问:
How to calculate the order of elements within these groups?
I'm struggling to understand how to calculate the order of specific elements within large groups.
For example if we have $(4,9)$ within $\Bbb Z_{18}\times \Bbb Z_{18}$ (apologies for the lack of use of mathjax).
I know if G is a finite group of order (size) n then, for any $g∈G$, the order (period) of $g$ is a divisor of $n$. Hence knowing the order of $\Bbb Z_{18}\times \Bbb Z_{18}=324=2^2. 9^2$, would this mean that the order of $(4,9)$ in this group must be $2$?

答:
Hint: Its a product group, so the order is the lcm of the orders of the components.

Here $4$ has order $9$ and $9$ has order $2$ in $\Bbb Z_{18}$.


问:
connected graph with a certain number of edges

Let $k\in \mathbb{Z}_+.$ Prove that if $G = (V,E)$ is a connected graph with $|E| > {k^k \choose 2},$ then $G$ has a vertex of degree at least $k$ or contains a path with $k$ edges.

It suffices to show that if $G$ does not have a vertex of degree at least $k$, then $G$ contains a path with $k$ edges. However, I'm not sure how to find this path. I tried considering a few small cases such as $k=1,2,$ but I'm not sure how to generalize the result. For instance, for $k=1, G$ clearly contains at least two vertices and a path with $1$ edge (the path between those two distinct vertices). If $k=2, |E| > 6$, and by connectedness of $G$, there must be a vertex of degree at least $k$; if all vertices are of degree one, since $G$ is connected, it can only have two vertices and one edge, but then the number of edges is not more than $6$, a contradiction. I should definitely use the connected property, but I'm not sure how; for $k=2,$ if $G$ is disconnected, the proposition clearly fails.

答:
I'll lay out an approach / hint in the first part, and then complete the proof in the second part of the post.

We can assume $k>1$.
If $G$ has a vertex of degree $k$, we are done. So we assume that the maximum degree of $G$ is some $\Delta \leq k-1$.

We let $u$ be a vertex of $G$, and assume for the sake of contradiction that $G$ does not contain any path of length $k$. In particular, this means that the diameter of $G$ is at most $k-1$ (since $G$ is connected and has no $k$-path), so every vertex of $G$ is within distance $k-1$ of $u$ (this is where connectivity is useful!).

If you just wanted a hint / way to approach the problem, stop reading here.

Now given the degree and distance constraints we have so far, let's see how many vertices $G$ can have.
We introduce the notation:

$$N_i(u) = \{v\in G : d(u,v) = i\}$$
The vertex $u$ has at most $(k-1)$ neighbours at distance $1$ from it (i.e., we have $|N_1(u)| \leq k-1$). Each of these in turn has at most $(k-2)$ neighbours besides $u$. So the set $N_2(u)$ of vertices distance $2$ from $u$ has $|N_2(u)| \leq (k-1)(k-2)$ (see the diagram).

In general, we get that $|N_i(u)| \leq (k-1)(k-2)^{i-1}$. So the total number of vertices $n$ is bounded above as follows:

$$n = |\{u\}| + |N_1(u)| + \dots + |N_{k-1}(u)|$$

And thus:

$$n \leq 1 + (k-1) + (k-1)(k-2) + \dots + (k-1)(k-2)^{k-2} = 1 + (k-1)\sum_{i=0}^{k-2}(k-2)^i$$

By using the formula for the sum of a geometric series, and a bit of inequality juggling, you can show that $n \leq k^k$. Thus the total possible number of edges that $G$ has is at most $\binom{n}{2} \leq \binom{k^k}{2}$. But this yields a contradiction, so $G$ must have a $k$-path.

ContentLicense: CC BY-SA 4.0


问:
How to establish the modus ponens inference rule in the LK sequent calculus for first-order logic?
How can the modus ponens inference rule
$$
\frac{\Gamma \vdash \varphi\hspace{1cm}\Delta\vdash \varphi\rightarrow\psi}{\Gamma,\Delta \vdash \psi}
$$
be established in the LK sequent calculus for first-order logic without resorting to the fact that this calculus is semantically complete?

答:
You can show that modus ponens is derivable in the sequent calculus LK.

In general, an inference rule

$$\dfrac{\Gamma_1 \vdash \Delta_1 \qquad \cdots \qquad \Gamma_n \vdash \Delta_n}{\Gamma \vdash \Delta}(*)$$

is derivable in LK if there is a LK derivation of the sequent $\Gamma \vdash \Delta$
starting from the premises $\Gamma_1 \vdash \Delta_1$, $\,\dots\,$, $\Gamma_n \vdash \Delta_n$.
Intuitively this means that the rule $(*)$ can be simulated by the inference rules of LK.

The following shows that modus ponens is derivable in LK.

$$\frac{\dfrac{\Gamma \vdash \varphi \qquad \dfrac{}{\psi \vdash \psi}\text{ax}}{\Gamma, \varphi \to \psi \vdash \psi}\to_L \qquad \Delta \vdash \varphi \to \psi}{\Gamma, \Delta \vdash \psi}\text{cut}$$

The idea is that modus ponens is an elimination rule for the connective $\to$, and LK can simulate it via the rules $\to_L$ and cut.


问:
Finding the value of $f(2001)$ given the functional equation $f\bigl(f(n)\bigr)= 3n$ and strict monotonicity of $f:\mathbb N\to\mathbb N$

Let $f:\mathbb N\to \mathbb N$ be a strictly increasing function such that $f\bigl(f(n)\bigr)= 3n\ \forall\ n\in \mathbb N$

Find the value of $f(2001).$

Now I tried to solve this by myself but I'm stuck somewhere in the middle of my solution. Please help me with my solution. Also my solution is a bit long as I'm writing almost every observation I made, so please be kind enough to bear with me.

My Approach:

We know that $f$ is strictly increasing.

Suppose for some $n_1$ and $n_2$, we have $f(n_1)=f(n_2)$. Thus $f\bigl(f(n_1)\bigr)=f\bigl(f(n_2)\bigr)\implies 3n_1=3n_2$ hence $n_1=n_2$.
 (I just realized that his step was actually not required.)

$\therefore$ $f$ is an injective strictly increasing function.

Suppose for some $n\in \mathbb N$, we have $f(n)\leq n$, then $f\bigl(f(n))\leq f(n)\leq n$ as $f$ is strictly increasing.

This gives us $3n\leq n$ which isn't true for any $n\in \mathbb N$.

$\therefore f(n)>n\ \forall\ n\in \mathbb N $

Now suppose $f(1)=l>1$. Thus $f\bigl(f(1)\bigr)=3=f(l)>f(1)=l\implies 1<l<3$ and since $l\in \mathbb N$, we know that $f(1)=2$.

This means that $f\bigl(f(1)\bigr)=f(2)=3$ and $f\bigl(f(2)\bigr)=f(3)=6$ and so on.

A few such values are:

$f(1)=2$

$f(2)=3$

$f(3)=6$

$f(6)=9$

$f(9)=18$

$f(18)=27$

$f(27)=54$

$f(54)=81$

Now here a pattern can be observed.

Claim: $f(3^n)=2\cdot3^n$

Proof: Suppose the above claim is true. Then $f\bigl(f(3^n)\bigr)=f(2\cdot3^n)=3^{n+1}$. Now $f\left(3^{n+1}\right)=f\bigl(f(2\cdot3^n)\bigr)=2\cdot3^n\cdot3=2\cdot3^{n+1}$

$\therefore$ $f(3^n)=2\cdot3^n$ and $f(2\cdot3^n)=3^{n+1}$

One more thing can be observed here that if $3^n<k<2\cdot3^n$, then $2\cdot3^n<f(k)<3^{n+1}$ and since there are exactly $3^n$ permitted values for both $k$ and $f(k)$ and $f$ is strictly increasing, the unique function satisfying the given condition can easily be found.

But unfortunately $2\cdot3^6<2001<3^7$, thus a unique function cannot be found using the observation stated above.

Now this is where I am stuck. Firstly, is this question solvable using my approach? If yes, then what should I add to my approach in order to reach the solution? Please help.

THANKS

答:
$f(f(7))=21$, but we have $f(12)=21$ from the "exactly $3^n$ permitted values" point, so $f(7)=12$. In the same way we can show $f(8)=15,f(19)=30$ and so on. In general
$$f(3^n)=2\cdot3^n$$
$$f(2\cdot3^n)=3\cdot3^n$$
$$f(3^n+k)=2\cdot3^n+k,0\le k\le3^n$$
$$f(2\cdot3^n+k)=3\cdot3^n+\mathbf{3k},\ 0\le k\le 3^n$$
Thus
$$f(2001)=f(2\cdot3^6+543)=3\cdot3^6+3\cdot543=3816$$
$f$ is OEIS A003605. The same question but asking for $f(1992)$ was problem 5 of BMO 1992.


问:
Limit multivarible
How do I solve this limit?

$$\lim_{x,y,z) \to (0,0,0)} \frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2+xyz} $$

This is equal to $$\frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2}\times\frac1{1+\frac{xyz}{x^2+y^2+z^2}}$$

The first one is a standard limit with value one, but I'n not sure about the other term.

答: $0 \leq | \frac  {xyz} {x^{2}+y^{2}+z^{2}}|\leq \frac  1 2 |z| \to 0$ since $|xy| \leq \frac 1  2{(x^{2}+y^{2})}$. Hence the given limit is $1$.

Title: Limit multivarible


问:
Noetherian ring question.
Let $R$ be a noetherian ring and $I$ a proper ideal of $R$. Show that $R/I$ is noetherian.

Note: This was the last assignment (due date already passed [April 30] so I'm not directly asking for hw answers) for my introduction to abstract algebra course. I'm posting it here because my professor decided not to post solutions for this last assignment since she'll be busy making the final exam. Still, I feel like these questions are very relevant and may pop up on the final.

Lastly:

Let $f = x^k+a_{k-1}x^{k-1}+ \cdots + a_0 \in \mathbb{Z}[x]$ be a polynomial and $p$ a prime integer. Suppose that $p$ divides $a_i$ for $0 \le i < k$ and $p^2$ does not divide $a_0$. Show that $f$ is irreducible.

答: Translating it into the language of modules you have that $ I$ is an ideal of $R$ iff it is a submodule of $R$ viewed as a module over itself. Then you can use the proposition which states "Let $ 0 \rightarrow I \rightarrow M \rightarrow N \rightarrow 0$ be an exact sequence of $R$ modules, then $M$ is noetherian iff  $I,N$ are both noetherian". Obviously in this case you consider $ 0 \rightarrow I \rightarrow R \rightarrow R/I \rightarrow 0$</p>

ContentLicense: CC BY-SA 3.0

Title: Noetherian ring question.


问:
Plane parallel to intersection of two lines
Determine the equation in normal form that contains the line $\left\{\begin{matrix}
x=1+t\\
y=3t\\
z=2t
\end{matrix}\right.$ and is parallel to the intersection between the planes $x+z=1$ and $y+z=2$.

$$x+z=1 \Rightarrow \overrightarrow{n}_{1}=(1,0,1)
\\y+z=2 \Rightarrow \overrightarrow{n}_{2}=(0,1,1)\\
(1,0,1)\times (0,1,1)=(-1,-1,1)\\$$

Now I need to figure out a normal vector that is parallel to (-1,-1,1). If I use dot product I get $(-1,-1,1)\cdot M_{1}=0 \Rightarrow M_{1}=0$ which doesn't seem right.

What's my next move?

答:
We have that a normal vector for the plane $n=(a,b,c)$ is

orthogonal to the direction vector for the line $(1,3,2)$

orthogonal to the direction vector for the intersection of the planes $(-1,-1,1)$

that is

$a+3b+2c=0$

$-a-b+c=0$

which can be solved assuming a free parameter, e.g. taking $a=5$ we obtain $n=(5,-3,2)$ and the plane equation is

$$5x-3y+2z+d=0$$

and form the condition that $P=(1,0,0)$ belongs to the plane we find that $d=-5$.


问:
How would you solve $\ln x = e^{\cos x}$ algebraically?
I am not sure where to start, of course, I can get; $$x = e^{e^{\cos(x)}}$$ or $$\ln(\ln(x)) = \cos(x)$$ however neither seem to be helpful in this case. I am aware that lambert W function may be used somewhere.

答:
Too long for a comment;

Very interested by @am301's approach, I noticed that the $[0,2]$ Padé approximant built around $x=2n\pi$ is
$$e^{e^{\cos (x)}}=\frac{2 e^e}{e (x-2n \pi )^2+2}$$ which is  close to the Lorentzian-shape function proposed in @am301's answer.

This approximation would lead to the estimates
$$\{2.12,5.07,7.19,12.14,12.92\}$$ which are not as good as @am301's ones.

In any manner, with the different proposed estimates, Newton method works like a charm. Doing it for the first root (to which corresponds the worse estimate)
$$\left(
\begin{array}{cccc}
 n & x_n \\
 0 & 2.124348  \\
 1 & 1.957533  \\
 2 & 1.969841  \\
 3 & 1.969924
\end{array}
\right)$$

Edit

All of this could be  improved (except for the first root) using the next Padé approximant which is
$$e^{e^{\cos (x)}}=e^e \frac{1-\frac{\left(31-60 e+15 e^2\right) }{30 (3 e-4)}(x-2 n\pi )^2 } {1+\frac{\left(30 e^2-31\right) }{30 (3 e-4)}(x-2 n\pi  )^2+\frac{\left(6 e+5
   e^3\right) }{40 (3 e-4)}(x-2 n\pi  )^4 }$$ which is of no use for this problem since leading to quintic equations.

The generated estimates would be
$$\{1.80,5.25,7.11,12.15,12.92\}$$ while the solutions are
$$\{1.97,5.24,7.11,12.15,12.92\}$$

Using the same idea, we could assume that
$$e^{e^{\cos (x)}}=a+\frac b {(x-2n \pi)^2+c}$$ and obtain the $(a,b,c)$ parameters minimizing
$$\Phi=\int_{-\pi}^\pi \left(a+\frac{b}{x^2+c}+e^{e^{\cos (x)}}\right)^2\,dx$$ This leads to
$$a=0.232893 \qquad b=8.39717\qquad c=0.550011$$ and then the estimates
$$\{2.03,5.22,7.10,12.17,12.90\}$$


问:
In which condition，the triangle have the maximum triangle area？
These trangles have the same perimeter 2q，in which condition，can we have the trangle with the maximum area?
I have tried to use the Heron's formula. $S=\sqrt{q(q-a)(q-b)(q-c)}$
I have sought  the partial derivative and tried to find the stationary point but the calculation  is complicated. Is there any strategy can we use to simply the calculation?

答:
For a triangle with semiperimeter $\rho$
and
the radii of the inscribed and circumscribed circles $r$ and $R$,
respectively,
the area
\begin{align}
S&=\rho\,r
,
\end{align}

so given $\rho$ fixed,
the maximal area corresponds
to the triangular shape with the largest incircle.

Since we are looking for the triangular shape with
the largest $r$ with respect to the given $\rho$,
we can consider $R=1$ without loss of generality.

It is well-known that in this case $\max r=\tfrac12$,
which corresponds to equilateral triangle.

So, the triangle of the maximal area with given $\rho$
is equilateral with the side length $\tfrac23\,\rho$,
and therefore
\begin{align}
\max S(\rho)&=
\tfrac{\sqrt3}9\,\rho^2
.
\end{align}


问:
What have I done wrong in solving this problem with indices rules?
The question asks to simplify:

$$\left(\dfrac{25x^4}{4}\right)^{-\frac{1}{2}}.$$

So I used $(a^m)^n=a^{mn}$ to get

$$\dfrac{25}{4}x^{-2} = \dfrac{25}{4} \times \dfrac{1}{x^2} = \frac{25}{4x^2} = \frac{25}{4}x^{-2}$$

However, this isn't the answer, and I can't see what I've done wrong.

This is what the mark scheme says:

$$\left(\dfrac{25x^4}{4}\right)^{-\frac{1}{2}} = \left[\left(\frac{4}{25x^4}\right)^{\frac{1}{2}} \text{ or } \left(\frac{5x^2}{2}\right)^{-1} \text{ or } \frac{1}{\left(\dfrac{25x^4}{4}\right)^{\frac{1}{2}}}\right] = \frac{2}{5}x^{-2}$$

To me, my answer looks more simple than theirs, and I can't see what I've done wrong.

Body:
$$(\dfrac{25x^4}{4})^{-\dfrac{1}{2}}$$
$$(\dfrac{25}{4})^{-\dfrac{1}{2}}(x^{-2})$$
$$(\dfrac{4}{25})^{\dfrac{1}{2}}(x^{-2})$$

$$(\dfrac{2}{5})(x^{-2})$$
$$\dfrac {2}{5x^2}$$


问:
Trouble understanding equation of an ellipse
I'm self-studying algebra by going through Sheldon Axler's algebra book. I'm a bit stuck on equation of an ellipse topic. Here's how it's explained in the book:

Example: Find an equation describing the ellipse in the xy-plane produced by stretching a circle of radius 1 centered at the origin horizontally by a factor of 5 and vertically by a factor of 3.
To find an equation describing ellipse, consider a typical point (u,v) on the circle of radius 1 centered at the origin. Thus $u^2 + v^2 = 1$. Stretching horizontally by a factor of 5 and stretching vertically by a factor of 3 transforms the point (u, v) to the point (5u, 3v). Rewrite the equation $u^2 + v^2 = 1$ in terms of this new point, getting $$\frac{(5u)^2}{25} + \frac{(3v)^2}{9} = 1$$

Shouldn't the resulting equation be $(5u)^2 + (3v)^2 = 1$? Where did 25 and 9 denominators come from?

The rest of the solution in the book:

Write the transformed point $(5u, 3v)$ as $(x,y)$, thus setting $x = 5u$ and $y = 3v$, getting $$\frac{x^2}{25} + \frac{y^2}{9} = 1,$$ which is the equation of the ellipse

答:
What you quote is not the full solution, is it?

The point $(u,v)$ on the circle satisfies

$$u^2+v^2=1$$

which we just rewrite in equivalent form:

$$\frac{(5u)^2}{25}+\frac{(3v)^2}{9}=1$$

and from this we see that the point $(x,y)=(5u,3v)$, which belongs to the ellipse, satisfies

$$\frac{x^2}{25}+\frac{y^2}{9}=1$$

and so this is the equation of the ellipse.

It would be much more natural to think about it the other way around. Take the point $(x,y)$ on the ellipse, note that $(x/5, y/3)$ (squeezing inverse to your stretching) belongs to the circle, and therefore

$$\left(\frac{x}{5}\right)^2+\left(\frac{y}{3}\right)^2=1$$

which gives the same answer.


问:
An inequality involving positive real numbers which satisfy $a^2+b^2+c^2 = 3abc$

Given that a, b, c are positive real numbers, show that
$$ \frac{a}{b^2c^2}+\frac{b}{a^2c^2}+\frac{c}{b^2a^2} \ge \frac{9}{a+b+c} $$
if $a^2 + b^2 + c^2 = 3abc$

My Attempt

I wrote the inequality as $$ (a^3+b^3+c^3)*(a+b+c) \ge (a^2+b^2+c^2)^2 $$
and after expanding and cancelling the terms we get
$$ \sum_{cyclic}\{a^3(b+c)-a^2(b^2+c^2) \ = \ a^2b(a-b) + a^2c(a-c) \} $$
now the sum is symmetric wrt a,b,c so if a $(b,a)$ term is there then there is a $(a,b)$ term so adding them separately gives the sum as
$$ \sum_{cyclic}\{ab(a-b)^2\} $$
so the inequality is true

But I was wondering whether there is an altenate way of proving this without using the right hand side of the inequality too much or any use of calculus is also good

Edit:
as @Isaac pointed out we can use Cauchy shwarz directly but is there any method which uses calculus

答:
Using the Cauchy-Schwarz inequality, we have
$$(a+b+c)\left(\frac{a}{b^2c^2}+\frac{b}{a^2c^2}+\frac{c}{b^2a^2}\right) \geqslant \left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)^2 = \left(\frac{a^2+b^2+c^2}{abc}\right)^2  =9.$$
Therefore
$$\frac{a}{b^2c^2}+\frac{b}{a^2c^2}+\frac{c}{b^2a^2} \geqslant \frac{9}{a+b+c}.$$


答:
Say equation of the plane $ax + by + cz = d$

Take points on the given line $(1, 0, 0), (2, 3, 2)$ for $t = 0, 1$. As the line is in the plane, these points are also in the plane.

From $(1, 0, 0)$, we get $d = a$ so equation of plane becomes $ax + by + cz - a = 0$.

From $(2, 3, 2)$, we get $a + 3b + 2c = 0$

From dot product of the normal vector with direction vector $(-1, -1, 1)$, we get $a + b - c = 0$.

So, $b = - \frac{3c}{2}, a = \frac{5c}{2}$

So equation of the plane is $5x - 3y + 2z = 5$.

Title: Plane parallel to intersection of two lines

Body:
Determine the equation in normal form that contains the line $\left\{\begin{matrix}
x=1+t\\
y=3t\\
z=2t
\end{matrix}\right.$ and is parallel to the intersection between the planes $x+z=1$ and $y+z=2$.

$$x+z=1 \Rightarrow \overrightarrow{n}_{1}=(1,0,1)
\\y+z=2 \Rightarrow \overrightarrow{n}_{2}=(0,1,1)\\
(1,0,1)\times (0,1,1)=(-1,-1,1)\\$$

Now I need to figure out a normal vector that is parallel to (-1,-1,1). If I use dot product I get $(-1,-1,1)\cdot M_{1}=0 \Rightarrow M_{1}=0$ which doesn't seem right.

What's my next move?


问:
$\sum_{k=1}^{n} k^m \mod 10$
With $S_{(n,m)}=\sum_{k=1}^{n} k^m$, I study $S_{(n,m)}\mod 10$ for n=2021.
More than just solving this specific exercise, I would like to know what general reasoning would be wise for dealing with such problems.
My first ideas are to :

search for common properties or links between $n \times(n+1) \over 2$, $n \times(n+1) \times (2 \times n +1) \over 6$, $\big({(n+1) \times (n+2) \over 2}\big)^2$ etc.

search for 2021 properties...

Can somebody give me some advice in order to build an accurate reasoning ?

Thanks in advance.

Romain

答:
I assume that $S_{n,m}$ is the sum, and that we are looking for $S_{n,m} \operatorname{mod} 10$.  If we know $S_{n,m} \operatorname{mod} 2$ and $S_{n,m} \operatorname{mod} 5$, the we can get $S_{n,m} \operatorname{mod} 10$ by the Chinese remainder theorem.

Since $k^m$ is even if and only if $k$ is even, modulo $2$ the value just depends on the number of odd positive integers $\leq n$.  This is odd when $n$ is if the form $4k+1$ and even for $n$ of the form $4k+3$.  Since $2021=4\cdot505+1$, $S_{2021,m}\equiv1\operatorname{mod} 2$ for all $m\geq1$.

To get $S_{n,m} \operatorname{mod} 5$, note that for all $k$ and $m$ we have $k^{m+4}\equiv k^m\operatorname{mod} 5$ by Fermat's little theorem, so we only need to compute the answers for $m=0,1,2,3$.  Since you know the values of $S_{n,m}$ for these values of $m$, and $2021\equiv1\operatorname{mod} 5$, it will be easy to finish.

I leave that up to you.

Title: $\sum_{k=1}^{n} k^m \mod 10$


答: Hint: $k^5 \equiv k \bmod 10$ for all $k \in \mathbb Z$.

Title: $\sum_{k=1}^{n} k^m \mod 10$

Body:
With $S_{(n,m)}=\sum_{k=1}^{n} k^m$, I study $S_{(n,m)}\mod 10$ for n=2021.
More than just solving this specific exercise, I would like to know what general reasoning would be wise for dealing with such problems.
My first ideas are to :

search for common properties or links between $n \times(n+1) \over 2$, $n \times(n+1) \times (2 \times n +1) \over 6$, $\big({(n+1) \times (n+2) \over 2}\big)^2$ etc.

search for 2021 properties...

Can somebody give me some advice in order to build an accurate reasoning ?

Thanks in advance.

Romain


问:
Logical notation in Ultraproducts Section of Book $\lambda i$. What does it mean?
Does anyone understand the notation $\lambda i$ circled in this image?

It is found in Logic and Structure By Dirk van Dalen, in the section on Ultraproducts. I can't find any clue as to what that notation is.

答: That looks like a lambda-abstraction in the lambda calculus that was introduced by Alonzo Church. It is a way of writing an anonymous function. For example $\lambda i. M$ is an abstraction that can be applied to an argument $N$, so that $N$ replaces every free occurrence of $i$ in $M$. That operation is called $\beta$-reduction. Have a look at https://en.wikipedia.org/wiki/Lambda_calculus as a start.


问:
The range of the derivative of a differentiable function
I read somewhere that, given a function $f$ differentiable on $[a,b]$, the range of $f'$ can be

(1) a closed interval or

(2) an open interval or

(3) a half-open interval or

(4) an unbounded interval

Can someone give an example for each one ?

答:
Let's see: For (1), any $f$ with continuous derivative works. The typical example for (4) is the continuous extension of $x^2\sin(1/x^2)$ to $[0,1]$.

An example for (2) is a bit harder. This comes from

Bernard R. Gelbaum, John M. H. Olmsted. Counterexamples in Analysis, Dover Books on Mathematics, 2003.

Differentiation is discussed in chapter 3. Consider the function
 $$ f(x)=\left\{\begin{array}{cc}x^4e^{-x^2/4}\sin(8/x^3)&x\ne0,\\ 0&x=0,\end{array}\right. $$
for $-1\le x\le 1$. Its derivative is
 $$ f'(x)=\left\{\begin{array}{cc}e^{-x^2/4}\left[\bigl(4x^3-\frac12x^5\bigr)\sin(8/x^3)-24\cos(8/x^3)\right]&x\ne0,\\ 0&x=0,\end{array}\right. $$
which has range $(-24,24)$ (in any neighborhood of zero). A proof is in their book (pp. 37-38). The reference they provide is

John M. H. Olmsted. Advanced calculus, Appleton Century Crofts, Inc.,
  New York, 1961.

I include the graph of $f'$ produced by WolframAlpha:

Once we have an example for (2), an example for (3) is easy: Simply pick some $t\ne0$ where $f'(t)=0$, say $0<t<1$, and replace $f$ to the right of $t$ with a continuous function $g$ such that $g(t)=f(t)$, $g'(t)=0$, and $g'$ has range $[-24,0]$.


问:
Find matrix $\mathbf{V}$ that gives constant value of $\mathrm{tr}(\mathbf{U}^T\mathbf{V}\mathbf{U})$ for degenerate eigenvectors $\mathbf{U}$
I have a symmetric matrix $\mathbf{A}\in\mathbb{R}^{n\times n}$ which has degenerate eigenvalues.
Let's say the degenerate eigenvectors of matrix $\mathbf{A}$ as $\mathbf{U}\in\mathbb{R}^{n\times m}$ (with $m < n$ as the number of degenerate eigenvectors for the same eigenvalue).
Please note that the degenerate eigenvectors matrix $\mathbf{U}$ can have multiple possible values while still being the valid eigenvectors of $\mathbf{A}$ because of its degeneracy.

How can I find a matrix/linear operator $\mathbf{V}\in\mathbb{R}^{n\times n}$ which gives the constant value of $\mathcal{L}=\mathrm{tr}(\mathbf{U}^T\mathbf{VU})$ for all possible values of $\mathbf{U}$?
Or what property does $\mathbf{V}$ has to have?

答:
If I have understood correctly, your question can be equivalently expressed as follows.

Let $S$ denote an $m$-dimensional subspace of $\Bbb R^n$ (which happens to correspond to a single eigenspace of a symmetric matrix $\mathbf A$). Characterize the matrices $\mathbf V \in \Bbb R^{n \times n}$ such that for every $\mathbf U$ whose columns form an orthonormal basis of $S$, the function $\mathcal L(\mathbf U) = \operatorname{tr}(\mathbf U^T \mathbf V\mathbf U)$ is constant.

In fact, this will be true for every matrix $\mathbf V$.

Let $\mathbf U_0$ denote a matrix with orthonormal columns whose columns span $S$. $\mathbf U$ will be another such matrix if and only if there exists an orthogonal matrix $\mathbf W$ such that $\mathbf U = \mathbf U_0 \mathbf W$. With that, we find that
$$
\operatorname{tr}(\mathbf U^T \mathbf V\mathbf U) = \\
\operatorname{tr}((\mathbf U_0 \mathbf W)^T \mathbf V(\mathbf U \mathbf W)) = \\
\operatorname{tr}(\mathbf W^T \mathbf U_0^T \mathbf V\mathbf U_0 \mathbf W) = \\
\operatorname{tr}(\mathbf W \mathbf W^T \mathbf U_0^T \mathbf V\mathbf U_0) =\\
\operatorname{tr}(\mathbf U_0^T \mathbf V\mathbf U_0)
$$


问:
Vector dot product equation
if $a\cdot b=a\cdot c$ and $a$ is not $0$,

Is $b = c$?

How should the solution be? I have to prove.

答:
If $\vec a\cdot \vec b=\vec a\cdot \vec c$, then $\vec a \cdot (\vec b-\vec c)=0$.  That means $\vec a\perp \vec b-\vec c$.

It does not necessarily mean $\vec b-\vec c=\vec 0$.

For example, $(0,1)\cdot(1,0)=(0,1)\cdot(2,0)$, but $(1,0)\ne(2,0)$.

Title: Vector dot product equation


答:
A normal vector to the plane has to be orthogonal to the directing vector to the line contained in the plane, $\vec u=(1,3,2)$, and to the directing vector of the intersection of the given planes, $\vec v=(1,1,-1)$.

An  obvious candidate is $\:\vec u\times \vec v=(5,-3,2)$.

Furthermore, the plane contains the point $(1,0,0)$, so the equation is
$$\color{red}{5x-3y+2z=}5\cdot 1-3\cdot 0+2\cdot 0=\color{red}5.$$


问:
Only one value for $\sin^{-1}|\sin\theta|$
I'm supposed to prove that $\cos^{-1}x=2\sin^{-1}\sqrt{\frac{1-x}{2}}$.

Substituting x for $\cos(2\theta)$, the right-hand expression simplifies like;
$$2\sin^{-1}\sqrt{\sin^2\theta}= 2\sin^{-1}|\sin\theta|$$

How does this work out to just $\theta$ when the sine's negative?

答: Here's an idea: choose the smallest range of $\theta\ge0$ you can that covers all $x\in[-1,\,1]$, then see what simplification that allows. In particular, you should find that not only can you drop the modulus signs, but that the $\arcsin$ then cancels the $\sin$.


问:
Finding the mean and standard deviation on a z-score
This is a problem that I don't know what to do. Our teacher didn't even teach us how to do this so I'm quite lost on how to explain this.

Here is the given question

In a given word problem, it is customary that all values are given. In the case of a word problem with normal distribution, it is common that the values of the score, the mean, and the standard deviation are given to compute for the value z. Likewise, it is also common that the value of z is already given. But what if in a given word problem concerning normal distribution, the values of the score (X) and the mean are missing and are required to be solved, how will you solve it? What if the word problem requires a percentage? How should you solve it?

答:
The usual statement for a standard score is $z=\frac{x-\mu}{\sigma}$ and you can solve for any one of the four given the other three with

$z=\frac{x-\mu}{\sigma}$ (assuming $\sigma > 0$)

$\sigma=\frac{x-\mu}{z}$  (if $z=0$ then $x=\mu$ and you cannot solve for $\sigma$; otherwise $z$ and $x-\mu$ must have the same sign)

$x = \mu+z \sigma $

$\mu = x- z \sigma$

If you also have a probability such as $p=\mathbb P(Z\le z)$ and a normal distribution then you can use the standard normal cumulative distribution function and its inverse to say $p = \Phi(z)$ or $z=\Phi^{-1}(p)$.  So knowing $p$ or $z$ implies you know the other.

You can then extend this for example to $\mathbb P(X \le x)  = \Phi\left(\frac{x-\mu}{\sigma}\right) $ or $x = \mu+\Phi^{-1}(p) \sigma $ or similar results.

In many problems, not knowing two of $x$ or $\mu$ or $\sigma$ or both of $z$ and $p$ could cause difficulties in being able to determine them.

Probability percentages may be easier to handle rewritten as decimal fractions: for example $95\% = 0.95$.


问:
Homology of Möbius Strip
How does one calculate the homology groups of the Möbius strip?

I'm thinking of two methods.

Use cellular homology. I tried to draw a delta-complex structure of the Möbius strip but I'm not sure if I'm right? I basically have a rectangle with opposite ends identified, so vertices (top-down) on the left are $a$ and $b$. On the right, they are $b$ and $a$. I know that there is an edge $e$ that joins from $b$ to $a$ and this is for both the left and right side of the rectangle. But this is not yet a delta complex structure, so I feel like I need to draw another edge across the rectangle from bottom left to top right, connecting $b$ to $b$. But how are these remaining 3 edges labelled? If I label the top and bottom edge, I think I'm creating another surface - the torus, RP2 or Klein bottle and so that's not the way to go. Does it mean this method is not possible?

Use $H_n(X^k,X)$ somehow, where $X^k$ is the $k$ skeleton of $X$. But I don't quite know how to proceed with this.

Any pointers in the right direction is greatly appreciated. Thanks!

答:
I've attached a model for the Möbius band. The vertices are $a$ and $b$. The ones that eventually get glued together are given the same letter. The edges are labeled $A$, $B$ and $C$. There are two edges labelled $A$, and they have arrows. The must be glued so that the arrows agree, i.e. you need to give a half twist before you glue. The arrows on $B$ and $C$ are only there because we must orient simplices. Finally, $\alpha$ is the one face. I don't know how to put a clockwise arrow around it, so please add one.

To find the homology groups, we must look at the images and the kernels of the boundary maps. Consider the series of maps $0 \to F \to E \to V \to 0$, where $F$ stands for faces, $E$ for edges and $V$ for vertices. In between each is a boundary map.

Consider $\partial : 0 \to F$. The image and the kernel are both $0$.

Consider $\partial : F \to E$. We have $\partial \alpha = 2A+B+C$ and so the image is non-empty. There was only one face, so the image is isomorphic to $\mathbb{Z}$. The only face had a non-zero image so the kernel is $0$.

Consider $\partial : E \to V$. We have $\partial A = b-a$, $\partial B = a-b$ and $\partial C = a-b$. Up to an integer, the images are all $a-b$ and so the image is one dimensional: $\mathbb{Z}$. There were three edges, and the image was one dimensional, so the kernel must be two dimensional: $\mathbb{Z}^2$.

Consider $\partial : V \to 0$. We have $\partial a = \partial b = 0$ and so the image is $0$. There are two vertices, and so the kernel must be $\mathbb{Z}^2.$

We can put all of this together. The group $H_2(M,\mathbb{Z})$ is given by the quotient of the kernel of $F \to E$ by the image of $0 \to F$, i.e. $0/0 \cong 0$.  The group $H_1(M,\mathbb{Z})$ is given by the quotient of the kernel of $E \to V$ by the image of $F \to E$, i.e. $\mathbb{Z}^2/\mathbb{Z} \cong \mathbb{Z}$. The group $H_0(M,\mathbb{Z})$ is given by the quotient of the kernel of $V \to 0$ by the image of $E \to V$, i.e. $\mathbb{Z}^2/\mathbb{Z} \cong \mathbb{Z}$. Hence:

\begin{array}{ccc}
H_2(M,\mathbb{Z}) &\cong& \{0\} \\
H_1(M,\mathbb{Z}) &\cong& \mathbb{Z} \\
H_0(M,\mathbb{Z}) &\cong& \mathbb{Z}
\end{array}

Title: Homology of Möbius Strip


答:
The principal range of $\arccos$ is $[0,\pi]$ and that of $\arcsin$ is $[\frac\pi2,\frac\pi2]$.

On letting $x = \cos(2\theta)$,  $-1\le x\le1 \Rightarrow 0\le2\theta\le\pi$

Now,  $0\le2\theta\le\pi \Rightarrow 0\le\theta\le\frac{\pi}{2}$ and in the first quadrant, $\sin$ is positive. So $|\sin\theta| = \sin\theta$

Title: Only one value for $\sin^{-1}|\sin\theta|$

Body:
I'm supposed to prove that $\cos^{-1}x=2\sin^{-1}\sqrt{\frac{1-x}{2}}$.

Substituting x for $\cos(2\theta)$, the right-hand expression simplifies like;
$$2\sin^{-1}\sqrt{\sin^2\theta}= 2\sin^{-1}|\sin\theta|$$

How does this work out to just $\theta$ when the sine's negative?


问:
Can a projection operator increase the $L_\infty$ norm?
Suppose $P$ is a $n\times n$ matrix of real values satisfying $P^2=P$ so that $P$ is a projection matrix. Is it possible that

$$\|Px\|_\infty > \|x\|_\infty$$

for some $x\in \mathbb{R}^n$? Here $\|y\|_\infty := \max_i |y_i|~.$

If possible, please give an example. If not possible, how can I prove it?

答:
Yes, in fact we can find an orthogonal projection with this property. If $u = (4/5,3/5)^T,$ $v = (1,1)^T$, and $P = uu^T$, then we find that
$$
Pv = u(u^Tv) = \frac 7{25}(4,3)^T  = (1.12,0.84)^T.
$$
We find that $\|Pv\|_\infty = 1.12 > \|v\|_\infty = 1$.

In fact, I would conjecture that if $x$ is a vector whose entries are of equal absolute value and $P$ is any orthogonal projection onto a subspace not containing any of the the standard axes, then we will necessarily have $\|Px\|_\infty > \|x\|_\infty$.


问:
Existence of a global solution to a matrix Riccati differential equation - indefinite & constant terms?
Consider the symmetric matrix Riccati differential equation :

$$
P'(t) = Q + X^\intercal P(t) +  P(t) X +  P(t)UP(t)
$$
Where everything is symmetric and the coefficients are constant (independent of time). The quadratic term U and the term Q are not definite, instead have a mix of positive and negative eigenvalues (say half half), in different bases.

If it can help, U can be written as :
$$
U = -U_1+\gamma U_2
$$
where $U_1$ is symmetric singular and $U_2$ symmetric semidefinite positive.

Is it possible to tackle the issue of finding conditions for boundedness of P ?

答:
This answer does not give a complete answer, but hopefully does help you gain some insights. Namely, it is possible to use a coordinate transformation in which the dynamics is linear, which might be easier to analyse. Namely, if $\bar{P}$ is a stationary solution to

$$
P'(t) = Q + X^\top P(t) +  P(t)\,X +  P(t)\,U\,P(t), \tag{1}
$$

so

$$
Q + X^\top \bar{P} +  \bar{P}\,X +  \bar{P}\,U\,\bar{P} = 0, \tag{2}
$$

then the transformed coordinates can be obtained using $P(t) = \bar{P} + Y^{-1}(t)$, which is equivalent to

$$
Y(t) = (P(t) - \bar{P})^{-1}. \tag{4}
$$

By using $(2)$ the dynamics of $Y(t)$ can be shown to simplify to

$$
Y'(t) = -\left(U + Y(t)\,(X^\top + \bar{P}\,U) + (X + U\,\bar{P})\,Y(t)\right). \tag{5}
$$

By using vectorization and the Kronecker product this can be written as the following system of linear first order ordinary differential equations in $x(t) = \text{vec}(Y(t))$

$$
x'(t) = M\,x(t) - \text{vec}(U), \tag{6}
$$

$$
M = -\left((X + U\,\bar{P}^\top) \otimes I + I \otimes (X + U\,\bar{P})\right), \tag{7}
$$

with $I$ the identity matrix of the same size as $P(t)$.

Solutions to $(2)$ can be found using the eigendecomposition of the following matrix

$$
Z =
\begin{bmatrix}
X & U \\ -Q & -X^\top
\end{bmatrix} = V\,\Lambda\,V^{-1}, \tag{8}
$$

with $\Lambda$ a diagonal matrix containing the eigenvalues of $Z$ and $V$ containing the corresponding eigenvectors. By defining $W$ as a matrix constructed from any selection of half of the columns of $V$, with

$$
W =
\begin{bmatrix}
W_1 \\ W_2
\end{bmatrix}, \tag{9}
$$

where $W_1$ and $W_2$ are of the same size as $P(t)$. Then a solution to $(2)$ can be found using

$$
\bar{P} = W_2\,W_1^{-1}. \tag{10}
$$

It can be noted that $V$ always has an even number of columns because the dimensions of $(8)$ are double that of $P(t)$.

In order for $P(t)$ to remain bounded requires that $Y(t)$ remains nonsingular, which I don't think can be shown easily, but the fact that it has linear dynamics hopefully makes it easier. Two things can be noted that if $M$ is Hurwitz $\lim_{t\to\infty} x(t) = M^{-1}\,\text{vec}(U)$ or if $M$ has at least one eigenvalue with positive real part $\lim_{t\to\infty} \|x(t)\| \to \infty$ for almost all initial condition with $x(t)$ dominated by the eigenmode of $M$ with the eigenvalue which has the largest positive real part. If transforming $M^{-1}\,\text{vec}(U)$ or the eigenvector of the dominant mode back into a square matrix is nonsingular implies that $\lim_{t\to\infty} Y^{-1}(t)$ should remain bounded. If this isn't the case then $\lim_{t\to\infty} P(t)$ should become unbounded. However, I am not sure this is a sufficient condition and there are probably a couple caveats, such as when there are multiple eigenvalues which have equal largest positive real part.


问:
Why are the forward error not computable?
I am quite new to Numerical Analysis and I was wondering if someone could guide me into understanding this concept presented below.

Thank you.

答:
Consider the problem of computing $y = f(x)$ where $f : \mathbb{R} \rightarrow \mathbb{R}$. Let $\hat{y}$ denote the computed value, i.e., the value returned by the computer. The forward absolute error $e$ is the difference between the target value and the computed value, i.e., $$e = y - \hat{y}.$$ If $y \not = 0$, then the forward relative error $r$ is defined and is given by $$r = \frac{e}{y} = \frac{y-\hat{y}}{y}.$$ Some texts are only interested in the absolute value of the errors, but from time to time it is useful to know if the computed value is too large or too small. Regardless, we cannot apply the definition and compute $e$ and $r$ unless we know the exact value of $y$.

The backward error is interesting because it allows us to bound the forward error using the relevant condition number. The connection between the relative backward error, the relative forward error and the relative condition number is discussed in this answer to a related question.

It is possible to say substantially more about this as there are several different kinds of condition numbers and the abstract definition of the backward error is not included in the reference. These could be topics for other questions.


问:
Hamming cube has two minimal cuts $S_1, S_2$ such that $E = S_1 \cup S_2$
Consider the d-dimensional Hamming cube $H_d$.

Show: There are two minimal cuts $S_1, S_2$ of $H_d$ such that $E = S_1 \cup S_2$.

Hint: Consider the case where $d=2$ and construct two minimal cuts for $H_3$. Then generalize the construction for all $d \ge 2$.

Thanks for your ideas/thoughts..

答:
Here's a rough outline of something to try.

Consider the following two cuts (red and blue) of $H_2$:

Each cut divides the cube into two equal pieces, that are symmetric if you swap all the $0$s with $1$s and vice-versa. This is easy to work with, so we try generalise this to cuts of $H_d$. We note that $H_{d+1}$ is the cartesian product $H_{d+1}= H_d \times K_2$, so we get $H_{d+1}$ by taking two copies of $H_d$ and joining the corresponding vertices with edges.

So to get $H_3$ with the right cuts, take two copies of $H_2$ (call them copy $0$ and copy $1$), reflect copy $1$ about the $y$-axis for ease of display, and swap the colours of the cuts in copy $1$:

Now you want to assign the edges between copy $0$ and copy $1$ to the two cuts in a way that gives you one big pair of cuts, that are again symmetric. This is easy enough, pick a  vertex $u$ in copy $0$ of $H_2$, and colour the edge $\{0u, 1u\}$ red. Then for every neighbour $v$ of $u$, colour the edge $\{0v, 1v\}$ blue, and so on, until all the in-between edges are coloured (such that no neighbours in a copy of $H_2$ have the same colour edge going between the two copies).

This procedure scales. So if you have these two minimal, symmetric cuts $S_1$ (red) and $S_2$ (blue) in $H_d$, you can get two minimal symmetric cuts for $H_{d+1}$ by taking the two copies of $H_d$, swapping the cuts in one, and giving the edges between the two copies 'alternating' colours. To see that this works in general, it helps a lot to notice that $H_d$ is bipartite.


答:
Apply $\cos$ to both sides. RHS gives $x$ and LHS too:

$$\cos\left(2\arcsin\sqrt{\frac{1-x}{2}}\right)=1-2\sin^2\left(\arcsin\sqrt{\frac{1-x}{2}}\right)=\\=1-2\left(\frac{1-x}{2}\right)=x$$


问:
Show that there is no finite ordered integral domain.

Show that there is no finite ordered integral domain.

I know the definition of integral domain and an ordered set, but how do I prove that no such integral domain can be finite?
Also, i think $\mathbb{Z}_2$ is a finite integral domain and as well as ordered,since $0$ is the smallest element. So how to prove this fact isn't clear to me. Can someone please help? Thanks in advance.

答:
I will use the definition of ordered integral domain in the book Number Systems and Foundations of Analysis by Elliot Mendelson.

First let $(D,+,\times)$ be a integral domain, then this integral domain with an binary relation $<$ on $D$ satisfying:

(O1): $x \not\lt x$ (Irreflexivity)

(O2): $[x \lt y \land y \lt z] \Rightarrow y\lt z$ (Transitivity)

(O3): $[x < y \lor x=y \lor y < x]$ (Trichotomy)

(O4): $x<y \Rightarrow x+z < y+z$

(O5): $[x<y \land 0<z] \Rightarrow x\times z < y \times z$

Then $(D,+,\times,<)$ is an ordered integral domain (with the stipulation that $1 \neq 0$).

Now we proceed by contradiction, assume that $D$ is a finite set, then it is impossible to form a strictly increasing sequence on $D$, from this follows that $D$ have some greatest element $m$ such that $(\forall x)([x \in D \land x \neq m] \Rightarrow x < m)$

We need to derive the simple fact that $x \neq 0 \Rightarrow 0<x^2$, thus first assume $x \neq0$, then by (O3) we have $0<x\lor x<0$.

Considering $0<x$ we can apply (O5) which gives $[0<x \land 0<x] \Rightarrow 0\times x < x \times x$, that is $0<x^2$.

Now considering $x<0$, apply (O4) using $-x$ as $z$, which gives $x+(-x) < 0+(-x)$, which is $0<-x$, now we apply (O5), $[0<-x \land 0<-x] \Rightarrow 0\times -x < -x \times -x$ from which follows again $0 < x^2$.

Since we stipulated that $1 \neq 0$, we have now that $0 < 1^2$, but as $1 \times 1 = 1$, it is just $0 < 1$, now apply (O4) with our greatest element $m$ in place of $z$ and we got $0+m < 1+m$ which is $m<1+m$, since $D$ is closed under addition we got $1+m \in D$ which contradicts the fact that $m$ is the greatest element.

Therefore, its possible to create a strictly increasing sequence with the element of $D$ since the set dont have a greatest element, then if an integral domain have an order relation satisfying those five properties it cant have a finite underlying set.


问:
Simplification of matrix operation
I’m struggling on how can I simplify the following equation such that I could calculate $\Psi^{-1}$ only once.

$$a^T\Psi^{-1}a - b^T\Psi^{-1}b$$.

Since $\Psi^{-1}$ is the inverse of a correlation matrix, $a$ and $b$ are vectors of same lengths that dimension of the correlation matrix.

Any hint on how can I simplify that?

答:
Here is a way to rewrite the expression that you might like. We have
$$
a^T\Psi^{-1}a - b^T\Psi^{-1}b = \pmatrix{a & b} ^T \psi^{-1} \pmatrix{a & -b}.
$$

Title: Simplification of matrix operation


问:
Why is the limit of $\frac{11^n n^4 + 9^n n^9} {7^{2n} +1}=0$?
$$\lim_{n\to+\infty}\frac{11^n n^4 + 9^n n^9} {7^{2n} +1}=0$$

I have used the fact exponentials grow faster than polynomials for this question but was wondering whether there was an algebraic way e.g. with the ratio lemma.

答:
Consider the sequence

$$s_n:=\frac{n^k}{a^n}$$ where $a>1$.

We have

$$\frac{s_{n+1}}{s_n}=\left(\frac{n+1}n\right)^k\frac1a.$$

Now if we take

$$n>\frac1{\sqrt[k]a-1},$$ we have $$\frac{s_{n+1}}{s_n}<\left(\frac{n+1}n\right)^k\frac1a=r<1$$ and by induction

$$s_{n+m}<s_nr^m\to0.$$

This proves that any exponential grows faster than any power and it is not big deal to generalize to any polynomial.

E.g. with

$$\frac{n^9}{7^{2n}},$$ as of $n\ge2$ the sequence is dominated by a geometric series of common ratio $\dfrac{19683}{25088}.$


问:
Understanding proof in Markov Chains and Mixing Times
Kindly help me to understand the proof of Lemma 4.10 from "Markov Chains and Mixing Times" (https://pages.uoregon.edu/dlevin/MARKOV/markovmixing.pdf).

The goal ist to prove $d(t) \le \bar{d}(t)$, with

$d(t) := \max\limits_{x \in S} \Vert P^t(x, \cdot) - \pi \Vert_{TV}$ and $\bar d(t):= \max\limits_{x,y \in S} \Vert P^t(x, \cdot) - P^t(y, \cdot) \Vert_{TV}$ and $P$ being the transition matrix and $S$ the state space and $\pi$ the stationary distribution of a Markov chain.

The proof starts with a reminder of the definition of the stationary distribution $\pi$,

$\pi(A) = \sum\limits_{y \in S}\pi(y)P^t(y,A)$.

Then the first equation says:

$P^t(x,A) - \pi(A) = \sum\limits_{y \in S} \pi(y) \big( P^t(x,A)-P^t(y,A) \big)$.

I understand that we can fill in the definition of $\pi(A)$ in $P^t(x,A) - \pi(A)$ what gives us

$P^t(x,A) - \pi(A) = P^t(x,A) - \sum\limits_{y \in S}\pi(y)P^t(y,A)$.

But I fail to see why I can multiple $P^t(x,A)$ with $\pi(y)$, sum over all $y \in S$ and then use the distriutive law to end up with $\sum\limits_{y \in S} \pi(y) \big( P^t(x,A)-P^t(y,A) \big)$.

I'm sure I just miss one step but would appreciate your feedback!

答: You mean the step you don't understand is why $P^t(x,A) - \sum_{y \in S} \pi(y) P^t(y,A) = \sum_{y \in S} \pi(y)(P^t(x,A) - P^t(y,A))$? This just works because $P^t(x,A)$ doesn't depend on $y$ so it can be pulled out of the sum of that term, and then of course the sum of $\pi(y)$ is $1$.


问:
Proving collinear points - geometry
Straight line $p$ contains the orthocenter $H$ of an acute triangle $ABC$, where $p$ intersects sides $AB$ and $AC$ with $p$ intersecting $AC$ in the point $P$. Straight line $q$ also contains $H$, is perpendicular to $p$ and intersects $AB$ and $BC$ with $q$ intersecting BC in the point $Q$. Straight line containing $A$, parallel to $q$ and a straight line containing $B$, parallel to $p$ intersect in the point $R$. Prove that $P,Q$ and $R$ are collinear.

A question I came across in a practice book, I am not sure even where to start... I'm allowed to use trigonometry, similar triangles, point-line-plane.

答:
Let $B'$ be a point on $AC$ such that $\angle AB'B = 90°$. Denote the intersection point of $AR$ and $BB'$ by $B''$.

Case 1: When $B''$ is out of $\overline {BH}$

Joint $AH$, $B'N$ and $PB''$. Let $M$ be the intersection point of $BR$ and $QH$, and $N$ be the intersection point of $AR$ and $HP$.

Since $\triangle BMH$ ~ $\triangle HNB''$, $BM \times NB'' = MH \times HN$.

Also, $\angle AB'H = \angle ANH = 90°$ implies $B'B''NP$ and $AHNB'$ are cyclic quadrilateral.

$\Rightarrow \angle B''PN = \angle HB'N = \angle HAN = 90° - \angle AHN$

$\Rightarrow B''P \perp AH$, $B''P$//$BC$

$\Rightarrow \triangle B''PN$ ~ $\triangle QBM$

$\Rightarrow PN\times MQ = BM \times NB'' = MH \times HN = RN \times MR$

$\Rightarrow \triangle NPR$ ~ $\triangle MRQ$

$\Rightarrow \angle QRM = \angle RPN$

$\Rightarrow P$, $Q$ and $R$ are collinear

Case 2: When $B''$ lies on $\overline {BH}$

A → B and B → A

As same as Case 1

Title: Proving collinear points - geometry


问:
Find the internal angle of the triangle between $\overrightarrow a \& \overrightarrow b $.
If $\overrightarrow a  = \hat j + \sqrt 3 \hat k;\overrightarrow b  =  - \hat j + \sqrt 3 \hat k;\overrightarrow c  = 2\sqrt 3 \hat k$ form a triangle then find  the internal angle of the triangle between $\overrightarrow a \& \overrightarrow b $.

My approach is as follow

Let $\overrightarrow a  + \overrightarrow b  = \overrightarrow c $

$\overrightarrow a  = \hat j + \sqrt 3 \hat k;\overrightarrow b  =  - \hat j + \sqrt 3 \hat k;\overrightarrow c  = 2\sqrt 3 \hat k$
$ \Rightarrow {\left( {\overrightarrow a  + \overrightarrow b } \right)^2} = {\left( {\overrightarrow c } \right)^2} \Rightarrow {\overrightarrow a ^2} + {\overrightarrow b ^2} + 2\overrightarrow a .\overrightarrow b  = {\overrightarrow c ^2}$

$ \Rightarrow \overrightarrow a .\overrightarrow b  = 2 \Rightarrow $

$\cos \theta  = \frac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|}} = \frac{1}{2} = \frac{\pi }{3}$

But official answer is $\frac{2\pi }{3}$ where I am making mistake

答: You have calculated the external angle. Draw vector signs on your diagram, it would be clear to you.</p>


问:
How to prove that $\frac{|x+y+z|}{1+|x+y+z|} \le \frac{|x|}{1+|y|+|z|}+\frac{|y|}{|1+|x|+|z|}+\frac{|z|}{1+|x|+|y|}$
I need to prove $\frac{|x+y+z|}{1+|x+y+z|} \le \frac{|x|}{1+|y|+|z|}+\frac{|y|}{1+|x|+|z|}+\frac{|z|}{1+|x|+|y|}$. I've tried to use triangle inequality or to explore the form of $(a+b+c)^2$ but it won't get me anywhere. I would be grateful for some suggestions.

答:
as $|x+y+z|\le |x|+|y|+|z|$

let $|x|=a,|y|=b,|z|=c$
it suffices to prove $$\sum \frac{a}{1+b+c}\ge \sum \frac{a+b+c}{1+a+b+c}$$

indeed by C-S/titu's lemma; $$\sum \frac{a}{1+b+c}=\sum \frac{a^2}{a+ba+ca}\ge \frac{{(a+b+c)}^2}{a+b+c+2(ab+bc+ca)}\ge \frac{a+b+c}{1+a+b+c}$$

Here we used $$\frac{2ab+2bc+2ca}{a+b+c}\le a+b+c$$ which is just $a^2+b^2+c^2\ge 0$


问:
If $ f_n\to f $ pointwise, and $f_n $ are increasing continuous functions, and $ f $ is continuous, then $f_n\to f $ uniformly
let $ f_{n}:[a,b]\to\mathbb{R} $ be a sequence of increasing and continuous functions (each function is continuous and increasing, it dosent mean that the sequence is increasing). And assume that exists $ f:[a,b]\to\mathbb{R} $ such that $ f $ is continuous and  $ f_{n}\to f $ pointwise.
I have to prove that the convergence is also uniformly.

I tried to  assume by contradiction that there is no uniformly convergence, but I got stuck. Now I have a new intuitio, since $ f_n $ and $ f $ are continious in $ [a,b] $ they are uniformly continuous. so actually for any $ x $ I can write:

$ |f_{n}\left(x\right)-f\left(x\right)|\leq|f_{n}\left(x\right)-f_{n}\left(x+\delta\right)|+|f\left(x\right)-f\left(x+\delta\right)|+|f_{n}\left(x+\delta\right)-f_{n}\left(x+\delta\right)| $

Where $ \delta $ would fit to some $ \varepsilon $ from the unifromly continuous definition, where the first two terms can be small as I want, I can continue with it untill $ x+k\delta $ "is close enough" to $ b $  where $ k \in \mathbb{R} $. But again, Im not sure how to do it since $ \delta $ would have to depend on $ n $.

Any hints would help.
Thanks in advance

答:
Let $\varepsilon>0$ and
$$
V_n=\{x\in [a,b] : |f_n(x)-f(x)|<\varepsilon \}, \quad n\in\mathbb N.
$$
Clearly, the $V_n$'s are open relative to $[a,b]$ and $\bigcup_{n\in\mathbb N}V_n=[a,b]$, and hence the $V_n$'s are an open cover of $[a,b]$.

Since $[a,b]$ is compact, there is a finite sub-cover
$$
[a,b]= V_{n_1}\cup\cdots\cup V_{n_k}
$$
where $n_1<n_2<\cdots<n_k$. But since $f_n$ is increasing, then $m<n\Longrightarrow V_m\subset V_n$, and hence
$$
[a,b]=  V_{n_k},
$$
i.e.,
$$
V_{n_k}=\{x\in [a,b] : |f_{n_k}(x)-f(x)|<\varepsilon \}=[a,b]
$$
and hence
$$
|f_{n_k}(x)-f(x)|<\varepsilon,
$$
for all $x\in [a,b]$, and since $f_n$ is an increasing sequence, then
$$
0 \le f(x)-f_{n}(x)<\varepsilon, \quad \text{for all $n\ge n_k$ and $x\in[a,b]$.}
$$


问:
Number of Jordan cells size $j+1$
Could you tell me how to prove that the number of Jordan cells of size $j+1$ is equal to $\dim V^j/V^{j-1}$(quotient space)?

$V^j = \ker (f-\lambda)^j$

I know that for $U_1, U_2, ..., U_s$ - Jordan subspaces of endomorphism $f: X \rightarrow X$ we have $X=U_1\bigoplus U_2\bigoplus ...\bigoplus U_s$

and $\ker (f - \lambda id_X)^p = \ker (f|_{U_1} - \lambda id_{U_1})^p \bigoplus ... \bigoplus \ker (f|_{U_s} - \lambda id_{U_s})^p$.

Can this be useful?

I would appreciate all your help.

Thank you.

答:
It seems the following.

Consider a Jordan form $\Lambda$ of the operator $f$. When you consider the sequence $\{(\Lambda-\lambda I)^j:j\in\mathbb N\}$ of powers, at $j$-th step the rank of each initial Jordan $\lambda$-cell of size $\ge j$ is decreasing by $1$. The sum of all these “1” should be exactly equal to the total decrement of rank, which is equal to $\dim V^j /V^{j−1}$.  Let $n_j$ denotes that the the number of Jordan $\lambda$-cells of size $j$. Then we have

$\dim V^1=n_1+n_2+\dots$

$\dim V^2/V^1=\dim V^2-\dim V^1=n_2+\dots$

$\dots$

So

$n_1=\dim V^1-(\dim V^2-\dim V^1) =2\dim V^1-\dim V^2$.

$n_2=(\dim V^2-\dim V^1)- (\dim V^3-\dim V^2) =2\dim V^2-\dim V^1-\dim V^3$.

$n_3=(\dim V^3-\dim V^2)- (\dim V^4-\dim V^3) =2\dim V^3-\dim V^2-\dim V^4$.

$\dots$

This formulae are different from $n_{j+1}=\dim V^j-\dim V^{j-1}$.  Maybe something is wrong.

Consider an example
$A=\left(\begin{matrix}2 & 1 & 0 & 0 & 0\\0 & 2 & 1 & 0 & 0\\ 0 & 0 & 2 & 0 & 0\\ 0 & 0 & 0 & 2 & 1\\ 0 & 0 & 0 & 0 & 2\end{matrix}\right)$

Let $\lambda=2$. Then $n_1=n_4=0$, $n_2=n_3=1$, $\dim V^0=0$, $\dim V^1=2$, $\dim V^2=4$, $\dim V^3=5$, and $\dim V^4=5$.

$n_1=0=2\cdot 2-4=2\dim V^1-\dim V^2$.

$n_2=1=2\cdot 4-2-5=2\dim V^2-\dim V^1-\dim V^3<>\dim V^1/V^0=2$.

$n_3=1=2\cdot 5-4-5=2\dim V^3-\dim V^2-\dim V^4<>\dim V^2/V^1=2$.

Maybe I did an error somewhere.

Title: Number of Jordan cells size $j+1$


问:
Partition of $[a,b]$ such that $f(x)$ is within $\frac{\epsilon}{2}$ on each subinterval.
Suppose that $f: [a,b] \rightarrow \mathbb{R}$ is continuous on [a, b].
Show that for every $\epsilon > 0$ there exists an ordered list of points $$x_0=a<x_1<x_2<...<x_n=b$$ such that the set $$E_k= \{f(x)|x \in [x_{k-1},x_k]\}$$ satisfies $$sup(E_k)-inf(E_k)< \epsilon$$ for $k=1,...,n$.

My attempt was starting with point a, we can find smallest $\delta_1$ such $f(a+\delta_1)$ equals to $f(a)+\frac{\epsilon}{2}$ or $f(a)-\frac{\epsilon}{2}$, whichever comes first given that f is continuous. Thus, the interval $[a, a+ \delta_1 ]$ satisfies the desired sup-inf condition. Then we can continue the construction off of point $a+\delta_1$ and get a set of $\{\delta_{i}\}$. Then I am stuck on how to show that the the set is finite such that we can reach point b in finite partitions. Any opinion on the viability of this approach? In retrospect, this seems a bit brute-forced and a lot of complications would arise (what I have now is mega-hand-wavy) in the end, so I am wondering if there would be a better approach to this?

答:
First of all, I am going to assume that this is an introductory problem, and that you haven't yet been introduced to the concept of uniform continuity, which would render this problem trivial.  (If you can use the fact that $ f $ is uniformly continuous on $ [a,b] $, that is the easy, correct solution.)

If you haven't yet learned of uniform continuity and want to write a proof "from scratch", your approach seems reasonable.  You are also correct in noting that it is not "obvious" that the $ \delta_i $ that you chose will reach $ b $ in a finite number of steps.  We can proceed by contradiction.  Suppose you don't reach $ b $.  Then you end up with a sequence of points $ x_1, x_2, x_3, \ldots $, with $ x_1 < x_2 < x_3 < \ldots < x^* = \lim x_i \leq b $.  Now you can demonstrate that $ f $ is discontinuous at $ x^* $, which is a contradiction.  If you want a hint about how to show discontinuity at $ x^* $, let me know, but give it a try yourself first.


问:
Prove $\| A(A^TA)^{-1}A^T\|_2 = 1$, when $rank (A)=n$
Given a matrix $A \in R^{m \times n}$, which is full ranked, prove $\| A(A^TA)^{-1}A^T\|_2 = 1$.
So my book already proved this by a nice geometrical approach.
But I wondered if we could use numerical methods to solve this.
I'm trying to use QR decomposition but well I couldn't proceed. Any ideas?

答:
This is a quick exercise if we use the SVD. Write $A = U \Sigma V^T$, where $U$ is orthogonal of size $m$, $V$ is orthogonal of size $n$, and $\Sigma$ is the "diagonal" matrix
$$
\Sigma = \pmatrix{\Sigma_0 \\ 0_{(m-n)\times n}}, \quad \Sigma_0 = \pmatrix{\sigma_1 \\ & \ddots \\ && \sigma_m\\
}.
$$
Because $A$ has full rank, all $\sigma_i$ are non-zero. Verify that the matrix $M = A(A^TA)^{-1}A$ can be written as
$$
M = U \pmatrix{I_n&0\\ 0&0}U^T.
$$
Because $U$ and $U^T$ are orthogonal, this is in fact an SVD of $M$, which means that the singular values of $A$ are the diagonal entries of the diagonal matrix in the middle. From there, $\|M\|_2$ is the largest singular value of $M$, which is $1$.


问:
Let N be a poisson random variable with parameter µ, and conditional on N, let M have a binomial distribution with parameters N and p. Then...
https://i.gyazo.com/3f6b7a71c5d32fa7c309a2e959610480.png

Help me understand what happens from line 2 to line 3. What rules are applied?

答:
Easy....

$$p^k(1-p)^{n-k}\mu^n=\underbrace{(\mu p)^k}_{\text{out of the sum}}\cdot[\mu(1-p)]^{n-k}$$


问:
Graphic Interpretation of the Linear Approximation for Vector Valued Functions
A linear approximation of a function $f:\mathbb{R} \to \mathbb{R}$ at the point $x=a$ draws a tangent line to $f(x)$ at $(a,f(a))$ and approximates values of $f(x)$ for $x$ near $a$ by using the function of the tangent line. The same thing is done for functions $f: \mathbb{R}^2 \to \mathbb{R}$, but with a tangent plane instead of a line. Here's my question:

I'm having trouble understanding what the linear approximation for a vector valued function $\vec{F}:D \subset \mathbb{R}^n \to \mathbb{R}^m$, which has the form $\vec{F}(\vec{x}) \approx \vec{F}(\vec{a}) + A(\vec{a})(\vec{x}-\vec{a})$ where $A_{ij}(\vec{a}) = \displaystyle{\frac{\partial F_i}{\partial x_j}(\vec{a})}$, repersents graphically. Any help is appreciated, thanks in advance :)

P.S. if you want to demonstrate it with an example, that's fine. Just please use a function in $\mathbb{R}^3$.

答:
To begin with, I would like to point out that the graphical interpretations of functions $f_1:\mathbb R \to \mathbb R$ and $f_2:\mathbb R^2 \to \mathbb R$ are just interpretations of the function using mathematical objects that never appear in the actual functions themselves.

For $f_1:\mathbb R \to \mathbb R$ we plot points in a Cartesian plane whose coordinates are in $\mathbb R^2.$ Where do you see $\mathbb R^2$ in the definition of $f_1$? It is nowhere in there. But by taking the Cartesian product of the domain and range we get $\mathbb R \times \mathbb R \sim \mathbb R^2,$
which we use as coordinates of a two-dimensional plane.
The set $\{(x,y) \in \mathbb R^2 \mid f_1(x) = y \}$ is a figure in that plane;
we call that figure the graph of $f_1,$
and if $f_1$ is differentiable at $x_0$ then part of that graph is a curve through the point with coordinates $(x_0,f(x_0))$ that has a tangent line at that point.

This is a very useful visualization for a real-valued function on the real numbers, but let's just be clear that it is something we invented by pulling numbers out of the function definition and recombining them in a way that the function definition never said we should do. (It also never said we could not do this, and it turns out that it works, so we do it.)

Likewise for $f_2:\mathbb R^2 \to \mathbb R$ we take each function value
$z = f_2(x,y)$ and plot it at the point $(x,y,z)$ in a three-dimensional Cartesian space with coordinates $\mathbb R^3,$
even though there is no three-dimensional space or the set $\mathbb R^3$ anywhere in the definition of $f_2.$

In short, we took the coordinates of the function's input and concatenated the function's output as a third coordinate in order to produce a graph.

You can do the same thing with $\vec F:D\subset\mathbb R^n \to \mathbb R^m.$
Take the $n$ coordinates of each $\vec x\in D$ and concatenate the $m$ coordinates of $\vec F(\vec x)$ to produce the coordinates of a point in an $(n+m)$-dimensional Cartesian space, $\mathbb R^{n+m}.$
The derivative of $\vec F$ at some input value $\vec x$ then corresponds to an $n$-dimensional hyperplane in that space.

This does not help much with visualization, however, since visualizing anything in more than three dimensions is very hard for most people.


问:
Problem with forces

A motor boat with the power to steer across a river at $30$ kmph is moving such that the bow is pointed in a northerly direction. The stream is moving eastward at $6$ kmph. The river is $1$ km wide. Where on the opposite side will the boat meet the land?

I wasted enough time on this problem already without getting any answer. Can I know how its
solution?

All I know is that the position vector of the boat, $t$ hours later, would be: $\vec{r}(t)= (6t)\hat{i}+(30t)\hat{j}$, but I'm not sure where to go from there. In other words, how to take advantage of the info given that the width of the river is $1$ km.

P.S. The book's answer just says: $200$m east of the initial point (and I have no clue what do they mean by east of initial point?)

Thanks in advance.

答: When the boat moves 1 km the stream takes it 6/30 km = 0.2 km downstream</p>

Title: Problem with forces


问:
How to establish the modus ponens inference rule in the LK sequent calculus for first-order logic?
How can the modus ponens inference rule
$$
\frac{\Gamma \vdash \varphi\hspace{1cm}\Delta\vdash \varphi\rightarrow\psi}{\Gamma,\Delta \vdash \psi}
$$
be established in the LK sequent calculus for first-order logic without resorting to the fact that this calculus is semantically complete?

答:
For future reference, I will generalize the result a little, and show that
$$
\dfrac{\Delta \vdash \Sigma, \varphi\rightarrow\psi\qquad\Gamma\vdash\Xi,\varphi}{\Delta,\Gamma\vdash\Sigma,\Xi,\psi}
$$
using a proof that is essentially the one given by Taroccoesbrocco, but a little more detailed:

$$
\dfrac{\Delta \vdash \Sigma,\varphi \to \psi\qquad\dfrac{\dfrac{\dfrac{}{\psi \vdash \psi}{\text{ax}}}{\dfrac{\Gamma,\psi\vdash\psi}{\dfrac{\psi, \Gamma \vdash \psi}{\dfrac{\psi, \Gamma \vdash \psi,\Xi}{\psi, \Gamma \vdash \Xi,\psi}{\text{x R}}}{\text{wk R}}}{\text{x L}}}{\text{wk L}}\qquad\dfrac{\Gamma \vdash \Xi,\varphi}{\dfrac{\Gamma \vdash \Xi,\varphi, \psi}{\Gamma \vdash \Xi, \psi, \varphi}{\text{x R}}}{\text{wk R}}}{\varphi \to \psi,\Gamma \vdash \Xi,\psi}{\to L}}{\Delta, \Gamma \vdash \Sigma, \Xi,\psi}{\text{cut}}
$$


问:
If $2 \sum_{j=0}^{n}a_jf(x_j)$ is a quadrature for $\int_{-1}^1f(x)dx$ exact for polynomials of degree $\le n$ and $x_j=-x_{n-j}$, show $a_j=a_{n-j}$.
This was an problem in today's exam in numerical methods and I would like to know if my solution is correct.

For approximating the integral $I(f) = \int_{-1}^{1} f(x) dx$ we take the quadrature formula
$$
Q(f)
= 2 \sum_{j = 0}^{n} \lambda_j f(x_j)
$$
with supporting points $x_0, \ldots, x_n \in [-1, 1]$, which is exact for polynomials with degree up to $n$.
Show that if $x_j = -x_{n - j}$ for $j \in \{ 0, \ldots, n \}$, we have $\lambda_j = \lambda_{n - j}$ for $j \in \{ 0, \ldots, n \}$.

What I've tried:
For an odd function $f$, we have
$$
Q(f)
= \sum_{j = 0}^{n} \lambda_j f(x_j) + \sum_{j = 0}^{n} \lambda_j f(x_j)
= \sum_{j = 0}^{n} \lambda_j f(x_j) + \sum_{j = 0}^{n} \lambda_{n - j} f(x_{n - j})
= \sum_{j = 0}^{n} (\lambda_j - \lambda_{n - j}) f(x_j).
$$
Now let $f$ be an odd polynomial of degree less than or equal to $n$.
Then the quadrature formula is equal to the integral and as $f$ is odd, we have
$$
0 = I(f)
= Q(f)
= \sum_{j = 0}^{n} (\lambda_j - \lambda_{n - j}) f(x_j).
$$
Since this holds for all odd polynomials of degree less than or equal to $n$, I want to conclude that the coefficients $\lambda_j - \lambda_{n - j}$ are zero, i.e. $\lambda_j = \lambda_{n - j}$, but I am not sure if this is valid.

答:
There are several ways to proceed:

Note that terms $(\lambda_j - \lambda_{n-j}) f(x_j)$ and $(\lambda_{n-j} - \lambda_{j}) f(x_{n-j})$ are in fact the same for odd $f(x)$. So the summation may be reduced to $Q(f) = 2 \sum_{j=0}^{\lfloor \frac{n-1}{2}\rfloor} (\lambda_j - \lambda_{n-j}) f(x_j)$. Since all $x_j$ in this sum are negative, one may choose arbitrary values for $f(x_j)$, mirror them for positive $x$ and take $f(x)$ as interpolating polynomial.

We might use Lagrange basis interpolating polynomials $\ell_j(x)$. One of the equivalent definitions of these polynomials is given as
$$
\deg \ell_j(x) = n\\
\ell_j(x_k) = \delta_{jk} \equiv \begin{cases}
1, &k = j\\
0, &k \neq j
\end{cases}
$$
It is easy to see now that $\ell_j(-x)$ is also an $n$-degree polynomial and at the nodes it satisfies
$$
\ell_j(-x_k) = \ell_j(x_{n-k}) = \delta_{j,n-k} = \delta_{n-j,k} = \ell_{n-j}(x_k).
$$
Due to uniqueness we conclude that $\ell_j(-x) = \ell_{n-j}(x)$.
Consider now
$$
f(x) = \ell_j(x) - \ell_{n-j}(x).
$$
It is clear now that $f(x)$ is odd. Plugging it to $Q(f)$ yields
$$
0 = Q(f) = (\lambda_j - \lambda_{n-j}) - (\lambda_{n-j} - \lambda_j) =
2(\lambda_j - \lambda_{n-j}).
$$
so $\lambda_j = \lambda_{n-j}$.

There is also different approach to the problem. If $Q$ is exact for all polynomials of degree $n$ then we might apply it to $\ell_k(x)$:
$$
\int_{-1}^1 \ell_k(x) dx = I(f) = Q(f) = \sum_{j=0}^n \lambda_j \ell_k(x_j) = \lambda_j.
$$
On the other hand
$$
\int_{-1}^1 \ell_k(x) dx = \int_{-1}^1 \ell_k(-x) dx =
\int_{-1}^1 \ell_{n-k}(x) dx,
$$
so it is necessary that $\lambda_j = \lambda_{n-j}$.


问:
Complex roots of unity in a matrix

Let $\omega $ be a complex cube root of unity with $\omega \ne 1 $ and $P=[p_{ij}]$ be a $n \times n $ matrix with $p_{ij}=\omega^{i+j}$ . Then $P^2 \ne0$ when $n=$

(A) 57

(B) 55

(C) 58

(D) 56

Official answer is $B,C,D$. Considering the values are greater than $50$ if we see the answer, based on my assumption, the value of $n$ is not a multiple of $3$.

We need to find the square matrix whose square is not equal to zero matrix.
How do we find the value of $n$?

Title: Complex roots of unity in a matrix

Body:
$P^2$=$\begin{bmatrix}w^2 &w^3 &w^4& .&.&.\\
  w^3 \\
w^4 \\
.\\
.\end{bmatrix}$ $\begin{bmatrix}w^2 &w^3 &w^4& .&.&.\\
  w^3 \\
w^4 \\
.\\
.\end{bmatrix}$
=$\begin{bmatrix}w^4+w^6+w^8+...& .&.&.\\
  . \\
. \\
.\\
.\end{bmatrix}$
=$\begin{bmatrix}w+1+w^2+...& .&.&.\\
  . \\
. \\
.\\
.\end{bmatrix}$=$\begin{bmatrix}0+...& .&.&.\\
  . \\
. \\
.\\
.\end{bmatrix}$

 We deduce that $n$ should not be a multiple of $3$. Doubt?


问:
What three numbers when multiplied together will have the same number arrangement?
I have tried to look for 3 numbers that when multiplied together will produce the same arrangement. eg( 2x3x4 = 234);

xyz = 100x+10y+z

z(xy-1) = 100x+10y

so xy must not in fraction form.

Let x=y-a=z-b for some b>a>0. If a and b are zero, then from equation 1, we'll have

x^3=111x

which will give a decimal number for x, thus x,y and z should not be the same number.

If a or b is zero, then

x(x)(x+b)=111x+b

and b will be in decimal number. So let x<y<z. We'll get that ab<111. I have tried to put all the constraint and using programming to solve for x, y and z but only get numbers that solved equation 1 yet does not have the same arrangement as the example. How should I approach this problem?

Body:
$$xyz < x (10)(10) = 100x \le 100x + 10 y + z = [xyz]_{10}$$

In general, for any base $b$:

$$xyz < x(b)(b) = b^2x \le  b^2x + by + z = [xyz]_b$$


问:
Terminology confusion: Why is $z$ considered a free variable in the set $z^*=\{x \in z: \exists u (x \in u \land u \in z)\}$
In my book "The Foundations of Mathematics" by Kenneth Kunen, the following sentence is written:

In Comprehension [referencing the Comprehension Schema axiom], $\varphi$ can even have $z$ free - for example, it's legitimate to form $z^*=\{x \in z: \exists u (x \in u \land u \in z)\}$

I am slightly confused as to why $z$ is considered free in this set. The Comprehension Schema reads as follows:

Comprehension Schema: $\forall z \Big ( \exists y \forall x (x \in y \leftrightarrow x\in z \land \varphi(x) \big) \Big)$.

Swapping out $\varphi (x)$ for $\exists u (x \in u \land u \in z)$ we would create:

$\forall z \Big ( \exists y \forall x \big (x \in y \leftrightarrow x\in z \land \exists u (x \in u \land u \in z) \big ) \Big)$

The $z$ in the $\varphi$ formula is in the scope of the first universal quantifier at the beginning of this sentence. So why exactly is $z$ "free", as stated by Kunen? Does "free" mean different things depending on the context?

Thank you~

Edit: I figured I would put this here in case anyone else had a similar question.

The terms "bound" and "free" are always relative to a particular formula.

Consider the following:

Let $\psi_1 := \forall z \Big ( \exists y \forall x \big (x \in y \leftrightarrow x\in z \land \exists u (x \in u \land u \in z) \big ) \Big)$. In $\psi_1$, all variables are bound (i.e. no variable is free).

Let $\psi_2: = \exists y \forall x \big (x \in y \leftrightarrow x\in z \land \exists u (x \in u \land u \in z) \big )$. In $\psi_2$, variables $y,x,$ and $u$ are all bound but variable $z$ is free.

Let $\psi_3: = \forall x \big (x \in y \leftrightarrow x\in z \land \exists u (x \in u \land u \in z) \big )$. In $\psi_3$, variables $x$ and $u$ are bound, but variables $z$ and $y$ are free.

Let $\psi_4: = x \in y \leftrightarrow x\in z \land \exists u (x \in u \land u \in z)$. In $\psi_4$, variable $u$ is bound, but variable $z, y,$ and $x$ are free.

答:
Variable $z$ is free in formula $∃u(x∈u ∧ u∈z)$ because there is no quantifier binding it, while the two occurrences of variable $u$ are bound because they are in the scope of the quantifier $∃u$

The general form of the axiom schema is $A = \{ x \mid \varphi(x) \}$ where $x$ must be free in the formula $\varphi(x)$ specifying the condition that defines the set.

Thus, Kunen's formula amounts to:

$A_z = \{x \mid x∈z \land ∃u(x∈u ∧ u∈z) \}$.

The formula $\varphi(x)$ is $(x∈z \land ∃u(x∈u ∧ u∈z))$ with $x$ free.

The formula has an additional free variables (a parameter): $z$.

This means that we are defining a "families" of sets $A_z$: one for each value of $z$.

The set-forming operator $\{ x \mid \varphi(x) \}$ maps a formula ($\varphi$) into a term (a "name") denoting a set.

It binds the variable $x$. Thus, if there are no other free variables in the formula, it is a closed term.

The key-property of the operator is:

$\forall z [z \in \{ x \mid \varphi(x) \} \leftrightarrow \varphi (z)]$.

Having said that, how to apply it to Kunen's example ?

We have:

$∀z(∃y∀x(x∈y ↔ x∈z∧∃u(x∈u ∧ u∈z)))$.

We have a "specifying condition": $\psi(x,z) := x∈z ∧ ∃u(x∈u ∧ u∈z)$, with two free variables.

The axioms schema asserts that there is a set $y$ with all and only those elements $x$ that satisfy the condition:

$y = \{ x \mid  x∈z ∧ ∃u(x∈u ∧ u∈z) \}$.

Thus, applying the above formula:

$\forall x [x \in y \leftrightarrow (x∈z ∧ ∃u(x∈u ∧ u∈z))]$.


问:
Basic book about mathematical proofs
I'm looking for an understandable book about mathematical proofs. </p>

<p>As an engineer/computer scientist major I'm used to do higher math on a daily base, but whenever I'm asked in an exercise to "show" something I get the "where should I start? and what exactly do they want me to do?" syndrome. </p>

答: There are books devoted to your question, e.g., Daniel Solow's book, How to Read and Do Proofs: An Introduction to Mathematical Thought Processes, http://www.wiley.com/WileyCDA/WileyTitle/productCd-EHEP000562.html but I'm not sure that's the way to go. It might be better to just pick up a good textbook in Number Theory (like Niven, Zuckerman, and Montgomery) or Abstract Algebra (like Fraleigh) or Discrete Math (maybe Grimaldi, or Brualdi), to see how people actually do proofs when they need them. </p>

Title: Basic book about mathematical proofs


问:
Normalized spaces. Prove that if one closed ball nested into other one, then $r_1 \leq r_2 - ||x_1-x_2||$
Let $(X, ||*||)$  normalized spase.

Prove that if one closed ball nested into other one ($\overline{B_{r_1}}(x_1) \subset \overline{B_{r_2}}(x_2))$, then $r_1 \leq r_2 -||x_1-x_2||$, where $r_1, r_2$ are radiuses of such balls and $x_1, x_2$ are their centers.

I tried to transform this inequality into: $||x - y|| \geq r_2 - r_1$, and make something with it. Because we know that: $||x - y|| \geq ||x|| - ||y||$.

答:
If $x_1 = x_2$, then the result is obviously true. So let's suppose that $x_1 \neq x_2$. Consider
$$x'' = x_1 + \frac{r_1}{||x_1-x_2||}(x_1-x_2)$$

You have
$$||x''-x_1|| = \frac{r_1}{||x_1-x_2||}||x_1-x_2||=r_1$$

so $x'' \in \overline{B}(x_1, r_1)$. Because $\overline{B}(x_1, r_1) \subset \overline{B}(x_2, r_2)$, you deduce that $x'' \in \overline{B}(x_2, r_2)$, i.e.
$$||x''-x_2|| \leq r_2 \quad \quad (*)$$

But $$||x''-x_2|| = \left|\left|x_1 + \frac{r_1}{||x_1-x_2||}(x_1-x_2)-x_2\right|\right|$$
$$=\left|\left|\left(x_1-x_2 \right) \left( 1 + \frac{r_1}{||x_1-x_2||}\right)\right|\right|$$
$$= \left( 1 + \frac{r_1}{||x_1-x_2||}\right) ||x_1-x_2|| = ||x_1-x_2|| + r_1$$

So the relation $(*)$ gives directly $$||x_1-x_2|| + r_1 \leq r_2$$

which is exactly what you asked for.


答:
The answer by @Yiorgos is the original proof. Here is another one I would like tu submit. As $(f_n)$ converges increasingly, $f \geqslant f_n$ for all $n$ and $|f(x) - f_n(x)| = f(x) - f_n(x)$. Let $M_n = \sup_{x\in [a,b]}f(x) - f_n(x) \geqslant 0$. The fact that $f$ is the uniform limit of $(f_n)$ is equivalent to the fact that $M_n$ converges to $0$.

By the increasing hypothesis of $(f_n)$, $M_n$ is a decreasing sequence. It is moreover positive, thus have a limit, say $m$. By continuity, there exists $x_n \in [a,b]$ such that $M_n = f(x_n)-f_n(x_n)$. By compacity of $[a,b]$, $(x_n)$ has a converging subsequence $(x_{\varphi(n)})$ with limit $x \in [a,b]$.

To conclude, notice that $M_{\varphi(n)}=f(x_{\varphi(n)}) - f_{\varphi(n)}(x_{\varphi(n)})$, and thus :
\begin{align}
m = \lim_{n\to\infty}M_{\varphi(n)} = \lim_{n\to \infty}f(x_{\varphi(n)}) - f_{\varphi(n)}(x_{\varphi(n)}) = f(x) - f(x) = 0
\end{align}

Title: If $ f_n\to f $ pointwise, and $f_n $ are increasing continuous functions, and $ f $ is continuous, then $f_n\to f $ uniformly

Body:
let $ f_{n}:[a,b]\to\mathbb{R} $ be a sequence of increasing and continuous functions (each function is continuous and increasing, it dosent mean that the sequence is increasing). And assume that exists $ f:[a,b]\to\mathbb{R} $ such that $ f $ is continuous and  $ f_{n}\to f $ pointwise.
I have to prove that the convergence is also uniformly.

I tried to  assume by contradiction that there is no uniformly convergence, but I got stuck. Now I have a new intuitio, since $ f_n $ and $ f $ are continious in $ [a,b] $ they are uniformly continuous. so actually for any $ x $ I can write:

$ |f_{n}\left(x\right)-f\left(x\right)|\leq|f_{n}\left(x\right)-f_{n}\left(x+\delta\right)|+|f\left(x\right)-f\left(x+\delta\right)|+|f_{n}\left(x+\delta\right)-f_{n}\left(x+\delta\right)| $

Where $ \delta $ would fit to some $ \varepsilon $ from the unifromly continuous definition, where the first two terms can be small as I want, I can continue with it untill $ x+k\delta $ "is close enough" to $ b $  where $ k \in \mathbb{R} $. But again, Im not sure how to do it since $ \delta $ would have to depend on $ n $.

Any hints would help.
Thanks in advance


问:
How do I mathematically model the changes in the shape of a circle projected onto non-flat surfaces at different angles?
Imagine I shine a flashlight straight down at my feet on a flat surface and it creates a perfect circle of radius 1. If I begin to move my arm forward the light will be projected on the ground at an angle and will draw an ellipse.

I want to know the math that dawns the ellipse at any angle. So at 90 degrees (straight down) its a perfect circle of x^2 + y^2 = 1. What would be the equation of the ellipse if I was shining the flashlight at the ground at 45 degrees? I'm just not sure how to approach this problem, so advice is appreciated. I took linear algebra years ago and I feel like there's some transformation matrix for this somewhere, but my search results havnt given me anything that I found useful yet.

答:
Here is a solution using analytic geometry. Let us consider that the flashline is at the height of the wrist $W$ situated at a unit height (i.e, for example at 1 meter from the ground). I will take a different convention:

$$\alpha=\text{angle between vertical and axis of the light}$$

Let us denote by $\beta$ the half aperture angle of the cone.

Let us express the equation of the light cone $(C)$ by the constraint:

$$M \in (C) \ \ \iff \ \ \dfrac{\vec{WM}.\vec{u}}{\|\vec{WM}\|\|\vec{u}\|}=\cos\beta$$

which becomes, with $M(x,y,z)$, $W(0,0,1)$, $\vec{u}(\tan \alpha,0,-1)$

$$\dfrac{x\tan \alpha - (z-1)}{\sqrt{1+(\tan \alpha)^2}\sqrt{x^2+y^2+(z-1)^2}}=\cos \beta\tag{1}$$

$$\cos \alpha (x\tan \alpha - (z-1))=\cos \beta \sqrt{x^2+y^2+(z-1)^2}\tag{2}$$

$$(x\sin \alpha - \cos \alpha (z-1))=\cos \beta \sqrt{x^2+y^2+(z-1)^2}\tag{3}$$

Intersection of the light cone with the ground is obtained by setting $z=0$ in (3) giving, squared:

$$(x\sin \alpha + \cos \alpha)^2=\cos^2 \beta (x^2+y^2+1)\tag{4}$$

We are able, using (4) to plot the corresponding ellipses.

We can also turn (4) into the canonical form:

$$\dfrac{(x-c)^2}{a^2}+\dfrac{y^2}{b^2}=1$$

with in particular simple forms for

$$a=\dfrac{1}{2}(\tan(\alpha+\beta)-\tan(\alpha-\beta))$$

$$c=\dfrac{1}{2}(\tan(\alpha+\beta)+\tan(\alpha-\beta))$$


问:
Homeomorphism between $k$-simplex and a product of $k$ unit intervals
How to prove that there is a homeomorphism between the $k$-simplex $[v_0, v_1, \dots, v_k]$ and a product of $k$ unit intervals?

This question is motivated by 'Lecture Notes on Elementary Topology' by I. Singer, where it is claimed that this "is not difficult to prove using barycentric coordinates" (Chapter 4, Page 71).

The $k$-simplex is defined as
$$
[v_0, \dots, v_k]
= \left\{\sum_{i=0}^{k} a_i v_i
\text{ such that } a_i \geq 0 \text{ and }
\sum_{i=0}^{k} a_i = 1
\right\} .
$$
$a_i$ are called barycentric coordinates.
(EDIT: $\{v_0, \dots, v_k\}$ is a set of $C$-independent vectors, namely the set $\{v_1-v_0, \dots, v_k-v_0\}$ is linearly independent.)

For example, given three noncolinear vectors, $[v_0, v_1, v_2]$ is the triangle with vertices located at $v_0$, $v_1$ and $v_2$.
According to the theorem that I am asking about, there should be a homeomorphism between that triangle and the unit square $[0,1]\times[0,1]$.

I do not know how to find that homeomorphism. My only idea is that the barycentric coordinates $(a_0, a_1, \dots, a_k)$, due to the condition $\sum_{i=0}^{k} a_i = 1$, are homeomorphic to a subset of $\Pi_{i=1}^{k} [0,1]$ by the map $f:[v_0, \dots, v_k] \rightarrow \Pi_{i=0}^{k}[0, 1]$ defined as
$$f(a_0, a_1, \dots, a_k) = (a_1, a_2, \dots, a_k).$$
But that is not a homeomorphism to $\Pi_{i=1}^{k} [0, 1]$: for example, the point $(1,1,\dots,1) \in \Pi_{i=1}^{k} [0, 1]$ has no nonempty inverse image.

I am a beginner in elementary topology and would appreciate very much any help.

答:
You do not tell us what $v_0,\ldots, v_k$ are, but certainly they are noncolinear vectors in some $\mathbb R^N$.

Let $\Delta^k = [0,e_1,\ldots,e_k] \subset \mathbb R^k$ be the standard $k$-simplex, where the $e_i$ are the standard basis vectors of $\mathbb R^k$. It is the set $\{ (x_1,\ldots,x_k) \in \mathbb R^k \mid x_i \ge 0, \sum_{i=1}^k x_i \le 1 \}$. The $x_i$ are the barycentric coordinates of $x = (x_1,\ldots,x_k)$ associated to $e_i$, $x_0 = 1 - \sum_{i=1}^k x_i$ is the barycentric coordinate associated to $0$.

First note that each $k$-simplex is homeomorphic to $\Delta^k$. In fact
$$h : \Delta_k \to [v_0,\ldots, v_k], h(x_1,\ldots,x_n) = (1 - \sum_{i=1}^k x_i)v_0  + \sum x_i v_i$$
is a homeomorphism. Observe that $[v_0,\ldots, v_k]$ is contained in any ambient $\mathbb R^N$ wit $N \ge k$.

This means that it suffices to show that $\Delta^k$ is homeomorphic to the cube $I^k \subset \mathbb R^k$.

Le $Q = \{ (x_1,\ldots,x_k) \mid x_i \ge 0 \text{ for all } i \}$. Consider the norms $\lVert (x_1,\ldots, x_k) \rVert_1 = \sum_{i=1}^k \lvert x_i \rvert$ and $\lVert (x_1,\ldots, x_k) \rVert_\infty = \max_{i=1}^k \lvert x_i \rvert$. Both are continuous real-valued functions on $\mathbb R^k$. Let $B_1$ and $B_\infty$ denote the closed unit balls with respect to these norms, i.e. $B_1 = \{ x \in \mathbb R^k \mid \lVert (x_1,\ldots, x_k) \rVert_1 \le 1\}$, similarly $B_\infty$. Then $\Delta^k = B_1 \cap Q$ and $I^k = B_\infty \cap Q$. Define
$$\phi : \Delta^k \to I^k, \phi(x) = \begin{cases} \dfrac{\lVert x \rVert_1}{\lVert x \rVert_\infty}   x \ne 0 \\ 0 & x = 0 \end{cases}$$
$$\psi : I^k \to \Delta^k, \psi(x) = \begin{cases} \dfrac{\lVert x \rVert_\infty}{\lVert x \rVert_1}   x \ne 0 \\ 0 & x = 0 \end{cases}$$
It is readily checked that $\psi \circ \phi = id$ and $\phi \circ \psi = id$, thus $\phi$ and $\psi$ are bijections which are inverse to each other. Both maps are obviously continuous in all $x \ne 0$. But they are also continuous in $0$ since
$$\lVert \phi(x) - \phi(0) \rVert_\infty = \lVert \phi(x) \rVert_\infty = \lVert x \rVert_1 = \lVert x -  0 \rVert_1 ,$$
$$\lVert \psi(x) - \psi(0) \rVert_1 =  \lVert x -  0 \rVert_\infty .$$
This means that $\phi,\psi$ are homeomorphisms.

Edited:

Usually $\mathbb R^n$ is endowed with the Euclidean norm $\lVert (x_1,\ldots, x_k) \rVert_2 = \sqrt{\sum_{i=1}^k x_i^2 }$ which generates the standard Euclidean topology. It is well-known that all norms on $\mathbb R^n$ are equivalent, i.e. the topology generated by any norm is the Euclidean topology. Therefore, to show that a function $D  = \mathbb R^n \to R  = \mathbb R^n$ is continuous, we may take any norm $\lVert - \rVert_D$ on the domain $D$ and any norm $\lVert - \rVert_R$ on the range $R$, similarly for maps $\mathbb R^n \to \mathbb R$. However, in the context of this answer we do not need the general norm equivalence theorem. Just note that
$$\lVert (x_1,\ldots, x_k) \rVert_\infty \le \lVert (x_1,\ldots, x_k) \rVert_2 = \sqrt{\sum_{i=1}^k x_i^2 } \le \sqrt n \max_{i=1}^k \lvert x_i \rvert = \sqrt n \lVert (x_1,\ldots, x_k) \rVert_\infty$$
$$\lVert (x_1,\ldots, x_k) \rVert_\infty \le \lVert (x_1,\ldots, x_k) \rVert_1 \le n \lVert (x_1,\ldots, x_k) \rVert_\infty$$


问:
Relation between eigenvector/values of A and R in QR factorization
It's intuitively clear that in QR factorization of given matrix, $A \in \mathbb{R}^{n\times n}$, i.e. $A =
 QR\,$ [where $Q \in \mathbb{R}^{n\times n}$ is orthogonal ($Q^T Q = I$) and $R \in \mathbb{R}^{n\times n}$ is upper triangular] because $R$ is undergoing an orthogonal transformation (via $Q$) the eigenvalues of A and R must be the same and that we would expect the eigenvectors of $A$ to be orthogonally transformed versions of those of $R$. However, I am not sure how to prove this. Any suggestions?

答:
The relationship that you suspect does not hold. As an example, consider
$$
A = \left(\begin{array}{cc} \frac{3}{5} & \frac{7}{5}\\ -\frac{4}{5} & -\frac{1}{5} \end{array}\right) \implies
Q = \left(\begin{array}{cc} \frac{3}{5} & \frac{4}{5}\\ -\frac{4}{5} & \frac{3}{5} \end{array}\right), \quad
R = \left(\begin{array}{cc} 1 & 1\\ 0 & 1 \end{array}\right).
$$
$R$ has $1$ as its only eigenvalue, and the associated eigenspace is spanned by $(1,0)$. $A$ has eigenvalues $\frac{1}{5}\pm\frac{2\sqrt{6}\,{}}{5}\mathrm{i}$, each with a corresponding complex eigenvector.


问:
Agree (or disagree) inequality $a\leq b$ then $a^n\leq b^n$ for all $n\geq 0, n\in\mathbb{R}$ and $a,b\in[0,1]$?

Let $a,b\in[0,1]$.  Agree (or disagree) if $a\leq b$ then $a^n\leq
 b^n$ for all $n\geq 0, n\in\mathbb{R}$?

I try to find example. Let $a=0.2$ and $b=0.5$. I try to plot the function as below. Red graph is $0.5^n$ and blue graph $0.2^n$. So, I can conclude if $a\leq b$ then $a^n\leq b^n$ for all $n\geq 0$, $n\in \mathbb{R}$.

Generally, does that inequality holds? I'm confused to conclude it, (maybe there are counterexamples)?

答:
If $a\leqslant b$, then either $b=0$, in which case $a=0$ and so $a^n\leqslant b^n$ indeed, or $b>0$, in which case $\frac ab\leqslant1$. And, for any $n\in\Bbb N$,$$\frac ab\leqslant1\implies\left(\frac ab\right)^n\iff\frac{a^n}{b^n}<1\iff a^n\leqslant b^n.$$

Note that I did not use at all the fact that $a,b\leqslant1$.


答:
If you want to get the full logic behind it, consider:

To travel 1 km (to the other side), the boat will take $\dfrac{1\ \text{km}}{30\ \text{km/h}} = \dfrac{1}{30} $ h.

In $\dfrac{1}{30}$ h, the stream would have carried the boat $\dfrac{1}{30}\ \text{h}\times6\ \text{km/h} = 0.2$ km eastwards.

Title: Problem with forces

Body:
A motor boat with the power to steer across a river at $30$ kmph is moving such that the bow is pointed in a northerly direction. The stream is moving eastward at $6$ kmph. The river is $1$ km wide. Where on the opposite side will the boat meet the land?

I wasted enough time on this problem already without getting any answer. Can I know how its
solution?

All I know is that the position vector of the boat, $t$ hours later, would be: $\vec{r}(t)= (6t)\hat{i}+(30t)\hat{j}$, but I'm not sure where to go from there. In other words, how to take advantage of the info given that the width of the river is $1$ km.

P.S. The book's answer just says: $200$m east of the initial point (and I have no clue what do they mean by east of initial point?)

Thanks in advance.


问:
Definition of free and bound variables in relation to specific formulas
Is it possible to say that a variable occurs free in a given formula, even though the formula is within the scope of a quantifier that binds the variable? For example, considering the statement $\exists y \forall x P(x, y)$, is it correct to say that the variable $y$ occurs free in the formula $\forall x P(x, y)$ and bound in the entire proposition $\exists y \forall x P(x, y)$? If not, is there a proper definition for the variable $y$ in $\forall x P(x, y)$? For instance, is it a constant in the formula or something like that?

答:
Yes, correct: In $\forall x P(x,y)$ as a subformula of $\exists y \forall x P(x,y)$, $y$ is free.

This is in fact sometimes used in definitions; for instance, you might run into something like "... if $\phi$ is of the form $\exists x \psi$, where $x$ occurs free in $\psi$..." to say that the quantifier is non-vacuous and actively binds occurrences of $x$ in $\psi$.


答:
Yes, it holds.
$$a\le b \Rightarrow \frac{a}{b}\le1 \Rightarrow \left(\frac{a}{b}\right)^n \le1 \Rightarrow a^n \le b^n$$ for all $$n \in \mathbb{N}.$$

Title: Agree (or disagree) inequality $a\leq b$ then $a^n\leq b^n$ for all $n\geq 0, n\in\mathbb{R}$ and $a,b\in[0,1]$?

Body:
Let $a,b\in[0,1]$.  Agree (or disagree) if $a\leq b$ then $a^n\leq
 b^n$ for all $n\geq 0, n\in\mathbb{R}$?

I try to find example. Let $a=0.2$ and $b=0.5$. I try to plot the function as below. Red graph is $0.5^n$ and blue graph $0.2^n$. So, I can conclude if $a\leq b$ then $a^n\leq b^n$ for all $n\geq 0$, $n\in \mathbb{R}$.

Generally, does that inequality holds? I'm confused to conclude it, (maybe there are counterexamples)?


问:
Check if $2^n = O(n^2)$ , prove it by using induction on $(2^n / n^2) > n$
I don't know how to solve by induction  $(2^n / n^2) > n$ to check if $2^n = O(n^2)$ . I know that
$2^n = O(n^2)$ will be false by using the big O definition, but I don't know how to prove it by using induction on  $(2^n / n^2) > n$. I don't even completely understand how the $(2^n / n^2) > n$ was created as it was already in the question.

答:
$2^n = O(n^2)$ would mean that $\dfrac{2^n}{n^2} \longrightarrow 0$ when $n \longrightarrow \infty$, by unwrapping the big $O$ definition.

So all you need to show is that this ratio does not got to $0$ at $\infty$.

Can you see why this follows from $\dfrac{2^n}{n^2} > n$ ?


问:
Prove $\int_{0}^{1}\!f(x)\cos(k\pi x)dx>0$
I have a positive continuous function $f(x)$ that is strictly monotonically decreasing on the interval $(0,1)$. That is, $f(x)>0$ and $f'(x)<0$ for $x\in(0,1)$. I want to show that the integral $$I = \int_{0}^{1}\! f(x)\cos(k\pi x)dx$$
is always positive, for all $k = 0,1,2,...$

Edit:
As pointed out in the comments, it's not possible to show this for general $f(x)$, but I have numerical eveidence to suggest that this is true for $$f(x) = {\frac {\left( 3\,{\epsilon}^{2}-\cos \left( \pi\,x
  \right) +1 \right) }{ \left( 2\,{\epsilon}^{2}
  -\cos \left( \pi\,(x) \right) +1 \right) ^{3/2}}}$$
where $0<\epsilon<1$.

Body:
For $k=1$:
$$I=\int_{0}^{1} f(x) \cos \pi x ~ dx ~~~(1)$$
Use: $\int_{0}^{a} f(x) dx=\int_{0}^{a} f(a-x) dx$, then
$$I=\int_{0}^{1} -f(1-x) \cos \pi x ~dx~~~~~(2)$$
Add (1) and (2), to gt
$$2I=\int_{0}^{1}[f(x)-f(1-x)] \cos \pi x ~dx~~~~(3)$$
$x <1-x, x \in (0,1/2)$, as $f(x)$ is decreasing function, we get
$$f(x)>f(1-x), x \in (0,1/2), ~f(x)< f)1-x), x \in (1/2,1)~~~~~~(4)$$
Then $$2I=\int_{0}^{1/2} [f(x)-f(1-x)] \cos \pi x~dx+\int_{1/2}^{1} [f(x)-f(1-x)] \cos \pi x ~dx~~~~(5)$$
$\cos \pi x <0, ~in~ (0,1/2)$ and $\cos \pi x >0 ~ in~ (1/2,1)$.
So it follows that $$2I>0$$


问:
What have I done wrong in solving this problem with indices rules?
The question asks to simplify:

$$\left(\dfrac{25x^4}{4}\right)^{-\frac{1}{2}}.$$

So I used $(a^m)^n=a^{mn}$ to get

$$\dfrac{25}{4}x^{-2} = \dfrac{25}{4} \times \dfrac{1}{x^2} = \frac{25}{4x^2} = \frac{25}{4}x^{-2}$$

However, this isn't the answer, and I can't see what I've done wrong.

This is what the mark scheme says:

$$\left(\dfrac{25x^4}{4}\right)^{-\frac{1}{2}} = \left[\left(\frac{4}{25x^4}\right)^{\frac{1}{2}} \text{ or } \left(\frac{5x^2}{2}\right)^{-1} \text{ or } \frac{1}{\left(\dfrac{25x^4}{4}\right)^{\frac{1}{2}}}\right] = \frac{2}{5}x^{-2}$$

To me, my answer looks more simple than theirs, and I can't see what I've done wrong.

Body: $$\Big(\frac{25x^4}{4}\Big)^{-\frac{1}{2}}=\Big(\frac{4}{25x^4}\Big)^{\frac{1}{2}}=\frac{4^{\frac{1}{2}}}{25^{\frac{1}{2}}(x^4)^{\frac{1}{2}}}=\frac{2}{5x^2}$$


问:
In which condition，the triangle have the maximum triangle area？
These trangles have the same perimeter 2q，in which condition，can we have the trangle with the maximum area?
I have tried to use the Heron's formula. $S=\sqrt{q(q-a)(q-b)(q-c)}$
I have sought  the partial derivative and tried to find the stationary point but the calculation  is complicated. Is there any strategy can we use to simply the calculation?

答: Assuming that a triangle of maximal area exists you can argue as follows: A triangle with vertices $A$, $B$ and a given perimeter has the third vertex on a certain ellipse with foci $A$, $B$. Choosing $C$ at a co-vertex of this ellipse makes the area largest and leads to an isosceles triangle. It follows that the triangle with largest area is isosceles with respect to all pairs of vertices, hence equilateral.


答:
Careful, you want to prove $\|x_1-x_2\| \le r_2-r_1$, not $\,\ge$.

Assume $x_1\ne x_2$ and define $y = (1-\lambda) x_1+\lambda x_2$.

If we set $\lambda = \frac{r_1}{\|x_1-x_2\|}$ then $$\|x_1-y\| = \lambda\|x_1-x_2\| = r_1$$ so $y \in \overline{B}(x_1,r_1)$. Therefore $y \in \overline{B}(x_2,r_2)$ as well so
$$\|x_1-x_2\| - r_1 = (1-\lambda)\|x_1-x_2\| = \|y-x_2\| \le r_2$$
which proves the desired claim.

Title: Normalized spaces. Prove that if one closed ball nested into other one, then $r_1 \leq r_2 - ||x_1-x_2||$

Body:
Let $(X, ||*||)$  normalized spase.

Prove that if one closed ball nested into other one ($\overline{B_{r_1}}(x_1) \subset \overline{B_{r_2}}(x_2))$, then $r_1 \leq r_2 -||x_1-x_2||$, where $r_1, r_2$ are radiuses of such balls and $x_1, x_2$ are their centers.

I tried to transform this inequality into: $||x - y|| \geq r_2 - r_1$, and make something with it. Because we know that: $||x - y|| \geq ||x|| - ||y||$.


问:
Can this problem have solutions for $N>3$:Find three $(N=3)$ positive integers a, b and c such that: $a+b+c=k^2$, $a+b=t^2$, $b+c=m^2$ and $a+c=n^2$
Can this question have solutions for N>3:

Find three ($N =3$) positive integers a, b and c such that: $a+b+c=k^2$, $a+b=x^2$, $b+c=m^2$ and $a+c=n^2$; $x, m, n∈\mathbb   N$

Solution:

Let $a+b+c=(x+1)^2=x^2+2x+1$

If $a+b=x^2$ then third number $c=2x+1$.

If $b+c=(x-1)^2=x^2-2x+1=m^2$ then :

$a=4x$, so $b=x^2-4x$

Now due to statement we must have $a+c=6x+1=n^2$

$n^2=6x+1$ can have infinite solutions such as:

$(x, n^2)=(20, 121=11^2), (60, 361=19^2), (140, 841=29^2)\cdot\cdot\cdot$

Which give:

$(a, b, c, k)=(80, 320, 41, 21), (240, 3360, 121, 61), (560, 19040, 281, 141),\cdot\cdot\cdot$

I tried to solve this problem for $N=4$, four numbers but no success. Now I have two questions:

1-Does this problem have solutions for N>3?

2-Any idea for better method?

答:
This answer is only for the case $N=4$. I fail to see how this method could help resolve the cases for larger $N$.

Consider the equations

$$a + b = u^2, c + d = v^2$$
$$a + c = w^2, b + d = x^2$$
$$a + d = y^2, b + c = z^2$$
$$a + b + c + d = k^2$$

Now the first three pairs of equations give

$$u^2 + v^2 = w^2 + x^2 = y^2 + z^2 = k^2$$

so a reasonable approach would be to investigate Pythagorean triples.

Not all Pythagorean triples generate positive integers solutions, but a solution could be generated if the pairs $(u,v), (w,x), (y,z)$ are "reasonably close". Consider

$$165^2+280^2 = 195^2+260^2 = 253^2+204^2 = 325^2$$

Take note of the parity as well (or multiply it by $2$ so you don't need to worry about it)

Solving the equations give

$$a = 11817, b = 15408, c = 26208, d = 52192$$

and more solutions can be generated this way.


答:
Partial answer: Your function $C:\ \Bbb{N}\ \longrightarrow\ \Bbb{N}$ can also be described as follows:
$$C(n)=\begin{cases}\varphi(n)&\text{ if $n$ is even}\\\sigma(n)&\text{ if $n$ is odd}\end{cases}.$$
Of course $\varphi(n)$ is even for every $n\geq3$, so for even $n$ we get $\varphi^{k}(n)\leq2$ for sufficiently large $k$. As $\varphi(2)=\varphi(1)=1$ this shows that every even number ends up at $1$.

Similarly $\sigma(n)$ is even for every odd $n$ that is not a perfect square, and so every odd number $n$ that is not a perfect square also ends up at $1$. So the question that remains is:

Is there an odd perfect square $n$ such that $\sigma^k(n)$ is an odd perfect square for all $k\in\Bbb{N}$?

This seems extremely unlikely, but I do not have a proof. The examples
$$\sigma(1^2)=1^2\qquad\text{ and }\qquad \sigma(9^2)=11^2,$$
show that it is possible for the divisor sum of an odd perfect square to be an odd perfect square again. Perhaps one can show that the number of prime factors is decreasing in any such sequence $(\sigma^k(n))_{k\in\Bbb{N}}$? Or perhaps there is a useful result on the related aliquot sequences, which have been studied in more detail?

Title: A Collatz like function or Eulers totient function vs. the sum of divisors?

Body:
Originally posted on MO, but it seems to be not a research level question:

Inspired by the Collatz function, I experimented with Euler's totient function $\phi$ vs. the sum of divisors function $\sigma$:

$$C(n) = \phi(n)\cdot\frac{(1+(-1)^n)}{2}+\sigma(n)\cdot\frac{(1-(-1)^n)}{2}$$

It seems that the only cycle is $1 \mapsto 1$ and that every number iterates to this cycle:

Experiment.

Example:

```<code>1
[1]
2
[2, 1]
3
[3, 4, 2, 1]
4
[4, 2, 1]
5
[5, 6, 2, 1]
6
[6, 2, 1]
7
[7, 8, 4, 2, 1]
8
[8, 4, 2, 1]
9
[9, 13, 14, 6, 2, 1]
10
[10, 4, 2, 1]
11
[11, 12, 4, 2, 1]
12
[12, 4, 2, 1]
13
[13, 14, 6, 2, 1]
14
[14, 6, 2, 1]
15
[15, 24, 8, 4, 2, 1]
16
[16, 8, 4, 2, 1]
</code>```

Can this be proven? (I am having trouble understanding why $\sigma(2n+1)$ where $2n+1$ is a square, will not be an odd square number?

Edit:
As per comment of @Servaes the question boils down to show the following:

There is no odd number $m$, besides $1$, such that:

$$\sigma^{(k)}(m^2), \forall k \ge 0$$

is for all $k$ an odd square number?

Here are some more examples:

```<code>81 3^4 121 11^2
>
1476326929 7^2 * 11^2 * 499^2 1891467081 3^2 * 7^2 * 19^2 * 109^2
>
10994571025 5^2 * 67^2 * 313^2 13884144561 3^2 * 7^2 * 31^2 * 181^2
>
61436066769 3^2 * 7^2 * 11^2 * 29^2 * 37^2 120776405841 3^2 * 7^2 * 13^2 * 19^2 * 67^2
>
98551417041 3^6 * 7^2 * 11^2 * 151^2 190189315449 3^2 * 7^2 * 19^2 * 1093^2
>
119582481249 3^4 * 7^2 * 11^2 * 499^2 228867516801 3^2 * 7^2 * 11^2 * 19^2 * 109^2
>
141447449025 3^2 * 5^2 * 25073^2 253358202409 13^2 * 31^2 * 1249^2
>
220545762129 3^2 * 7^2 * 11^2 * 19^2 * 107^2 433950245001 3^2 * 7^2 * 13^2 * 19^2 * 127^2
>
388895668225 5^2 * 191^2 * 653^2 485512123369 7^2 * 13^4 * 19^2 * 31^2
>
619568914129 11^2 * 163^2 * 439^2 686781810729 3^2 * 7^2 * 19^2 * 31^2 * 67^2
>
890560253025 3^4 * 5^2 * 67^2 * 313^2 1679981491881 3^2 * 7^2 * 11^2 * 31^2 * 181^2
>
970952066161 7^2 * 11^2 * 67^2 * 191^2 1266928085241 3^2 * 7^4 * 13^2 * 19^2 * 31^2
>
2580138650961 3^2 * 29^2 * 37^2 * 499^2 3974915450961 3^2 * 7^2 * 13^2 * 67^2 * 109^2
>
3076652813521 7^2 * 83^2 * 3019^2 3623803369641 3^2 * 7^2 * 13^2 * 19^2 * 367^2
>
3739640454225 3^2 * 5^2 * 13^2 * 47^2 * 211^2 7445874636369 3^2 * 13^2 * 31^2 * 37^2 * 61^2
>
4138876942929 3^6 * 151^2 * 499^2 6259388523129 3^2 * 7^2 * 109^2 * 1093^2
>
4758545225281 19^2 * 29^2 * 37^2 * 107^2 5396129223849 3^2 * 7^2 * 13^2 * 67^2 * 127^2
>
9262289646801 3^2 * 19^2 * 107^2 * 499^2 14281891581321 3^2 * 7^2 * 13^2 * 109^2 * 127^2
</code>```


问:
The curvature of line of intersection of 2 surfaces.
The problem is as follows.

Let $S_1$=$x^2+y^2-z^2=1, z>0$ and $S_2=z^2-2x+y=0$. And define $c$ is curve which is line of intersection of $S_1$ and $S_2$. Find the curvature of $C$ at $P=(1,1,1) $

I tried to parametrization of C by calculating $S_1$ and $S_2$ BUT it has the root on z-axis, i think it may have problem about regularity ...
In fact, i calculate $x= 1+$$ 3\over 2$$cost$ and $y=$$-1 \over 2$ $+$ $3 \over 2$ $sint$... please the full explanation the solution of this problem.

Body:
So far we found

$x(t) = 1 + \frac{3}{2} \cos t$

$y(t) = -\frac{1}{2} + \frac{3}{2} \sin t$

Substituting in any of $S1$ or $S2$, we get

$z(t)^2 = \frac{5}{2} + 3 \cos t - \frac{3}{2} \sin t$

So, $z(t) = \sqrt{\frac{5}{2} + 3 \cos t - \frac{3}{2} \sin t}$

$r(t) = (1 + \frac{3}{2} \cos t, -\frac{1}{2} + \frac{3}{2} \sin t, \sqrt{\frac{5}{2} + 3 \cos t - \frac{3}{2} \sin t} \,)$

$r'(t) = (-\frac{3}{2} \sin t, \frac{3}{2} \cos t, - \frac{3(\sin t + \frac{1}{2} \cos t)} {2 \, \sqrt{\frac{5}{2} + 3 \cos t - \frac{3}{2} \sin t}} \, \,)$

$r''(t) = (-\frac{3}{2} \cos t, -\frac{3}{2} \sin t, -\frac{3(\cos t - \frac{1}{2} \sin t)} {2 \, \sqrt{\frac{5}{2} + 3 \cos t - \frac{3}{2} \sin t}} - \frac{9(\sin t + \frac{1}{2} \cos t)^2} {4 \, (\frac{5}{2} + 3 \cos t - \frac{3}{2} \sin t)^{3/2}} \, \,)$

Using $r(t)$, we know that at point $(1, 1, 1), t = \frac{\pi}{2}$.

So, $r'(t) = (-\frac{3}{2}, 0, -\frac{3}{2})$

$|r'(t)| = \frac{3}{\sqrt2}$

$r''(t) = (0, -\frac{3}{2}, -\frac{3}{2})$

Curvature $k = \displaystyle \frac{|r'(t) \times {r''(t)}|}{|r'(t)|^3}$

Cross product of $r'(t)$ and ${r''(t)} = (-\frac{9}{4}, -\frac{9}{4}, \frac{9}{4})$

$|r'(t) \times {r''(t)}| = \frac{9 \sqrt3}{4}$

So, curvature $k = \frac{1}{\sqrt 6}$


问:
Limit of Sum related question

For $a\in R$ &  $a \ne 1$
$$\mathop {\lim }\limits_{n \to \infty } \frac{{{1^a} + {2^a} + ... + {n^a}}}{{{{\left( {n + 1} \right)}^{a - 1}}\left[ {\left( {na + 1} \right) + \left( {na + 2} \right) + .. + \left( {na + n} \right)} \right]}} = \frac{1}{{60}}$$, then find the value of $a$.

My approach is as follows:

$$\mathop {\lim }\limits_{n \to \infty } \frac{{{1^a} + {2^a} + ... + {n^a}}}{{{{\left( {n + 1} \right)}^{a - 1}}\left[ {\left( {{n^2}a + \frac{{n\left( {n + 1} \right)}}{2}} \right)} \right]}} = \frac{1}{{60}}$$

How to I convert it into limit of sum so that I can proceed with the integration?

答:
Dividing numerator and denominator of the given expression by $n^a$, we  get $$\mathop {\lim }\limits_{n \to \infty } \frac{\sum_{r=1}^n(\frac rn)^a}{\frac{(n + 1)^{a - 1}}{n^a}\sum_{r=1}^n\left( na + r\right)} = \frac{1}{{60}}$$
 $$\implies \mathop {\lim }\limits_{n \to \infty } \frac{(n+1)\sum_{r=1}^n(\frac rn)^a}{(\frac{n + 1}{n})^{a}\cdot n\sum_{r=1}^n\left( a + \frac rn\right)} = \frac{1}{{60}}$$
 Now, divide numerator and denominator by $n$ and you would be able to convert limit into integration. $a$ would come out to be $7, -\frac{17}2$. Doubt?

Title: Limit of Sum related question


问:
Negative Determinant and eigenvalues
How can I proof that a Matrix with a negative determinant hat at least a real eigenvalue ?</p>

答: Hint : Trigonalize in $\mathbb{C}$. If all the eigenvalues are non-real, then for each eigenvalue $\omega$, you also have $\overline{\omega}$ as an eigenvalue. What can you say of the determinant in that case ?

Title: Negative Determinant and eigenvalues