问:
Strategies for a game where players alternate placing $2\times 2$ squares and three-square "L" shapes on a $100\times100$ board
This is a question for a homework assignment that I need to solve: please only provide advice, I need to understand it. I also want to solve it on my own.

The question is as follows:

On a $100\times100$ board, two people, Art and Bart take turns putting pieces on the board.

Art's pieces are $2\times2$ squares.

Bart's pieces are "L" shapes made out of $3$ squares (a $2\times2$ square without one corner).

The player who can no longer fit one of their pieces on the board loses.

a) If Art starts, does either player have a winning strategy?

b) If Bart starts, does either player have a winning strategy?

答:
Part $1$:

If Bart starts, he can first place the L-shape tile in the center $2\times2$ area first with bottom right corner empty. (since the board is $100\times100$ it has a center $2\times2$).

Then whatever Art does, Bart always try to "mirror Art's move" with respect to the center point. Note that this is always possible, I will show some hints and let you prove the details.

Hint: try to break into two cases, one case where Art places the $2\times2$ tile that contains an "empty corner of an L shape"(think about the situation on the mirrored side when is this possible), and the other case otherwise.

Notice that, in the first case you also need to show that if Art places the $2\times2$ tile in the center area "bottom-right empty corner", Bart can also mirror his move.

There is a minor point that you need to prove Bart is always able to avoid the case where he places four L-shape tiles in a circular pattern (i.e. there is a $2\times 2$ space in the middle) by using only two orientations. I will let you think about why and how.

Once this is proven, Bart has his winning strategy because he always has a move no matter Art does.

Part $2$:

Partial Answer:

If Art starts anything other than filling the $2\times2$ center area with his first move, Bart can use similar strategy as in part $1$ to win. This means Art has to start by filling the center $2\times 2$ square.

I think Bart has a winning strategy by carefully leading the situation into the mirror game, but have thought for a while without being able to prove it. I hope this helps.


问:
calculate $ \lim_{k\rightarrow\infty}\sin\left(kx\right) $
if $ x \notin \pi\mathbb{N} $ why the limit

 $ \lim_{k\rightarrow\infty}\sin\left(kx\right)  $

does not exist?

答:
so when $x\rightarrow \infty $ what you see ? is there any existed value for f(x)

note thate $-1\leq sin(kx)\leq 1$ :$k\in R$

ContentLicense: CC BY-SA 3.0


问:
Using Euler's formula to evaluate $\int_0^{\infty} \sin(x^2)dx$
We let
$$
I = \int_0^{\infty} \sin(x^2)dx
$$
In other terms
$$
I = \int_0^{\infty} \Im\left(e^{ix^2}\right)dx = \Im \left(\int_0^{\infty} e^{ix^2} dx\right)
$$
We let
$$
J = \int_0^{\infty} e^{ix^2}dx
$$
This looks like the gaussian integral so using the standard technique of conversion to polar coordinates we have
$$
J^2 = \left( \int_0^{\infty} e^{ix^2}dx \right)\cdot\left( \int_0^{\infty} e^{iy^2}dy \right) = \iint_{\mathbb{R}^2} e^{i(x^2+y^2)}dA
$$
Converting:
$$
\begin{cases}
r^2 = x^2 + y^2 \\
dA = r \, dr d\theta
\end{cases}
$$
we obtain
$$
J^2 = \int_0^{2\pi} \int_0^\infty re^{ir^2} \, dr d\theta
$$
Evaluating and simplifying:
$$
J^2 = \frac{\pi}{i} \left[e^{ir^2} \right]_0^\infty
$$
Does this diverge? What can i do here, because I am pretty sure that $I$ converges? Thanks

答:
$$\cos \left(x^2\right)+i \sin \left(x^2\right)=e^{ix^2}$$
$$\int_0^{\infty}e^{ix^2}\,dx=\left[-\frac{(1+i) \sqrt{\pi } \text{erf}\left(-\frac{(1-i) x}{\sqrt{2}}\right)}{2 \sqrt{2}}\right]_0^{\infty}=\left(\frac{1}{2}+\frac{i}{2}\right) \sqrt{\frac{\pi }{2}}$$
$$\int_0^{\infty}\sin(x^2)\,dx=\text{Im}\left(\int_0^{\infty}e^{ix^2}\,dx\right)=\text{Im}\left[\left(\frac{1}{2}+\frac{i}{2}\right) \sqrt{\frac{\pi }{2}}\right]=\frac{1}{2}\sqrt{\frac{\pi }{2}}$$

ContentLicense: CC BY-SA 4.0


问:
Proof the three intersections are in the same line.
Some days ago I discovered a conclusion which is proved true:

Let $AX,AY,AZ$ be three rays (in which $A$ is the initial point) which any two of them form an acute angle, and $AY$ is inside $\angle XAZ$. Let
$$
\begin{align}
&A_1\in AX,\ B_1,B_2\in AZ(AB_2<AB_1),\\
&A_1B_1\cap AY=C_1,\ A_1B_2\cap AY=C_2,\ B_1C_2\cap AX=A_2,\ A_2B_2\cap AY = C_3,\\
&B_1C_2\cap B_2C_1=D_1,\ B_1C_3\cap B_2C_2=D_2,
\end{align}
$$
then $A,D_1,D_2$ are on a same line.

I've been working on it since a month ago, but I can't solve it in a geometric way. Could anyone provide a solution? Thanks!

(It's better that the solution is pure geometric, not analytic.)

答:
Here is a pedestrian way to show the result. (So we avoid Pappus, Desargues, et co. because we can extract an intermediate result which brings more light in the situation.)

Let $P$ be the intersection of the lines $A_1B_1$ and $A_2B_2$. Then the we have the following harmonic proportions, written as cross ratios equal to $-1$:
$$
(A_1,B_1;P,C_1)=(A_2,B_2;P,C_3)=-1\ .
$$
(For the first cross ratio, apply in $\Delta AA_1B_1$ Ceva w.r.t. the point $C_2$, and Menelaus w.r.t. the transversal $PA_2B_2$. For the second cross ratio do the same in $\Delta AA_2B_2$, or better, "move" the first configuration from the line $PA_1C_1B_1$ to the line $PA_2C_3B_2$ using the perspective point $A$.)

Now we look at $(A_1,B_1;P,C_1)=-1$ and move this harmonic proportion from the line $PA_1C_1B_1$ to the line $Ay=AC_3C_2C_1$ using the perspective point $B_2$. In other words, we build the lines $B_2A_1$, $B_2B_1$, $B_2P$, $B_2C_1$ (in this order, the one of the points in $(A_1,B_1;P,C_1)=-1$,) and intersect them with $Ay$. By perspectivity we get:
$$
(C_2,A;C_3,C_1)=-1\ .
$$
We isolate this relation, and state a simpler result which does not involve the points $A_1,A_2$. The same notations are used.

Proposition: Let $A,C_2;C_1,C_3$ be four points on a line in harmonic proportion. Let $B_1,B_2$ be two further points on a line through $A$. Construct $D_1,D_2$ as in the OP. Then $A,D_1,D_2$ are on a line.

Pedestrian proof:
We have:
$$
\begin{aligned}
+1 &=
\color{blue}{\frac{D_1C_2}{D_1B_1}}
\cdot \frac{B_2B_1}{B_2A}
\cdot \frac{C_1A}{C_1C_2}\ ,
\\
&\text{ Menelaus, $\Delta C_2B_1A$, secant $D_1B_2C_1$,}
\\[2mm]
+1 &=
\color{blue}{\frac{D_2C_2}{D_2B_2}}
\cdot \frac{B_1B_2}{B_1A}
\cdot \frac{C_3A}{C_3C_2}\ ,
\\
&\text{ Menelaus, $\Delta C_2B_2A$, secant $D_2B_1C_3$,}
\\[2mm]
&\text{ which gives}
\\[2mm]
\color{blue}{\frac{D_1B_1}{D_1C_2}}\cdot
\color{blue}{\frac{D_2C_2}{D_2B_2}}\cdot
\color{maroon}{\frac{AB_2}{AB_1}}
&=
\frac{B_2B_1}{B_2A}
\cdot \frac{C_1A}{C_1C_2}\ \cdot\
\frac{B_1A}{B_1B_2}
\cdot \frac{C_3C_2}{C_3A}
\ \cdot\
\color{maroon}{\frac{AB_2}{AB_1}}
\\
&=
-\frac{C_1A}{C_1C_2} \cdot \frac{C_3C_2}{C_3A}
\\
&=
-(C_1,C_3;A,C_2)
\\
&=-(-1)=+1\ .
\end{aligned}
$$
The reciprocal Menelaus in $\Delta C_2B_1B_2$ shows now the claimed colinearity.

$\square$

Note: The statement and the proof of the isolated proposition may be simpler to follow if we use a better notation, for instance $A'$ instead of $C_2$, and the remained points using maybe only indices $1,2$.


问:
Paralellogram based pyramid volume
Given three points of a paralellogram being
$
P(2,3,4) ; Q(3,1,4); R(2,5,3)
$
I've already calculated the fourth point with
$S(1,7,3)$

Further there's the tip point $Z(5,7,8)$

To get the volume of the pyramid I need to get the distance between the plane that the paralellogram spans and the the tip point.

From my lecture, the parameterform of a plane for the vectors p,v & w would be:

$E=p+ \mathbb{R}\cdot v +\mathbb{R}\cdot w$

Following this, what's the third vector for my example? So far it'd look like this

$$E=PQ+\lambda\cdot PS+ \dots$$

Further: once I have the plane of the parallelogram, is the rest of the way I described correct?

答: I'd take a cross product, say, $(P-Q)\times(R-Q)$ to get a vector orthogonal to the plane, normalize that to get a normal vector, then form the scalar product of $Z-Q$ with that normal vector and take the absolute value to get the distance of $Z$ from the plane.</p>

Title: Paralellogram based pyramid volume


问:
Pseudo metric space $d(x,y)= x+y$
Can I say that $d(x,y)= x+y$ is pseudo metric?

I am confused because if I put $d(1,1)=2$ not $0$ so it is not pseudo.

But some say that it is pseudo because I can put two distinct numbers and get $0$ like $d(-1,1)$ which is zero so it is pseudo but not metric.

Can you help me please?

答:
You make it sound like "pseudo" and "metric" are two independent properties a map $d\colon X\times X\to\mathbb R_{\ge 0}$ could have. That is not the case. There is a definition for what a metric is and a definition for what a pseudometric is.

As you can see from the definitions, the only difference is that a metric is required to satisfy $d(x,y)>0$ for $x\neq y$ while a pseudometric might have $d(x,y)=0$ for some $x\neq y$.

Both of them must satisfy $d(x,x)=0$ for all $x\in X$ so your candidate is neither a metric nor a pseudometric.

Note that every metric is a pseudometric but not every pseudometric is a metric.

Title: Pseudo metric space $d(x,y)= x+y$


问:
Pre-Algebra Fractional Exponent Question
Why does $t^{\frac{3}{2}} \cdot t^{\frac{1}{2}} = t^2$?

What I tried to do was multiply the exponents together $\frac{3}{2} \cdot \frac{1}{2} = \frac{3}{4}$ so my final answer was $t^{\frac{3}{4}}$ but according to my Wileyplus homework it is $t^2$.

Can someone please explain?

Body:
Just remember this:
$$a^b+a^c= a^{b+c}$$
so in your problem,
$$a=t, \, b=3/2, \, c=1/2$$
and $$ t^3/2+t^1/2= t^{1/2+3/2}= t^4/2=t^2$$

hope this helped!


问:
Why is my solution to $-xy(\frac{\partial}{\partial x}f(x,y))+(x-y^2)(\frac{\partial}{\partial y} f(x,y))=0$ incorrect?
I've been trying to solve this PDE: $$-xy(\frac{\partial}{\partial x}f(x,y))+(x-y^2)(\frac{\partial}{\partial y} f(x,y))=0$$

A condition is given in the form of: $f(x,0)=-x$

A suggestion with the following substitution is given, where $a$ is a "suitable" constant:

$$\begin{Bmatrix} u(x,y)=\frac{y^2}{x^2}+\frac{a}{x}\\v(x,y)= \frac{y}{x} \end{Bmatrix}$$

$$\frac{\partial}{\partial x}f(u(x,y),v(x,y))=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}=f_{u}^{'}(-\frac{ax+2y^2}{x^3})+f_{v}^{'}(\frac{y}{x^2}) \\
\frac{\partial}{\partial y}f(u(x,y),v(x,y))=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}=f_{u}^{'}(-\frac{y}{x^2})+f_{v}^{'}(\frac{1}{x})$$

So now we can substitue:

$$-xy(\frac{\partial}{\partial x}f(x,y))+(x-y^2)(\frac{\partial}{\partial y} f(x,y))=0\\ f_{u}^{'}\ \frac{(2+a)y}{x}+ f_{v}^{'} =0$$

If we let $a=-2$ we arrive at:

$$f_{v}^{'} =0 \\ f(u,v)= g(u)$$
From the condition we know: $-u=f(u,0)=g(u)$
So we should arrive at: $$f(x,y)=-\frac{y^2}{x^2}+\frac{2}{x}$$
Nevertheless this incorrect, but I cannot see where my mistake is.

答:
$$-xy(\frac{\partial}{\partial x}f(x,y))+(x-y^2)(\frac{\partial}{\partial y} f(x,y))=0$$
In fact a characteristic equation is :
$$-\frac{y^2}{x^2}+\frac{2}{x}=c$$
Thus the general solution is :
$$f(x,y)=\Phi\left(-\frac{y^2}{x^2}+\frac{2}{x}\right)$$
where $\Phi(X)$ is an arbitrary function (to be determined according to the specified condition).

Condition : $f(x,0)=-x$
$$f(x,0)=\Phi\left(\frac{2}{x}\right)=-x$$
Let $X=\frac{2}{x} \quad\implies\quad x=\frac{2}{X}$
$$\Phi(X)=-\frac{2}{X}$$
Now the function $\Phi(X)$ is determined. We put it into the above general solution where $X=-\frac{y^2}{x^2}+\frac{2}{x}$
$$f(x,y)=-\frac{2}{X}=-\frac{2}{-\frac{y^2}{x^2}+\frac{2}{x}}$$
$$f(x,y)=\frac{2x^2}{y^2-2x}$$


问:
Lattices in $\mathbb{C}$
I have the following assignment:

consider the map

$$|\cdot|:\mathbb{Z}[i]\longrightarrow \mathbb{N},\qquad |a+ib|:=a^2+b^2$$

1) Prove that $|\alpha|<|\beta|$ iff $|\alpha|\leq |\beta|-1$ and $|\alpha|<1$ iff $\alpha=0$

2) Let $\alpha,\beta\in\mathbb{Z}[i],\beta\neq 0$. Prove that the map $f:\mathbb{Z}[i]\longrightarrow\mathbb{Z}[i], f(\gamma):=\alpha-\gamma\beta$ is the composition of a dilatation by the factor $\sqrt{|\beta|}$, a rotation (angle?) and a translation.

3) Deduce that there exists $\gamma\in\mathbb{Z}[i]$ such that $|f(\gamma)|$ is strictly smaller than $|\beta|$.

$\textbf{Hint:}$ compare the size of a cell of the lattice $f(\mathbb{Z}[i])$ and the size of the set of points whose distance to $0$ is $\leq\sqrt{|\beta|}$.

What i did: point 1) is a trivial consequence of the fact that the norm takes integer non negative values. For point 2), I use complex multiplication of numbers which is: multiply absolute values and add angles. For point 3), i'm actually waiting for a miracle... I suppose i should prove that there exists a cell in $f(\mathbb{Z}[i])$ intersecting the open ball centered at the origin with radius $\sqrt{|\beta|}$, but i have no idea how to write down this. Only thing i noticed is that $f$ acts with a rotation, which does not affect distance from the origin, so that the only changes in $|\gamma|$ come from dilatation and by adding $\alpha$.

Could someone put me on the right direction?

答:
Hints (or not...)

1) This follows at once from the fac that $\;\forall\alpha\in\Bbb Z[i]\;,\;\;|\alpha|\in\Bbb N\,$

2) We have that

$$f=f_\alpha\circ f_\gamma\;,\;\;\text{with}\;\;f_\gamma(x):=\gamma x\;,\;\;f_\alpha(x)=\alpha-x$$

3) The closed unit circle centerd at the origin is compact and thus there're only a finite number of elements of any lattice (say, $\,\Bbb Z[i]\,$ ...?) intersecting it

Now you can dilate (shrink, in scientific talk...:)  ) elements as shown in (2) so that their inside that circle...

The basic assumption here is that $\,\beta\neq0\,$ , I presume...

Title: Lattices in $\mathbb{C}$


问:
The converse problem about duality of $L^p$

Let $Ω$ be a bounded open subset in $R^n$ and $f$ be measureable with respect to Lebesgue measure on $Ω$. If there exist $M＞0$ s.t.$$\left|\int fg\,\mathrm{d}x\right|≦M\left(\int |g|^q\mathrm{d}x\right)^\frac{1}{q}$$ for all continuous function $g$ on $Ω$ (the integral is on $Ω$), where $1＜q＜+∞$, $\frac{1}{p}+\frac{1}{q}=1$, then $f \in L^p(Ω)$ and $\|f\|_{L^p(Ω)}≦M$.

I find some references: the conclusion is true if $g$ can be substituted by all simple functions.

But how can we do this in the continuous case? Should we choose simple functions that approximate the continuous $g$ by using Lusin's theorem  (but I have some problem in seeing how to do this)? Could someone show me the details? Thank you!

My doubt: even though we can use a continuous $g$ to approximate simple functions in $L^1$ (i.e. in the left side integral), we need that to prove they are close enough also in $L^q$ (i.e. in the right side integral).

答:
$\newcommand\ds\displaystyle$

Since continuous functions with compact support are dense in $L^q(\Omega)$, the inequality
$$\label{1}\tag{1}
\bigg|\int_\Omega fg\bigg|\leq M\,\|g\|_q$$ holds for all $g\in L^q$ (proof at the end).

Suppose that $\int_\Omega|f|^p=\infty$. Let
$$
X_n=\{|f|\leq n\},\qquad n\in\mathbb N.
$$
Then $\Omega=\bigcup_nX_n$. Write $f(x)=|f(x)|\,e^{i\theta(x)}$. Define
$$
g_n=\frac{e^{-i\theta}\,|f|^{p-1}\,1_{X_n}}{\bigg(\ds\int_{X_n}|f|^p\Bigg)^{1/q}}.
$$
Then
$$
\|g_n\|_q^q=\frac{\ds\int_{X_n}|f|^{q(p-1)}}{\ds\int_{X_n}|f|^p }=1.
$$
And
$$
\int_\Omega f\,g_n=\frac{\ds\int_{X_n}|f|^p}{\bigg(\ds\int_{X_n}|f|^p\Bigg)^{1/q}}=\bigg(\ds\int_{X_n}|f|^p\Bigg)^{1-1/q}=\bigg(\ds\int_{X_n}|f|^p\Bigg)^{1/p}\xrightarrow[n\to\infty]{}\bigg(\int_\Omega|f|^p\Bigg)^{1/p}=\infty,
$$
contradicting your inequality. So $f\in L^p$.

Proof of \eqref{1}. Given $g\in L^q(\Omega)$ there exists a sequence $\{h_n\}$ such that $\|g-h_n\|_q\to0$ (proofs here and here).  Then there exists a subsquence, that we still name $\{h_n\}$ that converges pointwise almost everywhere. Then using Fatou's Lemma,
$$
\bigg|\int_\Omega fg\bigg|=\bigg|\int_\Omega \lim_nfh_n\bigg|\leq\liminf_n\bigg|\int_\Omega fh_n\bigg|\leq\liminf_n M\|h_n\|_q=M\|g\|_q.
$$
The last equality is due to $\big|\|g\|_q-\|h_n\|_q\big|\leq\|g-h_n\|_q$ by the reverse triangle inequality.


问:
Can I remove ε from an inequality because it is arbitrary?
I'm looking at this proof (relating to lower/upper sums and integrals) and near the end it says this:

U(f) ≤ U(f,P) < L(f,P) + ε ≤ L(f) + ε

And then it says: "Since ε > 0 is arbitrary, we must have U(f) ≤ L(f)".
I don't have any trouble understanding the rest of the proof but I don't understand how we get from the inequality above to U(f) ≤ L(f). Can someone explain to me the reason why we can remove ε and get to U(f) ≤ L(f)?

I've added a picture of the full proof to make it easier to understand my question.

答:
If you have two (real) quantities $A$ and $B$ such that
$$
A\le B+\epsilon
$$
for every $\epsilon>0$, then you must have
$$
A\le B
$$
since otherwise ($A>B$), you can find some $\epsilon_0>0$ such that
$$
A>B+\epsilon_0
$$

(For instance, you can take $\epsilon_0=\frac{A-B}{2}$.)

In your example $A=U(f)$ and $B=L(f)$.


问:
Why do I need Boltzano-Cauchy in proving Laplace transform of a derivative?
I'm studying Laplace transform right now and I'm trying to prove the following theorem:

Let $k\in$ ℕ, $f(x)$ has $k$ continuous derivatives and each of them is locally integrable and has its own constant $c_n$ such that $f^{(n)}e^{-c_nx}$ is integrable over (0, ∞). Then
$$
(f^{(k)})(p) = p^k(f)(p)\,-\,\sum_{j=0}^{k-1}p^jf^{k-1-j}_+(0)
$$
for $\text{Re}\{p\}>\text{max}\{ c_n\}$

The proof is a straight forward induction, starting with $f'$, where via per partes integration you arrive at

Now it would seem logical to me, that the leftmost term must be zero, for with $\text{Re}\{p\}>\text{max} c_n$ it is integrabe and no integrable function can go to any nonzero value at ∞ (otherwise the integral would be infinite). But in my materials this isn't the end of the proof at all! The author goes on to show that
$$
\psi_p := e^{-pt}f(t)
$$
has for a fixed $p$ large enough integrable derivative and then invokes

to show Boltzano-Cauchy condition holds. Why do I need that? How is it not always fufilled for integrable functions?

答:
Turns out there can be a function, that's nonzero on a countably infinite union of sets that get smaller as $x \rightarrow$ ∞ similar to this:

This function surely doesn't have a limit zero at infinity, but as the $x$ gets larger, only smaller and smaller intervals contribute to the integral. It then turns into basically an infinite sum and you surely can make the function such that they satisfy the theorem's requirements and that their individual integrals behive like, for instance $\sum \frac{1}{n^2}$


问:
Proving that if $AB=A$ and $BA=B$, then both matrices are idempotent
Let $A, B$ be two matrices such that $AB=A$ and $BA=B$, how do I show that $A\cdot A=A$ and $B\cdot B=B$?

Steps I took:

Let $A= \left[\begin{array}{rr}
    a & b \\
    c & d \\
   \end{array}\right]$ and let $B= \left[\begin{array}{rr}
    w & x \\
    y & z \\
   \end{array}\right]$

$\left[\begin{array}{rr}
    a & b \\
    c & d \\
   \end{array}\right] \cdot \left[\begin{array}{rr}
    w & x \\
    y & z \\
   \end{array}\right] = \left[\begin{array}{rr}
    aw+by & ax+bz \\
    cw+dy & cx+dz \\
   \end{array}\right] $ (Which is also equal to A)

$\left[\begin{array}{rr}
    w & x \\
    y & z \\
   \end{array}\right] \cdot \left[\begin{array}{rr}
    a & b \\
    c & d \\
   \end{array}\right] = \left[\begin{array}{rr}
    wa+xc & wb+xd \\
    ya+zc & yb+zd \\
   \end{array}\right] $ (Which is also equal to B)

At this point I am stuck. I don't know how to proceed and I imagine that I started off on the wrong track to begin with. I'd like a hint to guide me in the right direction.

My proof after consultation with answerers below:

Proof:

1) Since $AB=A$, we can say that: $(AB)A=AA$ which is equal to $A^2$

2) Then, (by associativity of matrix multiplication), we can say that $A(BA) = AB$ (since $BA=B$)

3) Then, $AB=A$ (since $AB=A$ was given)

4) Therefore, $AA$ is equal to $A$

5) Since $BA=B$, we can also say that: $(BA)B=BB$ which is equal to $B^2$

6) Then, (by associativity of matrix multiplication) we can say that $B(AB)=BA$ (since $AB=A$)

7) Then, $BA=B$

8) Therefore, $BB$ is equal to $B$

9) Thus, $AA$ is equal to $A$ and $BB$ is equal to $B$

Q.E.D.

答:
Since $AB=A$ and $BA=B$, to show that $A$ is an idempotent we have to show that $A=A^2$
Therefore $A^2=(AB)^2 = A^2B^{ } = AB = A(BA) = AA = A^2 = A$. Proven


问:
Is there any close solution for $\operatorname{argmin}_D ||X − DZ||^2_F$
Is there any closed solution or algorithm for solving the following optimization problem based on $Z$, $X$ as an input?

$D \leftarrow \operatorname{argmin}\|X − DZ\|^2_F$ s.t $\|d_k\|_2^2 = 1, \forall k \in \{1,...K\}.$

where $d_k$ is kth column of $D$ and $F$ stands for Frobenius norm.

$Z \in \mathbb{R}^{K\times n}$, $X \in \mathbb{R}^{m\times n}$, and $D \in \mathbb{R}^{m\times K}$.

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}\def\D{\operatorname{Diag}}\def\S{\operatorname{Sym}}$Here is an unconventional approach to your problem.

For algebraic convenience, use a colon to denote the trace, i.e.
$$A:B = {\rm Tr}(A^TB)$$
This Frobenius product acquires several nice properties from the underlying trace function, e.g.
$$\eqalign{
A:A &= \big\|A\big\|_F^2 \\
A:B &= B:A = B^T:A^T \\
CA:B &= A:C^TB = C:BA^T \\
}$$
The objective function is then
$$\phi(D) = \big(DZ-X\big):\big(DZ-X\big)$$
Next, introduce an unconstrained variable $U$ and use it to construct the $D$ matrix such that the constraint is automatically satisfied

$$\eqalign{
W &=\D\big(U^TU\big)^{-1/2}\quad\implies\quad W^2=\D\big(U^TU\big)^{-1}\\
D &= UW \quad\implies\quad I=\D(D^TD)\iff\big\|d_k\big\|^2_F=1  \\
}$$
where the function $\D(X)$ retains the main diagonal of its argument but sets all other elements to zero.

It will be useful to have the differentials of these quantities for later use (note that diagonal matrices behave much like scalars under differentiation)
$$\eqalign{
dW &= -\tfrac 12\,\D\big(U^TU\big)^{-3/2}\D(U^TdU+dU^TU) \\&= -W^3\D\big(\S(U^TdU)\big) \\
dD &= dU\,W + U\,dW \\&= dU\,W-UW^3\D\big(\S(U^TdU)\big) \\
}$$
where the function $\;\S(X)\doteq\tfrac 12(X^T+X)$

Now we can calculate the gradient $\left(\p{\phi}{U}\right)$, set it to zero, and solve for the optimal matrix.
$$\eqalign{
d\phi &= (DZ-X):dD\,Z \\
 &= (DZ-X)Z^T:dD \\
 &= M:dD \\
 &= M:dU\,W - M:UW^3\D\big(\S(U^TdU)\big) \\
 &= MW:dU - \D(W^3U^TM):U^TdU \\
 &= \Big(M - UW^2\D(U^TM)\Big)W:dU \\
\p{\phi}{U} &= \Big(M - UW^2\D(U^TM)\Big)W \;\doteq\; 0 \\
M &= UW^2\D(U^TM) \\
U &= M\Big(W^2\D\big(U^TM\big)\Big)^{-1} \\
}$$
$\big[\,$The fact that $W$ is diagonal (and symmetric) was used to simplify several of the steps.$\,\big]$

While this isn't a closed form solution, it does provide the basis for an iterative solution.

After initializing $\,k=0,\;U_0=$ (a random matrix), perform the following sequence of calculations until convergence is achieved
$$\eqalign{
W_{k+1} &= \D(U^T_kU_k)^{-1/2} \\
M_{k+1} &= U_kW_{k+1}ZZ^T-XZ^T \\
U_{k+1} &= M_{k+1}\Big(W_{k+1}^2\D\big(U^T_kM_{k+1}\big)\Big)^{-1} \\
k &= k+1 \\
}$$
The computation is very inexpensive since it's easy to invert (and extract roots of) a diagonal matrix. The desired solution is then
$$D=\arg\min(\phi) = \lim_{k\to\infty}U_k W_k$$


问:
Show that a vector that is orthogonal to every other vector is the zero vector
I have the following question, and I'd like to get some tips on how to write the proof. I know why it is, but I'm still not so great at writing it mathematically.

If $u$ is a vector in $\mathbb{R}^n$ that is orthogonal to every vector in $\mathbb{R}^n$, then $u$ must be the zero vector. Why?

I'm starting off like this, but I don't know if it's the right way to do it, or if it is and I just don't know how to continue.

\begin{align}
(\exists u\in\mathbb{R}^n)(\forall v\in\mathbb{R}^n)[\text{u is orthogonal to v}]&\iff u\cdot v=0\\
&\iff ?
\end{align}

From here, instinctively I want to divide both sides by $v$, but I don't know if there is such a thing as dividing a dot product.

答:
Perhaps this is slightly off topic but it's worth explaining slightly more intuitively why you can't divide by a vector.

The key idea is to think about information (or more precisely numbers of constraints/equations). A dot product takes two vectors, both with $n$ numbers in them, and combines them in a particular way (related to projecting then onto each other) to give you one number. So $2n\to 1$.

Lots of information is clearly not being used in this function. There must be many different  inputs giving the same output so it's hard to invert the process.  But that's not unfamiliar, multiplication (or even addition) does something similar, taking $2\to 1$. Indeed $4\times 1=2\times 2$.

The idea of division normally works by saying "Okay, so suppose I know one of the inputs as well as the output. Can I figure out the other input?" The answer is yes. Schematically, you're now given two pieces of information at the end, so $2\to 2$ pieces of information. Then you have enough to figure out what you started with! (So long as you weren't just given a zero for the input, in the case of multiplication. It works perfectly for addition.)

But what about the dot product? If someone told you what one of the vectors was as well as the  dot product, that's $n$ extra pieces of information. Therefore you now get $2n\to n+1$. But unless $n=1$ (normal multiplication) you still don't have enough data! The problem is everything in the vector you seek which is orthogonal to the input vector you were told is thrown away. You have a gap of $2n-(n+1)=n-1$ in your data. You have improved by just $1$ on your original ignorance.

But if you get another $k$ input vectors, and are told both them and their dot products with your mystery vector, you get $(k+1)n\to k(n+1)$. Clearly when $n=k$ you get a balance! Therefore, trying out dot products with $n$ different fixed vectors ought to be enough information (typically) to reverse the process.

So what is this magical new division? Well, you take $n$ vectors of $n$ entries, dot them all with the unknown vector, and collect the $n$ results. If you check it out, this is exactly matrix multiplication by an $n\times n$ matrix formed by taking rows of $A$ to be the vectors chosen. That is, you are given $A u=d$ where $d$ is the vector of dot products. So the division is actually just inverting the matrix $A$ to get $u=A^{-1} d$. Magic! And just as multiplication fails to be invertible when you times by zero, this fails if $A$ is not invertible; which is equivalent (as you might know) to the test input vectors all being linearly independent.

Since in this problem you know that all vectors dotted with $u$ give zero, you can choose any linearly independent set of test vectors and you'll find $u=A^{-1} \pmatrix{0 \\ 0 \\\vdots \\ 0}$ which vanishes. Hence $u$ is the zero vector.


问:
Definition of a limit when there are no cones?
A very pedantic question: If a category $\mathcal{C}$ admits no cones, then does it contains all limits vacuously? Or does the definition of a limit imply the existence of at least one limit?

答:
Every nonempty category admits at least one cone (not to say universal cone, but just a cone): a cone for any functor $X:\ast \to \mathscr{C}$ which picks out a fixed object $X$ of $\mathscr{C}$ and the identity morphism on $X$. To find a category $\mathscr{C}$ that admits no cones, you necessitate that the category be not nonempty.

The only category that has no cones is $\emptyset$. This category admits exactly one functor (the identity functor, which is the empty functor), and this has no cones because there do not exist any objects in $\emptyset$. In particular, a limit to $id_{\emptyset}$ is an object in $\emptyset$ so $\emptyset$ cannot admit any limits.


问:
Show that the equation has no integer solutions
I am trying to show that the equation $7x^5-23x^2+2x-8=0$ has no integer solutions i.e $x$ belonging to $\mathbb{Z}$.

Now I am aware calculus can be used to show there is no real root except the rational between $1$ and $2$ but I'd like to use modular arithmetic to show the lack of integer solutions.

What I've tried so far is:

Assume $k$ belonging to $\mathbb{Z}$ is a root

$k(7k^4-23k+2)=8$ which shows that $8$ must divide either $x$ or $7x^4-23x+2$

Take $k=8m$ and insert

?

Now this might be totally the wrong approach.

Thank you in advance.

答:
$k(7k^4-23k+2)=8$

Start from here, you first would notice if $k$ is integer, then $7k^4-23k+2$ is also integer.

Therefore $k$ can be $1,2,4,8,-1,-2,-4,-8$. You should check each of these case and calculate $7k^4-23k+2$ and confirm $k(7k^4-23k+2)$ is not $8$.


问:
find the remainder for $\color{purple}{TETRATION}$ such that $( {^{2021}2021}(\mod13))$
It is given that $ {^{2021}2021}$ is a tetration and I want to find the remainder when it is divided by $13$.

Briefly , I am looking for $ {^{2021}2021}\pmod{13}$

My work :

$\color{purple}{Firstly:}$ I found $2021\equiv 6\pmod{13}$

$\color{blue}{Secondly:}$ I thought that $6^{\color{red}{a}}\equiv 6^{\color{red}{b}}\pmod{13}$ should be satisfied.

$\color{orange}{Thirdly:}$ $a\equiv b\pmod{12}$ by Euler's phi function.Then , i said that $a\equiv 11$ and $b\equiv -1$

$\color{red}{Fourthly:}$ I said that $6^{11^m}\equiv 6^{(-1)^m}\pmod{13}$ and there left $2020$ superscript for tetration. Therefore , I said that it must be equal to $6^{(-1)^{2020}}\equiv6\pmod{13}$

$\therefore$ Answer is $6$

Is my solution correct , if not , can you share any trick or shortcut for finding remainders when we encounter tetrations..

$\color{red}{NOTE:}$  WHAT IS TETRATION $\rightarrow $  https://en.wikipedia.org/wiki/Tetration

$\color{red}{NOTE-2:}$ I would rather colorful answer  for the sake of @terasalisbon :)

$\color{GREEN}{THANKS...}$

答:
The only thing you need to know is that looking at a power $a^b$ modulo $k$ you can reduce the base $a$ modulo $k$ and the exponent $b$ modulo $\varphi(k)$.

Hence, when you consider the a tetration
$$
{}^{2021}2021 = \color{purple}{2021}^{\color{blue}{2021}^{\color{orange}{2021}^{\cdots^{\color{red}{2021}}}}},
$$
you can reduce the base modulo $13$ the first exponent modulo $\varphi(13)=12$, the third exponent modulo $\varphi(12)=4$, where already $2021\equiv 1 \pmod 4$.

Hence
$$
{}^{2021}2021 \equiv \color{purple}6^{\color{blue}5^{\color{orange}1^{\cdots^{\color{red}{2021}}}}} \equiv 6^5 \equiv 7776 \equiv 2 \pmod{13}.
$$


问:
The topological conjugacy $h$ guaranteed by the Hartman-Grobman theorem is generally not a lipeomorphism
Yesterday I asked a question here Topological conjugacy between linear maps and https://math.stackexchange.com/users/349785/paul-frost answered my question very well ! Thank him. Now I want to ask a very interesting question, at first I prefer to write detail again :

Let $A_{\alpha}: \mathbb{R} \to \mathbb{R}$ denote the linear map

\begin{align}
A_{\alpha}(x)=\alpha x
\end{align}
We proved that if $ 0 < \alpha < 1$ and $ 0 < \beta < 1 $ then $A_{\alpha}$ and $A_{\beta}$ are topologically conjugate.

Notice that two homeomorphisms $f:X \to X$ and $g:X \to X $ are topologically conjugate to each other if there is a homeomorphism $h:X \to X $ such that $hf=gh$

Now the question is this :

What if $\alpha \neq \beta$? In the book of Lan Wen I studied and found out that if $\alpha \neq \beta$ then there is no lipeomorphism $h:\mathbb{R} \to \mathbb{R}$ such that $hA_{\alpha}=A_{\beta}h$. Could anyone help me show this?

Note that We call a homeomorphism $f: E \to E^{'}$ a lipeomorphism if both $f$ and $f^{-1}$ is lipschitz.

答:
Assume that $hA_{\alpha}=A_{\beta}h$ with
$$
c|x-y|\le|h(x)-h(y)|\le d|x-y|.
$$
Then $hA_{\alpha}^n=A_{\beta}^nh$ for all integers $n$ and so
$$
c\alpha^n|x-y|\le|h(A_{\alpha}^nx)-h(A_{\alpha}^ny)|=|A_{\beta}^nh(x)-A_{\beta}^nh(y)|\le d\beta^n |x-y|.
$$
In particular, we have
$$
c\alpha^n|x-y|\le d\beta^n |x-y|.
$$
Now you just need to let $n\to\pm\infty$ to get a contradition.


问:
Chaos in finite field
Let's think about some finite field $\mathbb{F}$. Is it possible to construct a map

$x[n+1] = \mathcal{P}(x[n], x[n-1],...,x[n-k]), \ \ \  \forall x\in\mathbb{F} $

where $\mathcal{P}$ - polynomial with coeffs $\in\mathbb{F}$, so it's behavior  be non-periodic, but chaotic,

so x[n] be "jumping" randomly over $\mathbb{F}$?

(As $x[n+1] =1-2\cdot x[n]^2$ on [-1,1])

答:
Let $f:F^k \to F$ be a function on any finite set $F$, then the sequence defined recursively by $x_{n}=f(x_{n-1},x_{n-2},...,x_{n-k})$ can only take a finite number of values. So it must be periodic, maybe only after a certain rank.

But your question is not trivial, though. Assuming the field $F$ to be infinite but of finite characteristic, for example the p-adic numbers, can you define chaotic in such a topology?

Title: Chaos in finite field


答:
use modular arithmetic to show the lack of integer solutions:

$$7x^5-23x^2+2x-8\equiv x+x^2+2x+1\equiv x^2+1\not\equiv0\pmod3$$

Title: Show that the equation has no integer solutions

Body:
I am trying to show that the equation $7x^5-23x^2+2x-8=0$ has no integer solutions i.e $x$ belonging to $\mathbb{Z}$.

Now I am aware calculus can be used to show there is no real root except the rational between $1$ and $2$ but I'd like to use modular arithmetic to show the lack of integer solutions.

What I've tried so far is:

Assume $k$ belonging to $\mathbb{Z}$ is a root

$k(7k^4-23k+2)=8$ which shows that $8$ must divide either $x$ or $7x^4-23x+2$

Take $k=8m$ and insert

?

Now this might be totally the wrong approach.

Thank you in advance.


答:
If I am not mistaken, a (mod 3) argument will work here.

$7x^5 + 2x \equiv 0 \pmod{3},~$ precisely because $x^5 \equiv x\pmod{3}.$

Then the problem reduces to noticing that $-23x^2 \equiv x^2 \pmod{3}$, which can't be congruent to $2 \pmod{3}.$


问:
A probability question regarding combinatorics
From a class of $300$ students, three are selected at random to receive three identical prizes. Of the students, $200$ are from department $A$, $60$ from department $B$ and $40$ from department $C$.

Find the probability that the three winners come from different departments.

Find the probability that all three winners come from the same department.

Suppose that all three are from the same department. Compute the probability that they are all from department $A$.

Here is my attempt. I am not sure if this is correct.

$P(\text{different departments})=3!\binom{200}{1}\binom{60}{1}\binom{40}{1}$ or $3!\frac{200}{300}\times \frac{60}{299} \times\frac{40}{298}$?

$P(\text{same department})=\dfrac{\binom{200}{1}+\binom{60}{1}+\binom{40}{1}}{\binom{300}{3}}= \dfrac{3}{44551}$.

$\begin{align} P(\text{from department A}|\text{same department}) & = \dfrac{P(\text{from department A}) P(\text{same department})}{P(\text{same department})} \\
& = \dfrac{\binom{200}{1} \frac{3}{4451}}{\frac{3}{4451}} \\
&=\binom{200}{1}. \end{align}$

Thank you in advance.

答:
Note: I am interpreting the problem to mean that the three winners must be different people.

Unfortunately all three proposed solutions are incorrect.

There are ${300 \choose 3}$ subsets of size 3.  This will be the denominator for the first two questions.  In the first question, the numerator is ${200\choose 1}{60\choose 1}{40 \choose 1}$.  We want one of our three winners to be from department $A$, one from $B$, and one from $C$; these three choices are made independently, so we use the multiplication principle.

The numerator of the second question is ${200\choose 3}+{60\choose 3}+{40\choose 3}$.  There are three types of subsets, those entirely from $A$, those entirely from $B$, and those entirely from $C$.  These are disjoint subsets of ${300\choose 3}$, so we use the addition principle.  To count the subsets entirely from $A$, we need to pick 3 winners.

The denominator in the third question is the same as the numerator of the second question, since now we are restricting to only those triplets that are from a single department, which the numerator from the second question was counting.  The numerator of the third question is ${200\choose 3}$.


问:
$E[g(X,Y)|Y=y] = E[g(X,y)|Y=y]$ - Exercise $3.77$ Ross Probability Models.
How would you show that $E[g(X,Y)|Y=y] = E[g(X,y)|Y=y]$ in the continuous case? This problem is in Probability Models by Ross. No solution is provided, and no information is given about the random variables, altough I suppose we could assume the functions involved are absolute convergent or something.

The penultimate step in the discrete case is that all $p(X=i, Y=j|Y=y)$ are $0$ unless $j = y$, which I can't translate into the continuous case. My real analysis skills are non-existent, but you aren't supposed to know that for this book either way.

答:
Let $\mu_{X,Y}$ be the joint law of $(X,Y)$.  This is the Borel probability measure in $\mathbb{R}^{d} \times \mathbb{R}^{d}$ (assuming $X$ and $Y$ are both random vectors taking values in $\mathbb{R}^{d}$) such that, for any two Borel sets $A, B \subseteq \mathbb{R}^{d}$,
\begin{equation*}
\mu_{X,Y}(A \times B) = \mathbb{P}\{X \in A, \, \, Y \in B\}.
\end{equation*}
Similarly, let $\mu_{Y}$ denote the law of $Y$.

It is possible to show that there is a weakly $\mu_{Y}$-measurable, probability measure-valued function $y \mapsto \mu_{y}^{X}$ such that if $f$ is a bounded, continuous function on $\mathbb{R}^{d} \times \mathbb{R}^{d}$, then
\begin{equation*}
\int_{\mathbb{R}^{d} \times \mathbb{R}^{d}} f(x,y) \, \mu_{X,Y}(dx \otimes dy) = \int_{\mathbb{R}^{d}} \left[ \int_{\mathbb{R}^{d}} f(x,y) \, \mu_{y}^{X}(dx) \right] \, \mu_{Y}(dy).
\end{equation*}
I have seen the measures $\{\mu^{X}_{y}\}_{y \in \mathbb{R}^{d}}$ referred to as a regular family of conditional probabilities --- the term disintegration is also used to describe this process of decomposing $\mu$.

One now chooses a suitable $\mu_{Y}$-measurable set $E \subseteq \mathbb{R}^{d}$ with $\mu_{Y}(E) = 1$ and defines $\mathbb{E}(\cdot \, \mid \, Y = y)$ for $y \in E$ and bounded, continuous $f : \mathbb{R}^{d} \times \mathbb{R}^{d} \to \mathbb{R}$ by the formula
\begin{equation*}
\mathbb{E}(f(X,Y) \, \mid \, Y = y) = \int_{\mathbb{R}^{d}} f(x,y) \, \mu^{X}_{y}(dx).
\end{equation*}
With this formula in hand, it is now clear that $\mathbb{E}(g(X,Y) \, \mid \, Y = y) = \mathbb{E}(g(X,y) \, \mid \, Y = y)$ for $\mu_{Y}$-a.e. $y \in \mathbb{R}^{d}$ (in particular, for $y \in E$).


答:
Tetration is a repeated raising to a given power, a "tower" of exponents. Effectively this is worked right to left (or "top down" looking at the tower analogy).

So this is a massively tall tower of exponents, but oddly enough that doesn't matter; we'll only be interested in the bottom few elements to find the answer.

You start well to find that $2021\equiv 6 \bmod 13$.

Then from Fermat's little theorem we know that $2021^{12}\equiv 1 \bmod 13$ so $2021^a$ will cycle round values on a pattern of $12$ length (at most).

Checking that, we see $(6^i){\large|_{i=1}^{12}}\equiv (6,10,8,9,2,12,7,3,5,4,11,1) \bmod 13$ in that order.

So now stepping up the tower of exponents, consider $2021^a$ where $a=2021^b$. We know that we only need to consider the values of $a\bmod 12$ from the above cycle, so that implies that we need to look at $a = 2021^b = 5^b\bmod 12$. So how does $5$ cycle under an exponent $\bmod 12$? It's easy to find by calucaltion that $5^2 = 25\equiv 1\bmod 12$, so the cycle now is only length two - effectively, we only need to know whether $b$ is odd or even.

In the tower of exponents though we know that $b = 2021^c$, which is odd. So we have our result:

$${}^{2021}2021 \equiv 6^{5^{\text{odd}}} \equiv 6^5\equiv 2 \bmod 13$$

Title: find the remainder for $\color{purple}{TETRATION}$ such that $( {^{2021}2021}(\mod13))$

Body:
It is given that $ {^{2021}2021}$ is a tetration and I want to find the remainder when it is divided by $13$.

Briefly , I am looking for $ {^{2021}2021}\pmod{13}$

My work :

$\color{purple}{Firstly:}$ I found $2021\equiv 6\pmod{13}$

$\color{blue}{Secondly:}$ I thought that $6^{\color{red}{a}}\equiv 6^{\color{red}{b}}\pmod{13}$ should be satisfied.

$\color{orange}{Thirdly:}$ $a\equiv b\pmod{12}$ by Euler's phi function.Then , i said that $a\equiv 11$ and $b\equiv -1$

$\color{red}{Fourthly:}$ I said that $6^{11^m}\equiv 6^{(-1)^m}\pmod{13}$ and there left $2020$ superscript for tetration. Therefore , I said that it must be equal to $6^{(-1)^{2020}}\equiv6\pmod{13}$

$\therefore$ Answer is $6$

Is my solution correct , if not , can you share any trick or shortcut for finding remainders when we encounter tetrations..

$\color{red}{NOTE:}$  WHAT IS TETRATION $\rightarrow $  https://en.wikipedia.org/wiki/Tetration

$\color{red}{NOTE-2:}$ I would rather colorful answer  for the sake of @terasalisbon :)

$\color{GREEN}{THANKS...}$


问:
Understanding the solution to the differential equation $\dot{x} = x^2\cos{t^2}$
I am looking at the Haberman book on ODEs, chapter 1.3 problem 11. The problem is to find the solution of the separable first order ODE.

$$
\frac{dx}{dt} = x^2\cos{t^2}
$$

Now the posted solution uses the definite integral, where $\bar{x}, \bar{t}$ are the dummy variables of integration.

$$
\int_{1}^{x} \bar{x}^{-2} d \bar{x}=\int_{0}^{t} \cos \left(\bar{t}^{2}\right) d \bar{t}
$$

My question was how to make the argument about the integration bounds on the integral on the LHS, $\int_{1}^{x} \bar{x}^{-2} d \bar{x}$. I understand that as $x \rightarrow 0$, the value of $\frac{1}{x^2}$ will explode. But it still seems a bit odd that I did an operation on the left hand side that is asymmetric with the operation on the right hand side. I also understand that $cos(\bar{t})$ can only vary between 1 and -1, so that would also limit the range of the left hand side integral as well. Can anyone provide a more rigorous justification for the change in the bounds of the integral or the asymmetry in the bounds of the integration. Thanks.

答:
For this Problem Haberman gives some initial condition : $x(0)=1$. So that we have:
$$\frac{dx}{dt} = x^2\cos{t^2}$$
$$\frac{dx}{x^2} = \cos{t^2} \, dt$$
$$\int_{x(0)}^{x(t)}\frac{d\bar x}{\bar x^2} =\int_0^t \cos{\bar t^2} \, d\bar t$$
$$\int_{1}^{x(t)}\frac{d \bar x}{\bar x^2} =\int_0^t \cos{\bar t^2} \, d\bar t$$


问:
Triangle inequality infinite terms $|a-b|\leq \sum_{i=1}^\infty|x_{i}-x_{i-1}|$
While solving a problem, I thought of kind of "Triangle inequality infinite terms". Clearly for finite terms if $a=x_0$ and $x_n=b$ we have $$|a-b|\leq |a-x_1+x_1-x_2+\cdots+x_{n-1}-b|\leq \sum_{i=1}^n|x_{i-1}-x_{i}|  $$ It seems like, under the conditions  $\lim_{n\to \infty}x_n=b$ and $x_0=a$, this should be extended to $$|a-b|\leq \sum_{i=1}^\infty|x_{i-1}-x_{i}|$$

Is it right? Is not a surprising idea, but I think I could not find this usually on textbooks.

答:
Your result is correct. You can formalize the proof as follows: For every $n \in \Bbb N$ is
$$
 |a-b| = |(a-x_1)+ (x_1-x_2) + \cdots + (x_{n-1}-x_n) + (x_n - b)| \\
\le \left(\sum_{i=1}^n |x_{i-1}- x_i|\right) + |x_n - b| \, ,
$$
using the triangle inequality for finitely many terms.

If the series $\sum_{i=1}^\infty |x_{i-1}- x_i|$ converges then we can take the limit $n \to \infty$ and conclude that
$$
|a-b| \le \sum_{i=1}^\infty |x_{i-1}- x_i|
$$
since  $\lim_{n\to \infty}x_n=b$.

Otherwise $\sum_{i=1}^\infty |x_{i-1}- x_i|$ diverges to $+\infty$ and the same estimate holds (trivially) in the set of extended real numbers: $|a-b| \le +\infty$.


问:
Discrete random variable with zero variance
I was reading this

Why is the expected value $E(X^2) \neq E(X)^2$?

and this

https://stats.stackexchange.com/questions/228624/random-variable-with-zero-variance

and this led me to the following question.

If a discrete r.v. has $0$ variance does it mean that this r.v. takes on only a single value $\mu$ (its own mean)? I think that should be true but I cannot quite convince myself right now.

The opposite of this statement is obviously true.

But is it true in this direction too?

How can we prove this?

I think should be a trivial proof but I don't quite see it right now.

答:
Yes. Use the definition

$$0=\operatorname{Var}X=\mathbb E[(X-\mu)^2].$$
But $Y=(X-\mu)^2\geq0$. So if there is some $\omega$ with $\mathbb P(\omega)>0$ and $Y(\omega)>0$, then $$\mathbb E[Y]=\sum_{\omega\in\Omega}Y(\omega)\cdot\mathbb P(\omega)>0,$$ a contradiction. So we must have $\mathbb P(Y=0)=1$ i.e. $\mathbb P(X=\mu)=1$.


问:
Linear transformation and orthonormal basis
A transformation $T: V\to W$ transform m-dimensional space $V$ to n-dimensional space $W$ is given by $n\times m$ matrix $M$, such that we map vector $x$ from $V$ to $W$ as follows:

$\mathbf{x}^i = \mathbf{M} \mathbf{x}$

My question is that what is $M$? is it the orthonormal basis of $W$.

答: $M$ is the matrix corresponding to the linear transformation $T$, expressed in some (possibly orthonormal) bases for $V$ and $W$.


问:
$\mathbb{C}$ as the Lie algebra of $\mathbb{C}$
It is known that $\mathbb{C}$ is a linear algebraic group as the set of matrices $\begin{pmatrix} 0 & a \\ 0 & 0 \end{pmatrix}$ with $a \in \mathbb{C}$. I read in some notes that $Lie(\mathbb{C})$, the Lie algebra associated to $\mathbb{C}$, is isomorphic to $\mathbb{C}$, but I cannot see why. I only know that if $G$ is a linear algebraic group, $Lie(G)$ is the set of all $\phi : O(G) \to O(G)$ (where $O(G)$ is the set of regular functions on $G$) that satisfy Leibniz rule and such that $g \cdot \phi(f) = \phi(g \cdot f), \forall g \in G, \forall f \in O(G)$. It looks hard to use this definition, so I was thinking of using the fact that $Lie(\mathbb{C})$ is isomorphic to the tangent space, but I am not really sure how. Any hints?

答:
[In your setting both the group $G$ and the underlying field $k$ "are" the complex numbers (once, its additive group viewed as algebraic group, once as field), which can be confusing. So in the following I will use the notations $k$ and $G$. If you insert "$\mathbb C$" for both, see what a mess this would become!]

If you want to use the language in your question, you can also proceed by showing:

$O(G)$ is isomorphic to the polynomial ring $k[T]$ where further

the element $a \in G$ acts on a polynomial $p(T)$ via $p(T) \mapsto p(T+a)$

Then an element of the Lie algebra as defined in your post is a $k$-linear map $\phi: k[T] \rightarrow k[T]$ such that $\phi(f\cdot g) = \phi(f) \cdot g + f \cdot \phi(g)$. Show that this condition entails that $\phi$ is completely determined by $\phi(T)$, and that $\phi(c)=0$ for all constants $c$. Then, the compatibility with the $G$-action says that, if $\phi(T) = \sum c_iT^i$, we have

$$ \phi(T+a)= \sum c_i (T+a)^i$$

for all $a \in G$, but by $\phi$ being linear and our already established $\phi(a)=0$, the left hand side is just $\phi(T) = \sum c_i T^i$. Show that this is impossible unless $c_i = 0$ for all $i \ge 1$. Consequently $\phi(T) = c_0$ is a constant, and you can now show that $\phi$ must have been the differential operator $c_0 \cdot \dfrac{d}{dT}$ all along.

So $Lie(G)$, as a $k$-vector space, is the set $\{c_0 \cdot \dfrac{d}{dT} : c_0 \in k\}$ which obviously is of $k$-dimension $1$. You can now either use the fact that in any $1$-dimensional Lie algebra every Lie bracket is $=0$, or use whatever definition of the Lie bracket is given in your source to see that it is identically $0$ here. Probably all you need for that is that $G$ is abelian.

Worthwhile: Look at the two alternative definitions of the Lie algebra of an algebraic group in chapter 12b of Milne's online book draft, and see how they identify with our result above. (Most directly you should be able to identify, in 12.5, our $\phi$ with  Milne's $D$ once you see that in our setting, $I/I^2 = T \cdot k[T]/ T^2 \cdot k[T]$.) Note also that then your example is a special case of both 12.6. and 12.7.


问:
Non-periodic paths of length 4 in a graph and the Lovasz Local Lemma
Let $G$ be a graph with maximum degree $\Delta$ and consider colouring its edges, each with one of $k$ colours. A path $v_0v_1v_2v_3v_4$ in a $G$ is $\textit{periodically coloured}$ if $v_0v_1$ and $v_2v_3$ have the same colour and $v_1v_2$ and $v_3v_4$ have the same colour.

a) Prove that $G$ has a colouring with no periodically coloured paths (of length $4$) and at most $\lceil 5\Delta^{3/2} \rceil$ colours.

b) Give a sequence of graphs $G_n$ to show that even if $\Delta$ is fixed and $k > 100\Delta^{3/2}$, the probability that a random $k$-edge-colouring of $G_n$ has no periodically coloured path of length $4$ can be arbitrarily small.

In a) I can give a proof but with $5$ replaced by a larger constant. Consider applying Lovasz Local Lemma (in its symmetric form - see e.g. Theorem 2.3 in https://theory.stanford.edu/~jvondrak/MATH233A-2018/Math233-lec02.pdf) on the events $A_i =$ 'the $i$-th path is periodically coloured'. Clearly $\mathbb{P}(A_i) = \frac{k^2}{k^4} = \frac{1}{k^2}$ and, for any $i$, $d+1 \leq $ 'number of paths of length $4$ which have at least one edge in common with the $i$-th path of length $4$'. (In this number we include the path of $A_i$ itself and that's why we can have $+1$ on the left-hand side.)

Now here's how I estimate the latter number. If the first path is $v_0v_1v_2v_3v_4$, then for the second path there are $4$ choices for which edge of the first path to be common and $4$ choices for what its position in the second path be. After that, there are at most $\Delta^3$ choices for the other three edges from the second path.

So the Local Lemma shall be applicable if $\frac{e}{k^2}16\Delta^3 \leq 1$, but this requires $k\geq \lceil 4e^{1/2}\Delta^{3/2} \rceil$ and unfortunately, $4e^{1/2} \sim 6.59 > 5$. So is there a way to improve the above counting in order to take care of this multiplicative constant issue? Or I somehow need to use the general form of the Local Lemma (but with what real numbers)?

Update: Thanks to Misha Lavrov for the simple example for b).

Any help appreciated!

答: Actually "4  choices for what its position in the second path be'' could be replaced by 2, since with 4 we count each path twice. For example, if $w$ is the common edge, then in $e_1e_2we_3$ it is in third position, but in $e_3we_2e_1$ it is in second position - however, these two are the same path! With this replacement the required multiplicative constant works out as wanted.


问:
Prove that if $f \circ T \in E^\star$ for any $ f \in F^\star$, then $T$ is linear and continuous.
i'm stucked for a while in this problem:

Let $E, F$ be Banach spaces and suppose $T: E \longrightarrow F$ is an operator such that for
any $f \in \mathrm{F}^{\star},$ we have that $f \circ T \in E^{\star} .$ Prove that $T$ is linear and continuous.

I have some tips to prove this but, i didn't figured why T is linear. It seems to be simple to solve, but i feel that i'm in the wrong path.

答: Hint: if $x,y\in E$ and $\lambda,\mu\in\mathbb R$, then $f(T(\lambda x+\mu y)-\lambda T(x)-\mu T(y))=0$ for all $f\in F^*$.


问:
what value of K does the system have a unique solution
$\begin{cases}x_1 + kx_2 - x_3 = 2\\2x_1 - x_2 + kx_3 = 5\\x_1 + 10x_2 -6x_3= 1\\
\end{cases}$

I've been trying echelon form where i took $R_2 = R_2 - 2R_1$ and $R_3 = R_3-R_1$

So I have $\left[\begin{array}{ccc|c}1&K&-1&2\\2&-2&K&5\\1&10&-6&1\end{array}\right]$

I've been trying echelon form where i took $R_2 = R_2 - 2R_1$ and $R_3 = R_3-R_1$
and reduced it

So I have $\left[\begin{array}{ccc|c}1&K&-1&2\\0&-1-2K&K+2&1\\0&10-K&-5&-1\end{array}\right]$

But now I am not sure how i could remove $10-K$ with $-1-2K$ any help would be appreciated

答:
Let's solve this system:

$x_1+kx_2-x_3=2 \implies 2x_1+2kx_2-2x_3=4$

$2x_1-x_2+kx_3=5$

$x_1+10x_2-6x_3=1 \implies 2x_1+20x_2-12x_3=2$

Subtracting each pair of equations we have:

$(k+2)x_3-(2k+1)x_2=1$

$(k-10)x_2+5x_3=1$

$-21x_2+(k+12)x_3=3$

Finally :

$(3k+6)x_3-(6k+3)x_2=(3k-30)x_2+15x_3=(k+12)x_3-21x_2=3$

Which is easy to solve since from the two first ones:

$(3k+6)x_3-(6k+3)x_2=(3k-30)x_2+15x_3 \implies (3k-9)x_3=(9k-27)x_2 \implies x_3=3x_2$ (unless $k=3$)

By the second and the third ones:

$(3k-30)x_2+15x_3=(k+12)x_3-21x_2 \implies (3k-9)x_2=(k-3)x_3 \implies x_3=3x_2$ (again unless $k=3$)

And by the first and the last ones:

$(3k+6)x_3-(6k+3)x_2=(k+12)x_3-21x_2 \implies (2k-6)x_3=(6k-18)x_2 \implies x_3=3x_2$

So all pairs $(x,3x)$ will satisfy these equations and hence there are infinitely many answers.  But then there is only a unique $x$ such that these three creatures are exactly equal to $3$ so there is a unique answer. (to be more exact, you shall put $(x_2,x_3)=(\frac{1}{k+5},\frac{3}{k+5})$
The case $k=3$ can also be easily checked. (hint: it does not satisfy)


问:
One square per person
There are n persons.

Each person draws k interior-disjoint squares.

I want to give each person a single square out of his chosen k, so that the n squares I give are interior-disjoint.

What is the minimum k (as a function of n) for which I can do this?

NOTES:

For n=1, obviously k=1.

For n=2, obviously k must be more than 2, since with 2 squares per person, it is easy to think of situations where both squares of person 1 intersect both squares or person 2. It seems that k=3 is enough, but I couldn't prove this formally.

If we don't limit ourselves to squares, but allow general rectangles, then even for n=2, no k will be large enough, as it is possible that every rectangle of player 1 intersects every other rectangle of player 2. So, the sqauare limitation is important.

EDIT: The problem has two versions: in one version, the squares are all axis-aligned. In the second version, the squares may be rotated. Solutions to any of these versions are welcome.

EDIT: Here is a possibly useful claim, relevant for the axis-aligned version:

Claim 1: If two axis-aligned squares, A and B, intersect, then one of the following 3 options hold:

At least 2 corners of A are covered by B, and B is as large or larger than A;

One corner of A is covered by B, and one corner of B is covered by A,

At least 2 corners of B are covered by A, and A is as large or larger than B.

Thus, if A intersects B, then, out of the 8 corners of A and B, at most 6 corners remain uncovered.

答: Here is my counter-example to the case of $n=2$ and $k=3$. Provided of course that the squares need not be the same size. You will have to trust me that all shapes are squares. However, we can see that each square overlaps all 3 of the squares of the other color.

Title: One square per person


答:
You can also just do it as an indefinite integral and the integration constant will be determined by the initial/boundary conditions:
$$\int\frac{dx}{x^2}=\int \cos t^2dt$$
$$\frac1x=-\int\cos t^2dt+C_1$$
another way of thinking about it would be that you want to integrate both sides over the same region of $t$, so you could do:
$$\int_{x(t_0)}^{x(t)}\frac{d\bar{x}}{\bar{x}^2}=\int_{t_0}^t\cos \bar{t}^2d\bar{t}$$

Title: Understanding the solution to the differential equation $\dot{x} = x^2\cos{t^2}$

Body:
I am looking at the Haberman book on ODEs, chapter 1.3 problem 11. The problem is to find the solution of the separable first order ODE.

$$
\frac{dx}{dt} = x^2\cos{t^2}
$$

Now the posted solution uses the definite integral, where $\bar{x}, \bar{t}$ are the dummy variables of integration.

$$
\int_{1}^{x} \bar{x}^{-2} d \bar{x}=\int_{0}^{t} \cos \left(\bar{t}^{2}\right) d \bar{t}
$$

My question was how to make the argument about the integration bounds on the integral on the LHS, $\int_{1}^{x} \bar{x}^{-2} d \bar{x}$. I understand that as $x \rightarrow 0$, the value of $\frac{1}{x^2}$ will explode. But it still seems a bit odd that I did an operation on the left hand side that is asymmetric with the operation on the right hand side. I also understand that $cos(\bar{t})$ can only vary between 1 and -1, so that would also limit the range of the left hand side integral as well. Can anyone provide a more rigorous justification for the change in the bounds of the integral or the asymmetry in the bounds of the integration. Thanks.


问:
If ${\lvert{E(G)}\rvert}>{\lvert{V(G)}\rvert}+3$, graph $G$ has at least two disjoint cycles.
Assume that $G$ is a graph which ${\lvert{E(G)}\rvert}>{\lvert{V(G)}\rvert}+3$. Prove $G$ has two disjoint cycles.

Answer:

I tried to prove it when $G$ is connected. consider the smallest cycle of length $k$. It exists because ${\lvert{E(G)}\rvert}>{\lvert{V(G)}\rvert}-1$.
Now, I divided vertices into $2$ groups. Group $A$ consists of $k$ vertices in the smallest cycle and group $B$ consists of the other $n-k$ vertices.

Now, between  $n-k$ vertices are exactly $n-k-1$ edges. Because, If it is more then we have another cycle (the proof is done) and if it is less then $G$ is disconnected (contradiction). So there are at least $5$ edges between group $A$ and $B$.
Now, what should I do? should I write every possibility to prove it? Are there any easy ways to continue the proof?

What about if $G$ is disconnected? I think we should do it in each component. Am I right?

Thank you so much for your answers.

答:
Suppose the result is false; take a minimal counterexample $G$ with $n$ vertices and exactly $n+4$ edges (if there were more edges, we could delete some).

Then $G$ cannot contain a $3$-cycle or $4$-cycle. Otherwise, removing the edges of such a cycle, we have a graph with $n$ vertices and at least $n$ edges, which must still contain another cycle.

Also, $G$ has minimum degree $3$. If $G$ had a vertex $v$ of degree $1$, we could delete it and the edge out of it, getting a smaller counterexample (since the leaf was not part of any cycle). If $G$ had a vertex $v$ of degree $2$, with neighbors $w_1, w_2$, we could replace $v$ and its two edges with the single edge $w_1w_2$. This also gives a smaller counterexample, since any cycle containing $v$ would have had to use both edges, and now it can use the edge $w_1w_2$ instead. Also, because $G$ had no $3$-cycle, this still gives a simple graph.

Since $G$ has minimum degree $3$ and $n$ vertices, it must have at least $\frac 32n$ edges, but we know it has exactly $n+4$. Therefore $n+4 \ge \frac32n$, or $n \le 8$.

However, if we take a vertex $v$, we know it must have at least $3$ neighbors $w_1, w_2, w_3$. These have no edges between them (because $G$ has no $3$-cycles) and each has $2$ more neighbors (by minimum degree $3$) which are all distinct (because $G$ has no $4$-cycles), giving us at least $10$ vertices already. This is a contradiction, so the result must be true.


问:
Is Bayes' Theorem really that interesting?
I have trouble understanding the massive importance that is afforded to Bayes' theorem in undergraduate courses in probability and popular science.

From the purely mathematical point of view, I think it would be uncontroversial to say that Bayes' theorem does not amount to a particularly sophisticated result. Indeed, the relation
$$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(B\cap A)P(A)}{P(B)P(A)}=\frac{P(B|A)P(A)}{P(B)}$$
is a one line proof that follows from expanding both sides directly from the definition of conditional probability. Thus, I expect that what people find interesting about Bayes' theorem has to do with its practical applications or implications. However, even in those cases I find the typical examples being used as a justification of this to be a bit artificial.

To illustrate this, the classical application of Bayes' theorem usually goes something like this: Suppose that

1% of women have breast cancer;

80% of mammograms are positive when breast cancer is present; and

10% of mammograms are positive when breast cancer is not present.

If a woman has a positive mammogram, then what is the probability that she has breast cancer?

I understand that Bayes' theorem allows to compute the desired probability with the given information, and that this probability is counterintuitively low. However, I can't help but feel that the premise of this question is wholly artificial. The only reason why we need to use Bayes' theorem here is that the full information with which the other probabilities (i.e., 1% have cancer, 80% true positive, etc.) have been computed is not provided to us. If we have access to the sample data with which these probabilities were computed, then we can directly find
$$P(\text{cancer}|\text{positive test})=\frac{\text{number of women with cancer and positive test}}{\text{number of women with positive test}}.$$
In mathematical terms, if you know how to compute $P(B|A)$, $P(A)$, and $P(B)$, then this means that you know how to compute $P(A\cap B)$ and $P(B)$, in which case you already have your answer.

From the above arguments, it seems to me that Bayes' theorem is essentially only useful for the following reasons:

In an adversarial context, i.e., someone who has access to the data only tells you about $P(B|A)$ when $P(A|B)$ is actually the quantity that is relevant to your interests, hoping that you will get confused and will not notice.

An opportunity to dispel the confusion between $P(A|B)$ and $P(B|A)$ with concrete examples, and to explain that these are very different when the ratio between $P(A)$ and $P(B)$ deviates significantly from one.

Am I missing something big about the usefulness of Bayes' theorem? In light of point 2., especially, I don't understand why Bayes' theorem stands out so much compared to, say, the Borel-Kolmogorov paradox, or the "paradox" that $P[X=x]=0$ when $X$ is a continuous random variable, etc.

答:
You are correct that Bayes' theorem follows trivially from axioms of probability that everyone accepts. The difference between Bayesans and frequentists is a cultural one. The actual mathematical axioms they subscribe to are trivially homologous.

The cultural divide is a pretty stark one though.

Frequentists tend to think computation is a dirty word and they dont care to
analyse problems that they cannot approach analytically, so basically they would prefer to think that everything is a gaussian. Also some of them tend to do this
funny numerology thing where they fetishise numbers like 0.01 and
0.05

Bayesians think that if they write down a uniform prior as a formula
it looks more like real mathematics and less like a stupid assumption
that rarely applies (appeals to 'entropy' make them feel great too); and they delude themselves into thinking that
labelling part of their likelihood function a prior makes them
special; as if frequentists couldn't multiply different likelihood
functions together to get a joint one just fine.

Actual examples where a non-strawman version of either approach to the same problem yields a different result, do not actually exist. Because there are not actually any differences in the fundamental axioms they subscribe to. That being said it is not as if the language, computational tools, and modelling approaches you use are unimportant to guiding your thought process. Itd be better if teaching methods focussed more on said homology though.


问:
Evaluate $\int_{0}^{1} \frac{\ln(1 + x + x^2 + \ldots + x^n)}{x}\mathrm d x$

How to evaluate:
$$\int_{0}^{1} \frac{\ln(1 + x + x^2 + \ldots + x^n)}{x}\mathrm d x$$

Attempt:

$$\int_{0}^{1}\frac{\ln(1 + x + x^2 + \ldots + x^n)}{x} \mathrm dx
= \int_{0}^{1}\frac{\ln(1 -x^{n+1}) - \ln(1 - x)}{x}\mathrm d x$$

Any hints would be appreciated.

Edit: Testing it with different values of $n$, it seems like the integral evaluates to be $\frac{n \pi^2}{6(n+1)}$

Body:
I derive under the integral
$$f(n)=\int_{0}^{1}\frac{\ln(1 -x^{n+1}) - \ln(1 - x)}{x}\,dx$$
$$f'(n)=\int_{0}^{1}\frac{\partial}{\partial n}\frac{\ln(1 -x^{n+1}) - \ln(1 - x)}{x}\,dx=$$
$$=\int_0^1\frac{x^n \log x}{x^{n+1}-1}\,dx=\frac{\pi ^2}{6 (n+1)^2}$$
$$f(n)=\int\frac{\pi ^2}{6 (n+1)^2}\,dn=C-\frac{\pi ^2}{6 (n+1)}$$
As $$f(1)=\int_0^1 \frac{\log (x)}{x-1} \, dx=\frac{\pi ^2}{6}$$
then $C=\frac{\pi ^2}{6}$ and finally we have
$$f(n)=\frac{\pi ^2}{6}-\frac{\pi ^2}{6 (n+1)}=\frac{\pi ^2 n}{6 (n+1)}$$


问:
Identity involving matrices of binomial coefficients
Let $P = (p_{i,j})$ and  $Q = (q_{i,j})$ be the $n \times n$ lower triangular matrices defined by

$$p_{i,j} =
\begin{cases}
\dbinom{i}{j-1} & \text{for}\ 1\le j\le i \le n\\
\hfill 0 \hfill & \text{elsewhere}
\end{cases}$$

and

$$q_{i,j} =
\begin{cases}
\dbinom{i}{j}   & \text{for}\ 1 \le j \le i \le n\\
\hfill 0 \hfill & \text{elsewhere}
\end{cases}$$

For any square matrix $M=(m_{i,j})$, define $M^{alt} = (m_{i,j}^{alt})$ with entries $m_{i,j}^{alt} = (-1)^{i+j} m_{i,j}$.

The following appears to be true per Mathematica:

$$P(P^{-1})^{alt} = Q.$$

Any ideas how to go about proving it?

Thank you.

答:
$P^{-1}$ has a closed form in terms of Bernoulli numbers: $$(P^{-1})_{i,j}=\begin{cases}\displaystyle\frac{1}{i}\binom{i}{j}B_{i-j},& 1\leqslant j\leqslant i\\\hfill 0,\hfill& \text{otherwise}\end{cases}$$

A direct way to show it is to use the definition $\sum_{n=0}^\infty B_n x^n/n!=x/(e^x-1)$. Namely, $$\sum_{k=j}^i\binom{i}{k-1}\frac{1}{k}\binom{k}{j}B_{k-j}=\sum_{k=j}^i\frac{i!\ B_{k-j}}{(i-k+1)!j!(k-j)!}=\frac{i!}{j!}S_{i-j}$$ where, using the "coefficient-of" notation, $$S_n=\sum_{k=0}^n\frac{B_k}{k!}\frac{1}{(n+1-k)!}=[x^{n+1}]\frac{x}{e^x-1}(e^x-1)=\begin{cases}1,&n=0\\0,&n\neq 0\end{cases}$$ as expected. Similarly, $[P(P^{-1})^{alt}]_{i,j}=(i!/j!)T_{i-j}$ where, as expected again, $$T_n=\sum_{k=0}^n\frac{(-1)^k B_k}{k!(n+1-k)!}=[x^{n+1}]\frac{-x}{e^{-x}-1}(e^x-1)=\frac{1}{n!}.$$


问:
Derivatives for a piecewise defined function
$f$ is a function on real line and $f(x)=\begin{cases} x^2& x<0\\ 2x+x^2&x\ge 0\end{cases}$. Could any one tell me which of the following is/are true?

1.$f'(0)$ doesn't exist

2.$f'(x)$ exist other than at $0$

3.$f''(x)=2$

4.$f''(0)$ does not exist

I have myself checked that $1$ is true as $\lim x\uparrow 0 f'(x)=0\neq \lim x\downarrow f'(x)=2$ I mean to say lefthand derivative is $0$ but right hand derivative is $2$, but I am not getting how to find $f''$ please help

答:
Hints:

At each side of $\,x=0\,$ the function is polynomial and thus infinitely differentiable, which already answers (2) and almost-almost (really almost!) (3).

Now, just as $\,f'(x_0)\,$ cannot exist if $\,f\,$ isn't defined at $\,x_0\,$ , we get that (4) is true after we take a peek at what we did in (1) ...


问:
Show that the Lebesgue Outer Measure is Translation Invariant
I am currently proving the following theorem in my measure theory course:

Let $A \in \mathcal P(\mathbb{R})$ and $t \in \mathbb{R}$. Then $m^*(A+t)=m^*(A)$, where $m^*$ is the Lebesgue outer measure.

Here is the first half of my proof and also where I am unsure of a few details.

Define a sequence of open intervals $(I_n)_{n=1}^\infty$ that covers $A$. Therefore, $(I_n+t)_{n=1}^\infty$ is a sequence of open intervals that covers $A+t$. So we can see that $$m^*(A+t) \leq \sum_{n=1}^{\infty}l(I_{n}+t)=\sum_{n=1}^{\infty}l(I_{n})$$ Where $l(I_n)$ is the length of the intervals. So for every sequence of open intervals $(I_n)_{n=1}^\infty$ that cover $A$ we have $m^*(A+t) \leq \sum_{n=1}^{\infty}l(I_{n})$, therefore $m^*(A+t) \leq m^*(A)$.

I believe everything is correct but my reasoning for why we can conclude that  $m^*(A+t) \leq m^*(A)$. Is this due to the monotonicity and sub-additivity of $m^*$? More precisely, is it because $$m^*(A+t) \leq m^*(A) \leq m^*(\bigcup_{n \geq1}I_{n})\leq \sum_{n=1}^{\infty}l(I_{n})$$ since $A \subseteq \bigcup_{n \geq1}I_{n}$ and we concluded that $\sum_{n=1}^{\infty}l(I_{n}+t)=\sum_{n=1}^{\infty}l(I_{n})$?

Any feedback is welcome.

答:
You wrote

So for every sequence of open intervals $(I_n)_{n=1}^\infty$ that cover $A$ we have $m^*(A+t) \leq \sum_{n=1}^{\infty}l(I_{n})$, therefore $m^*(A+t) \leq m^*(A)$.

This follows immediately by definition of the outer measure as an infimum of a set of real numbers. Recall that if $E\subset \Bbb{R}$ is non-empty, then $\inf E$ is by definition an element of $[-\infty,\infty)$ such that two things are satisfied:

$\inf E$ is a lower bound for the set $E$: for every $x\in E$, we have $\inf E \leq x$.

For any lower bound $\beta$ of $E$, we have $\beta \leq \inf E$.

In your case, we're fixing a set $A\subset\Bbb{R}$, and letting $E$ be the set of all sums $\sum_{n\geq 1} l(I_n)$, where $\{I_n\}_{n=1}^{\infty}$ is a countable cover for $A$ consisting entirely of open intervals. What your proof shows is that $m^*(A+t)$ is a lower bound for $E$, hence by definition of infimum, $m^*(A+t)\leq \inf E =: m^*(A)$.


答:
How do you deine $m^*$?

If your definition is:
$$m^*(A)=\inf\{\sum_n l(J_n):\ A\subset\bigcup_n J_n,\, J_n\text{ are open intervals}\}$$
then you have proven that given ANY open cover (by intervals) of $A$,
$m*(A+t)\le \sum_n l(I_n)$.
You only have to take $\inf$ over all possible covers to get $m^*(A+t)\le m^*(A)$.

The reverse inequality is a consequence of the previous inequality:
$m^*(A)=m^*((A+t)-t)\le m ^*(A+t)$

Title: Show that the Lebesgue Outer Measure is Translation Invariant

Body:
I am currently proving the following theorem in my measure theory course:

Let $A \in \mathcal P(\mathbb{R})$ and $t \in \mathbb{R}$. Then $m^*(A+t)=m^*(A)$, where $m^*$ is the Lebesgue outer measure.

Here is the first half of my proof and also where I am unsure of a few details.

Define a sequence of open intervals $(I_n)_{n=1}^\infty$ that covers $A$. Therefore, $(I_n+t)_{n=1}^\infty$ is a sequence of open intervals that covers $A+t$. So we can see that $$m^*(A+t) \leq \sum_{n=1}^{\infty}l(I_{n}+t)=\sum_{n=1}^{\infty}l(I_{n})$$ Where $l(I_n)$ is the length of the intervals. So for every sequence of open intervals $(I_n)_{n=1}^\infty$ that cover $A$ we have $m^*(A+t) \leq \sum_{n=1}^{\infty}l(I_{n})$, therefore $m^*(A+t) \leq m^*(A)$.

I believe everything is correct but my reasoning for why we can conclude that  $m^*(A+t) \leq m^*(A)$. Is this due to the monotonicity and sub-additivity of $m^*$? More precisely, is it because $$m^*(A+t) \leq m^*(A) \leq m^*(\bigcup_{n \geq1}I_{n})\leq \sum_{n=1}^{\infty}l(I_{n})$$ since $A \subseteq \bigcup_{n \geq1}I_{n}$ and we concluded that $\sum_{n=1}^{\infty}l(I_{n}+t)=\sum_{n=1}^{\infty}l(I_{n})$?

Any feedback is welcome.


问:
How do I find a general formula for $\int_0^\infty e^{-(ax^2+bx+c)} dx$
Setting our integral to $I$ and squaring it:
$$
I = \int_0^\infty e^{-(ax^2+bx+c)} dx
$$
$$
I^2 = e^{-2c}\iint_{\mathbb{R}^2} e^{-(a(x^2+y^2) + b(x+y))}dxdy
$$
I want to convert to polar co-ordinates but I get an ugly expression and depending on the values of $a$ and $b$ we will not obtain circular symmetry:
$$
I^2 =e^{-2c} \int_0^{2\pi}\int_0^\infty e^{-ar^2 - br\sqrt{1+2\sin(\theta)}}rdrd\theta
$$
Are my bounds incorrect and if so how can I change them?

答:
I'm confused as to whether your bounds are $[0,\infty)^2$ or $(-\infty,\infty)^2$ so I will assume it is the former. We shall say:
$$I=\int_{0}^\infty \exp(-(ax^2+bx+c))dx$$
now notice that:
$$ax^2+bx+c=\left(\sqrt{a}x+\frac{b}{2\sqrt{a}}\right)^2+\left(c-\frac{b^2}{4a}\right)$$
now lets make the substitution:
$$u=\sqrt{a}x+\frac{b}{2\sqrt{a}}\Rightarrow dx=\frac{du}{\sqrt{a}}$$
so what we now have is:
$$I=\frac{\exp\left(\frac{b^2}{4a}-c\right)}{\sqrt{a}}\int\limits_{b/2\sqrt{a}}^\infty\exp(-u^2)du$$


Title: what value of K does the system have a unique solution

Body:
$\begin{cases}x_1 + kx_2 - x_3 = 2\\2x_1 - x_2 + kx_3 = 5\\x_1 + 10x_2 -6x_3= 1\\
\end{cases}$

I've been trying echelon form where i took $R_2 = R_2 - 2R_1$ and $R_3 = R_3-R_1$

So I have $\left[\begin{array}{ccc|c}1&K&-1&2\\2&-2&K&5\\1&10&-6&1\end{array}\right]$

I've been trying echelon form where i took $R_2 = R_2 - 2R_1$ and $R_3 = R_3-R_1$
and reduced it

So I have $\left[\begin{array}{ccc|c}1&K&-1&2\\0&-1-2K&K+2&1\\0&10-K&-5&-1\end{array}\right]$

But now I am not sure how i could remove $10-K$ with $-1-2K$ any help would be appreciated

Body:
If $k\neq -\frac12$
$$\left(
    \begin{matrix}
    1 & k & -1 &  | & 2\\
    2 & -1 & k &  | & 5 \\
    1 & 10 & -6 & | & 1 \\
     \end{matrix}
\right)  \xrightarrow[\text{$R_3=R_3-R_1$}]{\text{$R_2=R_2-2R_1$}}$$ $$\left(
    \begin{matrix}
    1 & k & -1 &  | & 2\\
    0 & -1-2k & k+2 &  | & 1 \\
    0 & 10-k & -5 & | & -1 \\
     \end{matrix}
\right) \xrightarrow[\text{$(2k+1\neq0)$}]{R_3=R_3-(\frac{10-k}{2k+1})R_2} $$
$$\left(
    \begin{matrix}
    1 & k & -1 &  | & 2\\
    0 & -1-2k & k+2 &  | & 1 \\
    0 & 0 & \frac{-k^2-2k+15}{2k+1} & | & \frac{-3k+9}{2k+1} \\
     \end{matrix}
\right) $$
Which has a unique solution $\iff \frac{-3k+9}{2k+1} \neq 0 \iff -3k+9\neq 0 \iff k\neq3$

If $k=-\frac12$ it is easy to show that the system has a unique solution.


问:
To show Galois extension is simple by using Galois correspondence
Galois correspondence can be proved by using primitive element theorem.

Here,I want to prove

Every Galois extension is simple extension, using galois
correspondence.

How can I do that with Galois correspondence theory?Thank you for your help.

答:
I am not sure what the question is exactly asking for, but here is an argument using the Galois correspondence:

Let $F/K$ be a finite Galois extension. Then by the Galois correspondence $F/K$ only has finitely many intermediate fields as the Galois group $G(F/K)$ is finite and hence has only finitely many subgroups. This implies that $F/K$ is simple. (In general an extension is simple and algebraic iff it has only finitely many intermediate fields, this is a variant of the primitive element theorem.)


问:
Show that $\operatorname{Vol}_{3}(\{0\leq x,y\leq1,\,z=2x+y\})=0$
I want to show that $\operatorname{Vol}_{3}=0$, where
$$D:=\{(x,y,z)\in\mathbb{R}^{3}\,:\,0\leq x,y\leq1,\,z=2x+y\}$$
by covering $D$ (for each $\epsilon>0$) with a finite number of rectangles $R_{i}$ such that $\sum_{i}\operatorname{Vol}_{3}(R_{i})<\epsilon.$

What sort of choice of rectangles will be suitable here?

I've done a few examples in $\mathbb{R}^{2}$, such as the line $y=x$ from $(0,0)$ to $(1,1)$, but I am having trouble extending this mindset in $\mathbb{R}^{3}$.

Any help is welcome!

答: HINT: Divide $[0,1]\times [0,1]$ into $n\times n$ squares. Over the $i$th square, you want a box $R_i$ of height $<\epsilon$ that encloses the portion of $D$ over that square. So you want to choose $n$ large enough so that the maximum of $2x+y$ minus the minimum of $2x+y$ on each square will be $<\epsilon$. Can you figure this out? (Spoiler: You will need $3/n<\epsilon$.)


问:
Write the general formula for this function
While playing with power series, I found this rather nice series from adding together series of functions.

$$\ln(1-x)+\dfrac{1}{1-x}=1+\dfrac{1}{2}x^2+\dfrac{2}{3}x^3+\dfrac{3}{4}x^4+\dfrac{4}{5}x^5...$$

$$1+\sum_{n=1}^{\infty}\dfrac{n}{n+1}x^{n+1}$$ seems to be true, it looks clumsy and not handy when I do convergence tests. This series misses an $x$ term and I don't know how to express it in the formula. Is there a way to write this formula in a more succinct form, including the $1$ inside the formula?

答: You can write it as $$1+\sum_{n=0}^\infty\frac{n}{n+1}x^{n+1}$$ or, if you prefer $$1+\sum_{n=1}^\infty\frac{n-1}{n}x^n$$


问:
Alexander polynomial of some 2-component links
I would like to understand whether the multivariate Alexander polynomial of a link $L$ does not vanish everywhere for a certain class of links; I don't know the links' diagrams in general but I have some information:

$L$ has two components, say $K_1$ and $K_2$;

each component is unknotted (considered singularly), and in particular $K_1$ is a simple curve;

$K_2$ is contained in the solid torus which is the complement of $K_1$;

the linking number of the components vanishes, i.e. $lk(K_1, K_2)=0$. So, $K_2$ is homotopically trivial in the exterior of $K_1$;

$L$ is non-split.

An example of such a link is the Whitehead link $W$, whose Alexander polynomial is $\Delta_W(t_1, t_2)=(1-t_1)(1-t_2)$.

The only general result I was able to find in the literature is that the Alexander polynomial $\Delta_L(t_1, t_2)$ evaluted in $(t_1, 1)$ is equal to zero, i.e. $\Delta_L(t_1, 1)=0$, unfortunately. My guess, just for the records, is that the Alexander polynomials of such links does not vanish everywhere. Any ideas or references are appreciated.

答:
I believe there are such examples where the Alexander polynomial is zero. Consider the link L11n244, pictured below.

Both the Knot Atlas and Link Info have it listed as having multivariable Alexander polynomial of zero. The picture above satisfies all of your conditions except that neither component is "simple" (which I'm taking to mean a diagram of the unknot with no crossings). But one can just isotope the diagram so that either of the components are simple. I've done that for the blue component below.

It occurs to me that maybe you meant something else with your third condition: $K_2$ is contained in the solid torus which is the complement of $K_1$. The reason I say that is if $K_1$ is an unknot, then its complement is a solid torus. So any two component link $L=K_1\cup K_2$ where $K_1$ is trivial would have $K_2$ contained in the solid torus complement of $K_1$.


问:
Elementary proof of Power Rule For differentiation $f'(x)=rx^{r-1}$ for $f(x)=x^r$
I could  not find any elementary proof of the Power Rule for differentiation:

Given $x,r \in R$ ,$x>0$ and a function $f(x)=x^r$, then its derivative is $f'(x)=rx^{r-1}$

Defintion:Let a sequence of rational numbers {$r_n$} tend to $r$ then $x^r=\lim_{r_n \to r}x^{r_n}$

All the proofs I had seen utilized the derivative of $logx$ but I dont think this is  elementary because one has to show that for $e=\text{Euler's number}$ , $t \in R$and $n \in N$ $\lim_{n \to \infty}(1+\frac{1}{n})^n=\lim_{t \to 0}(1+t)^\frac{1}{t}=e$ which can be proved using the Power Rule

Ofcourse the proof for the Power Rule is straight forward if one is familiar with the result of Power Rule if $r$ is a rational number and theorem  regarding  derivative of a sequence of derivatives that are uniformly convergent.

答:
You're not going to get an elementary proof for real $r$ in general. But you can go a long way by elementary means.

natural $r$:

Use induction starting at $f(x)=x$ and then use the product rule for the inductive step. For $r=0$ the result is trivial.

integer $r$:

Use that $x^{-r}=(x^r)^{-1}$, then apply the chain rule. This way you only need to prove $\frac{\mathrm dx^{-1}}{\mathrm dx}=-x^{-2}$. Which goes like this:

$$\begin{align}\frac{(x+h)^{-1}-x^{-1}}{h}&=\frac{\frac{x}{x(x+h)}-\frac{x+h}{x(x+h)}}{h}\\
&=\frac{-\frac{h}{x(x+h)}}{h}\\
&=-\frac{1}{x(x+h)},\end{align}$$

which goes to $-x^{-2}$ as $h\to0$.

rational $r$:

Use that $x^{\frac pq}=(x^p)^{\frac1q}$ and the chain rule, so you only need to prove the fact for $r=\frac1q,~q\in\mathbb N_{\geq2}$. Which follows from the rule for the derivative of the inverse function, since $x^{\frac1q}$ is the inverse of $x^q$. Exception: at $x=0$ the inverse function rule doesn't work since we would divide by $0$. But for $x=0$, the derivative is again elementary, since $\frac{h^q}{h}=h^{q-1}$ goes to $0$ as $h\to0$ for all $q>1$.

irrational $r$:

This is where elementary means won't help us anymore. But it can be done without logarithms. Let's say $r_n\to r$, where $r_n$ are all rational. We want to calculate the limit

$$\lim_{h\to0}\frac{\lim\limits_{n\to\infty}(x+h)^{r_n}+\lim\limits_{n\to\infty}x^{r_n}}{h}.$$

Using the limit laws we can manipulate this to get

$$\lim_{h\to0}\lim_{n\to\infty}\frac{(x+h)^{r_n}-x^{r_n}}{h}.$$

If we could exchange the two limits, we would get that this is equal to

$$\lim_{n\to\infty}\lim_{h\to0}\frac{(x+h)^{r_n}-x^{r_n}}{h}=\lim_{n\to\infty}\frac{\mathrm dx^{r_n}}{\mathrm dx}=\lim_{n\to\infty}r_n x^{r_n-1}=r x^{r-1},$$

where the last step uses elementary limit laws and your definition of irrational powers. But we need to be able to exchange the limits. There is a theorem which says that if both $\lim_{h\to0} a_n(h)$ and $\lim_{n\to\infty}a_n(h)$ exist and the latter limit is uniform if we consider $a_n$ as a sequence of functions of $h$, then we are allowed to exchange the limits. But I'd classify this theorem as similarly non-elementary as logarithms. Anyway, we now need to prove that

$$\frac{(x+h)^{r_n}-x^{r_n}}{h}$$

converges uniformly as a sequence of functions of $h$. I'm having a hard time doing so right now, but I'm pretty optimistic that it can be proved. You might want to try it as an exercise.


答:
Hint.

The set $D$ is the graph of a Riemann integrable function $f$ on a square $[0,1]^2$.

If you consider the sets of "3d rectangles" that give you the lower sum and the upper sum, the set difference between the "upper rectangles" and the "lower rectangles" contains the set $D$. Because of integrability, the difference between the lower sum and the upper sum can be arbitrarily small.

Lower dimension analogue:

Title: Show that $\operatorname{Vol}_{3}(\{0\leq x,y\leq1,\,z=2x+y\})=0$

Body:
I want to show that $\operatorname{Vol}_{3}=0$, where
$$D:=\{(x,y,z)\in\mathbb{R}^{3}\,:\,0\leq x,y\leq1,\,z=2x+y\}$$
by covering $D$ (for each $\epsilon>0$) with a finite number of rectangles $R_{i}$ such that $\sum_{i}\operatorname{Vol}_{3}(R_{i})<\epsilon.$

What sort of choice of rectangles will be suitable here?

I've done a few examples in $\mathbb{R}^{2}$, such as the line $y=x$ from $(0,0)$ to $(1,1)$, but I am having trouble extending this mindset in $\mathbb{R}^{3}$.

Any help is welcome!


Title: Write the general formula for this function

Body:
While playing with power series, I found this rather nice series from adding together series of functions.

$$\ln(1-x)+\dfrac{1}{1-x}=1+\dfrac{1}{2}x^2+\dfrac{2}{3}x^3+\dfrac{3}{4}x^4+\dfrac{4}{5}x^5...$$

$$1+\sum_{n=1}^{\infty}\dfrac{n}{n+1}x^{n+1}$$ seems to be true, it looks clumsy and not handy when I do convergence tests. This series misses an $x$ term and I don't know how to express it in the formula. Is there a way to write this formula in a more succinct form, including the $1$ inside the formula?

Body:
If $ |x|<1$, then

$$\frac{1}{1-x}=1+\sum_{k=1}^{+\infty}x^k$$

and by integration

$$\ln(1-x)=-\sum_{k=1}^{+\infty}\frac{x^k}{k}$$

Thus

$$\ln(1-x)+\frac{1}{1-x}=$$

$$1+\sum_{k=1}^{+\infty}(1-\frac 1k)x^k$$


问:
Accuracy of approximation using linear interpolation
The use of linear interpolation is shown (in textbook) together with interval bisection and Newton-Raphson process as an introduction to numerical methods. The basic idea is when a root of the function $f(x)$ lies in a given interval, we draw a straight line between those two points, and with aid of similar triangles, we can use the root of the straight line as our approximation.

The problem I have with this method is exactly at which stage can I obtain an approximation to the required degree of accuracy. As an example, the equation $$x^3+4x-9=0$$ with initial interval $[1,2]$ has subsequent intervals of:

$[1.363...,2]$,

$[1.443...,2]$,

$[1.460...,2]$, and

$[1.463...,2]$ respectively.

It is written that "Two successive approximations give the root as $1.5$ to one d.p." so we can stop there. However, I don't think it's obvious that further use of the method would not give a lower bound greater than $1.5$ (or more likely, greater than $1.46$ for $2$ d.p.). As opposed to the method of interval bisection where the lower bound and upper bound gives the same result when rounded to the required accuracy, here the upper bound is still $2$, and anything could happen for such a wide interval if we assume no algebraic understanding of the curve. Any insights?

答:
The polynomial that produces your values is $p(x)=x^3+4x-9$. Then indeed the regula-falsi or false position method, that is, the bracketing method using the secant root,

```<code>a,b=1,2
f=lambda x: x*(x*x+4)-9
fa,fb = f(a),f(b)
for k in range(2,20):
    c = b-fb*(b-a)/(fb-fa)
    fc = f(c)
    if fa*fc > 0:
        a,fa = b,fb
    if abs(c-b) < 1e-10: break
    b,fb = c,fc
</code>```

produces the points and values

step
x
f(x)

c[0]=a
1
-4

c[1]=b
2
7

c[2]
1.3636363636363638
-1.0097670924117192

c[3]
1.443860801050558
-0.21449107408405865

c[4]
1.46039514739571
-0.04375583788976023

c[5]
1.4637471747797475
-0.008851314007729982

...
...
...

c[15]
1.4645957011474617
-9.946194978738276e-10

c[16]
1.4645957012235364
-2.007709554163739e-10

As all the values found are negative, only the point with negative value gets changed, the other side of the interval with the positive value remains unchanged. This stalling behavior is rather typical of the "plain vanilla" regula falsi method.

There are anti-stalling modifications that shift the calculated mid-point more and more towards the stalling side so that eventually a point gets found that replaces that end of the interval. The easiest is the so-called Illinois variant that reduces the function value at the stalling side by a factor of 2 in each stalling step.

```<code>a,b=1,2
f=lambda x: x*(x*x+4)-9
fa,fb = f(a),f(b)
for k in range(2,20):
    c = b-fb*(b-a)/(fb-fa)
    fc = f(c)
    if fa*fc > 0:
        a,fa = b,fb
    else:              # this new branch is the only change
        fa *= 0.5
    if abs(c-b) < 1e-10: break
    b,fb = c,fc
</code>```

resulting in an iteration

step
x
f(x)

c[0]=a
1
-4

c[1]=b
2
7

c[2]
1.3636363636363638
-1.0097670924117192

c[3]
1.443860801050558
-0.21449107408405865

c[4]
1.475974727761858
0.11931191740107039

c[5]
1.4644961782313235
-0.0010384912181606865

c[6]
1.4645952254470451
-4.964985366839869e-06

c[7]
1.4645961724498568
4.9171042082463146e-06

c[8]
1.464595701242682
-9.841016890277388e-13

c[9]
1.4645957012427764
0.0

Note that the sign of the values alternates frequently, and apparently the convergence is super-linear, here as fast as the Newton method in terms of function evaluations.

See also Bracketing root-finding methods: my modified Illinois method, https://stackoverflow.com/questions/22273751/regula-falsi-algorithm

Using a multi-precision floating point type, one can continue the table above to ever smaller errors, where the period-3 pattern of the error progression becomes clearly visible

k
$c_k$
$c_k-c_{k-1}$
$f(c_k)$

8
1.464595701242682
-4.7120717493e-07
-9.8507967689e-13

9
1.464595701242776
9.4400382608e-14
-1.9544502434e-19

10
1.464595701242776
3.7459071605e-20
1.9544494679e-19

11
1.464595701242776
-1.8729532086e-20
-1.5413207939e-39

12
1.464595701242776
1.4770510950e-40
-1.2155181734e-59

13
1.464595701242776
2.3296674592e-60
1.2155181734e-59

14
1.464595701242776
-1.1648337296e-60
-5.9616556262e-120

15
1.464595701242776
5.7130676528e-121
-2.9239659746e-180

16
1.464595701242776
5.6040860040e-181
2.9239659746e-180

Body:
The polynomial that produces your values is $p(x)=x^3+4x-9$. Then indeed the regula-falsi or false position method, that is, the bracketing method using the secant root,

a,b=1,2
f=lambda x: x*(x*x+4)-9
fa,fb = f(a),f(b)
for k in range(2,20):
    c = b-fb*(b-a)/(fb-fa)
    fc = f(c)
    if fa*fc > 0:
        a,fa = b,fb
    if abs(c-b) < 1e-10: break
    b,fb = c,fc

produces the points and values

step
x
f(x)

c[0]=a
1
-4

c[1]=b
2
7

c[2]
1.3636363636363638
-1.0097670924117192

c[3]
1.443860801050558
-0.21449107408405865

c[4]
1.46039514739571
-0.04375583788976023

c[5]
1.4637471747797475
-0.008851314007729982

...
...
...

c[15]
1.4645957011474617
-9.946194978738276e-10

c[16]
1.4645957012235364
-2.007709554163739e-10

As all the values found are negative, only the point with negative value gets changed, the other side of the interval with the positive value remains unchanged. This stalling behavior is rather typical of the "plain vanilla" regula falsi method.

There are anti-stalling modifications that shift the calculated mid-point more and more towards the stalling side so that eventually a point gets found that replaces that end of the interval. The easiest is the so-called Illinois variant that reduces the function value at the stalling side by a factor of 2 in each stalling step.

a,b=1,2
f=lambda x: x*(x*x+4)-9
fa,fb = f(a),f(b)
for k in range(2,20):
    c = b-fb*(b-a)/(fb-fa)
    fc = f(c)
    if fa*fc > 0:
        a,fa = b,fb
    else:              # this new branch is the only change
        fa *= 0.5
    if abs(c-b) < 1e-10: break
    b,fb = c,fc

resulting in an iteration

step
x
f(x)

c[0]=a
1
-4

c[1]=b
2
7

c[2]
1.3636363636363638
-1.0097670924117192

c[3]
1.443860801050558
-0.21449107408405865

c[4]
1.475974727761858
0.11931191740107039

c[5]
1.4644961782313235
-0.0010384912181606865

c[6]
1.4645952254470451
-4.964985366839869e-06

c[7]
1.4645961724498568
4.9171042082463146e-06

c[8]
1.464595701242682
-9.841016890277388e-13

c[9]
1.4645957012427764
0.0

Note that the sign of the values alternates frequently, and apparently the convergence is super-linear, here as fast as the Newton method in terms of function evaluations.

See also Bracketing root-finding methods: my modified Illinois method, https://stackoverflow.com/questions/22273751/regula-falsi-algorithm

Using a multi-precision floating point type, one can continue the table above to ever smaller errors, where the period-3 pattern of the error progression becomes clearly visible

k
$c_k$
$c_k-c_{k-1}$
$f(c_k)$

8
1.464595701242682
-4.7120717493e-07
-9.8507967689e-13

9
1.464595701242776
9.4400382608e-14
-1.9544502434e-19

10
1.464595701242776
3.7459071605e-20
1.9544494679e-19

11
1.464595701242776
-1.8729532086e-20
-1.5413207939e-39

12
1.464595701242776
1.4770510950e-40
-1.2155181734e-59

13
1.464595701242776
2.3296674592e-60
1.2155181734e-59

14
1.464595701242776
-1.1648337296e-60
-5.9616556262e-120

15
1.464595701242776
5.7130676528e-121
-2.9239659746e-180

16
1.464595701242776
5.6040860040e-181
2.9239659746e-180


问:
Empty theory is a sequence of isolated points
I'm working my way through Henson and van den Dries Model Theory lecture notes doing odd-numbered exercises. Yet I am stuck on the exercise 4.17:

Show that if $T$ is the empty theory in the language $L$ of equality, then the space $S_0(T)$ consists of a sequence of points $(T_n | n \geq 1)$ that are isolated, together with a point $T_\infty$ to which this sequence converges.

Nowhere in the notes do they define what an "empty theory" is (I am only aware of an empty language). How would such a thing even look like? What are the consequences of it? In particular, wouldn't it imply everything?

Then the "isolated points" are also not defined anywhere. How could I possibly start proving the statement? It seems more of a topology question to me than model theory...

Any help or hints would be appreciated!

答:
Remember that a theory in this context is just a set of sentences. The empty theory is $\{\}$, the theory containing no sentences at all. This is the weakest possible theory (so the answer to your question "wouldn't it imply everything?" is "no"), and the only things it proves are the logical tautologies, that is, the sentences true in every structure. Remember that $\mathcal{M}\models S$ iff every sentence in $S$ is true in $\mathcal{M}$; if $S=\{\}$, then $\mathcal{M}\models\mathcal{S}$ is vacuously true for every structure $\mathcal{M}$, so the consequences of $\mathcal{M}$ are merely the sentences true in all structures.

Meanwhile, the notion of "isolated point" is indeed a topological notion, but that doesn't mean it's irrelevant to logic; indeed, the whole point of considering the type spaces is that there is a connection between logical properties of theories and topological properties of the associated space.

You should break the exercise down into three parts:

First, classify the models of $T$ up to isomorphism. (HINT: there's exactly one of each cardinality.)

Next, determine the complete consistent theories extending $T$. Note that multiple non-isomorphic models may have the same theory. (Indeed, there will only be countably many such theories.)

Note that so far, topology hasn't entered the picture; only the final bulletpoint involves topology:

Lastly, remember that points in $S_0(T)$ are exactly complete consistent extensions of $T$, so the previous bulletpoint tells you what they are; it only remains to determine the topology. Establishing isolatedness will be pretty easy (HINT: can you write down a sentence true in exactly those structures of cardinality $17$, for example?); understanding the non-isolated point is a bit more nuanced.


问:
Equation for a line through a plane in homogeneous coordinates.
For calculations in 2D space, there exist a few useful equations to compute general geometry with the vector dot product . and the vector cross product x when working with homogeneous coordinates (remember even though we are working in 2D space the homogeneous representation of lines and point are 3D vectors):

1) The point where two lines cross x = l1 x l2.

2) The line between two points l = x1 x x2

3) A point lies on a line if x . l = 0

In 3D space a plane can be described by a normal vector n = [n1, n2, n3] and a point on the plane x = [x1, x2, x3] both in Euclidean coordinates.

In homogeneous coordinated the plane can be defined as p = [n1, n2, n3, -(n1*x1 + n2*x2 + n3*x3)].

Is there a short equation for finding the point where a line passes through a plane? Intuitively it feels as it should be x = l x p in homogeneous coordinates, but this computation does not exist, since there is no cross product in 4 dimensions.

At the moment I am only able to compute the intersection by defining the line with the equation l(t) = a + b(t), where a is a point on the line and b is the direction of the line in Euclidean coordinates.

For a = [a1, a2, a3], b = [b1, b2, b3] and the plane in question p = [p1, p2, p3, p4], the point of intersection x = [x1, x2, x3]  can be obtained by substituting t in the line equation with t = -(p . [a1, a2, a3, 1])/(p . [b1, b2, b3, 0])

In summary, is there a elegant equation to find a point x where a line l crosses through a plane p preferably in homogeneous coordinates?

答:
To answer your question about the line and plane in $\mathbb R^3$, think of it this way: You want to find the line of intersection of a $2$-dimensional subspace $L$ and a $3$-dimensional subspace $P$ in $\mathbb R^4$. Let $\mathbf n$ be the normal vector of $P$; you want the line in $L$ orthogonal to $\mathbf n$. This is really the same as your approach.

If you want something more like your approach one dimension lower, you need two vectors orthogonal to $L$, i.e., a basis for $L^\perp$, say $\mathbf v_1$ and $\mathbf v_2$. Then compute the cross product $\mathbf n\times \mathbf v_1\times \mathbf v_2$, and, voilà! You compute the cross product of $n$ vectors in $\mathbb R^{n+1}$ exactly as in the case of $n=2$.


问:
Find tangent point on circle with external point

Given this image where we know P, O and the radius of the circle in 2D, how to find T?

(PMN is irrelevant to the problem)

There is an item rotating in 2D along the circle, when it reaches T, it should leave the circle and move along the red line (similar to the disc throwing concept).

I feel I should be using the law of sines and SOHCAHTOA but I fail to get a 2D point value for T.

EDIT: I'm thinking that since I know the length of PO and the radius of the circle on which the tangent should be placed, I can do:

```<code>alpha = sin(^-1)(r / PO)
theta = 180 - 90 - alpha
norm = PO / |PO|
v = norm.Rotate(theta)
v * r
</code>```

alpha is the angle at $P$ and theta is the angle at $O$.

I should be able to normalize $PO$, rotate by theta and multiply by r. (Since this is meant to happen in a 3D engine, I have access to quaternion and transform methods).

Though this feel like a long path for something I may be able to find out with trigonometry.

答:
Using Pythagoras theory we know that
$$
\overline{PT} = \sqrt{\overline{P0}^2 - \overline{0T}^2}
$$
Hence, we can find the point $T$ by calculatin the intersect between two circles,
$$
x = \frac{d^2 - r^2 + R^2}{2d}
$$


问:
close form for $\sum_{k=0}^n\binom{2n}{n} \binom{2n-2k}{n-k}$
Is it possible to find a closed form for the expression below?

$\sum_{k=0}^n\binom{2n}{n} \binom{2n-2k}{n-k}$

I think it is possible to use a former solved one below.

$\frac{1}{\sqrt{1-4x}} =  \sum_{n=0}^\infty \binom{2n}{n} x^n $

答:
Let $ x\in\mathbb{R} $, such that $ \left|x\right|<\frac{1}{4} $. Using cauchy product theorem, we have :

\begin{aligned}\sum_{n=0}^{+\infty}{\left(\sum_{k=0}^{n}{\binom{2k}{k}\binom{2n-2k}{n-k}}\right)x^{n}}&=\left(\sum_{n=0}^{+\infty}{\binom{2n}{n}x^{n}}\right)^{2}\\ &=\left(\sum_{n=0}^{+\infty}{\binom{-\frac{1}{2}}{n}\left(-4x\right)^{n}}\right)^{2}\\ &=\left(\frac{1}{\sqrt{1-4x}}\right)^{2}\\ &=\frac{1}{1-4x}\\ \sum_{n=0}^{+\infty}{\left(\sum_{k=0}^{n}{\binom{2k}{k}\binom{2n-2k}{n-k}}\right)x^{n}}&=\sum_{n=0}^{+\infty}{4^{n}x^{n}}\end{aligned}

Thus for any $ n\in\mathbb{N} $, we have : $$ \sum_{k=0}^{n}{\binom{2k}{k}\binom{2n-2k}{n-k}}=4^{n} $$