问:
Proof verification for $A$ $\times$ $B$ $\subseteq$ $\bigcup$ $\mathcal{F}$ $\otimes$ $\mathcal{G}$
Suppose $\mathcal{F}$ is a partition of $A$ and $\mathcal{G}$ is a partition of $B$. Let $\mathcal{F}$ $\otimes$ $\mathcal{G}$ be defined as follows: $$\mathcal{F} \otimes\mathcal{G} = \{Z \in \mathscr{P}(A\times B) \mid (\exists X)(\exists Y)(X\in \mathcal{F} \wedge Y \in \mathcal{G} \wedge Z = X \times Y)\}$$

Prove that $\mathcal{F}$ $\otimes$ $\mathcal{G}$ is a partition of $A$ $\times$ $B$.

My question is in regards to the proof of $A\times B\subseteq \bigcup\mathcal{F}\otimes \mathcal{G}$. My attempt is as follows:

Suppose $(a,b)$ $\in$ $A$ $\times$ $B$. Since $\mathcal{F}$ is a partition of $A
$, $a$ $\in$ $X$ for some $X$ $\in$ $\mathcal{F}$. Similarly, $b$ $\in$ $Y$ for some $Y$ $\in$ $\mathcal{G}$ Let $Z$ = $X$ $\times$ $Y$. Suppose $(i,j)$ $\in$ Z. Then, $i$ $\in$ X $\subseteq$ A and $j$ $\in$ Y $\subseteq$ B or $(i,j)$ $\in$ $A$ $\times$ $B$. Thus, $Z$ $\subseteq$ $A$ $\times$ $B$ or Z $\in$ $\mathscr{P}$($A$ $\times$ $B$). Now it can be seen that Z $\in$ $\mathcal{F}$ $\otimes$ $\mathcal{G}$ and $A$ $\times$ $B$ $\subseteq$ $\cup$ $\mathcal{F}$ $\otimes$ $\mathcal{G}$.  $\square$

Is my attempt valid? This proof differs from some other attempts I have seen online and I want to verify it.

答:
Let me suggest the most general version of this phenomenon. Consider a family $A$ of sets indexed by index set $I$. Agreeing to denote the set of all partitions of a given set $M$ by $\mathscr{Part}(M)$, let us consider a family $\mathscr{P} \in \displaystyle\prod_{i \in I}\mathscr{Part}(A_i)$ of partitions, which means that the component at each index $i \in I$ is a partition $\mathscr{P}_i \in \mathscr{Part}(A_i)$ on $A_i$.

We proceed to define the following set:
$$\bigotimes_{i \in I}\mathscr{P}_i \colon=\left\{\prod_{i \in I}X_i\right\}_{X \in \displaystyle\prod_{i \in I}\mathscr{P}_i}$$
and seek to prove that:
$$\displaystyle\bigotimes_{i \in I}\mathscr{P}_i \in \mathscr{Part}\left(\displaystyle\prod_{i \in I}A_i\right).$$

To prove this assertion, let us set $\mathscr{Q}\colon=\displaystyle\bigotimes_{i \in I}\mathscr{P}_i$ respectively $B\colon=\displaystyle\prod_{i \in I}A_i$. There are three statements we need to establish separately:

Any $U \in \mathscr{Q}$ is a nonempty subset $\varnothing \neq U \subseteq B$: by definition, we have the expression $U=\displaystyle\prod_{i \in I}X_i$ for a certain family $X \in \displaystyle\prod_{i \in I}\mathscr{P}_i$. Since $\mathscr{P}_i$ is a partition of $A_i$ we infer $X_i \subseteq A_i$ for every $i \in I$ and thus $U \subseteq B$ (by virtue of the monotony of the cartesian product operation). Likewise, since $X_i \in \mathscr{P}_i$ for each $i \in I$ it follows that $X_i \neq \varnothing$ and as cartesian products of nonempty sets remain nonempty we gather that $U \neq \varnothing$.

For any $U, V \in \mathscr{Q}$ we have $U \neq V \Rightarrow U\cap V=\varnothing$. Once again, let us consider expressions $U=\displaystyle\prod_{i \in I}X_i$ and $V=\displaystyle\prod_{i \in I}Y_i$, afforded by families $X, Y \in \displaystyle\prod_{i \in I}\mathscr{P}_i$. As $U \neq V$, there must exist an index $i \in I$ such that $X_i \neq Y_i$. However, since $X_i, Y_i$ are distinct members of the partition $\mathscr{P}_i$, this entails $X_i \cap Y_i=\varnothing$ and hence:
$$U \cap V=\left(\prod_{i \in I}X_i\right) \cap \left(\prod_{i \in I}Y_i\right)=\prod_{i \in I}\left(X_i \cap Y_i\right)=\varnothing,$$
bearing in mind that the cartesian product of a family of sets at least one component of which is empty is itself (the product) empty.

Finally, $B=\bigcup\mathscr{Q}$. Here we shall rely on the following fundamental:

Proposition. Let $J$ be a family of sets indexed by set $I$ and let furthermore $X$ be another family of sets, indexed by the disjoint union $\displaystyle\bigsqcup_{i \in I}J_i\colon=\displaystyle\bigcup_{i \in I}\left(\{i\} \times J_i\right)$. We then have the relation:
$$\prod_{i \in I}\bigcup_{j \in J_i}X_{ij}=\bigcup_{k \in \prod_{i \in I}J_i}\prod_{i \in i}X_{ik_i}.$$

We apply the general proposition above to the instance of family $\mathscr{P}$ of partitions playing the role of $J$ and family $M$ -- given by $M_{iX}=X$, for any $i \in I$ and $X \in \mathscr{P}_i$ -- playing the role of $X$, in order to infer that:
$$\begin{align}
B&=\prod_{i \in I}A_i\\
&=\prod_{i \in I}\bigcup\mathscr{P}_i\\
&=\prod_{i \in I}\bigcup_{X \in \mathscr{P}_i}X\\
&=\bigcup_{Y \in \prod_{i \in I}\mathscr{P}_i}\prod_{i \in Y}M_{iY_i}\\
&=\bigcup_{Y \in \prod_{i \in I}\mathscr{P}_i}\prod_{i \in I}Y_i\\
&=\bigcup_{U \in \mathscr{Q}}U\\
&=\bigcup\mathscr{Q}.
\end{align}$$

This concludes the argumentation and signifies the existence of a natural map:
$$\begin{align}
\prod_{i \in I}\mathscr{Part}(A_i) &\to \mathscr{Part}\left(\prod_{i \in I}A_i\right)\\
\mathscr{P} &\mapsto \bigotimes_{i \in I}\mathscr{P}_i.
\end{align}$$


问:
Prove the inequality: $(y+z)*x^4+(x+z)*y^4+(x+y)*z^4\le((x+y+z)^5)/12$ for positive reals.
The inequality is homogeneous, but I am not sure if that helps.
Also, I was thinking about considering a concave function and apply Jensen or Karamata on LHS.</p>

答:
Suppose $x \geqslant y \geqslant z$ and
$$f(x,y,z) = x^4(y+z)+y^4(z+x)+z^4(x+y).$$
We have
$$\begin{aligned}f(x,y,z)-f(x,y+z,0) & = y^4(z+x)+z^4(x+y)-x(y+z)^4\\ \\& = y^4(z+x)+z^4(x+y)-x(y+z)^4 \\ \\&=-yz\left[3x(y+z)^2+y^2(x-y)+z^2(x-z)\right] \leqslant 0.\end{aligned}.$$
Therefore $f(x,y,z) \leqslant f(x,y+z,0).$ Finally, we need to prove
$$f(x,y+z,0) = x^4(y+z)+x(y+z)^4\leqslant \frac{1}{12}(x+y+z)^5.$$
Setting $a = y +z,$ the inequality become
$$12ax(a^3+x^3) \leqslant (x+a)^5,$$
or
$$(a+x)(a^2-4ax+x^2)^2 \geqslant 0.$$
Which is true. The proof is completed.

ContentLicense: CC BY-SA 4.0


问:
Simplification of an expression
How do I simplify the following expression?
$$\displaystyle \frac{\int_q^1 w(s) \int_0^s e(\xi) d\xi ds}{2\int_q^1 w(s) ds} p$$
where $w(t)$ is nondecreasing $w(t)>0$ on $(q,1]$ , $e :(0,1)\rightarrow \mathbb{R}$ is continuous, $e\in L(0,1)$,
and $\displaystyle\frac12<p<q<1$

Please help me.  Thank you.

答:
Changing the order if integration gives
$$ \int_q^1 w(s) \int_0^s e(\xi) d\xi ds = \int_0^q e(\xi) \int_q^1 w(s)\, ds d\xi + \int_q^1  e(\xi)\int_{\xi}^1 w(s)\, ds d\xi .$$

Note that, the first integral on the right hand side can be written as

$$ \int_0^q e(\xi)d\xi \int_q^1 w(s)\, ds = \int_q^1 w(s)\, ds\int_0^q e(\xi)d\xi.  $$

Plot the region to see how the first equation derived.

Added: Plot the region $ \xi(s) = s $ where $ q \leq s \leq 1 $. Then to  change the order of integration, consider taking a horizontal strip and notice that the horizontal strip will be bounded below by two different functions, namely, $s=q$ and $s=\xi$. Accordingly, find the limits for $\xi$.

ContentLicense: CC BY-SA 3.0

Title: Simplification of an expression


问:
Distribution of a Standard Brownian Motion Function
Let $W(t)$ be a standard Brownian motion. What is the distribution of the following function?

$Y(t)=W(t)+W(t^2)$,

Note: We know that $W(t^2)$ is not a Brownian motion, since $R_{W(t^2)}(t_1,t_2)=min(t_1^2,t_2^2)$.
$R_{w}(.)$ stands for autocorrelation.

答: $W(t)$ and $W(s)$ are jointly normal for any $t$ and $s$.  Hence $Y(t)$ is normal. You only have to find the mean and the variance. The mean is, of course, $0$. Now $EY(t)^{2}=E[W(t)]^{2}+E[W(t^{2})]^{2}+2EW(t)W(t^{2})=t+t^{2}+2\min \{t, t^{2}\}$.


问:
Confusion whilst reading the proof of uniqueness of Poisson Equation
I wanted to show uniqueness to $\nabla^2 \phi=f$ subject to mixed boundary conditions; I was considering the following:

$\nabla^2u=0$ in a closed domain $D$ subject to $\alpha u+\beta\frac{\partial u}{\partial n}=0$ on $\partial D.$ (To avoid confusion, this is the general mixed condition. $\alpha =1, \beta =0$ is the Dirichlet, $\alpha =0, \beta =1$ is the Neumann.) Then the focus is to show if $u\equiv 0.$

Then I was presented this proposition that I had some trouble with.

Proposition: If $\alpha \neq0,\frac{\beta}{\alpha}\geq0$ then we have unicity.

Proof: Consider div$(u\nabla u)$, by simple identity div$(u\nabla u)=|\nabla u|^2+u\nabla^2u.$ Integrate over domain $D$ on both side and use divergence theorem we arrive at $$\int_D |\nabla u|^2dA=\int_{\partial D} u\frac{\partial u}{\partial n}ds.$$

Note from the above that $\frac{\partial u}{\partial n}=-\frac{\alpha u}{\beta}$ on the boundary and so by considering signs we deduce $|\nabla u|=0$ thus $u=const=0,$ as required.

My confusion:

Where did we use $\alpha \neq 0?$ It seems to me that we used $\beta \neq 0$ instead?

However, my suspect cannot be possibly right because if we have $\alpha=0, \beta=1$ this is just Neumann and I already knew this won't correspond to a unique solution.

Where did I go wrong? Many thanks in advance!

答:
(just promoting comment to answer)

The answer to your question comes from looking carefully at the steps and seeing exactly what is used where, and why.
In either case $\alpha\neq 0$ or   $\beta\neq0$, you can use the boundary condition:
$$ \int_D |\nabla u|^2 dA = \int_{\partial D} u\frac{\partial u}{\partial n} ds\overset{*}=\begin{cases}-\int_{\partial D} \frac\alpha\beta u^2 ds &\beta \neq 0 \\ -\int_{\partial D} \frac\beta\alpha (\frac{\partial u}{\partial n})^2 ds &\alpha \neq 0 \end{cases}$$
(The step marked $*$ is where the boundary condition is used.)
Considering the sign of the expressions as you said will lead to $u\equiv c$ for some $c\in\mathbb R$, and hence $\frac{\partial u}{\partial n}=0$ on $\partial \Omega$. Therefore the boundary condition reads
$$ \alpha u = 0 \text{ on } \partial \Omega.$$
It is at this point that you need $\alpha\neq 0$ to conclude that $u=0$ on $\partial \Omega$, and hence $u=c=0$ on $\Omega$.


问:
Is it possible to simplify this version of a geometric series? $f(x) = \sum\limits_{k=1}^n \frac{x^k}{k}$
Is it possible to simplify this series?

$$
f(x) = \sum\limits_{k=1}^n \frac{x^k}{k}
$$

If I derive it, I get the geometric series

$$
f'(x) = \sum\limits_{k=1}^n x^{k-1} = \sum\limits_{k=0}^{n-1} x^{k} = \frac{1-x^n}{1-x}
$$

But this seems to lead to nothing as

$$
f(x) = \int \frac{1-x^n}{1-x} dx
$$

can't expressed simpler as a comment in my previous question Antiderivative of $\frac{1-x^n}{1-x}$ mentioned.

I know, that $\lim\limits_{n \to \infty} \int \frac{1-x^n}{1-x} dx = -\ln{(|x-1|)}, \quad -1 < x < 1$

But what about without these restrictions of $n$ and $x$?

答:
Beside the  incomplete Beta function, you also have
$$f(x) = \sum\limits_{k=1}^n \frac{x^k}{k}=-x^{n+1}\, \Phi (x,1,n+1)-\log (1-x)$$ where appears the Lerch transcendent function.


问:
Compute the integral of $f=\frac{1}{z^2}$ along the given contour $T$.
My professor wrote the homework questions by hand so I posted screenshots for reference.

In office hours earlier today, my professor gave my friend the answer to b) claiming that it was 0. My friend then told me that the reason for this was $\int_{T}f = 0$ because the curve is closed. However I am finding it really hard to believe this answer.

What I ended up doing was using the fact that by Cauchy's theorem, the integral along $T$ is the same as the integral around 2 unit circles centered at $0$. Then I computed the contour integral and got a nonzero number.

I don't understand what theorem or property, or even intuitively looking at this picture why the answer would be 0. Any help/independent verification would be appreciated, thanks.

Edit: How I got my answer

Just FYI Our professor did NOT teach us about residues so that knowledge technically isn't known at this point.

By Cauchy's integral formula, we have

$$f(f(z)) = \frac{1}{2\pi i}\int_{T} \frac{f(u)}{u-z}du$$
where $f(u)$ is analytic on a region containing $T$ and its interior. Since the unit circle is contained in $T$, by Cauchy's theorem we can compute the integral along a double circle. This is what I attempted to do. Using the formula $$\int_{T}f(z)dz = \int_{a}^{b}f(z(t))z'(t)dt$$ where $z'(t) = rie^{it}$ and letting $C$ be the unit circle centered at $0$, I get

\begin{align*}
\int_{C}\frac{1}{z^2}z'(t)dt = \int_{0}^{2\pi}\frac{1}{r^2e^{2it}}rie^{it}dt = \frac{i}{r}\int_{0}^{2\pi}\frac{1}{e^{it}}dt = i\int_{0}^{2\pi}\frac{1}{z}dt = 2k\pi i^2
\end{align*}
where we know from class that $\int_{0}^{2\pi}\frac{1}{z}dz = 2\pi i$ and the winding number is $k=2$ I got $-4\pi$ as my answer.

答:
The function $\frac{1}{z^2}$ has a pole at $0$, so your gut feeling that using the closure of the curve is not sufficient for the integral to be $0$ is correct. In fact, for a different, but very similar function which also has a pole at $0$, the integral doesn't vanish: $\frac1z$.

In the comments it was said that $\frac{1}{z^2}$ has $0$ as a residue, so the integral vanishes. But philosophically speaking, that's the wrong way around: $\frac{1}{z^2}$ has $0$ residue because the integral along your given curve vanishes. If it didn't vanish, we'd have defined the residue differently, so it wouldn't be $0$.

The deeper reason why this integral vanishes is the fact that your function has an antiderivative. If a holomorphic function has an antiderivative on the entirety of its domain, then contour integrals only depend on the start and end points of the contour, which makes closed integrals vanish. And $\frac{1}{z^2}$ has an antiderivative: it's $-\frac{1}{z}$. In contrast, the example $\frac{1}{z}$ does not have an antiderivative on its entire domain. Such an antiderivative would have to be a holomorphic branch of the logarithm, but no such branch exists on the punctured plane $\mathbb C\backslash\{0\}$. That's why closed integrals over that function don't necessarily vanish.


问:
Existence of a global bounded solution of an IVP
I am having difficulties rigorously proving the following:

Show that the initial value problem
$$y'=e^{-y^2}(y^{2014}-y^{2013}), \quad
y(0)=1/2$$
possesses a bounded solution defined on the whole of $\mathbb{R}$.

Context:
I have managed to show that the solution is defined on $\mathbb{R}$ by inductively using the existence theorem. I have also sketched out the phase portrait, which allowed me to conjecture that the solution is bounded between $0$ and $1$ and that it is decreasing, since the function $f(x,y)=e^{-y^2}(y^{2014}-y^{2013})$ is nonpositive for all $y\in [0,1]$ and has zeros at $0$ and $1$. But I think my argument still has to be refined a bit.

Any help would be greatly appreciated.

答:
The IVP
$$
y'=\mathrm{e}^{-y^2}(y^{2014}-y^{2013}), \quad y(0)=\frac{1}{2} \tag{1}
$$
enjoys uniqueness, since its flux function $\,f(x,y)=\mathrm{e}^{-y^2}(y^{2014}-y^{2013})\,$ is $C^1$. (In fact, it is $C^\infty$.) If $\varphi: I\to\mathbb R, \,$ is a solution, where $I$ is an open interval, then
$$
0<\varphi(x)<1, \quad \text{for all $x\in I$}.
$$
For if $\varphi(x_0)=1$, for some $x_0\in I$, then $\varphi$ would also satisfy the IVP
$$
y'=\mathrm{e}^{-y^2}(y^{2014}-y^{2013}), \quad y(x_0)=1,
$$
which possesses as unique solution the $\psi(x)=1$, hence we would have that $\varphi(x)\equiv1$. We can similarly show that $\varphi(x)\ne 0$, for all $x\in I$.

Observe also that $\varphi$ is strictly decreasing.

Next, we use the fact that every IVP, which enjoys uniqueness, possesses a maximally defined solution. (That is, one which can not be extended further as a solution.) Assume $\varphi: I\to\mathbb R$ is the maximally defined solution, and its maximal domain is $I=(a,b)$. If $b<\infty$, let
$$
y_+=\lim_{x\to b}\varphi(x).
$$
Clearly, as $\varphi$ is decreasing, this limit exists, and
$$
0<\varphi(x)<1 \quad\Longrightarrow\quad 0\le y_+<1.
$$
If now $\theta: (b-\eta,b+\eta)\to\mathbb R$ satisfies the IVP
$$
y'=\mathrm{e}^{-y^2}(y^{2014}-y^{2013}), \quad y(b)=y_+,
$$
then
$$
\tilde\varphi(x)=\left\{
\begin{array}{ccc}
\varphi(x) & \text{if} & x\in(a,b),\\
\theta(x) & \text{if} & x\in [b,b+\eta),\\
\end{array}
\right.
$$
is also a solution of (1) which contradict the maximality of $(a,b)$. We act similarly if $a>-\infty$.


问:
Adjoint operator of a partial differential operator used to do the carleman estimate
I am reading Carleman estimate for parabolic equations and applications written by Masahiro Yamamoto.

Let $P$ be an operator defined by
$$
Pw=\partial_tw-\Delta w+2s\nabla \varphi\cdot\nabla w+(-s\partial_t \varphi-s^2|\nabla\varphi|^2+s\Delta \varphi)w.
$$
The article on page 4 says that the way is the decomposition of $P$ into the symmetric part $P_+$ and antisymmetric part $P_-$. To do this, he consider the formal ajoint operator $P^*$ to $P$:
$$
(Pw,v)_{L^2(D)}=(w,P^*v)_{L^2(D)},\quad v,w\in C_0^\infty(D).
$$
By integration by parts, the Green's theorem and $v,w \in C_0^\infty(D)$, he obtains
$$
P^*=-\partial_t w-\Delta w -2s\nabla\varphi\cdot\nabla w-(s\Delta \varphi+s^2|\nabla\varphi|^2+s(\partial_t\varphi))w.
$$
It can be seen that the term $s\Delta\varphi w$ change to $-s\Delta \varphi w$. I cannot figure out why this change happens. According to my understanding (may not be true),
$$
(s\Delta \varphi w,v)=(w,s\Delta \varphi v),
$$
so there must be something wrong with my thinking.

答:
You are right that the part $w\mapsto s(\Delta \phi)w$ is symmetric. However, there is additionally a contribution from the part $w\mapsto 2s\nabla \phi\cdot\nabla w$:
$$
\int v\nabla\phi\cdot\nabla w=-\int w\nabla\cdot(v\nabla \phi)=-\int w(\nabla v\cdot\nabla \phi+v\Delta \phi).
$$
So you end up with $s(\Delta \phi)w-2s(\Delta \phi)w=-s (\Delta \phi)w$ in the adjoint.


问:
Minimal degree affect on a tree subgraph.
Show that each graph G with minimal degree m contains any tree with m+1 vertices as a subgraph.
I have no idea how to start here, would appreciate help. Thanks!</p>

答:
Hint:

Induct on $m$.

First, can you show the $m=1$ case is true?

Then, assume the claim for graphs of minimal degree $m$. We want to show the claim for a graph of minimal degree $m+1$.

Fix a graph $G$ with minimal degree at least $m+1$. Fix your favorite tree $T$ with $m+2$ vertices. Then remove a leaf $u$ from $T$ to get a tree $T'$ with $m+1$ vertices.

Can you use the inductive hypothesis to show $G$ contains a copy of $T'$?

Can you use the fact that $G$ has minimal degree $m+1$ to find a way to reattach $u$ to $T'$ in $G$? This will give you a copy of $T$ in $G$. As desired.

I hope this helps ^_^


问:
How to Solve $\frac{\operatorname dy}{\operatorname dx}=\frac{{(x+y)}^2}{(x+2)(y-2)}$
I am a 12th  grader in India .
I am currently in my Differential Equations Course and
I am a beginner.
Can anybody help me in solving this Differential Equation.
I could nowhere find a solution on web .</p>

Body:
$$\frac{dy}{dx}=\frac{(x+y)^2}{(x+2)(y-2)}$$
Let $x-2=X, y-2=Y$, then
$$\frac{dY}{dX}=\frac{(X+Y)^2}{XY}$$
Let $$Y=VX \implies \frac{dY}{dx}=X\frac{dV}{dX}+V$$
$$\implies X\frac{dV}{dX}+V= \frac{(1+V)^2}{V} \implies X\frac{dV}{dX}=\frac{(1+V)^2}{V}-V  \implies \int \frac{VdV}  {1+2V}=  \int \frac{dX}{X}$$

$$\implies V=2\ln[CX(1+2V)^{1/4}], V=Y/X, X=x+2. Y=y-2$$                                                                                $$


问:
Better understanding of the meaning of implication in equations
While solving this equation
$$\sqrt{x^2+1}+x=1$$
I started to have some doubts about my real logical understanding of what I'm doing: I will show the steps to try to communicate better what I mean. We know that adding/subtracting functions that belong to the domain of the original equation gives an equivalent equation (meaning it has the same solutions of the first one), so if I want to add $-x$ both sides the correct logical operator is
$$\sqrt{x^2+1}+x=1\iff\sqrt{x^2+1}=1-x $$
Going on, I know that is true that $a=b\implies a^2=b^2$, so the next step is not an equivalence but an implication; meaning that the correct logic operator is
$$\sqrt{x^2+1}=1-x \implies x^2+1=(1-x)^2$$
Now since $(1-x)^2=x^2-2x+1$, I suppose that an equality is an equivalence (correct me if I'm wrong) so I get
$$x^2+1=(1-x)^2 \iff x^2+1=x^2-2x+1 \iff -2x=0 \iff x=0$$
So, since they are all equivalence and there is an implication, connecting the first and the last I get
$$\sqrt{x^2+1}=1-x \implies x=0$$
So the questions are:

a) since an equation is the problem to determine which unknowns, substituted in the equality, makes the equality true, it seems to me, of course wrongly, that we've obtained the converse, that is "if equation is zero then the unknowns are $*$" while, if I've not understood wrong, we are searching for what unknowns make the equation verified, so we're searching for the converse implication "if unknowns are $*$ these then the equation is verified". What am I missing here?

b) In the equation $\sqrt{x^2+1}=1-x$ I have on the left hand side a non negative quantity while on the right hand side a quantity that can be negative for some values of $x$; so, when arriving to $(2)$, I should add a condition $1-x \geq 0$ to have a real equivalence?

Thanks.

答:
There are two ways of dealing with this. One of them is to do what you did: to be aware of the fact that, at a certain step, what we have is an implication which is not an equivalence. So, at the end, when we get $x=0$, we have to check whether or not $0$ is really a solution of the original equation (it is).

The other way has to do with what you wrote in the final paragraph: when we have $\sqrt{x^2+1}=1-x$ we decide that, from now when, we will work only on $(-\infty,1]$; if $x>1$, then the equation has no solutions (since $\sqrt{x^2+1}\geqslant0$ and $1-x<0$. And, on that interval, we do have an equivalence:$$\sqrt{x^2+1}=1-x\iff x^2+1=(1-x)^2.$$At the end of the computations, we must eliminat all solutions that we got that do not belong to $(-\infty,1]$ (there will be none in this case).

Finally, I dont' see why is that that you assert that what we were supposed to be doing is “if unknowns are ∗ these then the equation is verified”; it's the other way around.


问:
Interesting infinite nested square roots of 2 for $2\cos1°$ and $2\sin1°$
It is interesting to note that any angle between 45° to 90° satisfying $1\over4$ < $p \over q$ <$1\over2$ where $ p \over q$ is of form $p = 2^n $ and $q$ is an odd number satisfying $2^{n+1} <q <2^{n+2}$ can be represented as cyclic infinite nested square roots of 2 ( Hereafter referred as $cin\sqrt2$)

Interestingly 64° falls between 45° and 90° and can be represented in radians as $16\pi \over 45$

Expansion of $2\cos\frac{16\pi}{45}$ happens as follows

$2\cos\frac{16\pi}{45} = \sqrt{2+2\cos\frac{32\pi}{45}} =\sqrt{2-2\cos\frac{13\pi}{45}} $
$=\sqrt{2- \sqrt{2+2\cos\frac{26\pi}{45}}} = \sqrt{2- \sqrt{2-2\cos\frac{19\pi}{45}}}$
$=\sqrt{2-\sqrt{2-\sqrt{2+2\cos\frac{38\pi}{45}}}} = \sqrt{2-\sqrt{2-\sqrt{2-2\cos\frac{7\pi}{45}}}}$
$=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+2\cos\frac{14\pi}{45}}}}}$
$=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+2\cos\frac{28\pi}{45}}}}}}$
$=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-2\cos\frac{17\pi}{45}}}}}}$
$=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+2\cos\frac{34\pi}{45}}}}}}}$
$=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-2\cos\frac{11\pi}{45}}}}}}}$
$=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+2\cos\frac{22\pi}{45}}}}}}}}$
$=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+2\cos\frac{44\pi}{45}}}}}}}}}$
$=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-2\cos\frac{\pi}{45}}}}}}}}}$....1

For the sake of simplicity last nested radical can be represented as $n\sqrt2[3-1+2-1+1-]$(nested square roots of 2 having $3-1+2-1+1-$)

$2\cos\frac{\pi}{45}$ is represented as $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+2\cos\frac{16\pi}{45}}}}}$ or simply as $n\sqrt2(4+2\cos\frac{16\pi}{45})$...2

Combining 1 & 2 will be single cycle of nested radical for $2\cos\frac{16\pi}{45}$ simply represented as  $n\sqrt2[3-1+2-1+1-4+]$

Now we can represent $2\cos\frac{16\pi}{45}$ as cyclic infinite nested square roots of 2 as
$cin\sqrt2(3-1+2-1+1-4+)$

As $64^\circ$ is $(2^6)^\circ$ taking half angle for 6 times will give noncyclic nested square roots of 2 as $n\sqrt2(6+)$

Therefore $2\cos1°$ can be represented as $n\sqrt2(6+)cin\sqrt2[3-1+2-1+1-4+]$ and

$2\sin1°$ can be represented as $n\sqrt2(1-5+)cin\sqrt2[3-1+2-1+1-4+]$

This opens the world of nested square roots of 2 for calculating $2\cos1°$ or $2\sin1°$ without any need  by taking root of cubic equation which involves imaginary component. And no need to know the value of$\pi$ to evaluate trig values as in Taylor series expansion.

Calculating single cycle itself provides result with accuracy of 7 digits after decimal point. (Six '2's in noncyclic part ten '2's in first cycle in infinite radical.)

With available scientific calculators and by programming, I have confirmed the results (for a long Post like this I feel it is difficult to incorporate those things)

My question is, is it possible to simplify above procedures by some other means?

答:
(Revising and extending some comments.)

We can generate a version of these representations thusly:

Let $\theta_0$ be the angle whose double-cosine we seek.

Let $\theta_{n+1}:=\min(2\theta_n,180^\circ-2\theta_n)$ and $\sigma_n := \operatorname{sgn}( 90^\circ-2\theta_n)$

Generate $\theta$- and $\sigma$-values, stopping either when $\sigma_N=0$ (the "terminating" case) or when $\theta_N=\theta_{m}$ for some $m<N$ (the "non-terminating" case).

Then either $\sigma(\theta_0) := [\sigma_0, \sigma_1, \ldots, \sigma_N=0|]$ (terminating) or $\sigma(\theta_0) := [\sigma_0, \sigma_1, \sigma_2, \ldots, \sigma_{m} \;|\; \sigma_{m+1}, \sigma_{m+2}, \ldots, \sigma_N ]$ (non-terminating) corresponds to a nested-radical representation of $2\cos \theta_0$: Each $\sigma_n$ gives the sign at the corresponding nesting level, as in $$2\cos \theta_0 = \sqrt{ 2 + \sigma_0 \sqrt{2 + \sigma_1 \sqrt{ 2+\sigma_2 \sqrt{\cdots} }}}$$ and the "$|$" separates a finite initial sequence from an infinitely-repeating cycle. (Caveat: The index arithmetic for my placement of the "$|$" may be off by one.)

For instance,
$$\sigma(45^\circ) = [0|] \qquad
\sigma(22.5^\circ) = [1\bar{0}|] \qquad
\sigma(64^\circ) = [|\bar{3}1\bar{2}1\bar{1}4] \qquad
\sigma(1^\circ) = [2|4\bar{3}1\bar{2}1\bar{1}]$$
where, for the sake of compactness, I'm collapsing strings of $k$ "$-1$"s to "$\bar{k}$" and strings of $k$ "$+1$"s to simply "$k$". Also, I'm writing a terminating "$0$" as either "$0$" or "$\bar{0}$", whichever contrasts with the "sign" of the preceding string; we can think of this as indicating a collapsed string of no "$+1$"s or no "$-1$"s. (BTW: Please pardon the potential confusion of the over-bar notation with repeating-decimal notation.)

The acute integer-degree angles separate into a few families of "doubled" angles (adjusted by the "min" step in the algorithm, if necessary) with related repeating cycles. For now, I've rendered the representations as Mathematica generated them, but one could make the cycles identical by augmenting the initial sequences with portions of them and manipulating appropriately; eg,
$$\begin{align}
\sigma(8^\circ) &= [| 3\bar{3}1\bar{2}1\bar{1}1] = [3\bar{3}1\bar{2}1\bar{1}1| 3\bar{3}1\bar{2}1\bar{1}1] = [3\bar{3}1\bar{2}1\bar{1}| 4\bar{3}1\bar{2}1\bar{1}] \\
\sigma(16^\circ) &= [| 2\bar{3}1\bar{2}1\bar{1}2] = [2\bar{3}1\bar{2}1\bar{1}2| 2\bar{3}1\bar{2}1\bar{1}2] = [2\bar{3}1\bar{2}1\bar{1}|4\bar{3}1\bar{2}1\bar{1}]
\end{align}$$

so that the last versions of $\sigma(8^\circ)$ and $\sigma(16^\circ)$ have the same repeating cycle as $\sigma(1^\circ)$, $\sigma(2^\circ)$, and $\sigma(4^\circ)$. It's not necessarily clear what the "canonical cycle" should be, but I might suggest that such a cycle would begin and end with different "signs" (ie, "$p\cdots\bar{q}$" or "$\bar{p}\cdots q$", but not "$p\cdots q$" or "$\bar{p}\cdots\bar{q}$").

Each table begins with an orbit of doubled-and-adjusted angles. It happens that these are all the angles whose corresponding $\sigma$s can be written with empty initial sequences.

Note that the formula $2\cos\theta=\sqrt{2+2\cos2\theta}$ implies that (some form of) $\sigma(\min(2\theta,180^\circ-2\theta))$ is obtained from (some form of) $\sigma(\theta)$ by reducing the first digit; eg, $[2\bar{3}\cdots]\to[1\bar{3}\cdots]\to [\bar{3}\cdots]$.

$$\begin{array}{rrcrrcrrcr}
\theta & \sigma(\theta) && \theta & \sigma(\theta) && \theta & \sigma(\theta) \\
\hline
4^\circ & [| 4\bar{3}1\bar{2}1\bar{1}] &\to&
\phantom{1}8^\circ & [| 3\bar{3}1\bar{2}1\bar{1}1] &\to&
16^\circ & [| 2\bar{3}1\bar{2}1\bar{1}2] \\
32^\circ & [| 1\bar{3}1\bar{2}1\bar{1}3] &\to&
64^\circ & [| \bar{3}1\bar{2}1\bar{1}4]  &\to&
52^\circ & [| \bar{2}1\bar{2}1\bar{1}4\bar{1}] \\
76^\circ & [| \bar{1}1\bar{2}1\bar{1}4\bar{2}] &\to&
28^\circ & [| 1\bar{2}1\bar{1}4\bar{3}] &\to&
56^\circ & [| \bar{2}1\bar{1}4\bar{3}1] \\
68^\circ & [| \bar{1}1\bar{1}4\bar{3}1\bar{1}] &\to&
44^\circ & [| 1\bar{1}4\bar{3}1\bar{2}] &\to&
88^\circ & [| \bar{1}4\bar{3}1\bar{2}1] &\to&
(4^\circ) \\
\hline
1^\circ & [2| 4\bar{3}1\bar{2}1\bar{1}] &\to&
2^\circ & [1| 4\bar{3}1\bar{2}1\bar{1}] &\to&
(\phantom{1}4^\circ) \\
\hline
 7^\circ & [2| 1\bar{2}1\bar{1}4\bar{3}] &\to&
14^\circ & [1| 1\bar{2}1\bar{1}4\bar{3}] &\to&
(28^\circ) \\
\hline
11^\circ & [2| 1\bar{1}4\bar{3}1\bar{2}] &\to&
22^\circ & [1| 1\bar{1}4\bar{3}1\bar{2}] &\to&
(44^\circ) \\
\hline
13^\circ & [2| \bar{2}1\bar{2}1\bar{1}4\bar{1}] &\to&
26^\circ & [1| \bar{2}1\bar{2}1\bar{1}4\bar{1}] &\to&
(52^\circ) \\
\hline
17^\circ & [2| \bar{1}1\bar{1}4\bar{3}1\bar{1}] &\to&
34^\circ & [1| \bar{1}1\bar{1}4\bar{3}1\bar{1}] &\to&
(68^\circ) \\
\hline
19^\circ & [2| \bar{1}1\bar{2}1\bar{1}4\bar{2}] &\to&
38^\circ & [1| \bar{1}1\bar{2}1\bar{1}4\bar{2}] &\to&
(76^\circ) \\
\hline
23^\circ & [1\bar{1}| \bar{1}4\bar{3}1\bar{2}1] &\to&
46^\circ & [\bar{1}| \bar{1}4\bar{3}1\bar{2}1] &\to&
(88^\circ) \\
\hline
29^\circ & [1\bar{1}| \bar{3}1\bar{2}1\bar{1}4] &\to&
58^\circ & [\bar{1}| \bar{3}1\bar{2}1\bar{1}4] &\to&
(64^\circ) \\
\hline
31^\circ & [1\bar{1}| \bar{2}1\bar{1}4\bar{3}1] &\to&
62^\circ & [\bar{1}| \bar{2}1\bar{1}4\bar{3}1] &\to&
(56^\circ) \\
\hline
37^\circ & [1\bar{1}| 1\bar{3}1\bar{2}1\bar{1}3] &\to&
74^\circ & [\bar{1}| 1\bar{3}1\bar{2}1\bar{1}3] &\to&
(32^\circ) \\
\hline
41^\circ & [1\bar{1}| 2\bar{3}1\bar{2}1\bar{1}2] &\to&
82^\circ & [\bar{1}| 2\bar{3}1\bar{2}1\bar{1}2] &\to&
(16^\circ) \\
\hline
43^\circ & [1\bar{1}| 3\bar{3}1\bar{2}1\bar{1}1] &\to&
86^\circ & [\bar{1}| 3\bar{3}1\bar{2}1\bar{1}1] &\to&
(\phantom{1}8^\circ) \\
\hline
47^\circ & [\bar{2}| 3\bar{3}1\bar{2}1\bar{1}1] &\to&
(86^\circ) \\
\hline
49^\circ & [\bar{2}| 2\bar{3}1\bar{2}1\bar{1}2] &\to&
(82^\circ) \\
\hline
53^\circ & [\bar{2}| 1\bar{3}1\bar{2}1\bar{1}3] &\to&
(74^\circ) \\
\hline
59^\circ & [\bar{2}| \bar{2}1\bar{1}4\bar{3}1] &\to&
(62^\circ) \\
\hline
61^\circ & [\bar{2}| \bar{3}1\bar{2}1\bar{1}4] &\to&
(58^\circ) \\
\hline
67^\circ & [\bar{2}| \bar{1}4\bar{3}1\bar{2}1] &\to&
(46^\circ) \\
\hline
71^\circ & [\bar{1}1| \bar{1}1\bar{2}1\bar{1}4\bar{2}] &\to&
(38^\circ) \\
\hline
73^\circ & [\bar{1}1| \bar{1}1\bar{1}4\bar{3}1\bar{1}] &\to&
(34^\circ) \\
\hline
77^\circ & [\bar{1}1| \bar{2}1\bar{2}1\bar{1}4\bar{1}] &\to&
(26^\circ) \\
\hline
79^\circ & [\bar{1}1| 1\bar{1}4\bar{3}1\bar{2}] &\to&
(22^\circ) \\
\hline
83^\circ & [\bar{1}1| 1\bar{2}1\bar{1}4\bar{3}] &\to&
(14^\circ) \\
\hline
89^\circ & [\bar{1}1| 4\bar{3}1\bar{2}1\bar{1}] &\to&
(\phantom{1}2^\circ) \\
\hline
\end{array}$$

$$\begin{array}{rrcrrcrrcrrcr}
\theta & \sigma(\theta) && \theta & \sigma(\theta) && \theta & \sigma(\theta) && \theta & \sigma(\theta) \\
\hline
12^\circ & [| 2\bar{2}] &\to&
24^\circ & [| 1\bar{2}1] &\to&
48^\circ & [| \bar{2}2] &\to&
84^\circ & [| \bar{1}2\bar{1}] &\to& (12^\circ) \\
\hline
 3^\circ & [2| 2\bar{2}] &\to&
 6^\circ & [1| 2\bar{2}] &\to& (12^\circ) \\
\hline
21^\circ & [2| \bar{1}2\bar{1}] &\to&
42^\circ & [1| \bar{1}2\bar{1}] &\to&
(84^\circ) \\
\hline
33^\circ & [1\bar{1}| \bar{2}2] &\to&
66^\circ & [\bar{1}| \bar{2}2] &\to&
(48^\circ) \\
\hline
39^\circ & [1\bar{1}| 1\bar{2}1] &\to&
78^\circ & [\bar{1}| 1\bar{2}1] &\to&
(24^\circ) \\
\hline
51^\circ & [\bar{2}| 1\bar{2}1] &\to&
(78^\circ) \\
\hline
55^\circ & [\bar{2}| 1\bar{1}1] &\to&
(70^\circ) \\
\hline
57^\circ & [\bar{2}| \bar{2}2] &\to&
(66^\circ) \\
\hline
69^\circ & [\bar{1}1| \bar{1}2\bar{1}] &\to&
(42^\circ) \\
\hline
87^\circ & [\bar{1}1| 2\bar{2}] &\to&
(\phantom{1}6^\circ) \\
\hline
\end{array}$$

$$\begin{array}{rrcrrcrrcr}
\theta & \sigma(\theta) && \theta & \sigma(\theta) && \theta & \sigma(\theta) \\
\hline
20^\circ & [| 2\bar{1}]  &\to&
40^\circ & [| 1\bar{1}1] &\to&
80^\circ & [| \bar{1}2] &\to&
(20^\circ) \\
\hline
 5^\circ & [2| 2\bar{1}] &\to&
10^\circ & [1| 2\bar{1}] &\to& (20^\circ) \\
\hline
25^\circ & [1\bar{1}| \bar{1}2] &\to&
50^\circ & [\bar{1}| \bar{1}2] &\to&
(80^\circ) \\
\hline
35^\circ & [1\bar{1}| 1\bar{1}1] &\to&
70^\circ & [\bar{1}| 1\bar{1}1] &\to&
(40^\circ) \\
\hline
65^\circ & [\bar{2}| \bar{1}2] &\to&
(50^\circ) \\
\hline
85^\circ & [\bar{1}1| 2\bar{1}] &\to&
(10^\circ) \\
\hline
\end{array}$$

$$\begin{array}{rrcrrcr}
\theta & \sigma(\theta) && \theta & \sigma(\theta) \\
\hline
36^\circ & [| 1\bar{1}] &\to&
72^\circ & [| \bar{1}1] &\to&
(36^\circ) \\
\hline
 9^\circ & [2| 1\bar{1}] &\to&
18^\circ & [1| 1\bar{1}] &\to& (36^\circ) \\
\hline
27^\circ & [1\bar{1}| \bar{1}1] &\to&
54^\circ & [\bar{1}| \bar{1}1] &\to&
(72^\circ) \\
\hline
63^\circ & [\bar{2}| \bar{1}1] &\to&
(54^\circ) \\
\hline
81^\circ & [\bar{1}1| 1\bar{1}] &\to&
(18^\circ) \\
\hline
\end{array}$$

$$\begin{array}{rrcrrcr}
\theta & \sigma(\theta) && \theta & \sigma(\theta) \\
\hline
60^\circ & [| \bar{1}] &\to&
(60^\circ) \\
\hline
15^\circ & [2| \bar{1}] &\to&
30^\circ & [1| \bar{1}] &\to& (60^\circ) \\
\hline
75^\circ & [\bar{1}1| \bar{1}] &\to&
(30^\circ) \\
\hline
\end{array}$$

$$\begin{array}{rr}
\theta & \sigma(\theta) \\
\hline
45^\circ & [0|] \\
\hline
\end{array}$$


问:
Number theory proof mod p
For a given prime $p$ and integer $k$,  $k∈<1, p-1>$ $S(k)$ is defined as a sum of remainders of $k, k^2, k^3, ..., k^{p-1}$ when divided by $p$. Prove that $S(k)=\frac{1}{2}p(p-1)$ is true for an odd number of different k.

I managed to prove that $S(p-1)=\frac{1}{2}p(p-1)$ but can't see how to do it further.

答: If $S(n)=\tfrac12p(p-1)$ then also $S(n^{-1})=\tfrac12p(p-1)$, where $n^{-1}$ denotes the inverse of $n$ mod $p$. This shows that solutions come in pairs, except those for which $n=n^{-1}$. Of course $n=n^{-1}$ holds only for $n=\pm1$, and a quick check shows that $S(1)=p-1$ and $S(-1)=\tfrac12p(p-1)$.

Title: Number theory proof mod p


问:
Calculus fundamental theorem proof rolles i i
Hi I have recently learned rolles theorem and I don't quite understand why the number c that we choose has to be in the open interval (a,b) why can it not be on the closed interval.for example a constant function such as y=2 has a f'(c)=0 when c =a or b which is in the closed interval.the same thing confuses me for the mean value theorem.some help would be greatly appreciated</p>

答: Rolle's theorem states that if $f\colon[a,b]\longrightarrow\Bbb R$ is continuous, if $f(a)=f(b)$, and if $f$ is differentiable on $(a,b)$, then there is some $c\in(a,b)$ such that $f'(c)=0$. So, we are not even assuming that $f$ is differentiable at $a$ or at $b$. And if we were, then yes, $f'$ might have a zero at $a$ or at $b$ (or both). But asserting that it has a zero on $(a,b)$ is a stronger statement than claiming that it has a zero on $[a,b]$. So, why would we claim a weaker statement when we can easily prove a stronger one?


问:
reason behind finding particular solution for non-homogeneous differential equation
I was wondering about the reason behind having to find a particular solution for a non-homogeneous linear differential equation and having to combine it with the general solution to get the whole solution. Thanks in advance.</p>

答:
It's essentially a result from linear algebra, using the fact that the functions in question form a real vector space and the derivative is a linear operator. So I'm going to start with a little bit of linear algebra:

Let $V,W$ be vector spaces over $\mathbb R$ or $\mathbb C$ (or any field, really, but for differential equations these two are of interest), and let $L:V\longrightarrow W$ be a linear map. Also let $b\in W$. If $v\in V$ is a solution of the equation $L(x)=b$, then the solution set of the equation is exactly $v+\ker L:=\{v+x~|~x\in\ker L\}$.

Proof: we first show that every element of $v+\ker L$ is a solution, and then we show that every solution is in $v+\ker L$.
The first part is easy: all elements of that set are of the form $v+x$, where $x\in\ker L$, that is $L(x)=0$. So
$$L(v+x)=L(v)+L(x)=b+0=b,$$
so we do in fact have a solution.
The second part is slightly trickier: if $v'$ is a solution of the equation, then $L(v')=b$. Since $L(v)=b$ as well, this gives us $L(v')=L(v)$, and by linearity we get $L(v'-v)=0$, so $v'-v\in\ker L$. But then $v'=v+(v'-v)$, which is $v$ plus an element of the kernel, so $v'\in v+\ker L$.

Now let's look at linear differential equations. They are of the form
$$\frac{\mathrm d}{\mathrm dt}y-Ay=b,$$
where $A$ is a matrix and $b$ some function (the inhomogenous part). Now the derivative and the matrix are linear, so we could instead define $L:=\frac{\mathrm d}{\mathrm dt}-A$ and write this as
$$L(y)=b.$$
Since $L$ is linear, we know by the theory above that the solution set is some arbitrary solution plus the kernel of $L$. The arbitrary solution is the particular solution. The kernel is the set of all functions satisfying $L(y)=0$, but that's just the homogenous equation, so $\ker L$ is the set of all solutions of the homogenous equations. You could say that it is the general solution to the homogenous equation. So we take a particular solution and add the general solution to the homogenous problem.


问:
Vector projection on space
There's a question i just can not understand and i need help with her:

we have 4 points, $(0,y1),(1,y2),(3,y3),(4,y4))$

and $ y=x−3$ is the line of the 4 points(that we received by least squares).

i need to calculate the projection of $\vec y=(y_1,y_2,y_3,y_4)$ on space

$W=SP\left\{\left(0,1,3,4\right),\left(1,1,1,1\right)\right\}$

To calculate normally i need the y values,but I can not find a way to find them.

Thank you

答:
This exercise is about understanding the line of best least-square fit as a projection problem.

If you denote $\vec{x}=\begin{pmatrix}0 \\ 1 \\3 \\ 4\end{pmatrix}$, $\vec{1}=\begin{pmatrix}1 \\ 1 \\1 \\ 1\end{pmatrix}$   and
$\vec{y}=\begin{pmatrix}y_1 \\ y_2 \\y_3 \\ a_4\end{pmatrix}$, then the line $y=ax+b$ of best least-square fit means

$$||\vec y - (a\vec x + b\vec 1)||_2^2 \text{ is minimal}$$

So, $a\vec x + b\vec 1$ is the orthogonal projection of $\vec y$ onto the span of $\vec x$ and $ \vec 1$.

Since $a=1$ and $b=-3$ are given, you get

$$a\vec{x}+b\vec{1}=\begin{pmatrix}0 \\ 1 \\3 \\ 4\end{pmatrix}-3\cdot\begin{pmatrix}1 \\ 1 \\1 \\ 1\end{pmatrix} = \begin{pmatrix}-3 \\ -2 \\0 \\ 1\end{pmatrix} $$

Title: Vector projection on space


问:
nonsingularity of a diagonally dominant matrix
I have a matrix which is sort of diagonally dominant but not quite.

Each row is a probability vector (nonngative and sums up to one).

I have for any $i$ and $j\neq i$:

$a_{ii} > a_{ij}$

$a_{ii} > a_{ji}$

I wonder whether it must be nonsingular.

答:
When $A$ is $n\times n$ for some $n\ge4$, the answer is negative. Here is a counterexample for the case $n=4$:
$$
A=A_4:=\frac{1}{8}\pmatrix{
3&1&2&2\\
1&3&2&2\\
2&2&3&1\\
2&2&1&3}.
$$
This doubly stochastic matrix is singular because the sum of its first two columns is equal to the sum of the other two.

When $n>4$, consider
$$
A=(1-t)\pmatrix{A_4\\ &I_{n-4}}+\frac{t}{n}E
$$
where $E$ is the matrix of ones and $t>0$ is sufficiently small. The matrix $A$ is doubly stochastic because it is a convex combination of two doubly stochastic matrices $A_4\oplus I_{n-4}$ and $\frac1nE$. It is singular because the sum of its first two columns is equal to the sum of the third and the fourth ones. It is easy to check that $a_{ii}>\max(a_{ij},a_{ji})$ for every $i\ne j$.


问:
Solving a functional equation over $\mathbb{R}^+$
How do we solve the functional equation $f : \mathbb{R}^{+}\rightarrow \mathbb{R}^+$ such that $f(x)f(y)+f(xy) = xy$?

答:
Notice that for $y=1$, one gets:

$$f(x)f(1)-f(x) = \left(x+2\right)\left(\frac{1}{x}+1\right) \Rightarrow f(x) = \frac{1}{f(1)-1}\left[\left(x+2\right)\left(\frac{1}{x}+1\right)\right].$$

Now, observe that:

$$f(1) = \frac{1}{f(1)-1}\left[\left(1+2\right)\left(\frac{1}{1}+1\right)\right] = \frac{6}{f(1)-1}.$$

Then:

$$f(1)(f(1)-1) = 6 \Rightarrow f^2(1) - f(1)-6 = 0 \Rightarrow f(1)= 3 \vee f(1) = -2.$$

Since $f:\mathbb{R}^+ \to \mathbb{R}^+$, then $f(1) = 3.$

Hence:

$$f(x) = \frac{1}{2}\left(x+2\right)\left(\frac{1}{x}+1\right).$$


答:
For $x=y=1$ you get $f(1)^2-f(1)=6\iff f(1)=3$, since the other possible solution is negative (namely, $-2$).

Now for any $x>0$, taking $y=1$ one gets
$$f(x)f(1)-f(x)=(x+2)(1/x+1)\iff f(x)=\frac{(x+2)(1/x+1)}{f(1)-1}=\frac{(x+1)(x+2)}{2x}$$

Title: Solving a functional equation over $\mathbb{R}^+$

Body: How do we solve the functional equation $f : \mathbb{R}^{+}\rightarrow \mathbb{R}^+$ such that $f(x)f(y)+f(xy) = xy$?


问:
Volume of solid of revolution around a line
What is the volume of the solid that results when the region bounded by the graphs
of $y=\sqrt{x}$, y = 0, and x=4 is revolved about the line x = 4.

My solution is
$$ \pi \int_0^2 (4-y^2)^2 \, dy = \frac{256\pi}{15} $$

I don't know if this correct. How can I integrate this in terms of x? In terms of x, is the result the same with my integration above?

答:
In terms of $x$, you can write as below -

$V = \displaystyle 2\pi \int_0^4 (4-x)\sqrt x \, dx$

The horizontal area of the circumference of thickness $dx$ is $2 \pi (4-x) dx \,$ where $(4-x)$ is the radius of the circumference. $y = (\sqrt x - 0)$ is the height.


问:
Is Euclidean Geometry complete and unique
Please help me understand this concept of completeness of geometry and set me on the right path.

This is my context: From wikipedia, a formal system is complete if every tautology is also a theorem. For me this means that a system of axioms is complete if every statement that is true can be proved from the axioms (the question is tautologies in what kind of logic, first order, second order,...?). Anyway, it seems then that the concept of completeness depend on the system of axioms.

So, in the case of Euclidean Geometry, its completeness depends on their axioms (For example Euclid's Axioms, Hilbert Axioms, Tarki's Axioms,etc). Comparing for example the axioms of Hilbert and the axioms given by Tarski, I can see that they are essentially different in that Hilbert uses second order logic and Tarski's only first order logic. Also Tarski proved that his system of axioms is complete.

Now, the incompleteness theorem of Gödel stated that the system of axioms of arithmetic is incomplete. So, when we consider $R^2$ and $R^3$ as a representation of Euclidean Geometry we are talking about a system that is not complete because we are using the real numbers. So, are we speaking of different Euclidean Geometries? I'm totally confused.

答:
It sounds like you might get a lot out of Greenberg's short paper Old and New Results in the Foundations of Elementary Plane Euclidean and Non-Euclidean Geometries.

In it he discusses how the ordinary theory of plane geometry is incomplete (pg 202), but why Tarski's geometry is complete (section 3.3, page 215).


问:
identity tensor proof
Can someone explain how can I prove the identity tensor multiplicating by a determinant of a $n\times n$ matrix A is again the determinant using Einstein Notation?

$$\det(AI)=\det(A)$$

答: $AI = A$ by the definition of the identity $I$. If $\det$ is a function (which it is), then we have to have $\det (AI) = \det (A)$.

Title: identity tensor proof


问:
Probability DIE question on Bayes Theoram
Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1,2,3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1,2,3 or 4 with the die ?

My Doubt?

Let E1 be the event that the outcome on the die is 5 or 6 and E2 be the event that the outcome on the die is 1,2,3 or 4. A - exactly one head.

I am not able to understand why the probability of 4 / 6 in {1,2,3,4} not getting multiplied with the probability of exactly one head i.e 1/2 to get P (A/E2)?

答:
The probability you mentioned is

$$P[E_2 \cap A] = P[E_2] \cdot P[A|E_2] = \left(\frac{2}{3}\right) \left(\frac{1}{2}\right) = \frac{1}{3}$$

However, the question is asking for $P[E_2|A]$, which, by Bayes' Theorem, is

$$\begin{align}
P[E_2|A] &= \frac{P[A|E_2] \cdot P[E_2]}{P[A]} \\
\\
&= \frac{P[A|E_2] \cdot P[E_2]}{P[A|E_1] \cdot P[E_1] + P[A|E_2] \cdot P[E_2] } \\
\\
&= \frac{\left(\frac{2}{3}\right) \left(\frac{1}{2}\right)}{\left(\frac{1}{3}\right) \left(\frac{3}{8}\right) + \left(\frac{2}{3}\right) \left(\frac{1}{2}\right)} \\
\\
&= \frac{8}{11}
\end{align}$$


问:
How much additional area does a ridge and furrow system create?
Various parts of Europe used a Ridge and furrow system for their fields. One side effect is it provides additional land area.

Approximating the ridges and furrows using parts of a circle with radius, r, and, given a depth / height of h from the mean height, how much additional area does it create in comparison to flat land with surface area, A?

E.g. flat land, where r is infinite and h is zero, would give a surface area of A.

At the other extreme, land with half circles for the ridges and furrows, where r = h, would give a surface area of A x (circumference of semi circle / distance across semi circle) = A x (πr / 2r) = Aπ/2.

What is the general solution?

Body:
Series expansion of YNK's answer in Mathematica

$$\frac{0.285763}{\sqrt{k}}+1.1321 \sqrt{k}-0.0615981 k^{3/2}+0.140062 k^{5/2}+0.0426328 k^{7/2}+0.017562 k^{9/2}+0.00772557 k^{11/2}+0.00350133 k^{13/2}+0.00161512 k^{15/2}+O\left(k^{17/2}\right).$$


问:
$\int_0^x f(t)dt = [f(x)]^2$ and f is never 0
Problem says:

Suppose that continuous $f(x)$ is such that:

$$\int_0^x f(t) dt= [f(x)]^2$$

and $f(x)$ is never zero. Then find $f(x)$.

I do the question thinking it say "$f(x)$ is not the zero function"  I find $f(x) = \frac12 x$ as solution.

But, I see from again reading again $f(x)$ can never be zero. Now my solution is not correct. $f(0) = 0$ for my solution.

From how I have came to solution I wonder:

Is this possible even to find?

In the way I have solved it seemed this was the only way possible solution.

答:
If you put $x=0$ then you immediately get $f(x) = 0$. I guess what they meant was "$f$ is not identically zero", since $f \equiv 0$ is another solution.

Also, if you follow J.M.'s advice then it's quite helpful to know that $f(x) \neq 0$, at least for $x \neq 0$. You have to assume that $f$ is differentiable, though.


问:
IMO 2010 N2: Find all pairs $(m,n)$ of nonnegative integers for which $m^2 + 2 \cdot 3^n = m\left(2^{n+1} - 1\right).$
Find all pairs $(m,n)$ of nonnegative integers for which $m^2 + 2 \cdot 3^n = m\left(2^{n+1} - 1\right).$

I am not able to start , however I got solutions $(m,n)$  as $(6,3), (9,3), (9,5),$ or $(54,5)$... Can one help me? Thanks!

答:
We note that $m^2-m(2^{n+1}-1)+2\cdot3^n = 0$,  now this equation is supposed to have integer solutions, so it's discriminant $d$ must be a perfect square. $$\implies d^2 = (2^{n+1}-1)^2-4\cdot2\cdot3^n \iff (2^{n+1}-1-d)(2^{n+1}-1+d)=2^3\cdot3^n.$$
Now, both the factors sum to $2(2^{n+1}-1)$ and both of them must be of the form $2\cdot3^a$ and $4\cdot3^b$, where $a,b$ are non-negative integers with $a+b=n.$
\begin{align*}
\implies2\cdot3^a+4\cdot3^b=2(2^{n+1}-1)\iff\boxed{3^a+2\cdot3^b=2^{a+b+1}-1}
\end{align*}

Motivation (Classic; for below). The only positive integers that satisfy $|3^x-2^y|=1$ are $(x,y)=\{(1,1);(2,3);(1,2)\}.$

Case 1: $\boxed{\min(a,b)\leqslant2}.$ For $a=0: 3^b=2^b-1$; there is not solution from above. For $a=1: 3^b=2^{b+1}-2$, since $3^b\equiv 1,3 \pmod 4$, and as $b\geqslant 1 \implies 2^{b+1}-2 \equiv 2 \pmod4.$ Hence, no solutions in this case as well. Now, for $a=2$ we have:
$$3^2+2\cdot3^b=2^{b+3}-1\iff3^b=2^{b+2}-5.$$
$3^2\equiv 1 \pmod 9 \implies 3^{2x}\equiv 1 \pmod 8$ and $3^{2x+1}\equiv 3 \pmod 8.$ Here, since $b=0$ doesn't work, for $b\geqslant1$, RHS $\equiv -5 \equiv 3 \pmod 8$, so $b$ has to be odd, and we can easily observe that only $b=1,3$ can satisfy, (also from here we get that $a$ has to be even), hence we're through with this case.

Case 2: $\boxed{\min(a,b)\geqslant3}.$ $a$ and $b$ are unequal from above
\begin{align*}
\implies\min(a,b)\leqslant v_3\left(3^a+2\cdot3^b\right)
&=v_3\left(2^{a+b+1}-1\right)
\\&=v_3\left(4^{\tfrac{a+b+1}{2}}-1\right)
\\&=1+v_3\left(\frac{a+b+1}{2}\right)-(*)
\\&\leqslant1+v_3(\max(a,b)).
\end{align*}
Since, $a,b$ are different, so $|a-b|\geqslant1$ and as $\max(a,b) = \frac{a+b+|a-b|}{2}$, so the desired in the last step above; and $(*)$ follows from LTE, as $3|4-1$, $3\nmid 4,1$ and $3\ne 2$.

This is another motivation for turning the problem towards $\frac{a+b+1}{2}$.

Now,
$$\implies \max(a,b) \geqslant 3^{\min(a,b)-1}.$$
Next, as $\max(a,b)+\min(a,b)=a+b$, so
$$2^{\max(a,b)+\min(a,b)+1}=2^{a+b+1}=3^a+2\cdot3^b+1>3^{\max(a,b)}$$
$$\implies2^{\min(a,b)+1} > \left(\frac 32\right)^{\max(a,b)} \geqslant \left(\frac 32\right)^{3^{\min(a,b)-1}}.$$
which isn't true for $\min(a,b) \geqslant 3.$ Thus, no solution in this case.

So now, the only possible solutions here are: $\boxed{(a,b)=(2,1);(2,3)}$ and $\boxed{n=3,5}.$
$$
\begin{cases}
n=3:m^2+2\cdot3^2=m(2^4-1)\implies m^2-15m+54=0\implies m=6\text{ or }9.\\
n=5:m^2+2\cdot3^5=m(2^6-1)\implies m^2-63m+162=0\implies m=9\text{ or }54.
\end{cases}
$$
So, the solutions are: $$\boxed{(m,n)=(6,3);(9,3);(9,5);(54,5).}$$

I did this problem with a nudge from v_Enhance, as I was stuck at the later part of the problem, where some relation between $\min(a,b)$ and $\max(a,b)$ helps to get out, through some contradiction etc; and rest of all the steps are pretty usual, just that formalizing in the form of cases $\le 2$, $\ge 3$, instead of randomly putting inputs and taking care of obtained values all the time.


问:
subdifferential of ReLU function composition with affine function
Let $f:\mathbb{R}^n\to\mathbb{R}^n$ be ReLU function, i.e., $f(x)=[\max(0,x_i)]_{i=1}^n$. Let also $g:\mathbb{R}^m \to \mathbb{R}^n$ be an affine function $g(x)=Ax+b$. What is the subdifferential of $f(g(x))$?

答:
The subdifferential of the composed function $h(x) := f(Ax+b)$ is simply $\partial h(x) = A^\top \partial f(Ax+b)$. Thus if suffices to find $\partial f(x)$.

For one-dimensional ReLU, $r(z) = \max(0,z)$ with $z\in\mathbb{R}$, it is straightforward to find its subdifferential (with a slight abuse of notation: a single element represents a singleton set containing that element):
$$\partial r(z) =
\begin{cases}
1 &\mbox{if } z > 0,\\
[0,1] &\mbox{if } z=0,\\
0 &\mbox{if } z < 0.
\end{cases}$$
Sketch of proof: $r(z)$ is differentiable every except at $z=0$, thus $\partial r (z) = r'(z) \ \forall z \neq 0$. For $z=0$ it is easy to show that $\partial r (z) = [0,1]$ based on the definition of subgradient.

For general $n$-dimensional ReLU ($\mathbb{R}^n\to \mathbb{R}^n$), $f(x) = \max(0,x)$ with $x\in \mathbb{R}^n$, and for vector-valued functions in general, there are different ways of generalizing the notion of subgradient. Once can use, for example, generalized inequalities based on convex cones (see this paper for example). Another solution is to use Clarke Jacobian (which is the Clarke subdifferential for vector-valued function). For the ReLU function, it can be shown that these two kinds of "subgradients" coincide (if one chooses the nonnegative orthant as the underlying cone in the first kind). I'll give a detailed derivation for the Clarke subdifferential because it's more common.

The Clarke subdifferential is the set of Clarke Jacobian matrices, defined as follows for a locally Lipschitz function $F:\mathbb{R}^n\to \mathbb{R}^m$:
\begin{equation}
D_CF(x) = \mathrm{co}\left\{\lim_{i\to\infty}DF(x_i): x_i\to x, x_i\not\in E\cup E_f \right\},\tag{*}
\end{equation}
where $E\in \mathbb{R}^n$ is any set of measure zero, $E_f$ is the set of points at which $F$ is not differentiable, $DF$ is the Jacobian matrix, and $\mathrm{co}$ denotes the convex hull. In other words: Take any sequence $x_i$ converging to $x$ while avoiding both $E$ and $E_f$, such that $DF(x_i)$ converges to a limit, then the convex hull of all such limits is $D_CF(x)$. Clearly, $D_CF(x)$ is a set of $m\times n$ matrices, and it is well-known that if $F$ is differentiable at $x$, then $D_CF(x) = \{DF(x)\}$.

Now let's compute $D_Cf$ for our ReLU function $f$. It is easy to see that $f$ is differentiable at any $x = (x_1,\dots,x_n)$ such that $x_i\neq 0 \ \forall i$, and the Jacobian at such point can be computed as
\begin{equation}
\partial f_i(x_j) =
\begin{cases}
0 &\mbox{if } i\neq j \mbox{ or } x_j < 0,\\
1 &\mbox{otherwise}.
\end{cases}
\end{equation}
(The Jacobian is thus a diagonal matrix where the diagonal elements are either $0$ or $1$). Clearly, we can partition $\mathbb{R}^n$ into $2^n$ subsets such that $f$ is differentiable on their interiors, and that $Df$ is constant over each subset. These $2^n$ subsets correspond to precisely the $2^n$ orthants of $\mathbb{R}^n$, and the $2^n$ Jacobians correspond to the $2^n$ diagonal matrices whose diagonals are from $\{0,1\}^n$. Now, for a point $x$ on the boundaries of the orthants (at which $f$ is clearly not differentiable), we can compute $D_Cf(x)$ according to $(*)$ as follows.

Suppose that $x$ lies in the intersection of $k$ orthants $S_1,\dots,S_k$, and that these orthants correspond to the (constant) Jacobians $G_1,\dots,G_k$. For each sequence $x_i$ that converges to $x$ (such that $x_i$ does not meet the boundaries), if $i$ is large enough then obviously $DF(x_i)$ can only take one of the values $G_1,\dots,G_k$, i.e., the sequence $DF(x_i)$ admits cluster points in the set $\{ G_1,\dots,G_k \}$. On the other hand, if we choose the sequence $x_i$ such that all of its elements belong to $S_j$ ($1\le j \le k$), then automatically $DF(x_i)$ converges to $G_j$ because $DF(x_i) = G_j \ \forall i$ (i.e., a constant sequence). We conclude that all the possible limits of in $(*)$ are $\{ G_1,\dots,G_k \}$ and therefore $D_Cf(x) = \mathrm{co} \{ G_1,\dots,G_k \}$.

It remains to determine $\{ G_1,\dots,G_k \}$. Because $x$ is on the boundaries, $x$ must contain at least one zero element. Suppose that $x$ contains exactly $p$ zero elements, then $x$ lies in the intersection of the $k = 2^{p}$ orthants corresponding to the $2^{p}$ Jacobians of the form $G=\mathrm{diag}(g_1,\dots,g_n)$, where
\begin{equation}
g_i =
\begin{cases}
0 &\mbox{if }  x_i < 0,\\
1 &\mbox{if }  x_i > 0,\\
\mbox{either } 0 \mbox{ or } 1 &\mbox{if }  x_i = 0.
\end{cases}
\end{equation}
The convex hull of these $2^{p}$ Jacobians is clearly the set of matrices $\mathrm{diag}(c_1,\dots,c_n)$ such that
\begin{equation}
c_i =
\begin{cases}
0 &\mbox{if }  x_i < 0,\\
1 &\mbox{if }  x_i > 0,\\
\mbox{any value in } [0,1] &\mbox{if }  x_i = 0.
\end{cases}
\end{equation}
For example, in $\mathbb{R}^3$, the point $x = (0, \color{blue}{2}, \color{red}{-7}, 0)$ lies in the intersection of the $4$ orthants whose Jacobians are $\mathrm{diag}(0, \color{blue}{1}, \color{red}{0}, 0)$, $\mathrm{diag}(0, \color{blue}{1}, \color{red}{0}, 1)$, $\mathrm{diag}(1, \color{blue}{1}, \color{red}{0}, 0)$, and $\mathrm{diag}(1, \color{blue}{1}, \color{red}{0}, 1)$, and thus
\begin{align}
D_Cf(x) &= \mathrm{co} \left\{\mathrm{diag}(0, \color{blue}{1}, \color{red}{0}, 0), \mathrm{diag}(0, \color{blue}{1}, \color{red}{0}, 1), \mathrm{diag}(1, \color{blue}{1}, \color{red}{0}, 0),  \mathrm{diag}(1, \color{blue}{1}, \color{red}{0}, 1)\right\} \\
&= \left\{\mathrm{diag}(c_1, \color{blue}{1}, \color{red}{0}, c_4): c_1\in [0,1], c_4\in [0,1]\right\}.
\end{align}

All the above just to arrive at the following simple rule for computing the subdifferential of the $n$-dimensional ReLU function $f$ at a point $x=(x_1,\dots,x_n)$:

Compute $g_i \in \partial \mathrm{ReLU}(x_i)$ (i.e., element-wise subgradient).

Then $\mathrm{diag}(g_1,\dots,g_n)$ is a Clarke Jacobian of $f$ at $x$.

If the reader find the above derivation hard to understand, I would recommend to apply it to $\mathbb{R}^2$, the idea will become clearer quickly.


问:
Is there a way to define a function such that the sets in the domain are the same size?
Let $A \subseteq \mathbb{R}^a$, $B \subseteq \mathbb{R}^b$ and $C \subseteq \mathbb{R}^c$.

I would like to define a function $f$ that can take any subset of $A$, $B$ and $C$ and return a real value $x$, but the function is only defined when these subsets are the same size. What is the correct notation for doing this? My, slightly hacky, attempt is:
$$f:\mathscr{P}(\mathbb{R}^a) \times \mathscr{P}(\mathbb{R}^b) \times \mathscr{P}(\mathbb{R}^c) \to \mathbb{R},(A,B,C) \mapsto x,\ \text{where $\mathscr{P}(X)$ is the power set of $X$, and |A|=|B|=|C|}.$$
But I don't think this really addresses the fact that the sets in the domain need to be the same size. Is there any way to capture this in the notation?

答:
What you did was almost the solution, you just have to translate mathematically what you wrote in words:
You introduce the set $$S = \{(A, B, C)\in \mathcal{P}(\mathbb{R}^a)\times \mathcal{P}(\mathbb{R}^b)\times \mathcal{P}(\mathbb{R}^c) : |A| = |B| = |C|\}.$$
This is a well-defined set, and then your function would be a function $f\colon S\to \mathbb{R}$.


问:
Traversing a maze from one cell to another
I have maze (10*10) as shown below:

Each cell has unique value ranging from [0,15] which actually defines its walls (see the image below). For example, value of first cell (r=0, c=0) is 7 and so on.

Now I need to travel from entry point to exit point. Entry point is always at the left side of maze which is created by removing any wall. Same is done on the right side of maze where the exit is. These points will be provided by indicating with an arrow in the image.

I know how we actually traverse from one point to another inside rectangular grid. It contains values in binary form where $1$ signify block to that cell and $0$ signify that path is possible from that cell.

PS: You will be only provided a 10*10 matrix with values filled accordingly and you will need to get row and column number as a path from entry point to exit point.

答:
One way to solve a general maze is to view it as a tree with two special vertices, a start and an end, and we need to find the unique simple path that connects them.

Define the start and end vertices as at least degree 2 by adding an edge to the 'outside'.

Any terminal vertices (degree 1) can all be removed, as it is impossible for them to be in the path.

This leaves a new tree, and again we can remove all of the terminal vertices.

This process leaves only the solution path between Start and End.

In your case, repeatedly replace all cells that are $7, 11, 13, 14$ with $15$ and recalculate the cell values.


问:
How many positive integers $\le 462$ are relatively prime to $462$?
How many positive integers less than or equal to $462$ are relatively prime to $462$?

My Approach
Well my approach is basically that we find $|A \cup B|$ where:

$A = \{\text{all prime numbers} \le 462\}$

$B = \{\text{every number which isn't a factor of } 462\}$

I just don't know how to calculate this so it is not an efficient method so please answer with a more efficient method. Thank you!

答: HINT: Euler's Totient function $\phi$ is multiplicative. In other words, if we have two positive integers $a$ and $b$ with $\gcd(a,b) = 1$, then $\phi(ab) = \phi(a)\phi(b)$. So, you can use prime factorization of $462$ to divide and conquer the problem.


问:
How many ways can the letters ABCDEFG be arranged such that A and B are never next to each other?
The problem (not homework) is: How many ways can the letters ABCDEFG be arranged such that A and B are never next to each other?

I couldn't find a quick formula for this, so I ended up doing it by intuition (I don't know if my answer is correct, hoping for some confirmation).

If the letter "A" is in the first spot, there are $5$ spots where "B" can go such that it's not next to "A". If "A" is in the second to sixth spots, there are $4$ possible spots each for "B" to be placed. $(5 \cdot 4) = 20$ total spots. If "A" is in the seventh spot, there are again $5$ possible spots for "B". So there are a total of $30$ rearrangements of "A" and "B" in $7$ spots where they aren't next to each other.

For every placement of "A" and "B", the $5$ other letters can be rearranged $120$ times ($5!$) such that they form different 'words' without moving the "A" and "B", so I multiplied the $30$ rearrangements of "A" and "B" by $120$ and got $3600$. Can anyone confirm or give a formula for this?

答: You can arrange CDEFG in $5!=120$ ways, with $6$ spaces in between them and at the ends. Choose one of these spaces for A and another for B; makes $120\cdot6\cdot5=3600$ in total.


问:
Why the zero vector always orthogonal to any other vector.
Im currently looking at inner products and was wondering why the inner product of any vector with the zero vector is equal to 0. I have researched on this and only found the information that the zero vector is orthogonal to all vectors but no proof alongside.

And hence I was wondering if anyone had any proof as to why this happens.

Thanks in advance.

Edit:

Thanks every for their comment, I was now wondering if anyone could help explain to me why the inner product of $\left \langle M \vec u,\vec v \right \rangle$ = $\left \langle \vec u,M^{T} \vec v \right \rangle$

where M is and $n*n$ square matrix.

答:
Let us denote the inner product by $( \cdot|\cdot).$

Then

$$(0|x)=(0x|x)=0(x|x)=0$$

for all $x$.


答:
For any fragment of euclidean geometry seems to be topologically and notionally equal.  R2 and R3 are different only in dimension.

One should be careful about what is being implied by Euclid's geometry.

Just as the curvature of space (the sum of angles of a triangle for example), leads to euclidean, hyperbolic and spherical geometries, there is a second statement for which any two, but not all three parts, can be true.  These are set in italics.

In a complete  and orientable space, lines cross once.

Euclid's geometry deals with a fragment of the plane, and never considers completeness.

The spheric geometries are necessarily complete, so one either rejects orientable to give elliptic, or once, giving spherical geometry.

So when one starts talking of the nature of the euclidean geometry and the horizon at infinity, one gets a presentation as a line (projective) or point (inversion).  Some people consider what happens on the other side of infinity, that is, they try to complet the euclidean plane.

A complete euclidean plane is perfectly modelled by the sphere at infinity.  This is the centre of a planes through a point, and thus must be a the centre of a plane centred on a real point.

The actual geometry is called "Möbius" geometry, but corresponds to the implemetation of 0/0 as undefined, that any circle is a straight line, and crosses other circles twice.  If one supposes that 'any point on the surface' is 'a point at infinity', and that straight lines contains the point at infinity, this folds down to the inversion model of the euclidean plane.

In short, Möbius geometry is oE, or orientable-complete-euclidean, whereas Euclid studied the geometry of orientable-fragment-euclidean.

Hyperbolic geometry is necessarily a fragment of space, even at 'infinity'.  This is because there is no meaning for a plane in 3D centred on a real point, unless you get Meta-infinities come to hand.

Title: Is Euclidean Geometry complete and unique

Body:
Please help me understand this concept of completeness of geometry and set me on the right path.

This is my context: From wikipedia, a formal system is complete if every tautology is also a theorem. For me this means that a system of axioms is complete if every statement that is true can be proved from the axioms (the question is tautologies in what kind of logic, first order, second order,...?). Anyway, it seems then that the concept of completeness depend on the system of axioms.

So, in the case of Euclidean Geometry, its completeness depends on their axioms (For example Euclid's Axioms, Hilbert Axioms, Tarki's Axioms,etc). Comparing for example the axioms of Hilbert and the axioms given by Tarski, I can see that they are essentially different in that Hilbert uses second order logic and Tarski's only first order logic. Also Tarski proved that his system of axioms is complete.

Now, the incompleteness theorem of Gödel stated that the system of axioms of arithmetic is incomplete. So, when we consider $R^2$ and $R^3$ as a representation of Euclidean Geometry we are talking about a system that is not complete because we are using the real numbers. So, are we speaking of different Euclidean Geometries? I'm totally confused.


问:
What does Fatou's lemma say about these $f_j$?
I know that Fatou's lemma states that if $\{f_j\}_{j\geq 1}$ is a sequence of nonnegative measurable functions and $f_j(x)\longrightarrow f(x)$ almost everywhere on a set E, then
\begin{eqnarray}
\int\limits_E f\leq \liminf\int\limits_E f_j\notag
\end{eqnarray}

Now I have this question: Consider $\mathbb{R}$ equipped with Lebesgue measure. For $j=1,2,\cdots,$ let
\begin{eqnarray}
f_j(x)=\begin{cases}1+\sin jx & \mathrm{if}~~ -\pi \leq x\leq \pi \\0 & \mathrm{if}~ ~ x<- \pi \\ 0 & \mathrm{if}~~  x> \pi. \end{cases}\notag
\end{eqnarray}
What does Fatou's lemma say about these $f_j$?

Can somebody assist me here please. I don't know how to begin.

答:
(This is essentially the same as the first comment on the question.)

Let $f = \liminf_{j} f_{j}$. Then by Fatou's lemma
$$
\int f \leq \lim_{j} \int f_{j}.
$$
You can easily calculate $\int f_{j}$ as well as the right-hand side above. For the left-hand side, you need to figure out what $f$ is. But without doing that, you know an upper bound on $\int f$ from the inequality above, and you can show certain properties of $f$. For example, $f$ cannot be equal to $2$ a.e. on $(-\pi,\pi)$.


问:
A property for subsets of real numbers with the condition $\mathbb{Q}^++A\subseteq A$
Consider the additive group $(\mathbb{R},+)$ and let $A$ be a subset
such that $\mathbb{Z}^++A\subseteq A$ (equivalently  $1+A\subseteq A$,
and if and only if
$$
\cdots \subseteq 3+A  \subseteq 2+A \subseteq 1+A \subseteq A
\subseteq -1+A \subseteq -2+A \subseteq -3+A \subseteq \cdots .
)$$
Then we have

$$
(*)\;\;\; (\mathbb{Z}^++A)\setminus (\bigcap_{i\in \mathbb{Z}} A+i)\subseteq
\mathbb{Z}^++(A\setminus (\mathbb{Z}^++A))
$$
Because, $1+A\subseteq A$ implies that $\mathbb{Z}^++A= 1+A$ and so
$$
(*)\Leftrightarrow$$ $$ (1+A)\setminus [(2+A)\cap(3+A)\cap(4+A)\cap \cdots]\subseteq [(1+A)\setminus
(2+A)]\cup [(2+A)\setminus (3+A)] \cup [(3+A)\setminus
(4+A)]\cup \cdots ,
$$
and the last one holds, clearly (even it is an equality).

Now, replacing $\mathbb{Z}^+$ by $\mathbb{Q}^+$ and $\mathbb{Z}$ by $\mathbb{Q}$, we would like to know that whether a similar relation to $(*)$ holds, that is: if $\mathbb{Q}^++A\subseteq A$, then is it true that
$$
(*')\;\;\; (\mathbb{Q}^++A)\setminus (\bigcap_{i\in \mathbb{Q}} A+i)\subseteq
\mathbb{Q}^++(A\setminus (\mathbb{Q}^++A))\;?
$$

if no, can one find infinitely many such non-subsemigroup-subsets $A$ satisfying $(*')$
(and of course $\mathbb{Q}^++A\subseteq A$)?, and some examples for
$$
(*'')\;\;\; \mathbb{Q}^++(A\setminus (\mathbb{Q}^++A)) \subsetneqq(\mathbb{Q}^++A)\setminus (\bigcap_{i\in \mathbb{Q}} A+i)
\;?
$$

Thanks in advance

答:
Define $A=\mathbb Q^++\alpha\mathbb Z$, where $\alpha$ is any irrational. Since $\mathbb Q^++\mathbb Q^+=\mathbb Q^+$, this set satisfies $\mathbb Q^++A=A$. In particular, the right side of $(*')$ is just $\mathbb Q^+$. However, since every element of $A$ can be uniquely represented as $x+\alpha y$ where $x\in \mathbb Q^+$ and $y\in \mathbb Z$, $A+i$ consists only of those values for which $x>i$, and so
$$\bigcap_{i\in \mathbb Q}A+i=\emptyset.$$
So, the left side of $(*')$ is
$$(\mathbb Q^++A)\setminus \emptyset = \mathbb Q^++A=\mathbb Q^++\alpha\mathbb Z,$$
which is not a subset of $\mathbb Q^+$.

Remark. This sort of proof does not work for $\mathbb Z^+$; while the intersection over $i\in\mathbb Z$ is still empty, $A\setminus(\mathbb Z^++A)$ is rather large, which is enough to encompass the factor of the first set.


问:
Is it possible to combine an arithmetic an geometric sequence, but first adding and then multiplying?
picture of solution for when you multiply first and add after

link of original discussion of the picture: Combined geometric and arithmetic series partial sum

In this picture, you can see a way of solving combined arithmetic and geometric sequence, however for this equation to work you need to multiply first and then add after. I want to be able to do it the other way around.

I know hoe to this somewhat by doing the following:

=((+)⋅+)⋅

You can just keep adding: {  ⋅+)  } to the equation to find the next term but this is still somewhat tedious, I want to find a way in which you can just find the nth term by plugging in its positional value into the equation. Meaning if I want to find the 5th term n=5 and so on.

 - the nth term, for example, the 5th term of the sequence

 - The original value, or the first value in the series

 - the amount added every time

 - the amount multiplied / common ration I think

If anybody knows how to do this that would be amazing Thank you!

答:
$$\begin{align}
a_n &= \bigg( \Bigl( \bigl( (a_1 + d) \cdot q + d \bigr) \cdot q + d \Bigr) \cdot q + d \bigg) \cdot q + \cdots \\
\\
&= \Bigl( \bigl( (a_1q + dq + d) \cdot q + d \bigr) \cdot q + d \Bigr) \cdot q + \cdots \\
\\
&= \bigl( (a_1q^2 + dq^2 + dq + d) \cdot q + d \bigr) \cdot q + \cdots \\
\\
&= (a_1q^3 + dq^3 + dq^2 + dq + d) \cdot q + \cdots \\
\\
& \;\; \vdots \\
\\
&= a_1q^{n-1} + dq^{n-1} + dq^{n-2} + \cdots + dq \\
\\
&= a_1q^{n-1} + dq \; (q^{n-2} + q^{n-3} + \cdots + 1) \\
\\
&= \boxed{ a_1q^{n-1} + dq \left( \frac{q^{n-1}-1}{q-1} \right) }
\end{align}$$


问:
Calculation of the expectation through conditioning
I'm stuck on the following problem:

Suppose we want to compute the expectation of X (number of failures until first success) ~ Geom(p).
We can introduce an event A – first trial was success.

Then, according to my textbook: $EX$ = $E(X|A)P(A) + E(X|A^c)P(A^c) = 0p + (1+EX)q$.

However, In my understanding $E(X|A) = \frac{E[XI_A]}{P(A)}$. So, $EX$ = $\frac{E[XI_A]}{P(A)}P(A) +\frac{E[XI_{A^c}]}{P(A^c)}P(A^c) = 0p + (EX+1)$.
 I don't get where this $q$ multiplier appeared from.
Could you please explain me that?

答:
Let's describe what each of the terms in the law of total expectation represents.

$\operatorname{E}[X \mid A]$ is the expected number of failures observed, given that the first trial is successful.  If the first trial is successful, that means no failures were observed, hence $X = 0$ is the only outcome and the expectation is $0$.  When you write $\operatorname{E}[X \mathbb 1_A]$, you are constructing another random variable $$X \mathbb 1_A = \begin{cases}X, & A \text{ occurs} \\ 0, & A \text{ does not occur}.\end{cases}$$  But as I have already pointed out, if $A$ occurs, that means the very first trial was successful, hence $X = 0$, so we have $X \mathbb 1_A = 0$ in both cases.

$\Pr[A]$ is simply the probability that the first trial is successful.  Since trials are independent and identically distributed Bernoulli random variables with probability of success $p$, we simply have $\Pr[A] = p$.

$\operatorname{E}[X \mid A^c]$ is the expected number of failures observed, given the first trial was a failure.  If the first trial fails, then we have observed $1$ failure, and all subsequent trials, being independent of what happened before, behave in the same way as if we had not performed the first trial.  That is to say, the expected number of additional failures after the first failure is also $\operatorname{E}[X]$.  Therefore, $\operatorname{E}[X \mid A^c] = 1 + \operatorname{E}[X]$.

$\Pr[A^c]$ is the probability that the first trial is a failure.  This is $q = 1 - p$.

Your use of indicator functions unnecessarily complicates the solution and suggests an over-reliance on formulas and definitions.  The expression $$X \mathbb 1_{A^c} = \begin{cases} X, & A^c \text{ occurs} \\ 0, & A^c \text{ does not occur} \end{cases}$$ does not provide any insight into how to evaluate the expectation.  Your claim that $\operatorname{E}[X \mathbb 1_{A^c}] = 1 + \operatorname{E}[X]$ is incorrect, because according to the above, we must have $\operatorname{E}[X \mathbb 1_{A^c}] \le \operatorname{E}[X]$.  This is because $X$ is a random variable with nonnegative support, so $0 \le X$ for all $X$, and $X \mathbb 1_{A^c} \le X$.


问:
How to graph an absolute value equation?
How would you graph:

$|x+y|=1$ ?
I  can do the normal $y=|x+1|$ and all that. But how would you do a question with two of these unknowns in the absolute value?

Any help would be greatly appreciated, thanks.

答:
It doesn't seem as though $|x+y|=1$ is a function, since for $x=2$, the values $y=-1$ and $y=-3$ both satisfy the equation.

The way to graph this would be to graph two individual equations: $x+y=1$ and $x+y=-1$ (these two equations result when we lift the absolute value). So we graph separately:
$$y=-x+1$$$$y=-x-1$$


问:
Can I move a constant out of a continuous function? When?
Can I move a constant out of a continuous function? When?

If I have $f(\frac{1}{k}g(x))$

Then is this (ever) same as

If I have $\frac{1}{k}f(g(x))$?

答:
Try with $g(x) = x$ and $f(x) = x^2$.

The property of $f$ that decides whether you can do that or not, is homogeneity. For example, linear functions, such as $f(x) = 21 x$ are homogeneous  (of degree one), but $f(x) = x^2$ is not.


答:
No. For example take $f(x)=x^2$ and $g(x)=x.$

Then

$$f(\frac{1}{k}g(x))=\frac{1}{k^2}x,$$

but

$$\frac{1}{k}f(g(x))=\frac{1}{k}x.$$

Body:
Can I move a constant out of a continuous function? When?

If I have $f(\frac{1}{k}g(x))$

Then is this (ever) same as

If I have $\frac{1}{k}f(g(x))$?


问:
$\overline{\mathrm{span} \{e_{2n} \ \mid \ n \}} \cap \overline{\mathrm{span} \{ e_{2n} + \frac{1}{n+1}e_{2n+1} \} } = 0$, but their sum is dense

Regarding this question:Show that the intersection of $\overline{\text{span }\{e_{2n}\}}$ and $\overline{\text{span }\{e_{2n}+\frac{1}{n+1}e_{2n+1}\}}$ is zero..

The above has an accepted answer, but there is no proof of the result from the title. I am looking for a proof for that result and also that
$$\overline{\overline{\mathrm{span} \{e_{2n} \ \mid \ n \}} + \overline{\mathrm{span} \{ e_{2n} + \frac{1}{n+1}e_{2n+1} \} }} = H. $$
I have not been able to do it myself.

For the first part, I have started by picking $x \in \overline{\mathrm{span} \{e_{2n} \}} \cap \overline{\mathrm{span} \{e_{2n} + \frac{1}{n+1} e_{2n+1} \}} .$

Then, we know that $$x = \sum_{n=1}^\infty \langle x, e_n \rangle e_n. $$
Moreover, there is a sequence $(x_k)_{k}$ with $x_k \in \mathrm{span} \{e_{2n} \}$ for every $k$ and such that $x_k \to x$.
Using the same reasoning, there is a sequence $(y_k)_k$ with $y_k \in \mathrm{span} \{e_{2n} + \frac{1}{n+1} e_{2n+1} \}$ and such that $y_k \to x$. However, I do not know how to proceed from here.

I do not know how to start the second part either.

答:
Regarding the density of the sum:

The sum contains the ONB vectors $e_k,k\in\mathbb N$, thus it also contains their finite linear combinations. These altogether constitute a dense subspace of $H$.

Regarding that the subspaces intersect trivially:

This is the subject of an older question which got
a fresh answer there.


答:
$462 = 2\times3\times7\times11$

let A be the set of integers from 1 to 462 divisible by 2

let B be the set of integers from 1 to 462 divisible by 3

let C be the set of integers from 1 to 462 divisible by 7

let D be the set of integers from 1 to 462 divisible by 11

$A\cup B\cup C\cup D$ is the set of all integers from 1 to 462 that are NOT relatively prime to 462.

So your answer will be

$462 - n(A\cup B\cup C\cup D)$

So you have to get $n(A\cup B\cup C\cup D)$. (that's just the number of elements in the union)

The pattern to get the number of elements in a union is simple. Start with $n(A)+n(B)+n(C)+n(D)$ then subtract the number of elements in the intersection of each pair... then add the number of elements in the intersection of each triple... then subtract the number of elements in the intersection of each quadruple. Described better here:

https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle

$n(A) = 462/2 = 231$... calculate all the others the same way.

Can you take it from here?

Totient function is an easier method, but I don't know if you're allowed to use that.

Title: How many positive integers $\le 462$ are relatively prime to $462$?

Body:
How many positive integers less than or equal to $462$ are relatively prime to $462$?

My Approach
Well my approach is basically that we find $|A \cup B|$ where:

$A = \{\text{all prime numbers} \le 462\}$

$B = \{\text{every number which isn't a factor of } 462\}$

I just don't know how to calculate this so it is not an efficient method so please answer with a more efficient method. Thank you!


问:
Paths in $\mathbb{R}$ are homotopic.
Let $f$ and $g$ be paths in $\mathbb{R}$. Show that $f$ is homotopic to $g$.

My attempt: To show that $f(x)$ and $g(x)$ are homotopic we must produce a continuous map $F:[0,1]\times [0,1]\rightarrow \mathbb{R}$ such that $F(x,0)=f(x)$ and $F(x,1)=g(x)$. Because $f$ and $g$ are paths, they are continuous so we can take $F$ to be $F(x,t)=(1-t)f(x)+tg(x)$. Is this a correct justification?

答:
This community wiki solution is intended to clear the question from the unanswered queue.

Yes, your approach works. You can easily verify that the same construction works for any two maps $f, g: X \to C$ where $X$ is an arbitrary topological space and $C$ is a convex subset of an arbitrary topological vector space $E$.

Title: Paths in $\mathbb{R}$ are homotopic.


问:
Integration of radial functions?
Let $f(|x|)$ be a integrable radial function in $\mathbb{R}^n$ ($|\cdot|$ denotes the euclidean norm as in convention). The following identity is used to simplify computations

$$\int_{\mathbb{R}^n}f(|x|)\mathrm{d}x = \omega_{n-1}\int_0^\infty f(r) r^{n-1} \mathrm{d}r,$$
where $\omega_{n-1}$ denotes the surface area of the $(n-1)$-sphere of radius 1.

But I have never found any full proofs on such topic. Does anyone have some references in mind, better some source which I can actually refer to?

updates

Recently I found this subject has been mentioned by Stein in the appendix from his book "Fourier Analysis", and it further refers to Buck's "Advanced Calculus", Folland's "Advanced Calculus" and Lang's "Undergraduate Analysis".
I have gone through the later two. It seems to me that Lang's treatment is pretty self-contained, though it is given in exercise hence details omitted.

Mentioned by TCL, Evan's book on measure theory is decent but I'm not sure if it's too heavy. It let me feel like shooting mosquito with bazooka.

答: Since $\mathbb{R}^n=\{0\}\cup (0,\infty)\times S_r^{n-1}$, its volume element $dV$ can be written as a the product of $dr$ times the volume element of the $n-1$ sphere of radius $r$. By dimensional grounds, the latter is equal to $r^{n-1}$ times the volume element of the unit $(n-1)$-sphere. For example, if $n=2$ one has $dV=r dr\wedge d\theta$, note that the integral of $d\theta$ over $S^1$ is $2\pi$, i.e. the "volume" of $S^1$; similarly for $n=2$, $dV=r^2 \sin\theta\, dr\wedge d\theta$ and $\int _{S^2} \sin\theta d\theta\wedge d\phi=4\pi$. Since your function is radial, you can factorize the integral into a radial and an angular part. The integral of the angular part, which you denote with $\omega_{n-1}$, is therefore the volume of the unit $(n-1)$-sphere. A nice way of calculating $\omega_{n-1}$ is presented here http://scipp.ucsc.edu/~haber/ph116A/volume_11.pdf</p>

Title: Integration of radial functions?


答: $$ \langle Mx,y\rangle =(Mx)^{T}y=x^{T}M^{T}y=x^{T}(M^{T}y)=\langle x,M^{T}y\rangle $$

Title: Why the zero vector always orthogonal to any other vector.

Body:
Im currently looking at inner products and was wondering why the inner product of any vector with the zero vector is equal to 0. I have researched on this and only found the information that the zero vector is orthogonal to all vectors but no proof alongside.

And hence I was wondering if anyone had any proof as to why this happens.

Thanks in advance.

Edit:

Thanks every for their comment, I was now wondering if anyone could help explain to me why the inner product of $\left \langle M \vec u,\vec v \right \rangle$ = $\left \langle \vec u,M^{T} \vec v \right \rangle$

where M is and $n*n$ square matrix.


问:
Find recurrence relation and general solution
I need help finding recurrence relation.

We will glue a decorative tile with a size of 1 × n squares. We have available

tiles of 4 colors with size 1x1 square

tiles of 5 colors with size 1x2 square

How much combinations is there to make decorative tile of size 1 x n squares? And what's the recurrence relation. I don't even know how to start solving this. Can you solve it or help me?

答:
Denoting  $u_n$ the number of possibilities, you have $u_0=1$, and $u_1=4$.

Then the first tile can be a $1x1$ square, with four possibilities, and then you pave the $n-1$ other squares, or it can be a $1x2$ block, with five possibilities, and then you pave the $n-2$ other squares, and all solutions are distinct.

To say it shorty, for $n \ge 2$, $u_n = 4u_{n-1}+5u_{n-2}$.

The roots of the characteristic polynomial $X^2-4X-5$ are $5$ and $-1$ so $u_n = \alpha 5^n + \beta (-1)^n$. Using $u_0=1$ and $u_1=4$, we get $\alpha=\frac{5}{6}$ and $\beta=\frac{1}{6}$ so $$u_n = \frac{5^{n+1}+(-1)^n}{6}$$

You can check it for $n=2$, where you have $4 \cdot 4$ possibilities for a tiling with two squares, and $5$ possibilities for a tiling with one block, so $u_2=21$. And $\frac{5^3 + (-1)^2}{6} = 21$.


问:
Conversion of Laplace distribution into Chisquare Distribution
I got a confusion regarding the following distribution conversion.
From wikipedia I got that
If $Xi$~$Laplace$($0,1/b$)

Then $|Xi|$~$exp$ with mean $b$

so 2*$sum_{i=1}^n |X_i|$/b~ Chisquare $(2n)$

Now again from wikipedia I got
If $Xi$~$Laplace$($a,1/b$) then
$2bsum_{i=1}^n |Xi-a|$~ Chisquare (2n)
Then according to this the above turns out to be
$2bsum_{i=1}^n |Xi|$~ Chisquare (2n) which is contradictory,can anyone please help me out to reach a firm conclusion about it

答:
... which is contradictory,can anyone please help me out to reach a firm conclusion about it

I understand: the results are both correct; the difference you see is due to a different parametrization of the Laplace.

In the first statement they consider the following density:

$$f_X(x)=\frac{1}{2b}e^{-\frac{|x|}{b}}$$

which transformed get $Y=|X|\sim Exp(\frac{1}{b})$ (neg exp with mean $b$, as they state).
In this situation (see proof of the second statement) the result is correct:

$$\frac{2}{b}\Sigma_i Y_i \sim Gamma\Big(n;\frac{1}{2}\Big)=\chi_{(2n)}^2$$

In fact, using the same parametrization for the Laplace, you get, for the second statement:

$$f_X(x)=\frac{1}{2b}e^{-\frac{|x-a|}{b}}$$

$$Y=|X-a|$$

with standard transformation you get

$$F_Y(y)=F_X(a+y)-F_X(a-y)=1-\frac{1}{2}e^{-\frac{a+y-a}{b}}-\frac{1}{2}e^{\frac{a-y-a}{b}}=1-e^{-\frac{y}{b}}$$

Thus

$$Y\sim Exp\Big(\frac{1}{b}\Big)=Gamma\Big(1;\frac{1}{b}\Big)$$

As known (easy to prove with MGF and independence)  you get $\Sigma_i Y_i \sim Gamma\Big(n;\frac{1}{b}\Big)$

and concluding (very easy to be proved with a simple monotonic transformation)

$$\frac{2}{b}\Sigma_i Y_i \sim Gamma\Big(n;\frac{1}{2}\Big)=\chi_{(2n)}^2$$

Now the two statement are coherent

I understand that they are ambiguous because saying Laplace(0;1/b) I mean the following density:

$$f_X(x)=\frac{b}{2}e^{-b|x|}$$

With this parametrization (that is the one they used in the second statement and correctly, in my opinion), the correct statement is

$$2b\Sigma_i Y_i \sim Gamma\Big(n;\frac{1}{2}\Big)=\chi_{(2n)}^2$$

Concluding: Case by case, do yourself all the calculations, expliciting well the parametrization you use and get your own result. It is a good exercise, not difficult but very useful


问:
Prove a dimension relation $\dim S+\dim T<\dim X+\dim(\overrightarrow{S}+\overrightarrow{T})$ when $S\cap T=\emptyset$
The proposition is:

Let $X$ be a finite-dimensional affine space and $S,T$ be subspaces of $X$. If $S\cap T=\emptyset$, then $$\dim S+\dim T<\dim X+\dim(\overrightarrow{S}+\overrightarrow{T}),$$ where
$\overrightarrow{S}$ and $\overrightarrow{T}$ represents the direction of $S$ and $T$, respectively.

My attempt is as follows. Since $\dim S=\dim\overrightarrow{S}, \dim T=\dim\overrightarrow{T}, \dim X=\dim\vec X$, if we could show $\dim(\overrightarrow{S}+\overrightarrow{T})>\dim(\overrightarrow{S}\cap\overrightarrow{T})$, the inequality would be proved by $\dim(\overrightarrow{S}+\overrightarrow{T})\le\dim\vec X$ and the Grassmann relation $\dim(\overrightarrow{S}+\overrightarrow{T})=\dim\overrightarrow{S}+\dim\overrightarrow{T}-\dim(\overrightarrow{S}\cap\overrightarrow{T})$. However, I cannot establish $\dim(\overrightarrow{S}+\overrightarrow{T})>\dim(\overrightarrow{S}\cap\overrightarrow{T})$ from condition $S\cap T=\emptyset$. Worse still, I later realized that even if $S\cap T=\emptyset$, their directions $\overrightarrow{S}$ and $\overrightarrow{T}$ could be the same if $S\parallel T$. Therefore $\dim(\overrightarrow{S}+\overrightarrow{T})=\dim(\overrightarrow{S}\cap\overrightarrow{T})$ in that special case. Could you please help me with the proof (you don't have to follow my unsuccessful attempt)? Thank you very much.

答:
My previous attempt is not in the right direction. Following is the proof I have just figured out.

The inequality is equivalent to $\dim S+\dim T-\dim(\overrightarrow{S}+\overrightarrow{T})<\dim X$ where the left side is equal to $\dim(\overrightarrow{S}\cap\overrightarrow{T})$ due to the Grassmann relation. If we can show $\dim(\overrightarrow{S}+\overrightarrow{T})<\dim X$, then $\dim(\overrightarrow{S}\cap\overrightarrow{T})\le\dim(\overrightarrow{S}+\overrightarrow{T})<\dim X$, which establishes the proposition.

To this end, we need to make use of two facts:

Let $a\in S$ and $b\in T$. Then $S\cap T\ne\emptyset$ is equivalent to $\overrightarrow{ab}\in\overrightarrow{S}+\overrightarrow{T}$.

Let $a\in S$ and $b\in T$. $\overrightarrow{\langle S\cup T\rangle}=\overrightarrow{S}+\overrightarrow{T}+K\vec{ab}$ where $K$ is the field of scalars.

Fact 1. is not hard to prove. A proof of Fact 2. is here.

From fact 1., we can infer that $\overrightarrow{S}+\overrightarrow{T}$ is independent of  $K\vec{ab}$. To show this, if $(\overrightarrow{S}+\overrightarrow{T})\cap K\vec{ab}$ contains an element $c\vec{ab}\ne 0$, then $c\ne0$, so we have $\vec{ab}\in (\overrightarrow{S}+\overrightarrow{T})$, which is forbidden by Fact 1., because $S\cap T=\emptyset$. Taking dimension on both sides in Fact 2., we have $\dim\overrightarrow{\langle S\cup T\rangle}=\dim\bigl((\overrightarrow{S}+\overrightarrow{T})+K\vec{ab}\bigr)$. The vector spaces in the parentheses on the right side are independent as we have just shown, so we have $\dim\overrightarrow{\langle S\cup T\rangle}=\dim(\overrightarrow{S}+\overrightarrow{T})+\dim K\vec{ab}=\dim(\overrightarrow{S}+\overrightarrow{T})+1$. $\dim\overrightarrow{\langle S\cup T\rangle}\le\dim\vec X$ in general, so we finally get $\dim(\overrightarrow{S}+\overrightarrow{T})<\dim\vec X$.


问:
If $f(n) \in O(g(n))$ then prove that $O(f(n) + g(n)) = O(g(n))$
I have to prove that if I have two functions $f$ and $g$ and $f \in O(g)$ then $$O(f(n) + g(n)) = O(g(n))$$ but I don't know how to do it. Any help will be appreciated :)

Thank you!

答:
as f(n) ∈ O(g) then ∃C1>0 ∃n0>0 ∀n≥n0:f(n)≤C1∗g(n)

note also that g(n) ≤ C2∗g(n) thus:
f(n) + g(n) ≤ C1∗g(n) + C2∗g(n) → f(n) + g(n) ≤ 2*max{C1,C2} *g(n)

and the claim follows.


问:
Skew line offset from cube center
How to find skew distance from the unit cube's ( of development below) center to straight line PQ in 3d?

Title: Skew line offset from cube center

Body:
Place the cube on x-y plane with corners of base at $(\pm 0.5, \pm 0.5,0)$.

Center $C$ is at $(0,0,0.5)$.

$P = (0,0.5,1)$. $Q = (-0.5,-0.5,0.3)$

Then distance of $C$ from $QP$ is $$d=\dfrac{|QP \times CP|}{|QP|}$$

$QP=(0.5,1,0.7)$. $CP=(0,0.5,0.5)$

$$\Rightarrow d=\dfrac{|(0.15,-0.25,0.25)|}{\sqrt{1.74}}$$

$$\Rightarrow d \approx 0.2912$$


问:
value of fraction
Can we find out the value of

$$\frac{1+i}{1-i}$$

I have tried to solve it by multiplying $(1+i)$ to both sides and in the end see that the result is still $i$. Am I correct,or is there a different solution?

答:
You divide by complex numbers, by multiplying by its conjugate, that is, replace in the divisor $+i$ by $-i$.  The product of $(a+ib) \cdot (a-ib) = a^2+b^2$, a real.
$$
\frac{1+i}{1-i}\frac{1+i}{1+i}=\frac{2i}2=i.
$$</p>

Title: value of fraction


问:
Why is $SU(2)$ diffeomorphic to $S^3$?
As the title,my question is why $SU(2)$ is diffeomorphic to $S^3$ ? How can I structure a diffeomorphic map between $SU(2)$ and $S^3$?

Please help me,and give me more details!

Thank you very much!

答:
Every $SU(2)$ matrix is a linear combination of Pauli matrices $\sigma_1, \sigma_2, \sigma_3 $ and the unit matrix I, which is a base of the associated Lie algebra of the SU2 group.
So every $SU(2)$ matrix  $X$ can be written a $X=x_1\sigma_1 +x_2\sigma_2 +x_3\sigma_3 +x_4.I$ and the condition that $det{X} = 1$ gives $x_1^2 + x_2^2 +x_3^2 +x_4^2 = 1$ which is the $S^3$ sphere. And every point on that sphere represents an $SU(2)$ matrix

Title: Why is $SU(2)$ diffeomorphic to $S^3$?


问:
Category of vector spaces is monoidal closed
How to show that the category $\textbf{Vect}_k$ of vector spaces with ground field $k$ is monoidal closed? I am aware that the right adjoint to the functor $-\otimes Y$ should be $[Y,-]: X\mapsto \textbf{Vect}_k(Y,X)$, with unit $d_X : X\to [Y,X\otimes Y]$ and counit $e_X: [Y,X]\otimes Y\to X$. What are these linear maps explicitly?

答:
You don't necessarily need the unit and counit to define an adjunction: it's enough to show that the isomorphism
$$[X,\, [Y,Z]]\ \cong\ [X\otimes Y,\,Z]$$
given by observing that both sides basically give the bilinear maps $X\times Y\to Z$, is natural in $X$ and $Z$.

Nevertheless, $d_X(x)=y\mapsto x\otimes y$, and

$e_X:[Y,X]\otimes Y\to X$ is $\ e_x(f\otimes y)=f(y)$.


问:
Checking if the induced mapping is well-defined
Let $\mathcal{M}$ be a $\sigma$-algebra of subsets of a nonempty set $T$, $X$ is Hausdorff LCTVS, $X'$ the topological dual of $X$, and $m: \mathcal{M}\to X$  a countably additive vector measure on $\mathcal{M}$.

Let $f: T\to \mathbb{R}$ be $\mathcal{M}$-measurable function. We say that $f$ is $m$-integrable if for each $x'\in X'$, $f$ is integrable with respect to the measure $x'\circ m:\mathcal{M}\to [0,+\infty]$ and for each $E\in \mathcal{M}$ there exists $x_E\in X$ such that for any $x'\in X'$, we have
$$x'(x_E)=\int_Ef\ d(x'\circ m).$$
We shall denote $x_E=\int_Ef\ dm.$

Suppose that $f$ is $m$-integrable. Define
$$n:\mathcal{M}\to X$$ by
$$n(E)=\int_Ef\ dm$$ for each $E\in \mathcal{M}.$ I need to show that the mapping $n$ is well-defined. How do we do this?

答:
Let $f \colon T \to \mathbb R$ be $\mathcal{M}$-measurable and assume that $f$ is $x'\circ m$-integrable for all $x' \in X'$. For every $E \in \mathcal M$ there is at most one point $x_E \in X$ such that
  $$
x'(x_E) = \int_{E} f\,d(x'\circ m) \quad\text{for all }x' \in X'. \tag{$*$}
$$

Suppose $x_E$ has the property $(\ast)$ and $y \neq x_E$. Since $X$ is Hausdorff, the Hahn-Banach theorem ensures that there is $x' \in X'$ such that $x'(y) \neq x'(x_E)$. But then
$$
x'(y) \neq x'(x_E) = \int_{E} f\,d(x'\circ m)
$$
shows that $y$ can't have the property $(\ast)$.

The definition of $m$-integrability of $f$ thus implies the existence of a unique $x_E$ having the property in $(\ast)$ and the map
$$
\begin{align*}
n \colon \mathcal{M} & \longrightarrow X \cr
 E & \longmapsto x_E
\end{align*}
$$
is well-defined. The point $x_E$ is the definition of the integral $x_E = \int_{E} f\,dm$.


问:
Compact connected surfaces such that non-null homotopic map exists from itself to $\mathbb S^1$.
First, I know that any compact connected surface is either $\mathbb S^2$, $\mathbb P^2 \# \cdots \# \mathbb P^2$ ($n$ copies), or $\mathbb T^2 \# \cdots \# \mathbb T^2$ ($n$ copies).

Also, suppose I know the following result:

-Given $X$ is a connected space with a contractible universal covering
space, $f : Y \to X$ is null homotopic if and only if the induced homomorphism $f_* : \pi_1(Y, y) \to \pi_1(X, f(y))$ is trivial for each $y \in Y$.

For the first case, I know that since the fundamental group of $\mathbb S^2$ is trivial, it must be that any continuous $f$ from $\mathbb S^2$ to $\mathbb S^1$ should be trivial.

For the second case, when $n$ is just one, since the fundamental group of the projective plane is just a group of order $2$.

However, I am not sure how to proceed with other cases, and show that there exists/does not exist non-null homotopic maps.

Also, what should I do to prove the fact I am assuming?

答:
You can use the Van Kampen theorem to determine the fundamental group of a connected sum, this should answer your question for the projective planes.

For the torus, what do you think, if $n=1$, of the projection $\mathbb T^2 \cong S^1\times S^1\to S^1$ ? What does it tell you for $n\geq 2$ ?

To prove the result you are assuming, you should use the lifting criterion: suppose $\tilde X\to X$ is a covering space, with say both $\tilde X$ and $X$ connected; then what is a criterion for when a map $Y\to X$ can be lifted to a map $Y\to \tilde X$ ?

What does it tell you when $\tilde X$ is contractible ?


问:
Percentage, Time & work
$A$ is $4$ times as good as $K$ and hence, $K$ takes $50$ more days than $A$ to complete a job. How long will it take to finish the job if they work together?

I tried and got:

$A=6$%/day

$K=1.5$%/day

$A+K=7.5$%/day therefore $13.\overline{3}$ days in all.

But the book has answer $20$ days.
What's wrong?

答:
To answer your final question: the book is.

I get the same as you. It's easy to see that even A alone will finish in less than $20$ days, since if A took $20$ and K took $70$, A would only be $3.5$ times as fast as K, and this ratio gets lower as you increase the times.

I'm not sure exactly what mistake has been made in the book. The easiest way to make it work would be to change $50$ to $75$, and then the numbers work out quite nicely with an exact number of days at every stage.

Title: Percentage, Time & work


问:
$ABCD$ is a rectangle with $AD = 2$ , $AB = 4$. $P$ is on $AB$ such that $AP : PB = 2 : 1$. $CE \perp DP$ at $E$. Find $CE$.

$ABCD$ is a rectangle with $AD = 2$, $AB = 4$. $P$ is on $AB$ such that $AP : PB = 2 : 1$. $CE \perp DP$ at $E$. Find $CE$.

What I Tried: Here is a picture :-

Since $AP : PB = 2 : 1$ and $AB = CD$, we have $AP = \frac{8}{3}$, $BP = \frac{4}{3}$.

Now :- $$[\Delta DPC] = \frac{[\square ABCD]}{2}$$
$$\Rightarrow [\Delta DPC] = 4$$

From Pythagoras Theorem :- $$(AP)^2 + (AD)^2 = (PD)^2$$
$$\Rightarrow PD = \frac{\sqrt{82}}{3}$$

So, $$[\Delta PDC] = \frac{1}{2} * PD * CE$$
$$\Rightarrow 4 = \frac{1}{2} * \frac{\sqrt{82}}{3} * CE$$
$$\Rightarrow CE = \frac{12 \sqrt{82}}{41}$$

But my answer was given as $\frac{12}{5}$ , so where did I go wrong?

Can anyone help me? Thank You.

答:
$PD=\color{red}{\dfrac{\sqrt{82}}{3}}$ is false.

$$AD=\dfrac{\color{blue}{6}}{3}, AP=\dfrac{\color{blue}{8}}{3} \Rightarrow PD=\dfrac{\color{blue}{10}}{3}$$


问:
Is $\frac{10}{459}$ a cube in $\Bbb{Q}_3$?
Is $\frac{10}{459}$ a cube in $\Bbb{Q}_3$, the 3-adic numbers?

My general strategy for showing if something is a cube in $\Bbb{Q}_3$ has been:

Use Hensel's Lemma to show it is (1)

If it isn't, suppose for contradiction it is a cube and equate 3-adic valuations to see if we can find a contradiction (2)

If it isn't, suppose for contradiction it is a cube and reduce modulo $3^k$ for some $k$ and try to arrive at a contradiction. (3)

(2) doesn't help in this instance, and I don't think I can apply either (1) or (3) as the number in question is a fraction.

How can I attempt to answer the question: "Is $x$ a cube in $\Bbb{Q}_3$?" in this specific case, when $x = \frac{10}{459}$, and I'm also interested in a general strategy for when $x$ is a fraction.

Thanks.

答:
First determine the valuation of the fraction by getting it for the numerator and denominator:

$v(10)=0, v(459)=v(17×27)=3, v(10/459)=v(10)-v(459)=-3=\text{a multiple of 3}$

Then since $v(10/459)$ is a multiple of $3$, we will have a cube if the integer obtained by shifting the "decimal point" so that the last nonzero digit is at the units place is a cube.  Multiply your fraction by $3^3=27$ to make this adjustment, setting $v$ to zero:

$(10/459)×3^3=10/17$

The resulting fraction will be a 3-adic cube, completing our test, if its residue $\bmod 9$ is $\pm1$.  Here $10\equiv1$ and $17\equiv-1$, so $(10/17)\equiv-1$ making it a $3$-adic cube, and then $(10/17)×(1/3^3)=10/459$ will be one also.


问:
Manipulation of summation index - power series
Problem

Consider the differential equation: $y''+ty'+2y=t^2e^t $

By setting $y(t)=\sum_{n=0}^{\infty}c_nt^n$ show that the differential equation can be rewritten as
$$\sum_{n=0}^{\infty} \big((n+2)(n+1)c_{n+2}+(n+2)c_n \big)t^n=\sum_{n=2}^{\infty}\frac{t^n}{(n-2)!} $$

An attempt

$$y''=\sum_{n=2}^{\infty}c_n(n-1)nt^{n-2}=\sum_{n=0}^{\infty}c_{n+2}(n+2)(n+1)t^n$$

$$ty'=t\sum_{n=1}^{\infty}c_nnt^{n-1}=t\sum_{n=0}^{\infty}c_{n+1}(n+1)t^n $$

$$2y=2\sum_{n=0}^{\infty}c_nt^n=\sum_{n=0}^{\infty}2c_nt^n $$

Combining all these terms on the left hand side gives

$$\sum_{n=0}^{\infty} \Bigg[c_{n+2}(n+2)(n+1)+tc_{n+1}(n+1)+2c_n \Bigg]t^n=t^2e^t$$

As you can see, I'm almost getting to the answer, but there are some things that don't match up. I'm also unsure how

$$t^2e^t=\sum_{n=2}^{\infty}\frac{t^n}{(n-2)!} $$

Can anyone help me out and perhaps spot if I have made a mistake?

答:
On  the left side observe that $ \sum\limits_{n=0}^{\infty} nc_nt^{n}$ is same as $\sum\limits_{n=0}^{\infty} (n+1)c_{n+1}t^{n+1}$ since $0c_0t^{0}=0$.

[You have to split $(n+2)c_n$ as $nc_n+2c_n$ to see why your answer and the given answer match].

Also, $\sum\limits_{n=2}^{\infty} \frac {t^{n}} {(n-2)!} =t^{2} \sum\limits_{n=2}^{\infty} \frac {t^{n-2}} {(n-2)!}=t^{2} \sum\limits_{n=0}^{\infty} \frac {t^{n}} {n!}=t^{2}e^{t}$.