!+\!\sqrt{x}}}{y}$$
$$\cdots$$
$$a_n\qquad\frac{\sqrt{x\!+\!\sqrt{\!x+\!\sqrt{\!\cdots\!+\sqrt{x}}}}}{y}$$

Is there a formula of an unconditional form to describe series $a_n$?

I thought of something along the lines of:

$$\sum _{k=1}^{n } \left(\sum _{j=1}^k \frac{\sqrt{x}}{y}\right)$$

but, I quickly realized that it was very incorrect; Then I thought of:

$$\sum _{k=1}^{n} \frac{\sum _{j=1}^k \sqrt{x}}{y}$$

which I also concluded as very incorrect...

I'm blank, but I would like to see an example of something along the lines of:

$$\sum _{k=1}^{n } \frac{\sqrt{x+\sqrt{x+\sqrt{\cdot\cdot\cdot+\sqrt{x}}}}}{y}$$

where each $\sqrt{x+\sqrt{\cdots}}$ addition, repeats $k$ times. (i.e $k=3 \Rightarrow \sqrt{x+\sqrt{x+\sqrt{x}}}$);
If it is possible...

Cheers!

答:
If all you are looking for is a compact representation, let
$$
s_{k}=\begin{cases}
0 & \text{if }k=0\\
\sqrt{a+s_{k-1}} & \text{if }k>0
\end{cases}.
$$
Then
\begin{align*}
S_n & =\left(\frac{\sqrt{a}}{b}\right)+\left(\frac{\sqrt{a+\sqrt{a}}}{b}\right)+\ldots+\left(\frac{\sqrt{a+\sqrt{a+\ldots+\sqrt{a}}}}{b}\right)\\
 & =\frac{1}{b}\left[\sqrt{a}+\sqrt{a+\sqrt{a}}+\ldots+\sqrt{a+\sqrt{a+\ldots+\sqrt{a}}}\right]\\
 & =\frac{1}{b}\sum_{k=1}^{n}s_{k}.
\end{align*}

Assume $a\in\mathbb{R}$ (you don't have to do this). We can show that the recurrence is
stable everywhere (weakly stable at $a=-\frac{1}{4}$). Particularly,
the fixed point is given by
$$
s^{2}-s-a=0,
$$
which has roots
$$
\frac{1\pm\sqrt{1+4a}}{2}.
$$
Particularly, the locally stable fixed point is the solution with $\pm$ is $+$.
So, for large enough $k$,
$$
s_k\approx\frac{1+\sqrt{1+4a}}{2}.
$$

This is as good an answer as you can hope for, save for error bounds on the above expression.


答:
The variables $X^2$ and $Y$ are independent (by definition) if for all $x,y$ we have $\Pr(X^2\leq x,Y\leq Y)=\Pr(X^2\leq x)\Pr(Y\leq Y)$. Clearly we can assume $x\geq 0$ here. So
$$\Pr(X^2\leq x,Y\leq Y)=\Pr(-\sqrt x\leq X\leq \sqrt x,Y\leq y)\\
=\Pr(X\leq \sqrt x,Y\leq y)-\Pr(X\leq -\sqrt x,Y\leq y)\\
=\Pr(X\leq \sqrt x)\Pr(Y\leq y)-\Pr(X\leq -\sqrt x)\Pr(Y\leq y)\\
=[\Pr(X\leq \sqrt x)-\Pr(X\leq -\sqrt x)]\Pr(Y\leq y)\\
=\Pr(X^2\leq x)\Pr(Y\leq y).$$

Title: If $X$ and $Y$ are two independent random variables, are $X^{2}$ and $Y$ independent too?

Body:
Let $X \sim \mathrm{Unif }[0,1]$ and $Y\sim \mathrm{Gamma}(2,r)$, with $r>0$, be two independent random variables.

I have to find $\mathrm{Cov}(U,V)$ where $U = X$ and $V = XY$.

From previous points, I have calculated that $$f_{U,V}(u,v) = \frac{r^{2}ve^{-r\frac{v}{u}}}{u^{2}} $$ with $0 < u < 1 $ and $ v > 0 $,

and$$f_{V}(v) = re^{-rv}$$ with $v > 0$, hence, $V \sim \textit{Exp}(r)$.

Thus,
\begin{align*}
\mathrm{Cov}(U,V) &= \mathbb{E}(U\cdot V) - \mathbb{E}U \cdot\mathbb{E}V \\
&= \mathbb{E}(X\cdot XY) - \mathbb{E}X\cdot\mathbb{E}(XY) \\
&\overset{indep.}{=} \mathbb{E}(X^{2}\cdot Y) - (\mathbb{E}X)^{2}\cdot\mathbb{E}Y
\end{align*}

If $X^{2}$ and $Y$ are independent, then $$ = \mathbb{E}Y\cdot \textit{Var}(X)$$ which would be pretty convenient. But also how should I prove this?

If they're not independent, then I'm wondering if there is an easy way to evaluate it, without having to find the density function of $UV$ which I predict would end up having the same "problem". I know that there is a way to approximate $\lvert\mathbb{E}(UV)\rvert$ (Cauchy-Schwarz inequality), that is,
$$\lvert\mathbb{E}(UV)\rvert \leq \sqrt{\mathbb{E}(U^{2})\cdot\mathbb{E}(V^{2})}$$ but I'm not sure this is the way to evaluate this.

(I've searched around for an answer, but they all involve Measure Theory which I have no knowledge of).

Any suggestion/help would be appreciated.


问:
Find the c.d.f. of X
John has an insurance policy on his car with a 200\$ deductible, meaning that if an accident occur, he would pay the cost of the repair up to 200\$ with the insurance policy paying the rest. So if he has an accident worth 123\$ he would pay 123\$ but if the accident is worth 345\$ he would only pay 200\$.
Assume that the cost of an accident is uniformly distributed over [50, 1000]. Denote X the amount that John pays.

Compute the c.d.f. of X

Is X discrete, continuous, or neither discrete nor continuous

What is the probability that $X=200$

My attempt:

I would think that X is continuous because it is a piece-wise function, where John pays whatever the price of the damage is until 200\$ and then he pays 200\$ up to 1000\$.

Let t=cost of damage and F(t)=amount that John pays

Then the c.d.f would look like:

\begin{equation}
F(t)=
\begin{cases}
0\quad\quad\quad\quad t<50\\
\frac{t-50}{200-50}\quad\quad 50\leq t<200\\
\frac{t-200}{1000-200}\quad\;\; 200\leq t<1000\\
1\quad\quad\quad\quad  t<1000\\
\end{cases}
\end{equation}

For the last question, I am not sure how to find a probability from a c.d.f. Can I also get verification on my last two answers? Thank you.

答:
Let $C$ be the cost of an accident. Then $C\sim U[50,1000]$. Now we can express $X$ in terms of $C:$ $$X=\begin{cases}
C, & 50\le C<200\\
200, & 200\le C\le 1000
\end{cases}$$
Note that $\mathbb P(200\le C<1000)=\frac{1000-200}{1000-50}=\frac{16}{19}$. So naturally $X=200$ with probability $\frac{16}{19}$.

Also note that $X$ is neither a discrete nor continuous r.v., as $X$ has a point probability at $X=200$ (for continuous r.v.s, $\mathbb P(X=x)=0\:\forall x$)

Regarding the cdf, we want the cdf to increase linearly up to $1-\frac{16}{19}=\frac{3}{19}$ on $[50,200)$, and then reach $1$ once $x$ approaches $200$. It is easy to see that $$F_X(x)=\begin{cases}
0, & x<50\\
\frac{x-50}{1000-50}=\frac{x-50}{950}, & 50\le x<200\\
1, & x\ge 200
\end{cases}$$

Title: Find the c.d.f. of X


问:
If $(f_n)$ is an orthonormal basis and $\lambda_n\to\infty$, is $T(t)x:=\sum_ne^{-λ_nt}\langle x,f_n\rangle f_n$ uniformly continuous?
Let $H$ be a complex Hilbert space, $(f_n)_{n\in\mathbb N}$ be an orthonormal basis of $H$ and $(\lambda_n)_{n\in\mathbb N}\subseteq(0,\infty)$ be nondecreasing with $\lambda_n\xrightarrow{n\to\infty}\infty$. It is easy to see that $$T(t)x:=\sum_{n\in\mathbb N}e^{-\lambda_nt}\langle x,f_n\rangle_Hf_n\;\;\;\text{for }x\in H\text{ and }t\ge0$$ is a well-defined contractive semigroup on $H$.

Are we able to show that

$(T(t))_{t\ge0}$ is uniformly continuous?

If $U$ is a complex Hilbert space compactly and densely embedded into $H$ and $$e_n:=\frac1{\sqrt\lambda_n}f_n\;\;\;\text{for }n\in\mathbb N$$ is an orthonormal basis of $U$, does it follow that $$S(t)u:=T(t)u\;\;\;\text{for }u\in U\text{ and }t\ge0$$ is a strongly/uniformly semigroup on $U$ as well?

Noting that $$\sup_{t\ge0}\left\|\sum_{i=1}^ne^{-\lambda_it}\langle x,f_i\rangle_Hf_i-T(t)x\right\|_H^2\le\sum_{i>n}\left|\langle x,f_i\rangle_H\right|^2\xrightarrow{n\to\infty}0\tag1$$ for all $x\in H$, we should be able to conclude that $(T(t))_{t\ge0}$ is at least strongly continuous.

答:
It is not uniformly continuous:
$$
\frac1t (T(t)f_i-f_i) = \frac1t (e^{-\lambda_i t}-1)f_i \to -\lambda_i f_i
$$
for $t\to0$. The operator $f_i \mapsto \lambda_i f_i$ is unbounded, so $T$ cannot be uniformly continuous. Or put differently, the convergence $\frac1t (T(t)f_i-f_i)  \to  -\lambda_i f_i$ is not uniform with respect to $i$, and $\frac1t\|T(t)-I\|$ does not exist.


问:
Prove a partial derivative change-of-variables identity
If $u=f(x,y)$ where $x=e^{3s}\cos(2t)$ and $y=e^{3s}\sin(2t)$, then show that

$$ \left(\dfrac{\partial u}{\partial x}\right)^2+\left(\dfrac{\partial u}{\partial y}\right)^2= g(s,t)\left(\dfrac{\partial u}{\partial s}\right)^2 + h(s,t) \left(\dfrac{\partial u}{\partial t}\right)^2$$

My attempt: I found $du/ds$ and $du/dt$, squared them, and added them. In the examples from class, something cancels out nicely but nothing did in this question. After some searching around, I found a midterm solution which is very similar to my question.

So far, I have $U_s^2 + U_t^2 = U_x^2(9x^2+4y^2)+10 U_s U_t xy+ U_y^2(9y^2+4x^2)$

Now I'm stuck because of the $9$ and $4$ in front of $x$ and $y$. Any help is greatly appreciated!

答:
From $u_s = u_x\cdot x_s+u_y\cdot y_s,\ u_t = u_x\cdot x_t+u_y\cdot y_t$:
$$u_s = 3e^{3s}(u_x\cos(2t)+u_y\sin(2t))$$
$$u_t = 2e^{3s}(u_y\cos(2t)-u_x\sin(2t))$$
Solve to get:
$$u_x =\frac{-2\cos(2t)u_s+3\sin(2t)u_t}{6e^{3s}}$$
$$u_y =\frac{2\sin(2t)u_s+3\cos(2t)u_t}{6e^{3s}}$$
Now add together $u_x^2$ and $u_y^2$ and see how all the the mixed terms ($\sim u_s u_t$) disappear.</p>


问:
How can I compute the volume integral of a parameterized set?

This is really odd to me. How can I compute the volume of this cup?

答:
Parametrized surface: $x = r \cos\varphi, \ y = r \sin \varphi - (1-\sqrt{4-z})^2, z$

$r \geq 0, 0 \leq \varphi \leq 2\pi, 3\sqrt r \leq z \leq 3$

As we can see from parametrization that for any given $z$, we have a circle of radius $r$ with center at $(0, - (1-\sqrt{4-z})^2, z)$.

Now $z \geq 3 \sqrt r \implies r \leq \frac{z^2}{9}$.

So using disk method, the volume can be found as

$\displaystyle \int_0^{2\pi} \int_0^{3} \int_0^{z^2/9} r \ dr \ dz \ d\varphi = \frac{3 \pi}{5}$

Using shell method,

$\displaystyle \int_0^{2\pi} \int_0^{1} \int_{3 \sqrt r}^3 r \ dz \ dr \ d\varphi = \frac{3 \pi}{5}$

(the upper bound of $r$ comes from $z = 3 \sqrt r = 3 \ $ or $r = 1)$

As Martin R said, this is same as the volume of a cone of same radius and height (Cavalieri's Principle)


问:
How do you respond to "I was always bad at math"?
Here in the U.S., it is my experience that over 75% of adults I meet socially will volunteer that phrase or a variation upon learning that I am a mathematician.  I find this frustrating, since almost nobody would  brag about being bad at history or English or really any other subject.   Normally I change the subject, or perhaps say they must have had a bad classroom experience, but that feels unsatisfying and turns the conversation in an ugly direction.

Surely some of the bright minds here on Math.SE have found a good response. Note that I'm not looking to  alienate a new acquaintance, so let's keep it positive.

答:
I am not a mathematician just a lay person who find abstractions easy to work with.

I reply and tell them that it's not their fault. The problem is that maths is about abstract thinking and all it takes it so get the right abstraction (or model) in the mind of the person and they'll get it.

I tell them that some people 'can visualise maths' easily, but others may need help in getting to the visualisation. It's down to the teacher to help the student 'see the maths' and that if they choose to ignore the fact that they suck as maths and imagine that they've just not had a map that lets them navigate that world easily, they could do so. It's just mid set.


问:
$K$ consecutive heads with a biased coin?
You toss a coin repeatedly and independently. The probability to get a head is $p$, tail is $1-p$. Let $A_k$ be the following event: $k$ or more consecutive heads occur amongst the tosses numbered $2^k,...,2^{k+1}-1$. Prove that $\mathbb P\left(A_k\hspace{1mm} \text i.\text o.\right)=1$ if $\displaystyle p\geq \frac{1}{2}$, $0$ otherwise.

I'd appreciate any help with this one! edit: Assuming this has to do with Borel-Cantelli, specifically a "$0/1$ law".

We know that if $\{A_n:n\geq 1\}$ is a sequence of independent events in a probability space, then either $\mathbb P\left(A\left(\text i.\text o.\right)\right)=0$, which is the $\mathbb E\left(N\right)<\infty$ case, or $\mathbb P\left(A\left(\text i.\text o.\right)\right)=1$, which is the $\mathbb E\left(N\right)=\infty$ case, where $N$ denotes the total number of $A_n$ to occur;

$\displaystyle N=\sum_{n=1}^{\infty}I_n$, where $I_n=I\{A_n\}$ denote the indicator rv for the event $A_n$.

$\displaystyle \mathbb{E}(N) = \sum_{k=1}^\infty \mathbb{P}(A_k)$, so we have to show that this sum converges for $\displaystyle p< \frac{1}{2}$, and diverges otherwise. (how?)

答:
I would like to offer a solution similar to what the OP started.

So we defined $A_k = \text{there were at least k consecutive heads between the tosses } 2^k \text{ and } 2^k+1$

Then for $p\geq0.5$:
$$\mathbb{P}\left(A_n^c\right)\leq\mathbb{P}\left(
\text{There weren't $n$ consecutive heads that started in the tosses numbered: } 2^n,2^n+n,2^n+2n,..\right)
\approx\left(1-{p^n}\right)^{2^n/n} = (*)
$$
because these are $\left(2^{n+1}-2^n\right)/n$ independent events.

We can the conclude that $(*)= {\left(1-{p^n}\right)^{\frac{1}{p^n}}}^{\frac{2^n\cdot p^n}{n}}$

Now, we obtained that for $p\geq0.5$ then $(*)\to0$ and hence $\sum{\mathbb{P}(A^c_n)}<\infty$ and by Borrell Cantelli $\mathbb{P}\left(\{A_n^c\;i.o\}\right)=0$ and hence $\mathbb{P}\left(\{A_n\;eventually\}\right)=1$ and that's why $\mathbb{P}\left(\{A_n\;i.o\}\right)=1$

For $p<0.5$:
$$\mathbb{P}\left(A_n\right)\leq\mathbb{P}\left(
\text{$\bigcup$ events where the $n$ consecutive heads started in the tosses numbered: } 2^n,2^n+1,2^n+2,..2^{n+1}-n\right)
=\left(2^{n+1}-2^n-n\right)\cdot p^n = (2^n-n)p^n
$$
And hence, $\sum\mathbb{P}\left({A_n}\right)\leq\sum{(2^n-n)p^n}<_{p<0.5}\infty$ and again by Borrell Cantelli we got that $\mathbb{P}(A_n\;i.o)=0$


问:
Three circles inside a square, finding maximum value of $r$

In following figure the the length of side of the square is $1$.
radius of each of three circles is equal to $r$. what is the maximum value
of $r$?

$$1)\frac13\quad\quad\quad\quad
 2)\frac14\quad\quad\quad\quad3)\frac{\sqrt2}6\quad\quad\quad\quad4)2\sqrt2-\sqrt6\quad\quad\quad5)\frac{\sqrt2}{1+2\sqrt2+\sqrt3}$$

From the image I can see lowest circle is tangent to only one side of the square.

I started from upper circle and draw a segment to connect center of this circle to center of right circle and also draw a tangent line from center of upper circle to upper section of right circle. then I made a right triangle with hypotenuse length $2r$ and horizontal length $r\sqrt3$(because the angle is $30^{\circ}$) so the length of square is $r+\sqrt3 r+r=1$ hence $r=\frac1{2+\sqrt3}=2-\sqrt3$. but this answer is not in the options.

答:
The equilateral triangle formed by the centers of three circles is enclosed by a shape which is necessarily a square. Then the two angles are equal and sum to $90^{\circ}-60^{\circ}=30^{\circ}$. The angles are then $15^{\circ}$ each.

So the equation is modified to
$$r(1+2\cos 15^{\circ} + 1)=1$$
$$\Rightarrow r =\frac{\sqrt2}{1+2\sqrt2+\sqrt3}$$


问:
PDE: solve IVP where $u_{t} +3u u _{x} = 0$, $u(0,x) = \left \{ \begin{matrix} 2, x < 1 \\ 0, x>1 \end{matrix} \right.$
I just started learning PDEs and I am at the very beginning of the book by Peter J. Olver. Now when I search in the stackexchange I found similar problems, all relating to Cauchy or Burger equations. The book so far has not spoken of these, so I guess I should be able to solve this without knowledge of them.

I am asked to solve the IVP where
$$u_{t} +3u u _{x} = 0, \quad \quad u(0,x) = \left \{ \begin{matrix} 2, x < 1 \\  0, x>1 \end{matrix} \right.$$

My reasoning (likely to be wrong) so far is the following: we have a non-linear transport equation, where the speed of the wave is dictated by $3u$ (i.e. the wave is moving faster the bigger it is). We also know the solution is constant along the characteristic curve. We can see the wave being constant at $x>1$ but moving to the right when $x<1$. This leads me to the following:

$$
\begin{align}
\frac{ \partial x}{\partial t} &= 3u \\
\frac{\partial u }{\partial t} &= 0 \\
u(0,x) &= f(x),
\end{align}
$$

Then $u(t,x) = f(x_{0})$ and now I am stuck. I do not know how to proceed.

Any suggestions or comments on my train of thought or on how to proceed would be much appreciated.

答:
After further reading I constructed the following:

We use a change of variables to find the characteristic curves and find
$$x = 3tf(y)+y.$$

We then have that $u(0,y)=2$ for $y<1$ and $u(o,y)=0$ for $y>1$. We can draw these lines in the $(x,t)$-plane and will find a triangular area of shock waves (i.e. multi valued solutions). Now by looking at two values of $y$ (one smaller and one larger than 1) we can easily derive

$$
u(t,x) = \left\{\begin{matrix}
2 & x-6t<1\\
0 & x>1
\end{matrix}\right.
$$


问:
Find all critical points of the function $f(x, y) =\vert x\vert + (x−1)^2−y^2$. Classify them using the second derivative test.
this is what I have so far:

$$\frac{\mathrm{d}f}{\mathrm{d}x} = x/\vert x\vert+2(x-1)=0$$
$$\frac{\mathrm{d}f}{\mathrm{d}y} = -2y=0$$

$y=0$

$x=2$ or $\vert x\vert = 1/2$

What do I do with $\vert x\vert$? Am I on the right track?

Thanks

答:
Hint:

The second derivatives are constant (except along $x=0$ where the function is not differentiable):

$$f_{xx}=2,f_{xy}=0,f_{yy}=-2.$$

As $\Delta<0$, you can conclude about the type.


问:
Triple integral $\iiint_{\mathrm{E}}(x^2+xy+y^2)\, dx\,dy\,dz$
The following function: $$\iiint_{\mathrm{E}}(x^2+xy+y^2)\, dx\,dy\,dz, $$ where $E$ is a globe with radius 2 and center in origin.

I have got the following limits using spherical coordinates $0 \le r \le 2$, $0 \le \theta \le \pi$ and $0 \le \phi \le 2\pi$.
I'm unsure if I've calculated correctly.

答:
Due to symmetry
$$\iiint_{\mathrm{E}}xy\, dx\,dy\,dz=0;\>\>\>
\iiint_{\mathrm{E}}x^2\, dx\,dy\,dz
=\iiint_{\mathrm{E}}y^2\, dx\,dy\,dz
=\iiint_{\mathrm{E}}z^2\, dx\,dy\,dz =I$$
Thus
$$\iiint_{\mathrm{E}}(x^2+xy+y^2)\, dx\,dy\,dz
=\frac23(3I)=\frac23\iiint_{\mathrm{E}}(x^2+y^2+z^2)\, dx\,dy\,dz\\
= \frac23 \int_0^{2\pi}\int_0^{\pi}\int_0^2r^2\>r^2\sin\theta drd\theta d\phi=\frac23\cdot2\pi\cdot2\cdot\frac{32}5=\frac{256}{15}\pi
$$


问:
Algebraic inequality $\sum \frac{x^3}{(x+y)(x+z)(x+t)}\geq \frac{1}{2}$
The inequality is
$$\frac{x^3}{(x+y)(x+z)(x+t)}+\frac{y^3}{(y+x)(y+z)(y+t)}+\frac{z^3}{(z+x)(z+y)(z+t)}+\frac{t^3}{(t+x)(t+y)(t+z)}\geq \frac{1}{2},$$
for $x,y,z,t>0$.

It originates from a 3-D geometry problem involving volumes of tetrahedra etc. Actually, it is equivalent with that problem (see Let ABCD be a tetrahedron of volume 1 and M,N,P,Q,R,S on AB,BC,CD,DA,AC,BD s.t. MP,NQ,RS are concurrent. Then the volume of MNRSPQ is less than 1/2.).

The three variables simpler case
$$\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+x)(y+z)}+\frac{z^2}{(z+x)(z+y)}\geq \frac{3}{4}$$
can be proved using the Cauchy-Schwarz inequality in "Engel's form".

I have tried variants of Holder type inequalities, until now unsuccessfully.

答:
This is a solution using Carlson inequality: if $x_{ij}(1\leq i\leq m,1\leq j\leq n)$ are non-negative reals, then
$$
\prod_{s=1}^m\sum_{t=1}^nx_{st}\geq\left(\sum_{t=1}^n\left(\prod_{s=1}^mx_{st}\right)^{\frac{1}{m}}\right)^{m}
$$

Proof. According to the Carlson inequality,
$$
\begin{align}
&\sum\frac{x^3}{(x+y)(x+z)(x+t)}\cdot\sum(x+y)\cdot\sum(x+z)(x+t)\\
\geq&\left(\sum x\right)^3\\
\because&\sum(x+y)=2(x+y+z+t),\sum(x+z)(x+t)=(x+y+z+t)^2\\
\therefore&\sum\frac{x^3}{(x+y)(x+z)(x+t)}\cdot2(x+y+z+t)\cdot(x+y+z+t)^2\geq(x+y+z+t)^3
\end{align}
$$
One can now easily observe that the original inequality is true.


问:
Is $\mathbb{S}^3 \backslash \mathbb{S}^1$ homotopic equivalent to $\mathbb{S}^1$?
Is $\mathbb{S}^3 \backslash \mathbb{S}^1$ homotopic equivalent to $\mathbb{S}^1$?

I am reading this from some notes on algebraic topology.

I am not even sure what does $\mathbb{S}^3 \backslash \mathbb{S}^1$ mean?

what is $\mathbb{S}^1 \in \mathbb{R}^3$? Is it any one particular circle? Can anyone please define this as a set?

How will $\mathbb{S}^2 \backslash \mathbb{S}^1$ look like? (homeomorphic to 2 open discs?)

答:
I assume that $S^1$ inside $S^3$ means some great circle of $S^3$, i.e. if

$$S^3=\big\{(x_0,x_1,x_2,x_3)\in\mathbb{R}^4\ \big|\ \sum x_i^2=1\big\}$$

then $S^1$ is for example $\{(x_0,x_1,x_2,x_3)\in S^3\ |\ x_2=x_3=0\}$ or any other combination of coordinates with two of them being $0$.

We most certainly cannot treat $S^1$ in $S^3$ as any homemorphic image, because the result then depends on the choice of such embedding.

If that's the case then we can apply the stereographic projection to $S^3\backslash S^1$ relative to some point in $S^1$, to obtain that this space is homeomorphic to $\mathbb{R}^3\backslash L$ with $L$ being a line, for example $L=\{(x,0,0)\ |\ x\in\mathbb{R}\}$.

And this space is homotopy equivalent to $S^1$. Indeed, we first construct a homotopy equivalence from $\mathbb{R}^3\backslash L$ to the tube $\mathbb{R}\times S^1$ via linear homotopy from the identity to $(x,y,z)\mapsto (x,c\cdot y, c\cdot z)$ map, where $c=1/\lVert (y,z)\rVert$. And then by contracting the $\mathbb{R}$ piece.


问:
Finite products of topological vector spaces
A finite product of an infinite Hilbert space is isometric to itself, since the isometry class of a Hilbert space is determined only by the cardinality of its dimension.
But what about less 'nice' topological vector spaces of infinite dimension? I am in particular interested in Hausdorff topological vector spaces, Hausdorff locally convex spaces, and Banach spaces (up to isomorphism in the respective category).</p>

答:
Nice question. I think that in the category of Banach spaces and linear contractions as morphisms (i.e., $\|f(x)\|\le \|x\|$) the space $c$ of all real convergent sequences with the sup-norm is an example: The unit ball of $c$ has two extreme points (the constant sequences with value $\pm 1$) but in the product $c\times c$ (in the mentioned category, this is endowed with the maximum of the norms of the two components) the unit ball has four extreme points.

However, in the usual category of Banach spaces with continuous linear maps as morphisms, $c\times c$ and $c$ are in fact isomorphic (because $c\cong \mathbb R \times c_0 \cong c_0\cong c_0\times c_0 \cong c\times c$, the first iso is $(x_n)\mapsto (x_\infty,x_n-x_\infty)$ where $x_\infty$ is the limit, and $c_0\times c_0\cong c_0$ by shuffeling two null sequences into a single sequence).

There might be simpler examples in this category, but I think the celebrated example of Argyros and Haydon of a Banach space $X$ where every continuous linear operator is of the form $\lambda id +K$ for a compact operator $K$ will be an example, because on $X\times X$ there are certainly other operators (the projections followed by the embeddings).


问:
Analysis question from a past exam paper
Assume that $f \in C^1 [a,b].$ Prove that $$\forall \epsilon \  \exists \delta > 0 : \\
x, y \in [a,b] , |x-y| < \delta \implies \left\vert\frac{f(x)-f(y)}{x-y} - f'(x) \right\vert < \epsilon $$

I did he following proof and I wonder if its entirely correct as the question itself would have carried 10 % of a 2hr exam.

Edit: edited the proof to make $\delta$ independent of $x$

Let $x,y \in [a,b]$

Since $f'$ is continuous on a compact set, then its uniformly continuous.
Hence
$\forall x,y \in [a,b], |x-y| < \delta \implies |f'(y)-f'(x)| < \epsilon $

By MVT, since $f \in C^1 [a,b], \ \exists c \in (x,y) : f'(c) = \frac{f(x)-f(y)}{x-y}$ .

Thus in particular, $|x-c|<|x-y| < \delta \implies |f'(c)-f'(x)| < \epsilon $ and we're done.

答:
Your proof is not correct as $\delta$ depends on $x$ in contradiction with what is requested. You can proceed as follows.

Choose $\epsilon \gt 0$. According to Heine–Cantor theorem it exists $\delta \gt 0$ such that $\left\vert f^\prime(t) - f^\prime(x) \right\vert \lt \epsilon$ for $\vert t - x \vert \lt \delta$.

To get the conclusion, pickup $x,y$ with $\vert x - y \vert \lt \delta$. According to the mean value theorem, for $x \neq y$, it exists $t \in (x,y)$ such that
$$\frac{f(x)-f(y)}{x-y} = f^\prime(t). $$ You have $\vert t - x \vert \lt \delta$ and therefore

$$\left\vert\frac{f(x)-f(y)}{x-y} - f^\prime(x) \right\vert < \epsilon $$


问:
Laplace transform of $u(t)=1/s$
I know that the Laplace transform of $u(t)$ is equal to $1/s$ (causal system). But the Laplace transform of the impulse response of the integration operation is also equal to $1/s$. Intuitively, could someone tell me how they are related? $u(t)$ is a constant for $t>0$. But an integrator's output is increasing linearly with respect to time.

答: I suppose I will answer my own question here but please do add additional comment if there is a better way to interpret it. So can we say that the inverse Laplace transform of 1/s is u(t) which exactly describes the integration operation because integration keeps its output forever? That is why it is a constant when we put an impulse into the system. The output, u(t), describes what the output looks like after a spike is applied to the input. It just remembers it.</p>

Title: Laplace transform of $u(t)=1/s$