问:
Finding points on $\vec r(t) = \vec At^3 + \vec Bt^2 + \vec Ct + \vec D$ where the tangent is parallel to the line $px + qy + k = 0$
Given a line defined by the equation:
$$px + qy + k = 0$$
and a parametric cubic curve defined by:
$$\vec r(t) = \vec At^3 + \vec Bt^2 + \vec Ct + \vec D$$
where both curves lie in 2D space, how can I find all points on the cubic curve $t$ where the line is parallel to the tangent in these points?

答:
Considering that $\vec r(t) = (x(t),y(t))$ and that the line

$$
p x + q y + k = 0
$$

can be written as $\vec r(t) = p_0 + \vec v t$ or $\cases{x = x_0 + v_x t\\ y = y_0 + v_y t}$ with the relationship between $p_0,\vec v, p, q, k$ which follows (assuming $v_x\ne 0$ and $v_y\ne 0$)

$$
\frac{x-x_0}{v_x}=\frac{y-y_0}{v_y}\Rightarrow v_y x- v_x y - x_0y_y+y_0 v_x = 0
$$

then $p = v_y, q = -v_x, k = - x_0y_y+y_0 v_x$

so we follow with $\vec r_1(t) = \vec At^3 + \vec Bt^2 + \vec Ct + \vec D$ and $\vec r_2(t) = p_0 + \vec v t$

Now the tangent vectors to $\vec r_1(t)$ are given by $\vec r_1'(t) = 3\vec At^2 + 2\vec Bt + \vec C$ and the tangent vectors to $r_2(t)$ are given by $\vec r_2'(t) = \vec v$. Finally if $\vec r_1'(t)$ and $\vec r_2'(t)$ are parallel then

$$
\vec r_1'(t) \times \vec r_2'(t) = \vec 0
$$

etc.


问:
Breaking down a conditional probability into a marginal probability
I was reading this blog post on using Bayesian statistics to obtain the probability of 2 heads in a row for a coin toss, given that 14 tosses resulted in 10 heads.

I am confused on how the highlighted step in the image was obtained on the page. If anyone could clarify or prove the equality, I'd really appreciate it.

Thanks!

答:
I am confused on how the highlighted step in the image was obtained on the page. If anyone could clarify or prove the equality, I'd really appreciate it.

First observe that

$$\mathbb{P}[HH,p|\text{data}]=\mathbb{P}[HH|p,\text{data}]\cdot\mathbb{P}[p|\text{data}]$$

now considering the assumption they made of conditional independence (w.r.t. the parameter  $p$) of the observations you get

$$\mathbb{P}[HH,p|\text{data}]=\mathbb{P}[HH|p]\cdot\mathbb{P}[p|\text{data}]$$

At this point, to eliminate $p$ in the left member you have to integrate both members in $dp$ obtaining

$$ \bbox[5px,border:2px solid black]
{
\mathbb{P}[HH|\text{data}]=\int_0^1\mathbb{P}[HH|p]\cdot\mathbb{P}[p|\text{data}]dp
\
}
$$

Observe that inside the integral you have

$$\underbrace{\mathbb{P}[HH|p]}_{=\text{model}}\cdot\underbrace{\mathbb{P}[p|\text{data}]}_{=\text{Posterior Distribution}}$$

ContentLicense: CC BY-SA 4.0


问:
Inductively prove that this sequence of integrals is bounded.
EDIT: I have an attempted solution to this in a post below, it is very long, but still incomplete.

EDIT:Alright, I've pretty much almost finished my solution, but my biggest problem is the 2nd integral, rightmost inequality. I cannot find a way to evaluate it and compare it to the n=1 case for the upper integral, which should give me what I need.

EDIT: still stuck!

EDIT: 10 days later, I still need help with this..

Can someone please help me with this? I'm not even sure where to go from here.

Here is the problem:

Prove that

$$\frac{1}{2}<\int_{0}^{\frac{1}{2}}\frac{dx}{\sqrt{1-x^{2n}}}\le0.52359,$$

for any integer $n\ge1$,

And that

$$\frac{1}{2}<\int_{0}^{1}\frac{dx}{\sqrt{4-x+x^{3}}}\le0.52359$$

Now I can do this for the $n=1$ case, as obviously this simply integrates to $\arcsin{\frac{1}{2}}$ giving $\frac{\pi}{6}$, but I'm completely stumped on any doing any further for this question.

All help very appreciated, thank you for your time.

答:
First note that the leftmost inequality is trivial, since $\frac{1}{\sqrt{1-x^{2n}}}\geq 1$.

Now note that for $x<1$, $x^{2(n+1)} = x^2x^{2n} \leq x^{2n}$. Hence $\frac{1}{\sqrt{1-x^{2(n+1)}}}\leq\frac{1}{\sqrt{1-x^{2n}}}$ for all $x<1$. Thus $\displaystyle\int_0^\frac{1}{2}\frac{\mathrm{d}x}{\sqrt{1-x^{2(n+1)}}} \leq \displaystyle\int_0^\frac{1}{2}\frac{\mathrm{d}x}{\sqrt{1-x^{2n}}}$. The rightmost inequality now follows by induction.

ContentLicense: CC BY-SA 3.0


问:
If $V$ is $k$-dimensional then show that $\dim \wedge^n V = \binom {k} {n}.$

If $V$ is $k$-dimensional then show that $\dim \wedge^n V = \binom {k} {n}.$

I don't understand this result. If $k \lt n$ then how does the quantity $\binom {k} {n}$ make sense? Can anybody please help me in this regard?

Thanks for your time.

EDIT $:$ Let $V$ be a vector space. Then for any $n \in \Bbb N$ we can identify $V^{\times n}$ by $V^{\{1,2, \cdots, n \}}.$ Since $\{1,2, \cdots, n \}$ is an $S_n$-set it follows that $V^{\{1,2, \cdots, n \}}$ is an $S_n$-vector space. For any $\sigma \in S_n$ and any function $f : \{1,2, \cdots, n \} \longrightarrow V$ the action of $\sigma$ on $f$ is given by $(\sigma \cdot f) (i) = f (\sigma^{-1} (i)).$ This shows that $V^{\times n}$ is a $S_n$-vector space. Let $\mu$ be the action of $S_n$ on $V^{\times n}$ then for each $\sigma \in S_n$ the map $\otimes_V \circ \mu (\sigma, \cdot) : V^{\times n} \longrightarrow V^{\otimes n}$ will be a bilinear map. Then by universal property of tensor products there exists a unique bijective linear map $\rho (\sigma) : V^{\otimes n} \longrightarrow V^{\otimes n}.$ This gives a group homomorphism $\rho : S_n \longrightarrow GL(V^{\otimes n}).$ Hence $V^{\otimes n}$ is also a $S_n$-vector space where for any element $\sigma \in S_n$ and for any $v \in V^{\otimes n}$ the action of $\sigma$ on $v$ is given by $\sigma \cdot v : = \rho (\sigma) (v).$ If $\rho (\sigma) (v) = \left (\text {sign} (\sigma) \right ) v,$ for all $\sigma \in S_n$ and for some $v \in V^{\otimes n}$ then $v$ is said to be an alternating/anti-symmetric tensor. The collection of all alternating/anti-symmetric tensors forms a subspace of $V^{\otimes n}$ which is denoted by $\wedge^n V$ or $V^{\wedge n}.$

答:
You could try constructing a basis for $\bigwedge^n V$, that is proving the following result:

If $\{e_i\}_{i=1}^k$ is a basis for a vector space $V$ over $\mathbb{F}$, then
\begin{align*}
    \{e_{i_1} \wedge e_{i_2} \wedge \ldots \wedge e_{i_n} : 1 \leq i_1 < i_2 < \ldots < i_n \leq k\}
\end{align*}
is a basis for $\bigwedge^n V$ (where $\wedge$ for now is just a formal symbol).

From this it follows that $\dim \bigwedge^n V = \binom{k}{n}$. If $k<n$, by the definition of the binomial coefficient, the dimension of $\bigwedge^n V = 0$.

This makes perfect sense as the basis for $\bigwedge^nV$ are all of the possible combinations of $k$ basis vectors (in increasing subscript order). If we try to make a combination of $k>n$ basis vectors, there are zero such combinations in $\bigwedge^n V$, as all vectors in $\bigwedge^n V$ are alternating (i.e. do not have repeated elemetns, which can be shown by your definition of $\rho$).


问:
Big Oh notation and inequality
Consider these expressions.
$B_{2}(x)=x\ln(2)+ O(\ln(x))$ and $\psi(x)\geq B_{2}(x)$

In the next step,
$ \psi(x)\geq x\ln(2)+ O(\ln(x)) ............(1)$

What does the expression (1) actually mean, mathematically? From what I know, when a big oh notation is involved, there is an equal to sign between LHS and RHS of the expression and the definition of big oh notation makes sense, but I am not to understand what it means when an inequality in involved like in (1).

答:
$O(f)$ is set of functions with well known definition(considering non-negative case): $h \in O(f)$ iif $\exists C>0, \exists$ neighbourhood for $x$, such that $h(x) \leqslant C f(x)$

So, equality means belonging: $B_{2}(x)=x\ln(2)+ O(\ln(x))$ is same with $B_{2}(x)\in x\ln(2)+ O(\ln(x))$ where in right side is again set $g+O(f)=\{g+\phi \colon \phi \in O(f) \}$.

Now, $\psi \geqslant g+O(f)$ we can understand as $\psi \geqslant h$ for $\forall h \in g+O(f)$.

Title: Big Oh notation and inequality


问:
$C_0(\Omega)$ is not dense in $L^{\infty}(\Omega)$.

Let $\Omega \subset \mathbb{R}^n$ be open set. Show that $C_0(\Omega)$ is not dense in $L^{\infty}(\Omega)$.

Notation :$f \in C(\Omega)$ and f has compact support then $f \in C_0(\Omega)$.

My attempt: Assume that $C_0(\Omega)$ is dense in $L^{\infty}(\Omega)$. Lets seek a contradiction. By the assumption for all $\varepsilon > 0$ there exist a function $g \in C_0(\Omega)$ such that $|f-g|_{\infty} < \varepsilon$. So we have $\mu(x \in \Omega: |f-g| > \varepsilon) = 0$. I tried to find an open subset of $\{x \in \Omega: |f-g| > \varepsilon\}$, but i failed. Any help would be apperciated.

答:
Let $f$ be any discontinuous bounded measurable function on $\Omega$. (For example, the characteristic function of an open disk inside $U$). If there is  sequence $(f_n) \subseteq C_0(\Omega)$ converging to $f$ in $L^{\infty}$ then $\|f_n-f_m\| \to 0$ which implies that $f_n$'s converge uniformly to $f$. But uniform limit of continuous functions is continuous.

[$L^{\infty}$ norm of a continuous function is same as its supremum norm].


问:
Let $(s_n)$ be a sequence with $s_1, s_2, ..., s_{100}$ being positive and $s_n < 0 $ for all $n > 100.$ Assume $\lim s_n = 0$
Let $(s_n)$ be a sequence with $s_1, s_2, ..., s_{100}$ being positive and $s_n < 0 $ for all $n > 100.$ Assume $\lim s_n = 0$

What is $\lim \frac{1}{s_n} $?

My thought:

Since $\lim s_n = 0$,

$\forall \epsilon > 0,$ there exists $N$ such that $\forall n > N,$   $\mid s_n - 0 \mid <\epsilon$

Then $\mid s_n \mid < 0$

$-\epsilon < s_n < \epsilon$

$-\epsilon < s_n$ (only care about lower bound)

$\frac{-\epsilon}{s_n} > 1$ (because $n$ should be large enough for $s_n$ to converge to $0$?? Then $s_n$ is negative.. question point)

$\frac{1}{s_n} < - \frac{1}{\epsilon}$

Therefore $\frac{1}{s_n}$ converges to $-\infty$.

I was wondering if that question point step works.. and also other steps.

答:
What you wrote is fine until (and including) $-\varepsilon<s_n<\varepsilon$.

At this point, you should use the fact that $n>100\implies s_n<0$ to deduce that, if $N$ is large enough, then $-\varepsilon<s_n<0$, which implies that $0<-s_n<\varepsilon$ and that therefore $-\frac1{s_n}>\frac1\varepsilon$, which is equivalent to $\frac1{s_n}<-\frac1\varepsilon$.

So, it is indeed true that $\lim_{n\to\infty}\frac1{s_n}=-\infty$: for any $M<0$, you know that $$\frac1{s_n}<\frac1{-1/M}$$if $n$ is large enough, and this is the same thing as asserting that $\frac1{s_n}<M$.


问:
Why the set of polynomials is not closed in $C[0,1]$
This question starts from the following theorem: "Every finite dimensional subspace $Y$ of
a normed space $X$ is closed in $X$" (Kreyzig 2.4-3)

So the given counterexample for an infinite dimension subspace, is that if we take $X=C[0,1]$ and $Y = span\{x_0,x_1,x_2...\}$, where $x_j = t^j$, then $Y$ is not closed.

I guess this is the case because for example a sine function can be expressed as the limit point of a sequence of polynomials that converge uniformly (Taylor series). But why is the sine function not considered as a polynomial generated by an infinite basis? After all, we are explicitly saying $Y$ is infinite dimensional. For an infinite dimensional basis case, I would expect to have a mapping between every sequence to a polynomial in $C[0,1]$. If this is not the case, what exacly does it mean for $span\{x_0,x_1,x_2...\}$ to be considered infinite dimensional? If I input an infinite basis, it seems like it is not longer a polynomial.

I am confused.

答:
To talk about convergence or closure, we first need to equip $C[0,1]$ with a topology. Which may be induced by a norm, though it's not mandatory. Here, since you are mentioning uniform convergence, define the norm on $C[0,1]$ by $||f||_\infty=\sup_{[0,1]}|f|$, so that $C[0,1]$ is now a Banach space. And convergence is defined as uniform convergence, i.e. $f_n\to f$ in $C[0,1]$ iff $\sup_{[0,1]} |f_n-f|\to 0$.

The subset $P$ of polynomials is closed in $C[0,1]$, iff for any sequence of polynomials $(p_n)_n$ that converges in $C[0,1]$, i.e. $p_n\to p$, then $p\in P$.

However, the Stone-Weierstrass theorem tells us that for any continuous function $f$ in $C[0,1]$, there is a sequence of polynomials $p_n$ such that $p_n\to f$, i.e. $||p_n-f||_\infty\to 0$. Take any $f$ in $C[0,1]$ that is not a polynomial, and you have a counterexample of the condition above, which proves that $P$ can't be closed in $C[0,1]$.

In fact, Stone-Weierstrass theorem means that with this definition of convergence (i.e. for this topology), $\overline P=C[0,1]$, that is, polynomials are dense in $C[0,1]$.

By the way we also have that $\mathring P=\emptyset$, since every polynomial $p$ is trivially the (uniform) limit of a sequence in $C[0,1]\backslash P$, for instance $f_n(x)=p(x)+\frac1n\exp(x)$. And $\mathring P\ne P$ proves that $P$ is not open.


问:
How to define summation on a general space
We have the notation of summation on the space of Real numbers. Now I want to define the notation of summation on a more general space (for example, space of real-valued function).

What is the general procedure to define such a notation ? Which set of criteria that I have to verify to ensure that my definition of summation is logically accepted ?

答: You are talking about 'group'. Just check it here: https://en.wikipedia.org/wiki/Group_(mathematics)</p>


问:
why is an annulus close to it's boundary when it's boundary curves are close?
This is the motivating question for the rather vague question here: when is the region bounded by a Jordan curve "skinny"?

Suppose we are given two Jordan curves in the plane, one inside the other and each contained in an epsilon neighborhood of the other. How can we conclude that the annular region between the two curves is itself contained in the two epsilon neighborhoods of the original curves?

Note: by epsilon neighborhood of a curve I mean the set of all points which lie within a distance epsilon from some point in the image of the curve.

Follow up: Since the statement above is false (see Hagen's answer) we might ask whether it is true if we assume point-wise closeness of chosen parametrizations of the curves.  This is answered in the affirmative by user7...

答:
I think this may be very similar to the counterexample that Hagen von Eitzen described in words, but I'm not entirely sure. So here's a fooplot.

If you assume that $|f(\theta)-g(\theta)|< 2\epsilon$ for all $\theta$ (where $f$ and $g$ are the two curves), then the conclusion is true. Indeed, suppose that the disk $|z-a|< \epsilon$ is contained in the annulus. We may and do assume $a=0$. Then the winding numbers of $f$ and $g$ around $0$ are different: one is $0$, the other is $1$. On the other hand, I claim that $$f_t(\theta)=(1-t)f(\theta)+tg(\theta)$$
is a homotopy between $f$ and $g$ in $\mathbb R^2\setminus \{0\}$.

Indeed, if $f_t(\theta)=0$ for some $\theta$, then the points $f(\theta),0, g(\theta)$ lie on a line, in the order listed. Hence $|f(\theta)-g(\theta)|=|f(\theta)|+|g(\theta)|\ge 2\epsilon$, which is a contradiction.


问:
A question about expectation of $L^{p}$ norms.
Let $X$ be a random vector, $x,y\in\mathbb{R}^{n}$ and $p\geq 1$. My question is the following. If we know that $\text{E}[\|X - x\|_{p}^{p}]\leq\text{E}[\|X - y\|_{p}^{p}]$ (provided that they exist), we can deduce that $\text{E}[\|X - x\|_{p}]\leq\text{E}[\|X - y\|_{p}]$ ?.

I try to use Jensen's inequality, but it doesn't work.

Thanks in advance.

答:
Consider the case $n=1$, and $p=2$ (note that the statement is always trivially true for $p=1$).

We choose $x=0$, and $y=-1$, and define $X$ as a random variables on $\{-1,1\}$ such that
$$
X = \begin{cases}
1 & \text{ with probability } 1/3\\
-1 & \text{ with probability } 2/3
\end{cases}
$$
Clearly, $\mathbb{E}[(X-x)^2]=\mathbb{E}[|X-x|]=1$. However, one can easily compute that
$$
\mathbb{E}[(X-y)^2] = \mathbb{E}[(X+1)^2] = \frac{4}{3} > \mathbb{E}[(X-x)^2]
$$
but
$$
\mathbb{E}[|X-y|] = \mathbb{E}[|X+1|] = \frac{2}{3} < \mathbb{E}[|X-x|]
$$


问:
Does every finite $CW$ complex have the homotopy type of a smooth manifold? How about infinite $CW$ complexes?
Let M be a smooth manifold. I was wondering if every finite or infinite CW complex has the homotopy type of a smooth manifold because it seems way easier to compute De Rham cohomology groups than homology groups.

If the answer to the questions are yes, is it hard to find a smooth approximation of a CW complex?

答:
This is a partial answer.

If $X$ is a finite CW-complex, then in general there does not exist a compact manifold which is homotopy equivalent to $X$.

It clearly suffices to consider connected CW-complexes because we can argue componentwise.

So let $X$ be a finite connected CW-complex of dimension $n$ such that its $n$-th homology group with coefficients in $\mathbb Z_2$ has the property $H_n(X;\mathbb Z_2) \ne \mathbb Z_2$. An example is $S^n \vee S^n$. Assume that $X$ is homotopy equivalent to a compact manifold $M$ of dimension $m$. It must clearly be connected.
A well-known fact (see e.g. Hatcher Theorem 3.26) is that $H_k(M;\mathbb Z_2) = 0$ for $k > m$ and $H_m(M;\mathbb Z_2) = \mathbb Z_2$. Hence $n \ge m$ is impossible. But also $n < m$ is impossible because in that case $H_m(X) = 0$.

This leaves open whether there always exists a non-compact manifold $M$ which is homotopy equivalent to $X$. In the example $X  = S^1 \vee S^1$ this is the case.

Update:

In Moishe Kohan's answer you see that the answer is "yes" for finite (connected) CW-complexes and non-compact (connected) manifolds. Note that if $X$ is homotopy equivalent to such $M$, then necessarily $\text{dim} M  >  \text{dim} X$ provided $H_{\text{dim} X}(X) \ne 0$. See Hatcher Proposition 3.29.


问:
Stability of a discrete-time closed-loop
I modeled a dynamic system like so:

$$
\dot{y} = au,
$$

i.e. as an integrator.

Every $T = 0.1$ seconds, the measurement is updated and remains constant in between. This motivated me to model it as a discrete-time system using zero-order hold:

$$
\int_t^{t+T} \dot{y} \; \text{dt} = y_{k+1} - y_k = au_kT= au(t) \int_{t}^{t+T} \text{dt},
$$

where $u(t) = u_k = \text{const.}$ if $t \in [t_k, t_k + T)$.

By means of looking at the input-output behavior, I determined parameter $a = 2$ (omitting engineering units).

Next, I normalized the system equation:

$$
y_{k+1} \cdot \frac{y_{ref}}{y_{ref}} = y'_{k+1} \cdot y_{ref} = y'_k \cdot y_{ref} + 0.2 u'_k \cdot u_{ref} = y_k \cdot \frac{y_{ref}}{y_{ref}} + 0.2 u_k \cdot \frac{u_{ref}}{u_{ref}}.
$$

By choosing the reference values ($y_{ref} = 3$, $u_{ref} = 1$), I end up with

$$
y'_{k+1} = y'_k + \frac{2}{30} \cdot u'_k.
$$

As control law, I choose a P-controller, i.e. proportional feedback

$$
u'_k = P \cdot (y'_{ref} - y'_k).
$$

Inserting the control law yields the closed-loop system dynamics

$$
y'_{k+1} = y'_k \cdot (1 - \frac{2}{30}P) + \frac{2}{30}Py'_{ref}.
$$

From the stability criterion of a discrete-time system (absolute value of coefficient of $y'_k$ smaller $1$), I would expect every choice of $P \in (0, 30)$ to render the system stable.

But it turns out that choosing $P=5$ exhibits oscillations with growing amplitudes caused by small initial disturbances, i.e. instable behavior:

While choosing $P=2.5$ attenuates disturbances, i.e. renders the system stable:

What causes the mismatch between theory and actual behavior?

答:
Following the guide of A.Γ., I took a second look at the simulation results of the linear model compared with measurements:

Apparently, there is a difference albeit not a large one. Mainly due to the linearization which ignores the change of rate over time of the physical system. While this definitely affects stability margins in the closed loop, it does not justify the large difference between observed destabilizing control parameter $P = 5$, and theoretical value $P = 30$.

But then I opposed simulation results to measurements for two times the input value of the previous one:

This reveals yet another nonlinearity: The rate of change of output does not only increase over time, it also increases unproportionally with input value.

Scaling the slope of the integrator with factor $5$ makes the two systems match better:

For $a=10$, the stability margin reduces when choosing $P=5$:

$$
|1 - \frac{10}{30}\cdot P| = |1 - \frac{1}{3} \cdot 5| = \frac{2}{3} < 1
$$

Theory still says it is stable, but I can live with that in the light of this very simple model and its rough approximations.

What still puzzles me is the strong dependence of system behavior on absolute input value. But this might be something for the materials section.


问:
What is the dimension of the "Julia set" generated by inverse iteration and why do I get numbers different from Hausdorff dimension
On the Julia set, the iterating the function $f:z\mapsto z^2+c$ generates a sequence of points  on the Julia set.  Because roundoff errors would be expected to cause the sequence to fall into an attractive cycle, I used inverse iteration $f_i:z\mapsto \pm (z-c)^{1/2}$, where a random number generator was used to pick the sign, to generate large numbers of elements for the (arbitrarily picked) value of $c = -1$.  The set looks like

I generated large (1,000,000) sets this way and For a set of random reference points in the set, I calculated the number of points within various radii of each reference point.  The following graph is a log-log plot of the result:

Where the x axis is the radius and the y axis is the number of points within the radius.  The radii ranged from 0.0001 to 0.0512 in factor of 2 steps.  The slope of the line is about 0.855, suggesting a dimension less than 1, which is contrary to my intuition for this Julia set.

My guess is that I am studying a chaotic dynamic system, which results in sets that have a different dimension.  From a naive statistical point of view, if I have N >>1 points, and the number of points in a ball of radius r goes as $r^{0.855}$ then the number of balls needed to cover the set should go as $r^{-0.855}$, suggesting that the Hausdorff dimension is 0.855.

答:
Lutz comment on the density obtained via simple inverse iteration is certainly correct. There are, though, ways to modify the inverse iteration algorithm to obtain a more uniform distribution. You can find a description of these techniques with an implementation using Mathematica in this paper. The main algorithm described in that paper is built into Mathematica as of version 10 and can be accessed via a command like

```<code>JuliaSetPlot[z^2 - 1, z, ColorFunction -> None, PlotStyle -> Black]
</code>```

The modified inverse iteration algorithm is also implemented in this web app, which I used to generate this image:

Performing box-counting on that image, I compute a fractal dimension of about 1.367.

Body:
Lutz comment on the density obtained via simple inverse iteration is certainly correct. There are, though, ways to modify the inverse iteration algorithm to obtain a more uniform distribution. You can find a description of these techniques with an implementation using Mathematica in this paper. The main algorithm described in that paper is built into Mathematica as of version 10 and can be accessed via a command like

JuliaSetPlot[z^2 - 1, z, ColorFunction -> None, PlotStyle -> Black]

The modified inverse iteration algorithm is also implemented in this web app, which I used to generate this image:

Performing box-counting on that image, I compute a fractal dimension of about 1.367.


问:
Deducing a result about entire functions
We need to show that for an entire function $f$ on $\mathbb{C}$, there are constants $a_1,...,a_n$ such that $\int^{2 \pi}_{0}|f(re^{i \theta})|^2 d\theta=2\pi\sum_{n=0}^{\infty}|a_n|^2r^{2n}$.

Thoughts so far are that we can find a power series expansion about $0$ by Taylor's Theorem. Given the result it seems natural to use $f(z)^2$ somewhere but I can't see how.

Any help appreciated.

答:
Let $f(z)=\sum\limits_{n=0}^\infty a_nz^n$. For $z=re^{i\theta} $ we have $$f(re^{i\theta })=\sum\limits_{n=0}^\infty a_nr^ne^{in\theta }$$
We have that $$|f(re^{i\theta})|^2=f(re^{i\theta})\cdot \overline{f(re^{i\theta})}=$$
$$\sum\limits_{n=0}^\infty a_nr^ne^{in\theta }\cdot \sum\limits_{m=0}^\infty \overline {a_m}r^me^{-im\theta }=$$
$$\sum\limits_{n=0}^\infty \sum\limits_{m=0}^\infty a_n \overline {a_m}r^{n+m}e^{i(n-m)\theta }$$
Therefore $$\int_0^{2\pi }|f(re^{i\theta})|^2d\theta =$$
$$\int_0^{2\pi}\sum\limits_{n=0}^\infty \sum\limits_{m=0}^\infty a_n \overline {a_m}r^{n+m}e^{i(n-m)\theta }d\theta =(*)$$
$$\sum\limits_{n=0}^\infty \sum\limits_{m=0}^\infty \int_0^{2\pi} a_n \overline {a_m}r^{n+m}e^{i(n-m)\theta }d\theta=(*)$$
$$\sum_{n=0}^\infty a_n\overline{a_n}r^{n+n}\cdot 2\pi= $$
$$2\pi\sum_{n=0}^{\infty}|a_n|^2r^{2n}$$
$(*)$Where we can use uniform convergence to switch integral and sums, and also note that $$\int_0^{2\pi}e^{i(n-m)\theta}d\theta=\begin{cases}0&n\neq m\\2\pi&n=m\end{cases}$$</p>


问:
Ratio of solutions of differential equation

Prove, that the ratio of two linearly independent solutions of
$$
x''+p(t)x'+q(t)x=0,
$$
where $p(t)$, $q(t)$ are continuous coefficients, can not have a local maximum.

I rewrote this as a system of equations:
$$
\begin{cases}
y_1'=y_2 \\
y_2'+p(t)y_2+q(t)y=0
\end{cases}
$$

I multiplied the second equation by $e^{\int pdt}$ and got:
$$
(y_2e^{\int pdt})'=-q(t)e^{\int pdt}y_1
$$
and after integrating both sides:
$$
y_2e^{\int pdt}=\int-q(t)e^{\int pdt}y_1dt
$$

Unfortunately, I don't know what to do next.

Also, I had an idea to assume one of the solutions to be $V=\begin{pmatrix} v_1\\ v_2 \end{pmatrix}$ and then try to get the oter one from the Liouville's formula, however I don't know whether that would work.

答:
Here is an outline of a solution.  Let $y_1, y_2$ be two linearly independent solutions and write $z= y_1/y_2$.  By symmetry, the sam argument can be applied to $y_2/y_1$.  Then
\begin{align*}
z' = \frac{y_1'y_2 - y_2'y_1}{y_2^2} \tag{1}\label{A}
\end{align*}
and, using the original differential equation
\begin{align*}
z'' &= -2y_2'\frac{y_1'y_2-y_2'y_1}{y_2^3}+\frac{y_1''y_2+y_1'y_2'-y_1'y_2'-y_2''y_1}{y_2^2} \\
&=-2y_2'\frac{y_1'y_2-y_1y_2'}{y_2^3}+\frac{(-p y_1'-qy_1)y_2-(-py_2'-qy_2)y_1}{y_2^2} \\
&=-2y_2'\frac{y_1'y_2-y_1y_2'}{y_2^3}-p\frac{y_1'y_2-y_2'y_1}{y_2^2} \tag{2}\label{B}
\end{align*}
so that
\begin{align*}
z'' + \left( 2\frac{y_2'}{y_2}-p \right) z' = 0
\end{align*}
and the function $z'$ satisfies a first order linear differential equation.  Now, if $y_1 / y_2$ has a local maximum or minimum then $(y_1/y_2)' = 0$ at some point, which is to say $y_1'y_2 - y_1y_2' = 0$ at that point.  The equation \eqref{A} for $z'$ implies $z' = 0 $ at the same point.  We now have $z'$ satisfies a first order linear differential equation with $z'=0$ at a point, which we can take to be the initial condition, so that the solution $z'$ must be identically zero.  The function $z$ is therefore constant and $y_1$ is a multiple of $y_2$, contradicting their linear independence.


问:
Counting the number of ways to get a sum of $2012$, given that any number of cards can be chosen

A person has $32$ cards out of which $10$ cards, each of blue, green and red colours, have
denominations as $\{2^1, 2^2,\cdots , 2^{10}\}$ and one black and one white each of whose value is unity. What is the number of ways in which the person can get a sum of $2012$ if he can choose any number of cards?

I couldn't think much in it, except that number of ways of choosing $\text{(blue,black,white)}$ is equal to if he chooses $\text{(green,black,white) or (red,black,white)}$.

So if suppose he chooses $\text{(blue,black,white)}$, the total sum of the cards would be $$1+1+2+2^2+\cdots+2^{10}=2^{11}=2048$$

So, if we remove the cards whose sum equals $36$, we will get $2012$. Thus, we need to find the number of ways in which the sum on cards equals $36$. I tried it, but there were so many cases, that I gave up. I also haven't taken the case if he chooses for instance $\text{(blue,green,white)}$ etc.

The answer to this is $1007^2$, which suggests some easy method must be there.

Any hints would be appreciated!

答:
Let's consider only the blue, green and white cards chosen. If the total sum of all numbers on these cards is $2012$ or less, than there is only a single way to choose remaining red and black cards containing numbers $\{2^0, 2^1, \dots, 2^{10} \}$ such that the total sum of all cards is $2012$. (Prove that)

Therefore, the number of ways (let's denote it by $N_1 (1006)$) in which the person can get a sum of $2012$ if he can choose any number of cards equals the number of ways in which the person can choose any number of blue, green and white cards such that the sum of numbers on cards is $2012$ or less.

By $N_2 (k) $ let's denote the number of ways to get a sum of $2k$ or less on green and blue cards, where $k \leq 1023 $.
By ${\rm Sum_{green}} (\leq j) = j+1$ let's denote the number of ways to get a sum of $2j$ or less on green cards only, $j \leq 1023$, by $ {\rm Sum_{blue}} ( m) = 1$ let's denote the number of ways to get a sum of exactly $2m$ on blue cards only, $m \leq 1023$.

Note that $\displaystyle N_2(k) = \sum\limits_{i=0}^k \left( {\rm Sum_{blue}} (i) \cdot {\rm Sum_{green}} (\leq k - i) \right) = \sum\limits_{i=0}^k (k - i + 1) = \frac{ (k+1)(k+2)}{2} $ .

The $N_1(1006)$ is the sum of the number of ways to chose green, blue and white cards such that a white card is present ($=N_2(1005)$) and the number of ways to chose green, blue and white cards such that a white card is absent ($=N_2(1006)$). Therefore

$\displaystyle N_1(1006) = N_2(1005) + N_2(1006) = \frac{1006 \cdot 1007}{2} + \frac{1007 \cdot 1008}{2} = 1007^2 $


问:
Maximum and minimum value of $\frac{8x - 3y}{\sqrt{4x^2+y^2}}$?
For real numbers $\,x, y\neq 0\,$ consider
$$\frac{8x - 3y}{\sqrt{4x^2+y^2}}\,.$$
How to find the maximum and minimum value?

I've already got the maximum by using the Cauchy–Schwarz inequality
$$\big[(2x)^2 + y^2\big]\big[4^2 + (-3)^2\big] \geq (8x - 3y)^2\\[3ex]
25 \geq \frac{(8x - 3y)^2}{(2x)^2 + y^2}\\[4ex]
5 \geq \frac{8x - 3y}{\sqrt{(2x)^2 + y^2}}$$

But I cannot get the minimum value.

答:
Let $r > 0$ and $0 \le \theta < 2\pi$, and $x = r \cos \theta$, $y = 2r \sin \theta$, so that $$4x^2 + y^2 = 4r^2,$$ hence $$\begin{align}f(x,y) &= \frac{8x-3y}{\sqrt{4x^2 + y^2}} \\ &= \frac{8r \cos \theta - 6r \sin \theta}{2r} \\ &= 4 \cos \theta - 3 \sin \theta \\
&= 5 \left(\frac{4}{5} \cos \theta - \frac{3}{5} \sin \theta \right) \\
&= 5 \left( \sin \psi \cos \theta - \cos \psi \sin \theta \right) \quad \quad \psi = \arcsin \frac{4}{5} \\
&= 5 \sin(\theta - \psi).
\end{align}$$  Now we can see this function depends only on the choice of angle $\theta$.  This is obviously maximized when $\theta = \psi + \pi/2$ and the maximum attained is $5$.

This also happens to get us the minimum for free, which occurs when $\theta = \psi + \frac{3\pi}{2},$ and the minimum is $-5$.


问:
Let $(X, d)$ be a metric space and $a \in X$. Show that the intersection of all neighborhoods of $a$ is the singleton $\{a\}$.

Let $(X, d)$ be a metric space and $a \in X$. Show that the intersection of all neighborhoods of $a$ is the singleton $\{a\}$.

Let $b \in X$ and $\varepsilon = |a-b|$. Now since $(a-\varepsilon, a+\varepsilon)$ is a neighborhood of $a$ I have that if $b \in (a-\varepsilon, a+\varepsilon) \Longrightarrow |a-b| < \varepsilon$ which would be a contradiction and would imply that $b$ doesn't belong in the collection of neighborhoods of $a$. What confuses me is that we're talking about general metric spaces and not $\mathbb{R}$, how should this be shown for a general metric space?

答: If $b\ne a$, then $b\notin B_{d(a,b)}(a)$. So, for every $b\in X\setminus\{a\}$, there is some neighborhood of $a$ to which $b$ does not belong. In other words, the only point of $X$ which belongs to every  neighborhood of $a$ is $a$ itself.


问:
Lipschitz modulus of quadratic loss function
I'm trying to figure out the Lipschitz modulus of quadratic function $\ell(\xi)=\xi^{\top} Q \xi+2 q^{\top} \xi + q_0$ with positive definite $Q$. According to the definition of Lipschitz modulus $\operatorname{Lip}(\ell)=\sup \left\{\|z\|_{*}: \ell^{*}(z)<\infty\right\}$, I need to find the supremum of the dual norm in the domain of the conjugate function of the quadratic, which is $\frac{1}{2}(z-q)^{\top}Q^{-1}(z-q) - q_0$. Since $z$ could be arbitrary vector, the supremum of dual norm is $\infty$. Am I right? Any hint would be appreciated.

答:
A quadratic function with positive definite matrix is strongly convex and for that reason, has unbounded Lipschitz modulus. First, notice that since $Q\in\mathbb{S}_{++}^n\Rightarrow \lambda_{\min}(Q)\geq m>0$, hence your quadratic function is strongly convex. Since it is $\sigma=\lambda_{\min}(Q)$ - strongly-convex, it implies that
$$ f(\mathbf{y})- f(\mathbf{x})\geq \nabla f(\mathbf{x})^T(\mathbf{y}-\mathbf{x})+\frac{\sigma}{2}\|\mathbf{y}-\mathbf{x}\|^2\overset{\textbf{C.S.}}{\geq}\|\mathbf{y}-\mathbf{x}\|(\frac{\sigma}{2}\|\mathbf{y}-\mathbf{x}\|-\|\nabla f(\mathbf{x})\|) $$
where the RHS goes to infinity since the coercive function $\|x-y\|\to\infty$ . This means that $|f(\mathbf{x})- f(\mathbf{y})|$ is unbounded, as you correctly suspected.


问:
How to take the directional derivative of an integral over $u$ from $(x,y)$ to $(1,1)$
Please consider the following function
$$
v(x,y) = \int_{(x,y)}^{(1, 1)}u\,dS
$$
$v(x,y)$ is the integral along the straight line from $(x,y)$ to $(1, 1)$. I'm wondering how I can take the directional derivative in the direction of the vector from $(x,y)$ to $(1, 1)$. Let's say that the vector going from the origin through $(x,y)$ and through $(1, 1)$ is called $\beta$. How do I take the directional derivative in the direction of $\beta$ at the point $(x,y)$? Intuitively I would say that
$$
\beta\cdot\nabla v(x,y) = - u(x,y)
$$
but I'm not sure how I can show this rigorously.

答:
Write $X$, $X_0$, and  $\underline 1$ for $(x,y)$, $(x_0,y_0)$ and $(1,1)$.   Unpacking the definition of the line integral $\int_{[X,\underline 1]}u ds$ of a scalar function (I assume $u$ is scalar but a similar calculation works for other cases) you want to differentiate $v(X)=|X-\underline 1|\int_0^{1} u(t\underline 1 + (1-t)X) dt $ in the direction $X_0=\underline1-X$. Note that the factor $|X-\underline 1|$ comes from the speed of the parameterisation I chose.  With no extra effort we can compute the derivative in all directions $X_0$. Setting $g(s)=|X +sX_0-\underline 1|$ and $f(s)=u(t\underline 1  + (1-t)(X + sX_0))$, you want the derivative of $g(s)\int^{1}_0 f(s,t)dt$ at $s=0$ which is $g'(0)\int f(0,t)dt +g(0)\int_0^1 \partial_s f(0,t)dt$.

$$g’(s)=\frac{X_0\cdot(X+sX_0-\underline1)}{|X+sX_0-\underline1|}$$ $$ f’(s)=X_0\cdot \nabla u(t\underline 1  + (1-t)(X + sX_0))$$

Put them together and you have the directional derivative at $X$ in direction $X_0$:
\begin{align}
 D_{X_0} v(X) &= \frac{X_0\cdot(X-\underline1)}{|X-\underline1|}\int_0^1 u(t\underline 1 +(1-t)X)dt \\&\quad + {|X-\underline1|}\int_0^{1}X_0\cdot\nabla u(t\underline 1  + (1-t)X ) dt
\\&= \frac{X_0\cdot(X-\underline1)}{|X-\underline1|^2}\int_{[X,\underline 1]} u ds + \int_{[X,\underline 1]} X_0\cdot \nabla u ds
\end{align}
Specialising now to $X_0=\underline 1-X$ one gets
$$ D_{\underline 1-X}v(X) = -  \int_{[X,\underline1]} u \,\text ds + \int_{[X,\underline1]} D_{\underline 1-X}u \,\text ds. $$

Since the answer is different from your guess, let me verify the answer  for a special case. If $u\equiv 1$ then $v(X)$ is just the unsigned distance of $X$ to $\underline 1$, $v(X) = |X-\underline 1|$. Its gradient at $X$ is $\frac{X-\underline 1}{|X-\underline 1|}$. Thus the directional derivative in the direction $\underline 1-X$ is $(\underline 1-X)\cdot \nabla v(X) = -|X-\underline 1|$ which is not $-u(X)=-1$.

However if you define the directional derivative in direction $X_0$ as $\frac{X_0}{|X_0|}\cdot\nabla $ then in this case one gets $-u(X)=-1$.


问:
Basis functions, functions spaces, inner products.
I am currently doing a Physics degree and am in my first year. I have been introduced to the concept of Fourier series, and a prerequisite of that was an explanation of the inner product between functions and a mention of 'infinite dimensional spaces' of functions.

I can't see how a function would form an infinite dimensional space. I am aware that there would be an infinite of coordinates defining points in some region, but this would surely form an infinite set instead?

Also (and I guess as a corollary of me not understanding the above), how exactly do functions form a 'basis set' for other functions? And, what exactly does an inner product between functions represent?

答:
A function does not form a infinite dimensional space. However, a collection of functions may very well form such a space. A basic example would be the space of all polynomials.

When asking what it means to have a basis, that's just linear algebra, except that in infinite dimensions we tend to tweek the definition a bit.

So just to combine the two above points: The space of polynomials, call it $\mathcal{P}$, is infinite dimensional. What does that even mean? Well, it means that for any finite set of polynomials $p_1,...,p_n$, the linear span of $p_1,...,p_n$ is not equal to all of $\mathcal{P}$. How would I prove such a thing?

Well, since there are finitely many of them, there is a number $d$ such that the degree of each $p_i$ is at most $d$. Consider now the polynomial $p(x)=x^{d+1}$. I want to argue that $p$ does not lie in the linear span of $p_1,...,p_n$. Indeed, if it did, there would exist constants $\alpha_1,...,\alpha_n$ such that $p(x)-\sum_{i=1}^n \alpha_ip_i(x)$ would be the zero-polynomial. But that can't be, since by elementary calculus, $\lim_{x\to \infty} p(x)-\sum_{i=1}^n \alpha_ip(x)=\infty$.

The conclusion is that $p$ does not lie in the linear span of $p_1,...,p_n$. Since these polynomials were arbitrary, $\mathcal{P}$ does not have a finite spanning set. Therefore, $\mathcal{P}$ is not finite dimensional.

Now, $\mathcal{P}$ admits the basis given by $p_i(x)=x^i$. That means that given a polynomial $q$, there exist unique constants $\alpha_i$ such that $\alpha_i=0$ for all but finitely many $i$ and $q(x)=\sum_{i=1}^{\infty} \alpha_i p_i(x)$. This is perfectly analogous to classic linear algebra.

However, usually, in the infinite dimensional setting, we don't want the above condition that the $\alpha_i$ are only non-zero for finitely many $i$. It somehow clicks wrong with the theory. However, the good infinite dimensional notion of basis of a function space would still be that a set $\{f_i|i\in I\}$ forms a basis if for every $g$ in that space, there are unique $\alpha_i$ such that $g(x)=\sum_{i\in I} \alpha_i f_i(x)$.

Finally: What should the inner product represent? Well... I'm assuming we're talking about the inner product $\langle f, g\rangle=\int \bar{f}(x) g(x)\textrm{d}x$. One way to think about this is that it's no different from the usual inner product. You take the coordinate-wise product of the two vectors and then you "sum" them. Only, when you're in a continuous setting, summing is replaced by integration.

Indeed, if $f$ and $g$ are continuous, you'd even get that if $a=x_0<x_1<...<x_n=b$ and the distance between each $x_i$ and $x_{i+1}$ is small, then
$$
\sum_{i=1}^n (\bar{f}(x_i)g(x_i))(x_{i+1}-x_i) \approx \int_a^b \bar{f}(x) g(x)\textrm{d}x
$$
and the left-hand side is just a renormalised inner-product between the more classical vectors $(f(x_i))_{1\leq i\leq n}$ and $(g(x_i))_{1\leq i\leq n}$.

Hope this helps somewhat.


答:
This is a good question as being able to expand 'any' function in a Fourier series isn't trivial at all, in fact it has been one of the biggest problems in mathematics having had many of the greatest mathematicians working on it in the past.

It all builds upon the concept of a basis. If you've taken linear algebra, you will have seen things like the vectors $e_1 = \begin{pmatrix}1\\
0\end{pmatrix}$ and $e_2 = \begin{pmatrix}0\\1\end{pmatrix}$ spanning all of $\mathbb{R}^2$, what that means is that we can get any element in $\mathbb{R}^2$ by taking linear combinations of $e_1$ and $e_2$, i.e. any vector in $\mathbb{R}^2$ can be written as
$$ae_1 + be_2$$ where $a, b$
are arbitrary constants. What makes this work is the fact that $e_1$ and $e_2$ are linearly independent.

Now what if we had a function instead of a vector in $\mathbb{R}^2$? It turns out that that the infinite set $$\{1/\sqrt2\, \sin(nx), \cos(nx) : n \in 1, 2, 3, \ldots\}$$ is a linearly independent set that span all piece-wise continuous and periodic functions, so in the same way as before, we can express any such function as a linear combination of the basis vectors
$$f(x)\sim \frac{a_0}{\sqrt{2}} + \sum_{n=1}^\infty (a_n \cos n x + b_n \sin nx)$$

where $a_0, a_n, b_n$ are constants determined by our function. This set is known as the Fourier basis.

As for what an inner product is, it is just an operation $\langle \cdot, \cdot \rangle$ we can equip our space with, in particular, it allows us to define angles between vectors through
\begin{align*}
    \cos{\theta} = \frac{\langle u, v\rangle}{\lvert u \rvert\lvert v \rvert}
\end{align*}
and thus we can now talk about orthogonality and orthogonal projections which can be helpful in determining the coefficients $a_n, b_n$ above.

Title: Basis functions, functions spaces, inner products.

Body:
I am currently doing a Physics degree and am in my first year. I have been introduced to the concept of Fourier series, and a prerequisite of that was an explanation of the inner product between functions and a mention of 'infinite dimensional spaces' of functions.

I can't see how a function would form an infinite dimensional space. I am aware that there would be an infinite of coordinates defining points in some region, but this would surely form an infinite set instead?

Also (and I guess as a corollary of me not understanding the above), how exactly do functions form a 'basis set' for other functions? And, what exactly does an inner product between functions represent?


答:
A function cannot possibly form an infinite dimensional space. But, say, the set of all continuous functions from $[0,2\pi]$ into $\Bbb R$ is an infinite dimensional space. The dimension is infinite because, for instance, it contains all functions of the form $x\mapsto x^n$, with $n\in\Bbb Z^+$. And you do nott need coordinates to work with elements from this vector space.

A natural inner product in this space is$$\langle f,g\rangle=\int_0^{2\pi}f(t)g(t)\,\mathrm dt.$$


答:
In terms of generating functions the problem is equivalent to finding the coefficient at $x^{k}$, with $k=2012$ in the product:
$$
(1+x)^2\prod_{i=1}^{10}\left(1+x^{2^i}\right)^3=\frac1{1+x}\prod_{i=0}^{10}\left(1+x^{2^i}\right)^3=\frac1{1+x}\left(\sum_{i=0}^{2^{11}-1}x^i\right)^3.\tag1
$$

Thus the problem essentially boils down to determine the coefficients $c_k$ at $x^k$ in the expression:
$$\left(\sum_{i=0}^{m-1}x^i\right)^n.
$$

This problem is equivalent to asking about the number of ways to distribute $k$ balls among $n$ bins with the capacity of each bin being $m-1$. In our case we however do not need to bother about capacity because $k<m$. So we need to find:
$$
[x^k:]\sum_{i\ge0}\binom{i+n-1}{n-1}x^i\sum_{j\ge0}(-x)^j=
\sum_{j\ge0}(-1)^j\binom{k+n-1-j}{n-1}.
$$

It remains only to prove that
$$
\sum_{j=0}^k(-1)^j\binom{k+2-j}{2}=\left(\frac k2+1\right)^2
$$
for even $k$ which I left as an exercise.

Title: Counting the number of ways to get a sum of $2012$, given that any number of cards can be chosen

Body:
A person has $32$ cards out of which $10$ cards, each of blue, green and red colours, have
denominations as $\{2^1, 2^2,\cdots , 2^{10}\}$ and one black and one white each of whose value is unity. What is the number of ways in which the person can get a sum of $2012$ if he can choose any number of cards?

I couldn't think much in it, except that number of ways of choosing $\text{(blue,black,white)}$ is equal to if he chooses $\text{(green,black,white) or (red,black,white)}$.

So if suppose he chooses $\text{(blue,black,white)}$, the total sum of the cards would be $$1+1+2+2^2+\cdots+2^{10}=2^{11}=2048$$

So, if we remove the cards whose sum equals $36$, we will get $2012$. Thus, we need to find the number of ways in which the sum on cards equals $36$. I tried it, but there were so many cases, that I gave up. I also haven't taken the case if he chooses for instance $\text{(blue,green,white)}$ etc.

The answer to this is $1007^2$, which suggests some easy method must be there.

Any hints would be appreciated!


问:
Question about the general case of $\sqrt{ab} \leq(a + b) / 2$
I'm an amateur working through Spivak's Calculus and one of the early exercises is:

given that $0 \le a \le b$, prove that a $ \sqrt{ab} \le \frac{a+b}{2}$

I eventually managed to barely puzzle through a proof for this that someone had posted on here. After playing around a bit, I've come to suspect that this could be generalized to imply that the $n$th root of $n$ numbers is less than or equal to the mean of the sum of those numbers. My understanding of the special case is admittedly shallow so I really don't know where to start with proving or disproving the following "conjecture":

Given that $x_1$, $x_2$, $x_3,\ldots x_n \in \mathbb R^+,$

$$\sqrt[n]{\strut x_1x_2x_3\cdots x_n} \le \frac{x_1+ x_2+ x_3+\cdots+x_n}{n}$$

My question is, am I correct? If so, why? Does this idea have a name?

答:
By induction:

Let us assume that $g$ and $a$ are the geometric and arithmetic mean of $n-1$ numbers. Then I claim that for any non-negative number $x$

$$g\le a\implies\sqrt[n]{g^{n-1}x}\le\frac{(n-1)a+x}n.$$

Indeed, this can be rewritten
$$\left(\frac gx\right)^{(n-1)/n}\le\frac{(n-1)\dfrac ax+1}n$$

and we can show the stronger

$$\left(\frac gx\right)^{(n-1)/n}\le\frac{(n-1)\dfrac gx+1}n$$

which derives from Bernouilli's inequality for $0<\alpha<1$:
$$t^\alpha\le\alpha(t-1)+1.$$


问:
Maintain unitary time evolution for a nonlinear ODE
I want to solve a nonlinear ODE of matrix $A(t)$
$$\mathrm{i}\dot A = A(t)M(t),\:\mathrm{with}\: M(t)=A^\dagger(t)H(t)A(t)$$ where $H(t)$ and hence $M(t)$ are Hermitian. Therefore, I presume the time evolution of $A(t)$ is unitary. Is there any algorithm that can maintain this unitarity? I've heard of Crank-Nicolson, but is it for linear ODE and $M$ is independent of $t$?

答:
You would like to get a formula $$A_{k+1}=A_kU_k,~~~U_k=(I-\tfrac12iS_k\Delta t)(I+\tfrac12iS_kΔt)^{-1},$$ where $S_k$ is symmetric/hermitean and close to $M(t_k,A_k)$. This construction ensures that $U^*=U^{-1}$ is unitary, so that if $A_0$ has this property, it is kept over the whole integration.

Now let's compare expansions to establish second order. Just setting $S_k=M_k$ would give a first order method. For the second derivative of the exact solution we get
$$
-\ddot A=i\dot A M+iA(\dot A^*HA+A^*\dot HA+A^*H\dot A)
=AM^2+A(-M^2+iA^*\dot HA+M^2)
$$
So on the one hand
$$
A(t_k+Δt)=A_k\Bigl(I-iMΔt-\tfrac12M^2Δt^2-i\tfrac12A^*\dot HAΔt^2+O(Δt^3)\Bigr)
$$
and on the other
$$\begin{align}
A_{k+1}&=A_k\Bigl(I-\tfrac12iSΔt)(I-\tfrac12iSΔt+(\tfrac12iSΔt)^2+O(Δt^3)\Bigr)
\\
&=A_k\Bigl(I-iSΔt-\tfrac12S^2Δt^2+O(Δt^3)\Bigr)
\end{align}$$
Now try an expansion $S=M+NΔt+O(Δt^2)$. Comparing one reads off $N=\frac12A^*\dot HA$.

From this information one can construct a semi-implicit method
$$\begin{align}
S&=A_k^*H(t_k+\tfrac12Δt)A_k\\
A_{k+1}&=A_k(I-iSΔt/2)(I+iSΔt/2)^{-1}
\end{align}$$


问:
Show that $P(A \cap B)\in[0.4, 0.6]$.
Let $P$ be a probability distribution on a sample space $\Omega$ and let $A, B \subseteq\Omega$ be events with $P(A) = 0.6$ and $P(B) = 0.8$. Suppose that the value $P(A \cap B)$ is unknown. Show that $P(A \cap B) \in [0.4, 0.6]$.

I really don't know where to start any help would be appreciated.

答:
We know that the cardinal number of the union of two sets is greater than or equal to the cardinal number of the set containing more elements, i.e.,  $P(A \cup B) \in [0.8,1]$

We also know that $P(A \cap B) = P(A) + P(B) - P(A \cup B)$

$\implies - P(A \cap B) + 0.6 + 0.8 = P(A \cup B) $

$\implies P(A \cap B) - 0.6 - 0.8 \in [-1, -0.8]$

$\implies P(A \cap B) \in [0.4,0.6]$

Title: Show that $P(A \cap B)\in[0.4, 0.6]$.


问:
Given $\lim_{x\to0}\frac{\sin^2x}{x}=0$, evaluate $\lim_{x\to0}\frac{1-\cos x}{x}$

Given that $\lim_{x\to0}\frac{\left(\sin x\right)^2}{x}=0$, find the limit $\lim_{x\to0}\frac{1-\cos x}{x}$.

a. $1\quad$ b. $0\quad$ c. $\frac12\quad$ d. The limit does not exist

New to Calculus, and these trig questions really trip me up. Thanks

答:
Once accordng to L'Hospital's rule
$$\mathop {\lim }\limits_{x \to c} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to c} \frac{{f'\left( x \right)}}{{g'\left( x \right)}}$$

Therefore, once
$$\frac{d}{dx}(1-\cos(x)) = \frac{d}{dx}(1)-\frac{d}{dx}(\cos (x)) = 0-(-\sin(x)) = \sin(x)$$

and

$$\frac{d}{dx}(x) = 1$$

So $\lim\limits_{x\to 0}\dfrac{1-\cos(x)}{x} = \lim\limits_{x\to 0}\dfrac{\sin(x)}{1} = 0$


答:
Observe that

$$P(A)+P(B)=0.6+0.8=1.4$$

thus as this probability cannot be greater than 1 the minimum value of $P(A\cap B)=0.4$

On the other hand, $P(A\cap B)$ cannot be greater than the minimum probability of the two events...this happens when $A \subset B$ that is $P(A\cap B)=P(A)$

concluding:

$$P(A\cap B) \in[0.4;0.6]$$

These are the two extreme situations

Title: Show that $P(A \cap B)\in[0.4, 0.6]$.

Body:
Let $P$ be a probability distribution on a sample space $\Omega$ and let $A, B \subseteq\Omega$ be events with $P(A) = 0.6$ and $P(B) = 0.8$. Suppose that the value $P(A \cap B)$ is unknown. Show that $P(A \cap B) \in [0.4, 0.6]$.

I really don't know where to start any help would be appreciated.


问:
Show that a solution to an equation set is also a solution to another equation
$y=\frac12x^2;\;
y^2+(x-3)^2=3^2$ (circle)

The function and circle intersect at two points, one is $(0,0)$ and lets say the other is $(x_0,y_0)$
Show that $x=x_0$ is also a solution to the equation $x^3+4x=24$.

I dont understand at all how to solve this. I tried solving
$(x^3+4x-24)/(x-x_0)$ and try and get a new expression for $x_0$(?) since there can't be any remainder, but then I just end up with the exact same equation, in terms of $x_0$ instead.

答:
If $y = \frac 1 2 x^2$ then we can substitute for $y$ in the equation of the circle to get:

$\frac 1 4 x^4 + (x-3)^2 = 9
\\ \Rightarrow x^4 + 4x^2 - 24x + 36 = 36
\\ \Rightarrow x(x^3 + 4x - 24) = 0
\\ \Rightarrow x=0 \text{ or } x^3 + 4x - 24=0$


答:
Recolour the cards of value  $1$ blue and green. Let $a, b,c$ be the sum of the numbers on the blue, green and red cards respectively. Making use of the unique representation of numbers using binary digits:-

The problem is equivalent to finding the number of ordered triples of non-negative integers $(a,b,c)$ such that $a+b+c=2012$, where $c$ must be even.

Considering all the possibilities for $c$, gives $$1+3+5+ ... +2013=1007^2.$$


答:
For finite complexes - yes, for infinite complexes in general - no.

Why yes: First, each finite CW  complex $W$ is homotopy-equivalent to a finite simplicial complex $C$. (This should be in Hatcher's "Algebraic Topology.") Next, each finite simplicial complex embeds as a subcomplex in a simplex which then embeds in some ${\mathbb R}^n$ as a part of a triangulation of the latter. Now, you have a subcomplex $X$ of a triangulated ${\mathbb R}^n$. Take the regular neighborhood $N_X$ of $X$ in ${\mathbb R}^n$: It will be homotopy-equivalent to $X$ and a compact (PL) submanifold with boundary. (Both should be in Rourke and Sanderson's "PL topology".)  Lastly, take the interior of $N_X$. This is your manifold.

With a bit more work, this proof extends to the case of countable CW complexes of finite dimension. (The main difference is that you first prove the existence of a finite-dimensional locally finite simplicial complex $C$ homotopy-equivalent to $W$, this is in Hatcher's book. Then the simplicial version of Whitney's theorem gives you a proper embedding as a subcomplex in ${\mathbb R}^n$. The rest is the same.)

Why no (even for countable complexes). Consider the wedge of spheres of all dimensions. The resulting CW complex $X$ has the property that
$$
H_k(X)\ne 0, \forall k\in {\mathbb N}.
$$
However, if $M$ is an $m$-dimensional manifold then $H_k(M)=0$ for all $k>m$. (This is a consequence of the Poincare duality, again is in Hatcher's book.)

As another, finite-dimensional, example, take $X$ to be an uncountable set with discrete topology, or (if you prefer a connected example), an uncountable wedge of circles. Then $H_1(X)$ has uncountable rank which is impossible for a manifold. (I am assuming that manifolds are required to be 2nd countable, which is a standard assumption, although some people disagree.)

Title: Does every finite $CW$ complex have the homotopy type of a smooth manifold? How about infinite $CW$ complexes?

Body:
Let M be a smooth manifold. I was wondering if every finite or infinite CW complex has the homotopy type of a smooth manifold because it seems way easier to compute De Rham cohomology groups than homology groups.

If the answer to the questions are yes, is it hard to find a smooth approximation of a CW complex?


问:
Adjoints preserve limits (resp. colimits) Do they preserves completeness (resp. cocompleteness)?
I know that left adjoints preserve colimits and right adjoints preserve limits.
So clearly if the limits (resp. colimits) in both categories exist, the adjoints map them to each other.

My question is, whether it's garantueed that the limit  exists?

To be more precise: Let $T\colon A\rightarrow B$ a functor that is a right (resp. left) adjoint of some functor and $D\colon I\rightarrow A$ a diagramn, such that it's limit (resp. colimit) (in $A$) exists. Does the limit (resp. colimt) $T\circ D$ always exist?

答: Yes, and this is precisely the statement that $T$ preserves limits! If $\{L \to D\}$ is a limit cone, and $T$ preserves limits, then $\{T(L) \to T \circ D\}$ is a limit cone.


问:
Quadratic variation - Solution check
The task at hand is a simple calculation of the quadratic variation of $(tB_t)_{t\geq0}$, where $B$ is Brownian motion. Technically I have a solution but get a wrong result after many recalculations.

I first apply Ito's time-dependent formula, after checking the integrability conditions for the existence of the stochastic integral:

$\forall t \geq 0:  t^2B_t^2=  \int_0^t2s^2B_sdB_s + \int_0^t (s^2 + 2sB_s^2)ds \tag{1}$

Now, since $(tB_t)_{t \geq 0} \in L^2$, the quadratic variation process exists and is uniquely determined as the continuous adapted process starting at $0$ for $t=0$ making $(t^2B_t^2 - \langle tB_t \rangle)_{t \geq 0}$  a martingale.

Thus, since $\int_0^t2s^2B_s^2dB_s$ is a martingale due to the integrability conditions, we conclude: the quadratic variation is $\langle \cdot B_\cdot \rangle_t = \int_0^t (s^2 + 2sB_s^2)ds \tag{2}$
However, when we take the expected value we obtain:

$\mathbb{E}[\langle \cdot B_\cdot \rangle_t] = \mathbb{E}[\int_0^t (s^2 + 2sB_s^2)ds] \stackrel{(Tonelli)}{=} \int_0^t s^2 + 2s^2 ds = t^3 \tag{3}$

On the other hand, it is well known that $\langle \cdot B_\cdot \rangle_t = t^3/3$ a.s.

Is there a mistake in the solution?

答: The problem is that you are wrongly assuming that $tB_t$ is a martingale, but it's not! In fact $E(tB_t| \mathcal F_s)=tB_s$, and hence you cannot expect the Doob-Meyer decomposition theorem to hold in this case.

Title: Quadratic variation - Solution check


问:
Extension of operator $T$ from a subspace $Y$ of $X$ to $\mathbb{R^n}$ (incomplete answer)
I came across the same question in Carothers' A Short Course on Banach Space Theory as another poster (Extension of operator $T$ from a subspace $Y$ of $X$ to $\Bbb R^n$ without increasing the norm) and was stumped.

The question is as follows:

Suppose $Y$ is a subspace of a Banach space $X$ and let $T\in B(Y,\mathbb{R}^n)$ the space of bounded linear operators from $Y$ to $\mathbb{R}^n$. I am asked to show that there exists an extension $\tilde{T} \in B(X, \mathbb{R}^n)$ such that the operator norm does not increase, i.e. $\|\tilde{T}\| = \|T\|$.

But I do not find the posted solution to be satisfactory - the accepted answer proves that we can proceed if we use the supremum norm. This is all well and good, but can we solve the problem for any norm on $\mathbb{R^n}$? The phrasing of the question seems to indicate that the answer is "yes", but I am not so sure. As the commenter for the accepted solution points out, it's not clear that this will hold for any other norm on $\mathbb{R^n}$.

Further, looking at related posts, such as this one (Bounding the norm of an extension of a linear function from a subspace of a normed space to a finite dimensional normed space) there is a comment about extending bounded linear operators "norm preservingly" (only?) when taking the supremum norm.

In another related post (Extension of linear operator), the accepted response states without proof that if the finite-dimensional space of interest is over $\mathbb{C}$ then the result does not hold for any norms other than the supremum norm.

However, I have not been able to come up with a counter example myself. Any further insight is appreciated.

答:
As you have found in many places around here, the result is true when the target space is equipped with the supremum norm. Let $W\subset X$ be an inclusion of subspaces and let $Y$ be a finite dimensional space equipped with an arbitrary norm $\|\cdot\|$. Let $T:W\to (Y,\|\cdot\|)$ be a bounded linear operator.

Now it is well-known that in a finite dimensional space all norms are equivalent. So if $\|\cdot\|$ is any norm on $Y$ there exist constants $c_1,c_2>0$ so that $\|\cdot\|\leq c_1\|\cdot\|_\infty$ and $\|\cdot\|_\infty\leq c_2\|\cdot\|$. Actually, the best such constants are interpreted as follows: Take the identity operator $I:(Y,\|\cdot\|_\infty)\to(Y,\|\cdot\|)$. Then $c_1=\|I\|$ and $c_2=\|I^{-1}\|$.

Now consider the operator $S:W\to (Y,\|\cdot\|_\infty)$ defined by $S=I^{-1}\circ T$. Then $S$ is a bounded operator and as we explained this extends to an operator $S':X\to(Y,\|\cdot\|_\infty)$ s.t. $\|S'\|=\|S\|$. Take $T'=I\circ S'$. Then $T':X\to(Y,\|\cdot\|)$ extends $T$. Now
$$\|T\|\leq \|T'\|\leq c_1\|S'\|=c_1\|S\|\leq c_1c_2\|T\|$$
So if $c_1c_2=1$, the extension is isometric (note that $1\leq\|I\|\cdot\|I^{-1}\|=c_1c_2$). I think that this is the best one can say "for free" without getting more specific about the norm that $Y$ has.

Anyway, I believe that one should not really worry about this; the main point that the exercise is trying to make is that bounded operators towards finite dimensional spaces can be extended from subspaces to the entire space. Now usually a finite dimensional space appears in an auxilliary way, so the norm we consider on it can be chosen freely, because all norms are equivalent, so we simply choose the supremum norm if we want to make our extension isometric. As one of the linked answers state, the result fails for the $\ell^p$ norms on $Y$, so one cannot expect to prove that the extension is isometric in general.

Sure though, it would be more appropriate to explicitly state that $\mathbb{R}^n$ is equipped with the supremum norm.


问:
Prove unitary evolution for an ODE
Consider a nonlinear ODE of square matrix $A(t)=(a_1(t),\cdots,a_n(t))^T$

$$\mathrm{i}\,\dot A(t) = A(t) M(t)$$

with $$M(t) = A^\dagger(t) H(t) A(t)$$

where $H(t)$ and hence $M(t)$ are Hermitian matrices. My question is whether the time evolution of $A(t)$ is unitary. Here by unitary I mean the inner product of the vectors $$\langle a_i(t),a_j(t)\rangle$$ is invariant. The case for a constant Hermitian $M$ is straightforward, but how about this more general case?

答:
Consider the function $G(t)=A(t)A(t)^†$. Then its derivative is
$$
\dot G=\dot AA^†+A\dot A^†=(-iAM)A^†+A(-iAM)^†=-iAMA^†+iAM^†A^†=0
$$
So if $G(0)=I$, then this is also true for all $t$.

Title: Prove unitary evolution for an ODE


问:
Separating a compact set from a closed set in $T_2$ and locally compact
Let $(X,T)$ be a $T_2$ and locally compact topological space, let $C$ be a compact subspace of $X$ and $U$ an open subset of $X$ containing $C$. Prove that there exists a continuous function $f:(X,T)\to ([0,1],T_u|_{[0,1]})$ such that $f(C)=\{0\}$ and $f(X\smallsetminus U)=\{1\}$.

Here's what I've tried:

I know the space is $T_{3a}$ because it's $T_2$ and locally compact, and I know the set $C$ is closed because all compact subsets of a $T_2$ space are closed. So I've tried using the functions from the definition of completely regular spaces with no success.

If my space was $T_4$, I could use Uryshon's Lemma to find the function I'm looking for, but the only way I've been able to come up with to use this is to say that, because it is $T_2$ and locally compact, I can find a compactification by one point, lets call it $X'$. $X'$ will be compact and $T_2$, so it will be $T_4$, so I can find a continuous function $f:(X',T)\to ([0,1],T_u|_{[0,1]})$ such that $f(C)=\{0\}$ and $f(X\smallsetminus U)=\{1\}$. My problem is that I'm not sure that $f|_X$ is still continuous and, if it always is, doesn't that meen that if a space is $T_2$ and locally compact I can always find this function with any given pair of closed sets? Wouldn't that meen by Uryshon's Lemma that it is $T_4$? And I've only used the compactness of $C$ to say it's closed.

答:
Here is a more elementary approach, which clarifies the importance of compactness. Let $X$ be completely regular and $C,F\subseteq X$ be disjoint sets with $C$ compact and $F$ closed. We want to construct a continuous function $f\colon X\to[0,1]$ with $f(F)=0$ and $f(C)=1$.

Since our space is completely regular we can find, for every $c\in C$, a continuous function $f_c\colon X\to [0,1]$ with $f(c)=1$ and $f(F)=0$. Consider the open sets $U_c=f^{-1}((1/2,1])$, those are an open cover of $C$ so we can find, by compactness, finitely many $c_i\in C$ such that $\{U_{c_i}\}$ covers $C$.

Now the function $f'(x)=\sum_if_{c_i}(x)$ is still continuous, being the sum of finitely many continuous functions, has value $0$ on $F$ since every $f_{c_i}$ does, and has value $>1/2$ on $C$, so the function $f(x)=2\min\{f'(x),1/2\}$ is as required (it is continuous being the min of finitely many continuous functions).

To see why compactness is important in your argument (which is correct) note that you cannot just apply Urysohn's lemma to $C$ and $X\setminus U$ as subspaces of $X'$,because those won't necessarily be closed. In particular $X\setminus U$ won't be closed, and it's closure will be $X'\setminus U$ (I'm assuming that $X$ is not compact to begin with to avoid trivial cases), while $C$ will be already closed, because it is compact in $X$, so you can indeed apply Urysohn's lemma to their closures since those are disjoint. This argument won't work for arbitrary closed sets, any two unbounded disjoint closed sets in $\Bbb R^2$ will have the point at infinity in common in their closure in the one point compactification for example.


问:
Injections of Sobolev Spaces
I am studying Sobolev Spaces and appear the following question, that look simple, by I do not know how answer it:

If $T>0$ and $k\geq 2$, with $k\in \mathbb{N}$, then $ H^{k-1}(0,T) \subset W^{k,\infty}(0,T)$.

Why it is true ?

答: It is not true. Counterexample: For $x>0$, let $\mathscr I f(x) = \int_0^xf(y)dy$ and consider $G:=\mathscr I^{k-1}g$ where $g(x)=\frac1{x^{1/4}}$. Note $G^{(k-1)}=g\in L^2((0,T))\setminus L^\infty((0,T))$. Easy to check that $G^{(j)}\in L^\infty$ for all $j\le {k-2}$. So $G$ is a function that is in $H^{k-1}((0,T))$ for every $T$, but not even in  $W^{k-1,\infty}((0,T))$, so its not in $W^{k,\infty}((0,T))$.

Title: Injections of Sobolev Spaces


问:
Let $(s_n)$ be a sequence with $s_1, s_2, ..., s_{100}$ being positive and $s_n < 0 $ for all $n > 100.$ Assume $\lim s_n = 0$
Let $(s_n)$ be a sequence with $s_1, s_2, ..., s_{100}$ being positive and $s_n < 0 $ for all $n > 100.$ Assume $\lim s_n = 0$

What is $\lim \frac{1}{s_n} $?

My thought:

Since $\lim s_n = 0$,

$\forall \epsilon > 0,$ there exists $N$ such that $\forall n > N,$   $\mid s_n - 0 \mid <\epsilon$

Then $\mid s_n \mid < 0$

$-\epsilon < s_n < \epsilon$

$-\epsilon < s_n$ (only care about lower bound)

$\frac{-\epsilon}{s_n} > 1$ (because $n$ should be large enough for $s_n$ to converge to $0$?? Then $s_n$ is negative.. question point)

$\frac{1}{s_n} < - \frac{1}{\epsilon}$

Therefore $\frac{1}{s_n}$ converges to $-\infty$.

I was wondering if that question point step works.. and also other steps.

答:
You didn't quite complete the last part of the proof, although you were close. And I also cleaned up what you have done.

Let's start with the following:

Lemma: Given $\ r<0,\ $ there exists $\ \varepsilon>0\ $ such that $-\frac{1}{\varepsilon} = r.\ $ Proof: Let $\ \varepsilon = \frac{-1}{r}.$

Now, let's begin the main proof. Let $\ \varepsilon>0.$

Since $\lim s_n = 0\quad $ and $\quad s_n < 0\quad \forall\ n \geq 101,\ $ there exists $\ N\geq 101\ $ such that $\ -\epsilon < s_n < 0\quad \forall\ n \geq N.$

$s_n<0\ $ and $\varepsilon>0 \implies \frac{-1}{\varepsilon s_n} > 0,\ $ so multiplying both sides of $-\varepsilon < s_n\ $ by $\ \frac{-1}{\varepsilon s_n}\ $ gives:

$\frac{1}{s_n} < - \frac{1}{\varepsilon} \quad \forall\ n \geq N \geq 101\quad.$

$\varepsilon>0\ $ is arbitrary in all of this, so using the Lemma, given any negative number $r<0,\ $ we can choose $\varepsilon>0\ $ so that $\frac{1}{s_n} < r = - \frac{1}{\varepsilon} \quad \forall\ n \geq N \geq 101.$

This matches a definition of $\frac{1}{s_n}$ diverges to $-\infty$.


问:
If $H_1 \stackrel d{\hookrightarrow} H_2$, $X \subset H_1$ and $Y \subset H_2$ then $X \stackrel d{\hookrightarrow} Y$?
Let $H_1=(H_1, (\cdot, \cdot )_1)$ and $H_2=(H_2, (\cdot, \cdot )_2)$ be Hilbert spaces. Suppose that $H_1$ is continuously and densely embedded in $H_2$. Simbolically, $H_1 \stackrel  d{\hookrightarrow} H_2$. Let $X \subset H_1$ and $Y \subset H_2$ be closed subspaces such that $X \subset Y$, with $X,Y \neq \emptyset$. I know that
$$
X=(X, (\cdot, \cdot )_1) \quad \text{and} \quad Y=(Y, (\cdot, \cdot )_2) \tag{1}
$$
are Hilbert spaces.

Question. Is $X \stackrel  d{\hookrightarrow} Y$?

Its clear for me that, since $X \subset Y$ and due to $(1)$ that $X$ is continuously embedded  in $Y$. But I don't know (and I couldn't prove) that $X$ is dense in $Y$. I tried using the Hahn-Banach theorem, but without success.

答: Of course not, the question fails trivially in this generality: take any Hilbert space $H$  and let $H_1=H_2=H$. Let $X,Y$ be closed subspaces with $X\subset Y$ and $X\neq Y$. If $X$ embeds densely in $Y$ that means that the identity function $i:X\to Y$ maps $X$ to a dense subspace of $Y$, i.e. $X$ is dense in $Y$, i.e. $\bar{X}=Y$. But $X$ is closed, so $X=Y$, a contradiction since $Y\neq X$.


问:
Integrating factor of the form $x^ay^b$ for a non-exact ODE
The task is to find $a$ and $b$ such that $u=x^ay^b$ is an integrating factor (IF) for the ODE $ydx+x(xy-1)dy=0$.

Attempt: For $u$ to be an IF for the given ODE, the multiplication of ODE by the IF must be an exact ODE. That is,
$$ x^ay^{b+1}dx + (x^{a+1}y^b)(xy-1)dy=0 $$
must be exact. This necessitates the condition $M_y=N_x$ which translates into
$$(b+1)x^ay^b = (a+1)x^ay^b(xy-1) + x^{a+1}y^{b+1}.$$
Expanding the expression on the right and equalizing the coefficients with the left side leads to the conditions
$a+b=-2$ and $a = -2$ which in turn gives $b=0$.

The solution manual however disagrees and says $a=-3$ and $b=1$. I'd be grateful if you could help me figure out why my answer is wrong.

答:
You can also approach from the other direction. With this type of integrating factor, the total degree of coefficient and differential changes in the same way for all terms. To get complete differentials, one can group terms of the same degree structure. Here this results in
$$
0=[y\,dx-x\,dy]+x^2y\,dy=-x^2\,d(y/x)+\frac12x^2\,d(y^2).
$$
From this one can see directly that dividing by $x^2$ removes all obstacles to integration.


问:
Embeddings and Imbeddings
I am studying Sobolev Spaces and I am very confused about the following:

What are the difference between the Embeddings and Imbeddings of spaces?

If I have that: $W^{k,p}\to W^{m,q} $  is a compact imbedding, so it means that $W^{k,p}\subset W^{m,q} $ or not ?

答:
AFAIK Gae. S. in comments is correct: embeddings and imbeddings are alternate spellings of each other.

A compact im/embedding $\iota : X\to Y$ between Banach spaces, say (which already means $\iota (X) \subset Y$) is a map satisfying the following:

(embedding property) $\iota$ is injective,

(Continuity) we have the norm bound $ \|\iota(u)\|_Y \le C \|u\|_X$ for some $C>0$,

(Compactness) if $u_n$ is a bounded sequence of $X$, then $\iota(u_n)$ has a convergent subsequence in $Y$.

So what you said is true: $\iota(W^{k,p})\subset W^{m,q}$, but you have even a quantitative control on norms and the ability to extract convergent subsequences in the 'larger' space. Note that it is common practice to not give a symbol to the embedding map, and say that $X$ is a subset of $Y$ (this is the identification of $X$ with its embedding in $Y$).

Title: Embeddings and Imbeddings


问:
What are the chances of rain on particular days
I study maths as a hobby and am on to elementary theory of probability.

I have come across this problem:

Find the probability of events A, B and C given it rained on exactly 2 days last week.

A: it rained Monday and Tuesday

B: it rained on 2 consecutive days

C: it rained neither Monday or Tuesday.

My first thought was to work from the assumption that there was  a 2/7 chance of rain over the week and work out the probability for any particular day.
But then I thought I should start by working out the total number of outcomes for 2 days rain in one week, which I take to  be $\binom {7}{2} = 21$

Now the answers given in the book are $\frac{1}{21}, \frac{2}{7}, \frac{10}{21}$

I can see that the probability of any particular 2 days having rain is $\frac{1}{21}$ so can see how the the first answer is correct, but even then I have doubts. As for the 2nd and 3rd answers, I cannot see where these come from.

答:
I study maths as a hobby

me too....

"Given that it rained exactly 2 days last week" means that the combination of the two rainig days are exaclty

$$\binom{7}{2}=21$$

A: it rained Monday and Tuesday: this is one of the 21 possible combination, thus

$$\mathbb{P}(A)=\frac{1}{21}$$

B: how many combinations are there with 2 consecutive raining days in a week?

$1100000$

$0110000$

$0011000$

$0001100$

$0000110$

$0000011$

thus the result is

$$\mathbb{P}(B)=\frac{6}{21}=\frac{2}{7}$$

C: how many combinations are there in a week with 2 raining days where the beginning sequence is $00$? Exactly $\binom{5}{2}=10$ ( you can even count them manually...)

thus the answer is $\mathbb{P}(C)=\frac{10}{21}$


答:
The key observation here, as you have mentioned, is the number of total outcomes, which is $\begin{pmatrix}7\\2\end{pmatrix}=21$. Equipped with this fact, you can proceed by finding out what is the number of outcomes that fulfills each part.

Yes, the first part is just $\frac{1}{21}$, taking one of the 21 possible combinations.

On the second part, notice that there are just 6 out of 21 possible outcomes that makes the raining days consecutive (day 1 and 2, 2 and 3, ..., 6 and 7). Therefore, the possibility is just $\frac{6}{21}=\frac{2}{7}$.

As for the third part, you would want to see how many combinations in them are not on two specific days. That would mean those combinations are only chosen from the other five days. So that would be $\begin{pmatrix}5\\2\end{pmatrix}=10$. Therefore there are 10 out of 21 possibilities that fulfill the requirement. The probability is $\frac{10}{21}$.

Title: What are the chances of rain on particular days

Body:
I study maths as a hobby and am on to elementary theory of probability.

I have come across this problem:

Find the probability of events A, B and C given it rained on exactly 2 days last week.

A: it rained Monday and Tuesday

B: it rained on 2 consecutive days

C: it rained neither Monday or Tuesday.

My first thought was to work from the assumption that there was  a 2/7 chance of rain over the week and work out the probability for any particular day.
But then I thought I should start by working out the total number of outcomes for 2 days rain in one week, which I take to  be $\binom {7}{2} = 21$

Now the answers given in the book are $\frac{1}{21}, \frac{2}{7}, \frac{10}{21}$

I can see that the probability of any particular 2 days having rain is $\frac{1}{21}$ so can see how the the first answer is correct, but even then I have doubts. As for the 2nd and 3rd answers, I cannot see where these come from.


问:
Maximum and minimum value of $\frac{8x - 3y}{\sqrt{4x^2+y^2}}$?
For real numbers $\,x, y\neq 0\,$ consider
$$\frac{8x - 3y}{\sqrt{4x^2+y^2}}\,.$$
How to find the maximum and minimum value?

I've already got the maximum by using the Cauchy–Schwarz inequality
$$\big[(2x)^2 + y^2\big]\big[4^2 + (-3)^2\big] \geq (8x - 3y)^2\\[3ex]
25 \geq \frac{(8x - 3y)^2}{(2x)^2 + y^2}\\[4ex]
5 \geq \frac{8x - 3y}{\sqrt{(2x)^2 + y^2}}$$

But I cannot get the minimum value.

答:
You are almost there: Note that by Cauchy-Schwarz you  did
$$25 \geq \frac{(8x - 3y)^2}{(2x)^2 + y^2}\,.$$
Thus
$$\left|\frac{8x-3y}{\sqrt{4x^2+y^2}}\right|\le 5\tag {*}\\[5ex]
\implies -5\le \frac{8x-3y}{\sqrt{4x^2+y^2}}\le 5$$
The important part was in step (*). The modulus was crucial as $8x-3y$ could be negative too.


问:
Books on computational complexity
Can anyone recommend a good book on the subjects of computability and computational complexity? What are the de facto standard texts (say, for graduate students) in this area?

I've heard a thing or two on these subjects from "CS people" back at the university. With lots of hand-waving they talked about the halting problem, about P versus NP, and so on.

What I would like to find now is an actual mathematics textbook (as opposed to programming and/or popular textbooks). With rigorous definitions and proofs. I'm no stranger to mathematical logic, including things like partial recursive functions and Godel's theorem. I suspect I would benefit from reading a standard graduate-level text on computational complexity. All I need now is some references to such texts. Any suggestions?

答: You should take a look at Computational Complexity: A Modern Approach by Arora and Barak if you haven't already. I have also heard good things about Goldreich's Computational Complexity: A Conceptual Perspective but I haven't read it.</p>

Title: Books on computational complexity


答:
Your idea using the one-point compactification $\alpha X$ of $X$ can work fine:

Let $C \subseteq X$ be closed and $K \subseteq X$ be compact and disjoint, while  $X$ is a locally compact Hausdorff space.

Then $K$ is still closed in $\alpha X$, (compact subspaces are closed in Hausdorff spaces), while $C$ is a subset of the closed set $C’=C \cup \{\infty\}$ of $\alpha X$ (its complement in $\alpha X$ is the same as $C$’s complement in $X$, which was open, and so is still open) and $C’ \cap K = \emptyset$ still.

As $\alpha X$ is $T_4$ (compact and Hausdorff implies this), we have a Urysohn function $f: \alpha X \to [0,1]$ separating $K$ and $C’$ and then $f\restriction_X$ is still continuous (restrictions of continuous functions always are) and separates $C$ and $K$. QED.

You’re applying this to $K=C$ and $C=X\setminus U$ in your original notation.

Title: Separating a compact set from a closed set in $T_2$ and locally compact

Body:
Let $(X,T)$ be a $T_2$ and locally compact topological space, let $C$ be a compact subspace of $X$ and $U$ an open subset of $X$ containing $C$. Prove that there exists a continuous function $f:(X,T)\to ([0,1],T_u|_{[0,1]})$ such that $f(C)=\{0\}$ and $f(X\smallsetminus U)=\{1\}$.

Here's what I've tried:

I know the space is $T_{3a}$ because it's $T_2$ and locally compact, and I know the set $C$ is closed because all compact subsets of a $T_2$ space are closed. So I've tried using the functions from the definition of completely regular spaces with no success.

If my space was $T_4$, I could use Uryshon's Lemma to find the function I'm looking for, but the only way I've been able to come up with to use this is to say that, because it is $T_2$ and locally compact, I can find a compactification by one point, lets call it $X'$. $X'$ will be compact and $T_2$, so it will be $T_4$, so I can find a continuous function $f:(X',T)\to ([0,1],T_u|_{[0,1]})$ such that $f(C)=\{0\}$ and $f(X\smallsetminus U)=\{1\}$. My problem is that I'm not sure that $f|_X$ is still continuous and, if it always is, doesn't that meen that if a space is $T_2$ and locally compact I can always find this function with any given pair of closed sets? Wouldn't that meen by Uryshon's Lemma that it is $T_4$? And I've only used the compactness of $C$ to say it's closed.


答:
Assuming any day of the week is equally likely to have rain than any other day, and that the probability of rain on a given day is independent of the probability of rain on any other days, then what we need to do is first enumerate the number of outcomes in which exactly two out of the seven days had rain.  There are $7$ ways to pick the first day of rain.  Then there are $6$ ways to pick the second day (which cannot be the same as the first).  However, we double-count this way, since for instance if we pick as the first day "Tuesday" and the second as "Friday," this is the same as if we picked the first day as "Friday" and the second as "Tuesday."  So the total number of outcomes is $7(6)/2 = 21$.  This is also just $\binom{7}{2} = 21$, using the "combinations" formula.

Next we observe that each of these $21$ outcomes in which it rains on exactly two days of the week are equally likely.  And since there is exactly one outcome corresponding to rain on Monday and Tuesday, it follows the desired probability for the first part is $1/21$.

For the second part, how many of the outcomes had rain on consecutive days?  Well, there are $6$ such outcomes:  Monday/Tuesday, Tuesday/Wednesday, etc., until Saturday/Sunday (if you count Monday as the beginning of the week; otherwise we count Sunday/Monday, Monday/Tuesday, etc., until Saturday/Sunday--the result is the same).  So the desired probability is $6/21 = 2/7$.

For the final part, again count the outcomes that exclude Monday and Tuesday.  There are $5$ remaining days from which to choose the $2$ rainy days, hence $\binom{5}{2} = 10$ such outcomes out of $21$, so the probability is $10/21$.


问:
What invertible matrices $A$ satisfy $A^T C A = C$?
Let $C$ be an invertible, symmetric, pos-def matrix. These properties might not be very relevant to my question, but in the application I have in mind, $C$ is a full rank correlation matrix.

What invertible square matrices $A$ satisfy $A^T C A = C$?

答:
See $C$ as the matrix, in the canonical basis, of a quadratic form $q$. Then the matrices $A$ you are looking for are exactly the matrices of isometries w.r.t $q$, i.e.
$$\lbrace A \text{ such that } A^TCA=C \rbrace = O(q)$$

Indeed, let $\varphi$ be the bilinear form associated to $q$. Then for any endomorphism $u$ whose matrix in the canonical basis is $A$, it is easy to see that the matrix of the bilinear form $\varphi(u(.),u(.))$ is exactly $A^TCA$. So $A^TCA=A$ means exactly that $\varphi(u(.),u(.))=\varphi(.,.)$, i.e. that $u$ is an isometry for the bilinear form $\varphi$ (or equivalently for the quadratic form $q$).

For example, if $C$ is the identity matrix, then you can see it as the matrix of the quadratic form given by the usual dot product ; then you get all the isometries (in the usual sense, i.e. w.r.t. the usual norm). This general to any quadratic form as I explained.


问:
Find the sum: $\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n$

Find the sum:
$$\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n$$

My try:

I played a bit with the coefficient to make it look easier/familiar:

First attempt:
$$\begin{align}
\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n
&= \sum_{n=0}^\infty \frac{n!}{2^n(2n-1)!!}x^n \\
&= \sum_{n=0}^\infty \frac{n!}{(2n-1)!!}\left(\frac x2\right)^n
\end{align}$$
Second attempt:
$$\begin{align}
\sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}x^n
&= \sum_{n=0}^\infty \frac{n!\cdot n!}{(n+n)!}x^n \\
&= \sum_{n=0}^\infty \frac{1}{{2n \choose n}}x^n
\end{align}$$
However, I could not proceed with any of them. Also, I have figured out that the convergence radius is $4$.

My research:

I have also found the same sum has been discussed at AoPS, which unfortunately uses Beta function that my course has not covered yet.

Entering the sum to Wolphram Alpha, I got the following output for the partial sum:

$$\sum_{n=0}^k\frac{(n!)^2x^n}{(2n)!}=\frac{4\sqrt{x}\left(\sin^{-1}\left(\frac{\sqrt{x}}{2}\right)-\frac{2^{2k}k!(k+1)!B_\frac{x}{4}\left(k+\frac{1}{2},\frac{3}{2}\right)}{(2k)!}\right)}{(4-x)^{3/2}}+\frac{4}{4-x}.$$

My background:

As I have already mentioned, I cannot use Gamma, Beta or similar functions. I only know about the convergence theorems on functional series and operations on them. So, I'm looking for some method that uses quite elementary tricks.

Thanks in advance.

Body:
More than tricky ! (and done using a CAS).
$$\sum_{n=0}^p \frac{(n!)^2}{(2n)!}x^n=\frac d {dx}\Bigg[\sum_{n=0}^p \frac{(n!)^2}{(n+1)\,(2n)!}x^{n+1}\Bigg]$$
$$\sum_{n=0}^p \frac{(n!)^2}{(n+1)\,(2n)!}x^{n+1}=x \, _3F_2\left(1,1,1;\frac{1}{2},2;\frac{x}{4}\right)+\color{red}{\text{monster}}$$ $$\color{red}{\text{monster}}=\frac{((p+1)!)^2 x^{p+2} \,
   _3F_2\left(1,p+2,p+2;p+\frac{3}{2},p+3;\frac{x}{4}\right)}{(p+2) (2 (p+1))!}$$
$$\frac d {dx}\Big[x \, _3F_2\left(1,1,1;\frac{1}{2},2;\frac{x}{4}\right)\Big]=\frac{4}{4-x}+\frac{4 \sqrt{x} \sin
   ^{-1}\left(\frac{\sqrt{x}}{2}\right)}{(4-x)^{\frac 32}}$$ and
$$\frac d {dx}\big[\color{red}{\text{monster}}\big]=\frac{x^{p+1} \Gamma (p+2)^2 \,
   _2F_1\left(1,p+2;p+\frac{3}{2};\frac{x}{4}\right)}{\Gamma (2 p+3)}$$ which, in the radius of convergence, tends very quickly to $0$.


问:
What is a linear transformation $T:\mathbb{R}^3\longrightarrow\mathbb{R}^4$, whose null space is generated by $(0,1,-3)$ and $(0,-3,4)$?
What is a linear transformation to $T:\mathbb{R}^3\longrightarrow\mathbb{R}^4$, whose null space is generated by $(0,1,-3)$ and $(0,-3,4)$?

So, I know that the given pair of vectors generate the subspace $\{(0,y,z):\ y,z\in\mathbb{R}\}$, as they are linearly independent. By the Rank-nullity theorem, the image space in $\mathbb{R}^4$ is of dimension $3-2=1$.

But I can't seem to go further than this. Any help will be appreciated.

答: Complete $\{(0,1,-3),(0,-3,4)\}$ to a basis.  One easy way is to add the first standard basis vector $e_1=(1,0,0)$.  Then define $T((0,1,-3))=T((0,-3,4))=0$ and set $T((1,0,0))=(1,0,0,0)$, say.  Then just demand that $T$ be linear.


问:
Matrix equation with diagonal matrix
Let $A \in\mathbb{R}^{n\times n}$ be a given Positive-Semi Definite matrix ,$\theta\in\mathbb{R}^n$ a vector, $x\in\mathbb{R}$ an unknown variable, and $a\in\mathbb{R}^+$ a positive constant. I have the following equation
$$
\theta^TA(A+xI)^{-1}(A+xI)^{-1}A\theta  = a
$$
where $I$ is the identity matrix.

Is it possible to solve this equation for $x?$

What I have tried:

$$
\text{trace}(\theta^TA(A+xI)^{-1}(A+xI)^{-1}A\theta)  = \text{trace}(a)
\\\theta^TAA\theta  \cdot\text{trace}((A+xI)^{-1}(A+xI)^{-1})  = a
\\ \sum_{i = 1}^n\frac{1}{(x+s_i)^2}=\frac{a}{\|\theta\|_{A^2}^2}
$$

where $s_i$ are the singular values of the $A$. Is it correct? How can i proceed from here?

答:
Because $A$ is symetric psd matrix, we can diagonalize $A$ as: $A = UDU^T$ where $U$ is a orthogonal matrix ($U^T = U^{-1}$) and $D$ is a diagonal matrix.
Hence, we have
\begin{align}
(A+xI)^{-1} &= (UDU^T+xUU^T)^{-1} \\
&= (U(D+xI)U^{-1})^{-1} \\
&= U(D+xI)^{-1}U^{-1} \\
\end{align}
Then
\begin{align}
a & =\theta^TA(A+xI)^{-1}(A+xI)^{-1}A\theta  \\
& =\theta^TUDU^T \left(U(D+xI)^{-1}U^{-1}\right) \left(U(D+xI)^{-1}U^{-1}\right)UDU^T\theta  \\
& =\theta^TUD \left((D+xI)^{-1}\right)^2DU^T\theta  \\
\end{align}

The matrix $ \left((D+xI)^{-1}\right)^2$ is a diagonal matrix of $\frac{1}{(d_i+x)^2}$ with $(d_1,...,d_n)$ is the diagonal of the matrix $D$.
Let's denote $\eta =(\eta_1,...,\eta_n) = DU^T\theta \in \mathbb{R}^n$, we have

$$a = \sum_{i=1}^n\frac{\eta_i^2 }{(d_i+x)^2}$$

Title: Matrix equation with diagonal matrix


问:
Constructing a proof for neglible function
I need to write a proof for neglible functions in cryptography, i have this definition:

```<code>Let negl1 andnegl2 be negligible functions. Then:

 1. The function negl3 defined by negl3(n) =negl1(n)+negl2(n) is negligible.

 2. For any positive polynomial p, the function negl4 defined by negl4 (n) =p(n)·negl1(n) is negligible.
</code>```

Now, I need to prove the second part of this definition, relying on this definition:

```<code>A function f from the natural numbers to the non-negative real numbers is negligible if for
every positive polynomial p there is an N such that for all integersn > N it holds that f(n) < 1 / p(n).
</code>```

So I thought I would start with the following:

```<code>negl_4(n) = p(n) * negl_1(1)
= p_i(n) * 1/p_j(n)
</code>```

where the two p's are two differnt polynomials. Is there some way that I could convert the last statement here to 1 / p(n), with some new polynomial function?

答:
To show $\textrm{negl}_4$ is negligible:

Let $q(n)$ be any arbitrary polynomial in $n$.

Apply the definition of negligibility of $\textrm{negl}_1$ to the new polynomial $p(n)\cdot  q(n)$. We conclude there is some $N$ such that for all $n >N$ we have

$$\textrm{negl}_1(n) < \frac{1}{p(n)q(n)} \implies \textrm{negl}_4(n)= p(n) \cdot \textrm{negl}_1(n) < p(n) \cdot \frac{1}{p(n)q(n)} = \frac{1}{q(n)}$$

And as $q(n)$ was arbitrary, we are done.


问:
Is this ODE a well-known and studied one?
I came across this ordinary differential equation
$$x \,(y')^2+a=0\,,$$
where $a$ is a real number, $x$ is a real variable and $y$ is a real algebraic function of $x$. I was wondering whether this is a well-known and studied ODE or not.

答: If $a<0$ and $y(x):= 2 \sqrt{-a} \sqrt{x}$, then $y$ is a solution of differential equation on $(0, \infty).$


问:
Determine if $\Bbb Z_2 \times \Bbb Z_3$ is isomorphic to $\Bbb Z_6$.

Determine if $\Bbb Z_2 \times \Bbb Z_3$ is isomorphic to $\Bbb Z_6$.

Computing the direct product I have that $\Bbb Z_2 \times \Bbb Z_3 = \{(x,y) \mid x \in \Bbb Z_2, y \in \Bbb Z_3 \} =\{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)\}$ this has $6$ elements as does $\Bbb Z_6 = \{0,1,2,3,4,5\}$ I used this to conclude that the two groups are isomorphic, but I have a feeling that this doesn't count as a way to determine this. It seems that this could be done by looking at the orders of each group? I know that the order of a $g \in G$ is the smallest $n$ for which $g^n = e$, but I don't see how this can be used to show isomorphism?

答: By the Chinese remeinder Theorem (https://en.wikipedia.org/wiki/Chinese_remainder_theorem), you have  $\mathbb Z_p \times \mathbb Z_q \cong \mathbb Z_{pq}$ if and ony if $\gcd(p,q)=1$.


答:
Yes, they are isomorphic, but having the same number of elements is not enough for that. Afer all $S_3$ also has $6$ elements, but it's not isomorphic to any of those two groups, which are abelian.

Simply define$$\begin{array}{rccc}\varphi\colon&\Bbb Z_6&\longrightarrow&\Bbb Z_3\times\Bbb Z_2\\&[n]&\mapsto&(n\pmod 3,n\pmod 2)\end{array}$$(that is, $\varphi(0)=(0,0)$, $\varphi(1)=(1,1)$, $\varphi(2)=(2,0)$, …). Now, prove that $\varphi$ is an isomorphism.

Body:
Determine if $\Bbb Z_2 \times \Bbb Z_3$ is isomorphic to $\Bbb Z_6$.

Computing the direct product I have that $\Bbb Z_2 \times \Bbb Z_3 = \{(x,y) \mid x \in \Bbb Z_2, y \in \Bbb Z_3 \} =\{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)\}$ this has $6$ elements as does $\Bbb Z_6 = \{0,1,2,3,4,5\}$ I used this to conclude that the two groups are isomorphic, but I have a feeling that this doesn't count as a way to determine this. It seems that this could be done by looking at the orders of each group? I know that the order of a $g \in G$ is the smallest $n$ for which $g^n = e$, but I don't see how this can be used to show isomorphism?


问:
Lebesgue measure nullsets
I have a question concerning nullsets of the Lebesgue measure $\lambda^d$ on $\mathbb{R}^d$.
I think the set
 $$\{(x_1,\ldots,x_n)\in\mathbb{R}^n: \sum_{i=1}^n x_i=0\}$$
is a nullset of the n-dimensional Lebesgue measure. Any advice how prove this? I have my difficulties.

Regards,
Analyst77

答:
Hint

1.This is a hyperlane in $\mathbb R^n$. For each $\varepsilon >0$ can you cover it with cubes with total volume less than $\varepsilon$?

2.As we can split a straight line in compact line segments, split the hyperplance in compact subsets. Prove that each comact subset is a null set. Then the hyperplance is the countable union of these sets, therefore a null set.

Title: Lebesgue measure nullsets


问:
$C_0(\Omega)$ is not dense in $L^{\infty}(\Omega)$.

Let $\Omega \subset \mathbb{R}^n$ be open set. Show that $C_0(\Omega)$ is not dense in $L^{\infty}(\Omega)$.

Notation :$f \in C(\Omega)$ and f has compact support then $f \in C_0(\Omega)$.

My attempt: Assume that $C_0(\Omega)$ is dense in $L^{\infty}(\Omega)$. Lets seek a contradiction. By the assumption for all $\varepsilon > 0$ there exist a function $g \in C_0(\Omega)$ such that $|f-g|_{\infty} < \varepsilon$. So we have $\mu(x \in \Omega: |f-g| > \varepsilon) = 0$. I tried to find an open subset of $\{x \in \Omega: |f-g| > \varepsilon\}$, but i failed. Any help would be apperciated.

答:
Consider the constant function $f : \Omega \rightarrow \mathbb{R}$ such that $f(x)=1$ for every $x \in \Omega$. Its belongs to $L^{\infty}(\Omega)$.

Then for every $g \in \mathcal{C}_0(\Omega)$, there exists on open subset $\omega \subset \Omega$ such that for every $y \in \omega$, one has $g(y)=0$ (because $g$ has a compact support), so
$$||g-f||_{\infty} \geq 1$$

Therefore there doesn't exist any sequence $(g_n)_{n \geq 0}$ of functions of $\mathcal{C}_0(\Omega)$ such that
$$\lim_{n \rightarrow +\infty} ||g_n-f||_{\infty} =0$$

i.e. $f$ cannot be written as the limit of a sequence of elements of $\mathcal{C}_0(\Omega)$.


答:
There is an isomorphism $\phi:\mathbb{Z}_2\times \mathbb{Z}_3\to \mathbb{Z}_6$.

Send $([1]_2,[1]_3)\mapsto [1]_6$ and $([0]_2,[0]_3)\mapsto [0]_6$, you can find out the mapping in detail from here, and then verify this is an isomorphism.

Title: Determine if $\Bbb Z_2 \times \Bbb Z_3$ is isomorphic to $\Bbb Z_6$.


问:
Find orthogonal matrix missing values
I was solving this problem, where I need to find the value x, which is missed in the orthogonal matrix A.

$$
A =
\begin{pmatrix}
x&0.5&-0.5&-0.5\\
x&0.5&0.5&0.5\\
x&-0.5&-0.5&0.5\\
x&-0.5&0.5&-0.5\\
\end{pmatrix}
$$

One of the properties of orthogonal matrix is that the dot product of orthogonal matrix and its transposed version is the identity matrix:

$$I=A.A^T$$

$$
A^T =
\begin{pmatrix}
x&x&x&x\\
0.5&0.5&-0.5&-0.5\\
-0.5&0.5&-0.5&0.5\\
-0.5&0.5&0.5&-0.5\\
\end{pmatrix}
$$

So I figured that $$x=0.5$$

That would result in identity matrix I:

$$
I =
\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1\\
\end{pmatrix}
$$

As the result I had to provide the sum of missing values, which $0.5\times4 = 2$. The grader (program) returned the error. Is there something I am missing (there most likely is, I guess)?

Thank you in advance!

答: $A$ is orthogonal $ \iff x= \pm \frac{1}{2}.$

Title: Find orthogonal matrix missing values


问:
How to sum the series $\sum_{n=4}^{\infty}\frac{1}{(2n-7)(2n-6)}$
I have been tinkering with telescoping sum and I've encountered by chance this series:

$$\sum_{n=4}^{+\infty}\dfrac{1}{(2n-7)(2n-6)}=\log(2)$$

I use Wolfram to sum this series because I am not sure if it is a telescoping sum. I wonder how can you sum this series? By converting the infinite sum to an improper integral?

I integrate this function:

$$\int_{4}^{\infty}\dfrac{1}{(2n-7)(2n-6)}dx=\dfrac{\log(2)}{2}$$

I am new to improper integration and numerical infinite series. I thought the sum of the series and the result of the integral should match, but they don't.

It seems that I miss something very basic and crucial. I am learning myself Cal 3 so I may miss some crucial theorems. What are conditions for the sum of the series and the limit of the integral to be the same?

Come to think of it, summation is just a discrete version of integration, so the values naturally cannot be the same.

Is there a way to sum this series and obtain its sum? This doesn't seem to be a telescopic series in my humble opinion.

答:
Firstly, note that
$$\sum_{n=4}^{\infty}\dfrac{1}{(2n-7)(2n-6)}=\sum_{n=1}^\infty\dfrac{1}{(2n-1)(2n)}$$
Also, we can use partial fraction decomposition to find that
$$\frac{1}{2n(2n-1)}=\frac{1}{2n-1}-\frac{1}{2n}$$
So our sum is equivalent to
$$\sum_{n=1}^\infty\left(\frac{1}{2n-1}-\frac{1}{2n}\right)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$$
But recall that the Maclaurin series for $\ln(1+x)$ which converges for $-1<x\leqslant1$ is
$$x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$
So simply plugging in $x=1$ we have
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots=\ln(1+1)=\ln 2$$
as required.

I hope that was helpful. If you have any questions please don't hesitate to ask :)


答: There are two solutions: $\frac12$ and $-\frac12$.

Title: Find orthogonal matrix missing values

Body:
I was solving this problem, where I need to find the value x, which is missed in the orthogonal matrix A.

$$
A =
\begin{pmatrix}
x&0.5&-0.5&-0.5\\
x&0.5&0.5&0.5\\
x&-0.5&-0.5&0.5\\
x&-0.5&0.5&-0.5\\
\end{pmatrix}
$$

One of the properties of orthogonal matrix is that the dot product of orthogonal matrix and its transposed version is the identity matrix:

$$I=A.A^T$$

$$
A^T =
\begin{pmatrix}
x&x&x&x\\
0.5&0.5&-0.5&-0.5\\
-0.5&0.5&-0.5&0.5\\
-0.5&0.5&0.5&-0.5\\
\end{pmatrix}
$$

So I figured that $$x=0.5$$

That would result in identity matrix I:

$$
I =
\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1\\
\end{pmatrix}
$$

As the result I had to provide the sum of missing values, which $0.5\times4 = 2$. The grader (program) returned the error. Is there something I am missing (there most likely is, I guess)?

Thank you in advance!


问:
$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}+\frac{d}{1-d}+\frac{e}{1-e}\ge\frac{5}{4}$
I tried to solve this inequality:
$$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}+\frac{d}{1-d}+\frac{e}{1-e}\ge\frac{5}{4}$$
with
$$a+b+c+d+e=1$$
I am stuck at this. I don't want the full solution, a hint would be enough.</p>

答:
HINT:

$$\frac a{1-a}=-1+\frac1{1-a}$$

Assuming  $a,b,c,d,e$ to be positive real numbers, $1-a>0 $

Using AM HM inequality on $1-a$ etc

$$\frac{\sum (1-a)}5\ge \frac5{\sum \frac1{1-a}}\implies \sum \frac1{1-a}\ge\frac{5^2}{5-\sum a}=\frac{25}{5-1}$$


问:
Can I get a hint for this rates of change question?
A ball of radius $24cm$ has its volume decreasing at a rate of three times its surface area per second. What is the time taken for the volume to be reduced to one-eighth of the initial size?

Initial volume is $18432π$ and one with of this is $2304π$. I tried to integrate as follows

$$\int _{2304\pi }^{14832\pi }\:\frac{dV}{dt}$$
Where $\frac{dV}{dt}=4\pi r^2\cdot\frac{dr}{dt}$

Also from the given statement $\frac{dV}{dt}=-3\:\cdot4\pi r^2$ implying that $\frac{dr}{dt}=-3$

I couldn't solve it this way so I assume it is wrong. How do I do it?

答:
Nice job obtaining $\dfrac {dr}{dt} = -3$. This means that the radius is decreasing at a constant rate of $3$ cm per second.

Initial radius of the ball is $24$cm. At one-eighth of its volume, the radius would be halved, i.e. $12$cm. This takes:

$$\frac {24-12}{3} = 4 \text{ seconds}$$


答: You don't say how you found $x$. If you actually compute $A^TA$ and compare with $I$, you get that $|x|=1/2$, so $x$ is either $1/2$ or $-1/2$.

Title: Find orthogonal matrix missing values


问:
Is $p^{2} \overset{8}{\equiv} 1$
I was working on an assignment on finite fields, and I think it boiled down to prove the following, for every prime $p>2$:

$p^{2} \overset{8}{\equiv} 1$

I checked it directly for small primes, and it seems to hold. Is this statement true? If it is, I'd appreciate help in reaching a general proof.

Many Thanks!

答:
Not only for prime numbers greater that 2, but also it's true for all odd numbers. For any odd integer $k$, the following is true: $k^2\overset{8}{\equiv}1$

The reason is that $k^2\overset{8}{\equiv}1 \iff 8\mid k^2-1=(k-1)(k+1)$

And because $k$ is odd, $(k-1),(k+1)$ are both even. Furthermore, one of two consecutive even numbers is also divisible by $4$ and since $(k-1),(k+1)$ are two consecutive even numbers, one of them is divisible by $2$ and the other is divisible by $4$ so their product is divisible by $8$.

Title: Is $p^{2} \overset{8}{\equiv} 1$


问:
Understanding the proof that countable product of second countable space is second countable under the product topology
This is a very short proof but I dont understand the reason the proof was done this way and why we have a countable basis for the product topology.

"A countable product of second countable spaces is second countable."

Suppose that for each $n$ ,$X_n$ is second countable, say $\mathcal{B}_n$ is a countable basis. Then $$\{\prod_n B_n : B_n = X_n \text{ for all but finitely many } n; \text{ if } B_n \neq X_n, \text{ then } B_n \in \mathcal{B}_n \}$$ is a countable basis for $X = \prod_n X_n$, so  $X$ is second countable.

答:
Let $\mathcal{B}_{n}:=\left\{N_{i}^{n}: i \in \mathbb{N}\right\}$ be a countable basis for $\tau_{n}$ for each $n \in I$.
Let $\mathrm{\pi}_{n}$ denote the projection of $X$ onto $X_{n}$.
Let $\mathcal{L}_{n}:=\left\{\operatorname{\pi}_{n}^{-1}\left(N_{i}^{n}\right): N_{i}^{n} \in B_{n}\right\}$.

Let $\mathcal{K}_{J}=\left\{\bigcap_{n \in J} L_{n}: \forall n\in J \text{ we have } L_{n} \in \mathcal{L}_{n}\right\}$ for $J \subset I,|J|<\infty$

From the definiton of basis of the product topology,
$\mathcal{B}=\bigcup_{J \subset I,|J|<\infty} \mathcal{K}_{J}$ is a basis of the product space.
Now, each of the $\mathcal{L}_{n}$'s is countable.
Since the $\mathcal{K}_{J}$'s can be identified with a finite product of countable sets, they are countable.

Since countable union of countable sets is countable, $\mathcal{B}$ forms a countable basis of $X$.


答: A short test is that $\pm 1=\det(A)=-2x$. This leaves both possibilities, namely $+1$ and $-1$ for the determinant, so $x=1/2$ and $x=-1/2$. Indeed, one checks that both cases yield an orthogonal matrix.

Title: Find orthogonal matrix missing values


答:
Actually, every odd number satisfies this property, since
$$1^2=1 \equiv 1 \,[8]$$
$$3^2=9 \equiv 1 \,[8]$$
$$5^2=25 \equiv 1 \,[8]$$
$$7^2=49 \equiv 1 \,[8]$$

Title: Is $p^{2} \overset{8}{\equiv} 1$

Body:
I was working on an assignment on finite fields, and I think it boiled down to prove the following, for every prime $p>2$:

$p^{2} \overset{8}{\equiv} 1$

I checked it directly for small primes, and it seems to hold. Is this statement true? If it is, I'd appreciate help in reaching a general proof.

Many Thanks!