问:
Out of the box & WOW effect Probability
It is a common knowledge, that problems from graph theory/combinatorics can be often solved via short and
astonishing probabilistic reasoning. In contrast to this, it is much harder to find such "out of the box"/"WOW effect" solutions for analysis, topology or algebra problems. So, what are your favorite problems like this, how do you prove them?

This question was inspired by a very nice analysis problem:

evaluate $ \ \lim_{n\to \infty} e^{-n} \cdot \sum_{k=0}^{n}\frac{n^k}{k!}$.

with the following, extremely beautiful solution:

This is $P[N_n\leqslant n]$ where $N_n$ is a random variable with Poisson distribution of parameter $n$. Hence each $N_n$ is distributed like $X_1+\cdots+X_n$ where the random variables $(X_k)$ are independent and identically distributed with Poisson distribution of parameter $1$.

By the central limit theorem, $Y_n=\frac1{\sqrt{n}}(X_1+\cdots+X_n-n)$ converges in distribution to a standard normal random variable $Z$, in particular, $P[Y_n\leqslant 0]\to P[Z\leqslant0]$.

Finally, $P[Z\leqslant0]=\frac12$ and $[N_n\leqslant n]=[Y_n\leqslant 0]$ hence $P[N_n\leqslant n]\to\frac12$, QED.

Both, question and solution come from Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$.

I think  that such a list might be helpful for educational purposes. Such problems act on the imagination of students/listeners and much more likely raise their interest.

答:
Here are two nice (and somewhat related) applications of probabilistic proofs. They require more sophisticated machinery than your example though.

Fundamental Theorem of Algebra: Every non constant polynomial $p\in\mathbb C[X]$ has a root in $\mathbb C$.

Consider a Brownian motion $B_t$ on $\mathbb C$, and suppose $p$ had no complex roots. Then $f=\frac{1}{p}$ is analytic and bounded. As $f$ is continous and non-constant, we can find reals $\alpha<\beta$ such that $\{\Re f(z)\leq\alpha\}$ and $\{\Re f(z)\geq\beta\}$ both contain (disjoint) open discs.

Then $\Re f(B_t)$ is a bounded martingale, so by the martingale convergence theorem, it converges almost surely. However, Brownian motion is recurrent on $\mathbb C$, so $B_t$ visits $\{\Re f(z)\leq\alpha\}$ and $\{\Re f(z)\geq\beta\}$ infinitely many times. So
$$\lim\inf\Re f(z)\leq\alpha<\beta\leq\lim\sup\Re f(z),$$
which contradicts martingale convergence.

If $f:\mathbb Z^2\to[0,\infty)$ satisfies
$$f(x,y)=\frac{f(x+1, y)+f(x, y+1) + f(x-1, y) +f(x, y-1)}{4}$$
for all $x,y\in\mathbb Z$, then $f$ is constant.

Let $X_n$ be a simple symmetric random walk on $\mathbb Z^2$. The condition implies that $M_n=f(X_n)$ is martingale. As $f\geq0$, we know by the martingale convergence theorem that $M_n$ converges almost surely.

But $(X_n)$ is irreducible and recurrent, so (almost surely) visits every point in $\mathbb Z^2$ infinitely often. So $M_n$ hits every value in the image of $f$ infinitely often. This contradicts the convergence of $M_n$ unless $f$ is constant.

Away from probabilistic arguments, I will always be partial to the Gaussian integral. I particularly like two proofs: the first is the standard change of variables to polars. The second is applying the Feynman trick to $$F(t)=\left(\int_0^te^{-x^2}\mathop{}\!\mathrm{d}x\right)^2.$$
I also like Axler's proof that every linear operator $T$ on a finite-dimensional vector space $V$ over $\mathbb C$ has an eigenvalue. He doesn't define need to define determinants/characteristic polynomials, and instead just considers $v, Tv,\dots, T^nv$, where $n=\dim V$. These vectors are linearly independent, so take a linear combination that equals $0$, and you are done by the fundamental theorem of algebra.

Title: Out of the box & WOW effect Probability


问:
Surface area (cylindrical coordinates)
I'm having troubles when calculating the area of part of surface bonded with:$$x^2+y^2+z^2=a^2, x^2+y^2=ax, x^2+y^2=-ax.$$ The other two cylinders should give me the bounds for integral, so after I set cylindrical coordinates $x=r\cos\phi, y=r\sin\phi$ for first cylinder I get $-\frac{\pi}{2}\le \phi\le\frac{\pi}{2}$ and for the other cylinder $\frac{\pi}{2}\le \phi\le \pi$. Now not sure what of those two to use when calculating the surface area.

答:
well: $x^2+y^2+z^2=a^2$ is a sphere of centre $(0,0,0)$ and radius $a$ then:
$$x^2+y^2=ax\Rightarrow (x-a/2)^2+y^2=a^2/4$$
so it is a cylinder where the cross- section is centre $(a/2,0,z)$ and the radius is $a/2$
$$x^2+y^2=-ax\Rightarrow (x+a/2)^2+y^2=a^2/4$$
so this is a cylinder where the cross-section is centre $(-a/2,0,z)$ and the radius is $a/2$.

It is useful to look at the first by looking down on the $x-y$ plane where we see 3 circles, where the big overlaps the others at $z=0$ and as we move towards $z=a/2$ they only touch at a single point.

We will have $-\pi/2\le\theta\le\pi/2$ since we want to go the whole way around the $z$-axis so now you need to find a nice expression for $r$ which will most likely be dependent on both $z,\theta$

ContentLicense: CC BY-SA 4.0

Title: Surface area (cylindrical coordinates)


问:
Sums of integer powers similar to Prouhet–Tarry–Escott problem
Recall Prouhet–Tarry–Escott problem. Its solution shows that certain sums of powers of integers can be made to vanish simultaneously if their signs are chosen to follow the Thue–Morse sequence, e.g.
$$\sum_{k=0}^{2^4-1}(-1)^{\sigma_2(k)}\,k^n=0\quad\text{for}\;n<4,\quad\text{i.e.}\\
\small
\begin{array}{l}
 0^1+3^1+5^1+6^1+9^1+10^1+12^1+15^1=1^1+2^1+4^1+7^1+8^1+11^1+13^1+14^1 \\
 0^2+3^2+5^2+6^2+9^2+10^2+12^2+15^2=1^2+2^2+4^2+7^2+8^2+11^2+13^2+14^2 \\
 0^3+3^3+5^3+6^3+9^3+10^3+12^3+15^3=1^3+2^3+4^3+7^3+8^3+11^3+13^3+14^3, \\
\end{array}$$
where $\sigma_2(k)$ is the sum of digits of $k$ in base-2.

I am interested in sums of powers of a similar form, but with larger exponents (such that the sum does not vanish anymore). I was able to make a few conjectures:
$$\begin{align}
&\sum _{k=0}^{2^n-1} (-1)^{\sigma _2(k)}\, k^{n+2}\stackrel{\color{gray}?}=\frac{2^{\binom{n}{2}}}{36}\, (-1)^n\, \left(2^n-1\right)\, \left(5\times2^n-4\right)\, (n+2)!\\
\\
&\sum _{k=0}^{2^n-1} (-1)^{\sigma _2(k)}\,k^{n+3}\stackrel{\color{gray}?}=\frac{2^{\binom{n}{2}}}{72}\,(-1)^n \,\left(2^n-1\right)^2\,\left(2\times2^n-1\right)\,(n+3)!\\
\\
&\small\sum _{k=0}^{2^n-1} (-1)^{\sigma_2(k)}\,k^{n+4}\stackrel{\color{gray}?}=\frac{2^{\binom{n}{2}}}{32400}\,(-1)^n\,\left(2^n-1\right)\,\left(143\!\times\! 8^n-307\!\times\!4^n+193\!\times\! 2^n-32\right)\,(n+4)!\end{align}$$
This pattern seems to continue within increasingly complicated polynomials of $2^n$ on the right-hand side. Can you suggest an approach to prove these, and to find a more general formula for an arbitrary exponent $k^{n+p}$?

Body:
The exponential generating function for $S_{n,m}=\displaystyle\sum_{k=0}^{2^n-1}(-1)^{\sigma_2(k)}k^m$ (w.r.t. $m$) is $$\sum_{m=0}^\infty S_{n,m}\frac{x^m}{m!}=\sum_{k=0}^{2^n-1}(-1)^{\sigma_2(k)}e^{kx}=\prod_{k=0}^{n-1}(1-e^{2^k x})=2^{n(n-1)/2}(-x)^n\exp\sum_{k=0}^{n-1}f(2^k x),$$ where $$f(x)=\log\frac{e^x-1}{x}=\sum_{m=1}^\infty\frac{B_m^+}{m}\frac{x^m}{m!}$$ using the Bernoulli numbers (look at $f'(x)$ to see the expansion). Thus,
\begin{align*}
S_{n,n+p}&=(-1)^n 2^{n(n-1)/2}(n+p)!\ A_p(2^n),
\\A_p(a)&=[x^p]\exp\sum_{m=1}^\infty\frac{a^m-1}{2^m-1}\frac{B_m^+}{m}\frac{x^m}{m!}.
\end{align*}

This gives an algorithmic way to compute $S_{n,n+p}$ for any fixed $p$. Continuing your table,
\begin{align*}
  A_0(a)&=1
\\2A_1(a)&=a-1
\\36A_2(a)&=(a-1)(5a-4)
\\72A_3(a)&=(a-1)^2(2a-1)
\\32400A_4(a)&=(a-1)(143a^3-307a^2+193a-32)
\\64800A_5(a)&=(a-1)^2(2a-1)(19a^2-24a+2)
\\85730400A_6(a)&=(a-1)(5765a^5-19372a^4+22670a^3-10405a^2+1355a+32)
\\342921600A_7(a)&=(a-1)^2(2a-1)(1166a^4-2850a^3+1715a^2+60a-1)
\end{align*}

Observe that $A_p(a)$ is always divisible by $a-1$, and even by $(a-1)^2(2a-1)$ if $p>1$ is odd (because of $B_m^+=0$ for odd $m>1$, and [the g.f. of] $A_p(1/2)$ in closed form).


问:
Is there a $\triangle ABC$, with $D$ on $BC$, such that $AB$, $BC$, $CA$, $AD$, $BD$, $CD$ are all integers, but only one is even?
This is just a question I came up with while working through another problem.

If we have a triangle $ABC$ and we draw a cevian from $A$ to $D$ on side $BC$, is it possible to assign integer values to lengths $AB$, $BC$, $AC$, $AD$, $BD$ or $CD$ such that only one of the listed lengths has an even valued length? If it is possible, what are these lengths?

I have done some parts of this problem on my own. I figured out that this even length could only be $BC$, $CD$, or $BD$, with $BD$ and $CD$ being the same case.

With $BC$ being the only even length I computed for $BC$ using Stewart's Theorem and got
$$BC = \frac{(AC^2)(BD)+(AB^2)(CD)}{(BD)(CD)+(AD)^2}$$ where every length but $BC$ is odd. The equation seems to work but I don't know how to obtain its integer solutions.

Similarly with $BD$/$CD$ case, I applied Stewart's theorem and got $$BD=\frac{(AB^2)(CD)-(BC)(AD^2)}{(CD)(BC)-(AC)^2}$$
Once again this seems to work, but I don't know how to obtain its integer solutions.

Pretty much all I need help is verifying whether or not my work is right, and if it  is how to find the integer solutions of my provided equations. If there is a better way or my work has flaws please let me know.

Body:
According to Heron's formula, area and semi-perimeter are given as follows.

Sides a,b and c.

$Area=\sqrt{s(s-a)(s-b)(s-c)}$

$s=\frac{a+b+c}{2}$

We consider the case of $AB=AD$.

Let $a=AB, b=AC, c=BC, a=AD, c1=BD, c2=DC$.

Let apply Heron's formula to triangle $(ABC, ABD, ADC)$.

Triangle $ABC$:

$s=\frac{a+b+c}{2}$

$Area=\sqrt{s(s-a)(s-b)(s-c)}$

$c2=c-c1$

Triangle $ABD$:

$s1=\frac{2a+c1}{2}$

$Area1=\sqrt{s1(s1-a)(s1-a)(s1-c1)}$

Triangle $ADC$:

$s2=\frac{a+b+c2}{2}$

$Area2=\sqrt{s2(s2-a)(s2-b)(s2-c2)}$

$Area=Area1+Area2.$

\begin{split}
 \sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)} &= c1\sqrt{(2a+c1)(2a-c1)} \\
                                     &+ \sqrt{(c1-a-b-c)(a-b-c+c1)(a-b+c-c1)(a+b-c+c1)}
 \end{split}

By simple algebra, we know that equation $Area-Area1-Area2=0$ has a factor $-cc1+a^2-b^2+c^2$.

Thus problem was reduced to finding the integer solutions of $-cc1+a^2-b^2+c^2=0$.

We found the integer solutions of $-cc1+a^2-b^2+c^2=0$ with $(a,b,c)<30$.

Jack D'Aurizio's solutions are included.

Similarly, we can apply this method to general case.

            [a  b  c  c1 c2]

            [ 3  7  8  3  5]
            [ 5  7  8  5  3]
            [ 5  9  8  1  7]
            [ 5 13 16  7  9]
            [ 5 23 24  3 21]
            [ 7 15 16  5 11]
            [ 7 17 16  1 15]
            [ 7 19 24 11 13]
            [ 9 11  8  3  5]
            [ 9 15 16  7  9]
            [ 9 17 16  3 13]
            [ 9 21 24  9 15]
            [11 13 16 13  3]
            [11 17 24 17  7]
            [11 19 16  1 15]
            [11 23 24  7 17]
            [11 25 24  3 21]
            [13 15  8  1  7]
            [13 17 24 19  5]
            [13 23 24  9 15]
            [13 25 24  5 19]
            [15 21 24 15  9]
            [15 27 24  3 21]
            [17 19 24 21  3]
            [17 23 16  1 15]
            [17 29 24  1 23]
            [19 21 16 11  5]
            [19 23 24 17  7]
            [19 25 24 13 11]
            [23 29 24 11 13]
            [25 29 24 15  9]
            [27 29 16  9  7]


问:
conditioning of the monomial basis
It is well-known that the Vandermonde matrix is ill-conditioned when all nodes, i.e., the points that generate it, are real. I have read the following on a paper:

The ill-
conditioning of the Vandermonde basis has an elementary explanation
in cases where the points xj are unequal in size. If $|x_j |$ varies with j, then the powers
$|(x_j)^k|$ vary exponentially. This means that function information associated with smaller
values of $x_j$ will only be resolvable through exponentially large expansion coefficients,
and accuracy will quickly be lost in floating point arithmetic

I have not understood the explanation because he terms "function information" and "expansion coefficients" are not defined in this context on the paper. Could somebody provide an explanation on why the Vandermonde matrix is ill-conditioned?.

答:
By "function information," I think they mean function evaluations $(y_j)_{0\leq j\leq J}$ and by "expansion coefficients" they mean $(c_i)_{0\leq i\leq k}$ below.

Here's the basic idea: let $(x_j)_{j\in\{0,1,\ldots,J\}}$ be your nodes, and for every $j\in\{0,1,\ldots, J\}$ let $y_j=f(x_j)$ be your function evaluations. Then expansion in the monomial basis, of rank up to $k$, would be attempting to determine the coefficients $(c_0,\ldots,c_k)$ by solving the system

$$
\begin{pmatrix}
1 & x_0 & \cdots & x_0^k\\
1 & x_1 & \cdots & x_1^k\\
1 & x_2 & \cdots & x_2^k\\
\vdots & & \ddots & \\
1 & x_J & \cdots & x_J^k\\
\end{pmatrix}
\begin{pmatrix}
c_0 \\ c_1 \\ c_2\\\vdots\\c_k
\end{pmatrix}
=
\begin{pmatrix}
y_0 \\ y_1 \\ y_2\\\vdots\\y_J
\end{pmatrix}.
\tag{*}
$$

The point the article is  trying to make is that, if your nodes have drastically different magnitudes, then the matrix in ($*$) becomes horribly ill-conditioned, which (roughly speaking) means that if its input is changed a small amount, its output can be drastically different. They claim that due to this, your only hope (without changing your basis) is to increase $k$ in order to get better accuracy. But, due to the magnitude differences in the matrix (due to increasing $k$), you suffer from massive amounts of floating point roundoff error, so in general this is just a bad idea.

Here is a brief illustration using octave:

```<code>nodes = [0.1; 0.5; 0.0001; 10; 100; 1000];
max_order = 10;
cond_list = zeros(1,max_order);
for order = 1:max_order
  cond_list(order) = cond(vander(nodes,order));
end
</code>```

$$ \begin{matrix}
\textrm{Order}\;(k) & \textrm{Condition number of the matrix in }(*)\\
1 & 1.0000e+00\\
2 &  4.5975e+02 \\
3 & 5.1510e+05 \\
4 & 5.9490e+08  \\
5 & 2.9047e+12 \\
6 & 3.8906e+16 \\
7 &  1.1060e+21 \\
8 & 1.5428e+24 \\
9 &  2.5684e+24 \\
10    &     \texttt{Inf}
\end{matrix}
$$

Title: conditioning of the monomial basis

Body:
By "function information," I think they mean function evaluations $(y_j)_{0\leq j\leq J}$ and by "expansion coefficients" they mean $(c_i)_{0\leq i\leq k}$ below.

Here's the basic idea: let $(x_j)_{j\in\{0,1,\ldots,J\}}$ be your nodes, and for every $j\in\{0,1,\ldots, J\}$ let $y_j=f(x_j)$ be your function evaluations. Then expansion in the monomial basis, of rank up to $k$, would be attempting to determine the coefficients $(c_0,\ldots,c_k)$ by solving the system

$$
\begin{pmatrix}
1 & x_0 & \cdots & x_0^k\\
1 & x_1 & \cdots & x_1^k\\
1 & x_2 & \cdots & x_2^k\\
\vdots & & \ddots & \\
1 & x_J & \cdots & x_J^k\\
\end{pmatrix}
\begin{pmatrix}
c_0 \\ c_1 \\ c_2\\\vdots\\c_k
\end{pmatrix}
=
\begin{pmatrix}
y_0 \\ y_1 \\ y_2\\\vdots\\y_J
\end{pmatrix}.
\tag{*}
$$

The point the article is  trying to make is that, if your nodes have drastically different magnitudes, then the matrix in ($*$) becomes horribly ill-conditioned, which (roughly speaking) means that if its input is changed a small amount, its output can be drastically different. They claim that due to this, your only hope (without changing your basis) is to increase $k$ in order to get better accuracy. But, due to the magnitude differences in the matrix (due to increasing $k$), you suffer from massive amounts of floating point roundoff error, so in general this is just a bad idea.

Here is a brief illustration using octave:

nodes = [0.1; 0.5; 0.0001; 10; 100; 1000];
max_order = 10;
cond_list = zeros(1,max_order);
for order = 1:max_order
  cond_list(order) = cond(vander(nodes,order));
end

$$ \begin{matrix}
\textrm{Order}\;(k) & \textrm{Condition number of the matrix in }(*)\\
1 & 1.0000e+00\\
2 &  4.5975e+02 \\
3 & 5.1510e+05 \\
4 & 5.9490e+08  \\
5 & 2.9047e+12 \\
6 & 3.8906e+16 \\
7 &  1.1060e+21 \\
8 & 1.5428e+24 \\
9 &  2.5684e+24 \\
10    &     \texttt{Inf}
\end{matrix}
$$


问:
Proof that Effros Borel space is standard
I have difficulty understanding the proof of Theorem 12.6 in Kechris's Classical Descriptive Set Theory that if $X$ is Polish then the Effros Borel space of $F(X)$ is standard. $F(X)$ consists of all closed sets in $X$, and I am not giving the definition of Effros Borel space since it is probably not so related to my confusion. The proof proceeds as follows:

Let $\overline{X}$ be a compactification of $X$.

Identify $F(X)$ with a subset $G$ of $K(\overline{X})$, the collection of compact sets in $\overline{X}$.

Prove that $G$ is $G_\delta$ in $K(\overline{X})$, hence Polish.

Carry the topology on $G$ back to $F(X)$. Prove that Effros Borel space coincides with the topology.

I guess in 3 the author implicitly assumes that $K(\overline{X})$ is Polish, which seems to depend on the metrizability of $\overline{X}$. But why is it metrizable? Certainly not any compactification works, for example the Stone–Čech compactification of $\mathbb{N}$ is not even first countable; I think the one-point compactification of $\mathbb{R}^{\mathbb{N}}$ is not first countable either. Can every Polish space be embedded into some compact metric space?

答:
The citations below refer, as in the OP, to Kechris' book.

This is what Theorem $4.14$ does:

Every separable metrizable space is homeomorphic to a subset of the Hilbert cube.

This gives the desired embedding result. Note that Kechris uses a nonstandard (in my experience) notion of "compactification," in Definition $4.15$: a compactification of a separable metrizable space $X$ is a compact metrizable space $Y$ such that $X$ is homeomorphic to a dense subset of $Y$. So the "pick a compactification" language in the proof in question is unproblematic ... if we grant Kechris' use of the term.


问:
How many individual bags are picked from when picking balls at random?
Assume that we have 100 balls, each identified by a number of 1 to 100. The balls are distributed equally in 10 bags, so we have 10 balls in each of the 10 bags. We know which balls are in which bag.

Now, 10 random numbers are choosen and we pick those from the bags. On average, from how many individual bags to we have to pick?

Keep in mind that we know where each ball is located, so no guessing is involved.

It seems like it doesn't matter if the balls are distributed randomly or not, so I didn't specify this.

Here's what I got so far:

When picking once:

We will always have to pick from one bag. The answer is $1$.

When picking twice:

We will always have the first bag to pick from. Additionally, the second pick will be from a new bag in 9 out of 10 cases, which means in those cases we will have picked from a second bag.
That means, on average when picking twice we will have to pick from the following number of bags.

$$\frac{1+(9*2)}{10} = 1.9$$

I think this is correct, but I don't the general way of solving this. In my real-world problem I have variable values for all variables (number of balls, number of bags, number of picks).

答:
One relatively straightforward way to solve this problem is to make use of the principle of linearity of expectation.  That is, the expected value of a sum, is the sum of the individual expected values.  The additional "trick," if one wishes to call it that, is to recognize that the probability of a trial being successful can be interpreted as the expected number of successes in that trial.

Suppose we have $b$ bags, each with $n$ numbers in it (so there are $bn$ distinct numbers altogether).  We select $k$ numbers.  What is the probability that at least one of the numbers is in the first bag?  It is one minus the probability that none of the numbers is in the first bag.  Assuming that all selections of $k$ numbers are equally likely, the probability of avoiding the first bag altogether is

$$
P(\text{no number in first bag}) = \frac{\binom{(b-1)n}{k}}{\binom{bn}{k}}
$$

Therefore,

$$
P(\text{at least one number in first bag}) = 1 - \frac{\binom{(b-1)n}{k}}{\binom{bn}{k}}
$$

As alluded to above, we observe that we can interpret this as

$$
E(\text{number of first bags containing a selected number}) \equiv
P(\text{at least one number in first bag})
$$

This is just an example of the basic principle that the expected number of successes of a single Bernoulli trial is equal to the probability of success.

Now, each bag has an equal probability of containing a selected number (equivalently, each bag has an equal expected number of successes), and the principle of linearity of expectation tells us that the expected number of bags containing at least one selected number is equal to the sum of the individual expected values.  That is, finally,

$$
E(\text{number of bags containing selected numbers})
    = b\left[ 1 - \frac{\binom{(b-1)n}{k}}{\binom{bn}{k}} \right]
$$

This expression is only applicable where $k \leq (b-1)n$, for otherwise, all of the bags must contain a selected number.  For the case in your problem, where $b = n = k = 10$, that condition is satisfied, and we have

$$
E(\text{number of bags containing selected numbers})
    = 10\left[ 1 - \frac{\binom{90}{10}}{\binom{100}{10}} \right]
    \approx 6.69524
$$

If we go by what I mistakenly said in the comments (which effectively assumes that the numbers are chosen with replacement), then the expected value is instead about $6.51322$.


问:
Convergence of $ \int^\infty \int^\infty \cdots \int^\infty \frac{dx_1 dx_2 \cdots dx_n}{(x_1^2+x_2^2+\cdots+x_n^2)^a}$, where $a > \frac{1}{2} n$.
I want to show that $ \int^\infty \int^\infty \cdots \int^\infty \frac{dx_1 dx_2 \cdots dx_n}{(x_1^2+x_2^2+\cdots+x_n^2)^a}$ converges, where $a > \frac{1}{2} n$. This problem is from Cambridge Math Tripos 1904. I am not sure why there is no lower bound for this integral. Anyone knows what is the default value for the empty lower bound? I can see if $n = 3$, we can use the spherical triple integrals to solve the problem. But how do we show that the general case for nth repeated integrals?

答:
I think one nice way to visualise this would be to use hyperspherical coordinates. We would have to first set up the definitions:
$$r^2=\sum_{i=1}^nx_i^2$$
$$1\le r<\infty,\,0\le\varphi_i\le \pi$$
and:
$$\mathrm{d}x_1\,\mathrm{d}x_2\cdots\,\mathrm{d}x_n=r^{n-1}\,\mathrm{d}r\prod_{i=1}^{n-1}\sin^{i-1}{\varphi_{i}\,\mathrm{d}\varphi_{i}}$$
and so with all of this combined we get:
$${\int_{1}^{+\infty}\cdots\int_{1}^{+\infty}\int_{1}^{+\infty}\frac{\mathrm{d}x_{1}\,\mathrm{d}x_{2}\cdots\,\mathrm{d}x_{n}}{\left(\sum\limits_{i=1}^{n}{x_i^2}\right)^a}}=\left(\int_1^\infty{\frac{r^{n-1}}{r^{2a}}\,\mathrm{d}r}\right)\left(\prod_{i=1}^{n-1}{\int_{0}^{\pi}{\sin^{i-1}\varphi_{i}\,\mathrm{d}\varphi_i}}\right)$$
Now if you look at the product of the integrals of sines you will notice that the maximum value an integral can take is $1$, so $n-1$ of these integrals as a product cannot be greater than one, and so:
$$I\le \int_1^\infty\frac{\mathrm{d}r}{r^{1+2a-n}}$$
and now finally, for this integral to converge we know the exponent of the integral must be greater than $1$ and so:
$$2a-n>1$$
$$a>\frac{n+1}{2}$$
I must have made a mistake somewhere because we have this extra factor of $1/2$ but I find this method quite nice


问:
Are there any practical implications of Infinite Monkey Theorem?
I've recently come across the Infinite Monkey Theorem which loosely states that if you gave a monkey a typewriter and an infinite amount of time, it would almost surely type out any given finite text (say, a Shakesperian sonnet).

I understand also that the "monkey" is a placeholder for some notion of an abstract machine that continually produces random strings of symbols (say, letters and numbers).

What are the implications of this theorem? I find the theorem fascinating, but am really not sure how (if at all) it has contributed to other branches of mathematics, real world applications, or academia in general. Any insight would be appreciated!

答:
Suppose that the monkeys' typewriters were limited to the 26 letters of the English alphabet and the space bar for a total of 27 symbols. The number of sequences of length $k$ that could be formed by typing at random (and for simplicity let's assume the random distribution is a uniform distribution: all 27 keys have the same probability of being struck, namely $1/27$) is $27^k$.

If we set $k:=80$, which would be reasonable for a single line of monkey prose, then there are $27^k$ lines possible, all of them equally likely given our assumption that the underlying distribution for each character chosen is uniformly random. There are roughly $10^{97}$ particles in the universe, and $27^{80} > 10^{97}$, so there are more 80 character monkey line combinations than there are particles in the universe. Most of them won't even parse in English, let alone be Pulitzer prize-winning prose. So, in that specific instance at least, not very practical.

Nevertheless, my specific example is just that - an example. Thus is doesn't conclusively disprove the monkeys' futility in general, so I'm curious to hear what others think.


问:
Time it takes to unroll a spool at a fixed RPM
I would like to know how to calculate the time it takes to unroll a roll at a fixed RPM

Variables:  Thickness of material - .125", RPM - 2000 RPM, Starting diameter - 10", Finish diameter - 3.3"

Application is a veneer lathe that peels logs into veneer that makes plywood.

Much appreciated
/D

答:
If you shave off a thickness of $1/8$" then each revolution you'll take $1/4$" off of the diameter.

This will take $4\times(10-3.3) = 27$ revolutions (rounding up) to shave off everything. This takes $27/2000$ minutes or about $0.81$ seconds.


问:
Proof: F is isometric if and only if its matrix is orthogonal/unitary
I'd like to show that $F \in End(V)$ isometric $<=> M_{\beta \beta} (F)$ orthogonal/unitary

But it seems as if I still have some trouble doing that ;/

"=>"
$<v_i, v_j> = <f(v_i),f(v_j)> = <K_{\beta}^{-1} (f(v_i)), K_{\beta}^{-1} (f(v_j))> $.
So, the rows of $M_{\beta \beta}$ form an orthogonal basis of $K^n$.

"<="
The rows of the matrix form an orthogonal basis, so we have $\delta_{ij} = <f(v_i), f(v_j)>$ = ??

答:
The answer provided by dineshdileep is only half-true. for the converse part you have to prove the norm of every column is equal to  1, and that the columns are perpendicular to each other.

Hints:

1.for the norm part you can use vectors like $x^T$ = $[1 \: 0 \: ...  \: 0]$

2.to show each two columns are perpendicular to each other you can use vectors like $x^T$ = $[1 \: 1 \: ... \: 0]$

(the vector I just mentioned can be used to show columns $q_1$ and $q_2$ are perpendicular.)

ContentLicense: CC BY-SA 3.0


问:
$T: V \rightarrow V$ be a normal linear operator on a finite dimensional inner product space.
I did an undergrad in mathematics but I've been working in the financial sector for a few years and it seems my math skills have all but deteriorated... I'm going problems in my old linear algebra textbook for fun and I've been trying to figure out these three factoids for awhile now... Would appreciate some help, thanks

Let $T: V \rightarrow V$ be a normal linear operator on a finite dimensional inner product space. I'm trying to understand why the following things are true:

$(i):$ $||T(x)|| = ||T^*(x)||$ for all $x \in V$

$(ii):$ If $\lambda$ is an eigenvalue for $T$ with corresponding eigenvector $x$, then $T^*(x) = \bar{\lambda}x$, where $\bar{\lambda}$ is the complex conjugate of $\lambda$.

$(iii):$ If $\lambda,\mu$ are distinct eigenvalues, with corresponding eigenvectors $x,y$, then $\langle x,y \rangle = 0$

My thoughts:

$(i)$: I was thinking that this should be true, since the $^*$ just takes the transpose and the conjugate, so the norm should remain the same... Would obviously like to understand this deeper though.

$(ii):$ Seems obvious but I don't know how to show it...

$(iii)$ I have no clue

If somebody could help me out with this and provide some details that'd be great, my math skills arn't what they used to be...

答:
$\|Tv\|^2 = \langle Tv, Tv \rangle = \langle v, T^* T v \rangle$. Now use the fact that $T$ is normal to show that the right-hand side equals $\|T^* v\|^2$.

Show that $T-\lambda I$ is also normal (note that its conjugate transpose is $T^* - \bar{\lambda} I$). Then note $0=\|(T-\lambda I) x\|^2$ and use the result in (i).

$\bar{\lambda} \langle x, y \rangle = \langle \lambda x, y \rangle = \langle Tx, y\rangle = \langle x, T^* y \rangle = \langle x, \bar{\mu} y \rangle = \bar{\mu} \langle x, y \rangle$, where we have used the result in (ii).


问:
How to solve two-body system equation that find $L$ and $u$ simultaneously make $(Lu^{\prime})^{\prime} + Lu = \frac{1}{L}$ satisified?
The dynamical system of two-body system is written as (the mass and other constants is assumed to be 1) :

$\ddot{r} - r{\dot{\theta}}^{2} = \frac{1}{r^2} \tag{1}$

Consider the two-body system is under the influence of torque $\Gamma$:

$ \frac{dL }{dt} = \frac{d r^2 \dot{\theta}}{dt}  = \Gamma \tag{2}$

where the angular momentum is $L = r^2 \dot{\theta}$.

With the substitution $u = \frac{1}{r}$ and following calculation

$\dot{r} = -\frac{\dot{u}}{u^2} = -L\frac{d u}{d\theta}$

$\ddot{r} = -\frac{L^2}{u^2} \frac{d^2u}{d\theta^2} - L\frac{dL}{d\theta}\frac{du}{d\theta}$

$\dot{\theta}^2r = L^2u^3 $

Then Equation.1 turns to be:

$(Lu^\prime)^\prime + Lu = \frac{1}{L} , ^\prime = \frac{d}{d\theta} \tag{3}$

which describes the evolution of two-body system. More information can be found in

https://iopscience.iop.org/article/10.1088/2399-6528/ab9c30,

https://en.wikipedia.org/wiki/Sturm–Liouville_theory

https://en.wikipedia.org/wiki/Two-body_problem.

Thank you for sharing ideas and inspiration.

答:
Assume angular momentum L is a constant that no torque is at present. Then equation.3 turns to be

$Lu^{\prime\prime} + Lu = \frac{1}{L}$

$u^{\prime\prime} + u = \frac{1}{L^2}$

Conic sections solutions are:

$ u = \frac{1}{L^2} + \epsilon cos(\theta+\phi)$

where $\epsilon$ is eccentricity.

Assume angular momentum is not a constant (some torque is present). Further assume $u$ is of the form $u = c_1 e^{\alpha_1 \theta}$ and angular momentum $L$ is of the form $L=c_2 e^{\alpha_2 \theta}$ :

For equation.3, we have

$(c_2 e^{\alpha_2 \theta} (c_1 e^{\alpha_1 \theta})^\prime)^\prime + c_2 e^{\alpha_2 \theta} c_1 e^{\alpha_1 \theta} = \frac{1}{c_2 e^{\alpha_2 \theta}}$

$(\alpha_1 c_1 c_2 e^{(\alpha_1 + \alpha_2) \theta} )^\prime + c_1 c_2 e^{(\alpha_1 +\alpha_2) \theta} = \frac{1}{c_2} e^{-\alpha_2 \theta}$

$\alpha_1 (\alpha_1 + \alpha_2) c_1 c_2 e^{(\alpha_1 + \alpha_2) \theta} + c_1 c_2 e^{(\alpha_1 +\alpha_2) \theta} = \frac{1}{c_2} e^{-\alpha_2 \theta}$

$(\alpha_1 (\alpha_1 + \alpha_2) +1) c_1 c_2 e^{(\alpha_1 + \alpha_2) \theta} = \frac{1}{c_2} e^{-\alpha_2 \theta}$

Then the equation requests

$\alpha_1 = -2 \alpha_2$ and $(1+\frac{\alpha_1^2}{2})c_1 c_2^2 = 1$.


问:
Max possible probability of exponential distribution for given $\lambda$?

The maximum possible probability an exponential distribution with an
arbitrary parameter can give to the interval $[3,6]$ is :

$(a) \ \ \dfrac{1}{2}$

$(b) \ \ \dfrac{1}{3}$

$(c) \ \ \dfrac{1}{4}$

$(d) \ \ \dfrac{2}{3}$

My input:
$X \sim exp(\lambda)$

$f_X(x)=\lambda e^{-\lambda x} ; x\ge0$

$F(6)-F(3)=e^{-3 \lambda}-e^{-6 \lambda}$

I started by putting different values of lambda and calculating probability. Which is very time consuming for the exam I am preparing for.

If I take $\lambda=\dfrac{1}{3}$ I get $0.232$

But there must be some other way to figure this out which I am missing.

答:
Hint: You're trying to maximize the quantity
$$h(\lambda):=e^{-3\lambda}-e^{-6\lambda}$$
over all $\lambda>0$. This can be solved with calculus: take the derivative, set to zero, check you have a maximum.


问:
Out of 6 apples, 5 mangoes and 4 bananas how many selection of fruits can be made such that at least one fruit from each type is always included
According to me the answer should be $(6)(5)(4)-1= 119$ but given answer is $120$

I subtracted the 1 to eliminate the case where none of the fruits are picked.

Is my answer correct?

答: $120$ is the correct answer. The number of ways to pick at least $1$ apple is $6$, because we can pick from $1$ to $6$ apples. Similarly, there are $5$ ways to pick at least $1$ mango, and $4$ ways to pick at least $1$ banana. Thus, the total number of selections in which at least $1$ fruit of each type is picked is $6\cdot 5\cdot 4 = \boxed{120}$.


问:
handshaking lemma and Erdos-Gallai theorem
The conditions for a sequence to be the degree sequence of a simple graph are given by the Erdos-Gallai theorem in addition to the handshaking lemma. Is there an example of a degree sequence where the handshaking lemma is satisfied, but the Erdos-Gallai theorem is not satisfied and thus, the sequence is not graphic (or vice versa).

Or, does satisfying one of these conditions ensure the other is always satisfied?

答:
Let $n=3$ and $d_1=d_2=d_3=1$. Then

$$\begin{align*}
&\sum_{i=1}^1d_i=1\le 2=1\cdot0+\sum_{i=2}^3\min\{d_i,1\}\,,\\
&\sum_{i=1}^2d_i=2\le 3=2\cdot1+\sum_{i=3}^3\min\{d_i,1\}\,,\text{ and}\\
&\sum_{i=1}^3d_i=3\le 6=3\cdot2+\sum_{i=4}^3\min\{d_i,1\}\,.
\end{align*}$$

Thus, the sum of the degrees is odd, but the sequence satisfies the inequality in the Erdős-Gallai theorem.

Obviously one cannot satisfy the full hypotheses of the theorem without also satisfying the handshaking requirement that the sum of the degrees be even, since those hypotheses include the requirement that the degrees be even. However, that requirement is made solely to ensure that the handshaking lemma is satisfied. The meat of the theorem is the inequality, and this example shows that it is possible for a sequence to satisfy that without satisfying the handshaking lemma. (Examples that satisfy the handshaking lemma but not the inequalities are not hard to find.)


问:
Difference between system of differential equations and maps in the context of dynamical systems
In Stephen Wiggins' "Introduction to Applied Nonlinear Dynamical Systems and Chaos," a distinction is made between systems of equations in the form
$$\dot{x}=f(x,t)$$
and
$$x \mapsto{} g(x)$$
Quite frankly, I don't understand what the latter expression means other than the fact that it is called a map. My understanding is that the first expression, for let's say $x \in \mathbb{R}^{2}$, is equivalent to the system
$$ \dot{x}_1 = f_1(x_1, x_2, t) $$
$$ \dot{x}_2 = f_2(x_1, x_2, t) $$

There is a clear derivative (the $\dot{x}$) in the first expression but I don't see any derivatives in the second. So my question: what does the second method of describing a dynamical system, called a map, mean?

答:
A dynamical system defined by a map uses discrete rather than continuous time steps. That is all. The dynamics are given by
$$ x_{t+1} = g\left( x_t \right) $$


问:
A small question concerning the correctness of induction
I think I've read enough about the Peano axioms and I also checked the similar posts here on StackExchange, but all I want to know is why induction can't be validated for all n by writing down the finite logical chains. We don't really need the well-ordering principle...

If we assume to the contrary that there exists an m st. P(m) is false, we may reach the clear contradiction by writing out the corresponding logical chain P(1)->P(2)->...P(m), and so P(m) is true.

What's wrong with this?

答:
Why should there in fact exist such a chain starting from $1$ and reaching all the way up to $m$? This uses an assumption about $\mathbb{N}$, which is ... exactly the well-ordering principle (or something equivalent to it).

This can feel very slippery at first. I think the best way to clarify the situation is to look at a structure which is like the natural numbers except for induction. Since induction fails in such a structure, there must be something special about the natural numbers which lets us prove that induction holds of them.

Specifically, I want to look at a discrete ordered semiring with a simple failure of induction. Here's an example I quite like:

Consider the set $\mathfrak{X}$ of polynomials in a single variable $t$ with integer coefficients, which are either zero or have positive leading term.

There are obvious notions of addition and multiplication of elements of $\mathfrak{X}$ (just the usual operations on polynomials). Moreover, $\mathfrak{X}$ carries a natural ordering: we set $p<q$ iff $\lim_{a\rightarrow \infty}p(a)<\lim_{a\rightarrow\infty}q(a)$. (It's a good exercise to give an equivalent definition of this ordering in terms of coefficients alone.) This turns $\mathfrak{X}$ into a discrete ordered semiring, just like $\mathbb{N}$.

However, induction fails in $\mathfrak{X}$: consider for example the property

$P(a)\equiv$ "For every $b<a$ there is some $c$ such that either $c+c=b$ or $c+c+1=b$."

That is, "Every number $<a$ is either even or odd." In any discrete ordered semiring whatsoever this is trivially satisfied by $0$ and satisfies $\forall x(P(x)\rightarrow P(x+1))$. However, $\forall x(P(x))$ is not true in $\mathfrak{X}$: consider the monomial $t$.

So there is something special about $\mathbb{N}$ which doesn't come from its discrete ordered semiring structure alone but needs to be considered separately, and this is exactly captured by the well-ordering principle. Put another way:

Since induction holds in $\mathbb{N}$ but fails in $\mathfrak{X}$, any argument that induction holds in $\mathbb{N}$ must use some property of $\mathbb{N}$ which doesn't hold of $\mathfrak{X}$. This is what the well-ordering principle provides us.


问:
Find a lower bound using Cauchy-Schwarz inequality
Let $B$ a square positive definite matrix. Define
$$\mu=\frac{(y^TB^{-1}y)(s^TBs)}{(y^Ts)^2}$$
How can I prove that $\mu\ge 1$? So far I get
$$\frac{(y^TB^{-1}y)(s^TBs)}{(y^Ts)^2}\ge \frac{(y^TB^{-1}y)(s^TBs)}{\lVert y \rVert^2\lVert s \rVert^2},$$
but, how can I use C-S inequality in the numerator to get the desired bound? I was thinking in the fact that $y^TB^{-1}y\ge -\lVert y\rVert \lVert B^{-1}y\rVert$ and the same for $s^TBs$, but I cannot see anything else. Even if I use the positivity of $B$, I get
$$ (y^TB^{-1}y)(s^TBs)\ge -\lVert y\rVert \lVert B^{-1}y\rVert (s^TBs),$$
but I do not see any useful expression. Any help will be appreciated

答:
$$y^\top s = y^\top B^{-1/2} B^{1/2} s = (B^{-1/2} y)^\top (B^{1/2} s).$$
Now apply the Cauchy-Schwarz inequality.


问:
Let $T: \mathbb{C}^2 \rightarrow \mathbb{C}^2$ be the linear operator defined by $T(z_1,z_2) = (z_1+5iz_2,(3+2i)z_1+3z_2)$.
Let $T: \mathbb{C}^2 \rightarrow \mathbb{C}^2$ be the linear operator defined by $T(z_1,z_2) = (z_1+5iz_2,(3+2i)z_1+3z_2)$.

I'm trying to find the matrix of $T$ relative to the standard basis (using the standard inner product) on $\mathbb{C}^2$ and then use this to determine the linear transformation $T^*$ (not just the matrix of $T^*$).

Thanks

答: If you find the matrix, you have found $T^*$.  I will assume you are working over $\Bbb C$, so the dimension is $2$.  Then applying to the standard basis, $\{(1,0),(0,1)\}$  we get the columns $(1,3+2i)^t$ and $(5i,3)^t$.  So the matrix is $\begin{pmatrix}1&5i\\3+2i&3\end{pmatrix}$.


问:
there is no bounded linear functional on $ H$
let $ H= L^2[0,1]$ and $ C^1 $ be the set of all continuouse functions on $ [0,1] $ that have continuouse derivative.Let $ t \in [0,1] $ and define $ L: C^1 \longrightarrow F $ by $ L (h)= h'(t) $. show that there is no bounded linear functional on $ H $ that agrees with $ L $ on $ H $ .thanks</p>

答: Hint: Try to find a sequence of functions $\{f_n\}$ in $C^1$, such that $f_n$ is bounded in $H$ (i.e., the function values should not be large), but $f_n'$ unbounded in $F$ (i.e., the derivative is large).</p>


问:
The role of the rewiring probability in small world networks
I am generating some network samples by using small-world package provided by igraph in R. When I keep the number of nodes and the neighborhood parameter the same, the number of edges do not change when changing the rewiring  probability. I was wondering if someone could explain me the exact role of this parameter. For instance, what does exactly happen if I constantly increase the rewiring  probability? It is obvious that the density of the graph does not change.

答:
Small-world networks start out with nodes that can be visualized as lying on a circle in the plane. Each node is initially connected to some number $k$ of nearest neighbors. Then each edge is considered for re-wiring.  With some probability (the re-wiring probability) each edge is disconnected from one of its ends and then connected to a random node in the network instead. These “long-distance” edges give the network its small-world behavior (short paths between arbitrary nodes).

The details can be found in a famous paper by Watts and Strogatz or on Wikipedia.


问:
separation of variables to find solution to heat equation (cylinder)
I'm trying to model heat flow in a cylinder using the heat equation PDE where heat flow is only radial:

$$ \frac{\partial u}{\partial t} = \frac{1}{r} \frac{\partial u}{\partial r} + \frac{\partial^2 u}{\partial r^2}
$$
The I.C for $0<r<1$:
$$
u(r,0) = 0
$$

The B.Cs for $t>0$:
$$
\lim_{r \rightarrow 0} \frac{\partial u}{\partial r} = 0
$$
$$
u(1,t) = 1
$$
Now, we use separation of variables:
Assume that the solution to the PDE is given by:
$$
u(r,t) = U(r)T(t)
$$
Then by rearranging, we have:
$$  \frac{T'(t)}{T(t)} = \frac{1}{r} \frac{U'(r)}{U(r)} + \frac{U''(r)}{U(r)} = -\lambda = -m^2
$$
Here $m>0.$
The first ODE is given by:
$$
T'(t) = -m^{2}T(t)
$$
which has the following solution:
$$
\sum_{n=1}^{\infty} C_{n} \exp(-m^2t)
$$
The second ODE is given by:
$$
U''(r) +\frac{1}{r}U'(r) +m^{2}U(r) = 0
$$
According to the literature, this is a second order ODE that would result in Bessel's functions.
How do you then find the solution to this ODE and then apply the initial and boundary conditions to then find $u(r,t)$?

答:
If you represent the equation in the standard form of Bessel's equation, like @Александр Пальма showed for you in the comment, then you will find a general solution
$$
U(r)=c_1 J_0(m r)+c_2 Y_0(m r)
$$
Then after applying two properties of Bessel's functions, 1)
$$
\frac{d }{d r} J_0(m r)=-m J_1(m r),\qquad
\frac{d }{d r}Y_0(m r)=-m Y_1(m r)
$$
and 2) $Y_1(m r)$ has a singularity at $r=0$,
you can consider the boundary condition to fix the constants.

The answer: $c_2$ is obviously zero, while $c_1=1/J_0(m)$.


问:
Differentiability of functions of several variables.
I've just read the proof of a theorem which states that if a function of several variables(two in this case) has partial derivatives in some neighborhood of a point (x,y) and these derivatives are continuous at (x,y) then the function is differentiable at that point. The proof uses the mean value theorem and ,analytically, I understand the proof but I can't feel why it is necessary to have these derivatives in a neighborhood not just in a point. Can someone experienced share his intuitive view on this question and show me some intuitive reasons for this being necessary?</p>

答:
Limits are far more subtle when you have more than $1$ variable. Consider the standard counterexample function
$$f(x,y) = \begin{cases} \frac{xy}{x^2+y^2}\,, & (x,y)\ne (0,0) \\ 0\,, & (x,y)=(0,0)
\end{cases}\,.$$
This function is not continuous at the origin, but both its partial derivatives are $0$ at the origin. Thus, if we were to have a tangent plane, it would be horizontal. On the other hand, along the line $y=(\tan\theta) x$, the function has value $\frac12\sin 2\theta$ everywhere except the origin. So existence of partials doesn't even give you continuity.

A continuous variant of this is given by
$$g(x,y) =  \begin{cases} \frac{x^2y}{x^2+y^2}\,, & (x,y)\ne (0,0) \\ 0\,, & (x,y)=(0,0)
\end{cases}\,.$$
The tangent plane would still be horizontal, if it existed, but the function now has directional derivative $\frac12\sin2\theta$ in direction $\theta$. Thus, our alleged tangent plane doesn't contain tangent lines to the surface other than in the coordinate directions.

What you should try to find is a function all of whose directional derivatives at the origin are $0$ but which is nevertheless not differentiable (e.g., not continuous).


问:
conditionally convergence of $\{\cos(n)/\sqrt{n}\}$
The problem is basically the title: prove that $\sum_{n=1}^\infty \frac{\cos(n)}{\sqrt{n}}$ is conditionally convergent.

I proved the convergence of the sum by Dirichlet's test, but i couldn't prove that $\sum_{n=1}^\infty \frac{|\cos(n)|}{\sqrt{n}}=\infty$.

Do you have any hint?

答:
Here is a simple argument for divergence of $\sum_{n=1}^{\infty} \frac{|\cos(n)|}{1/\sqrt n}$.

Since cos(x) is zero only at multiplicities of $\frac{2k+1}{2}\pi$ and since these zeros are $\pi$ apart one of the three consecutive numbers 3k, 3k+1, 3k + 2 must be "far" from such zero. Say, $|cos(n)| >= c> 0$ for such n. Hence
$$\sum_{n=1}^{\infty} \frac{|\cos(n)|}{1/\sqrt n} = $$
$$\sum_{k=1}^{\infty} ( \frac{|\cos(3k)|}{1/\sqrt 3k} +  \frac{|\cos(3k + 1)|}{1/\sqrt (3k +1)} + \frac{|\cos(3k+2)|}{1/\sqrt (3k+2)} ) > = $$
$$\sum_{k=1}^{\infty} ( \frac{c}{1/\sqrt (3k+2)}$$
and since the series $\sum \frac{1}{1/\sqrt (3k+2)}$ diverges the original is divergent as well.


答:
You write about conjugate transposes, but on an abstract inner product space there is no such thing as a "transpose" since linear maps are not matrices.  At least I would not speak of a "conjugate transpose". I'd speak of an adjoint: $T^*$ is the adjoint of $T$.

The most important condition to use here is how a linear operator interacts with the inner product: $\boxed{\langle Tv,w\rangle = \langle v,T^*w\rangle}$ for all $v, w \in V$. This formula, quantified over all $v$ and $w$ in $V$, is how you define the mapping $T^* \colon V \to V$. That is, for each $w \in V$, the function $\mathbf C^n \to \mathbf C$ given by $v \mapsto \langle Tv,w\rangle$ is $\mathbf C$-linear and therefore it must be pairing against a specific vector in $V$ that we define to be $T^*w$:  $\langle Tv,w\rangle =  \langle v,T^*w\rangle$ for all $v \in V$.

In the special case $V = \mathbf C^n$ and $\langle \cdot,\cdot\rangle$ is the standard inner product on $\mathbf C^n$ such that $\langle \sum z_je_j,\sum w_je_j\rangle = \sum z_j\overline{w_j}$ for standard basis $e_1,\ldots, e_n$ of $\mathbf C^n$, you can check that
$$
\langle M(\mathbf v),\mathbf w\rangle = \langle \mathbf v,\overline{M}^\top(\mathbf w)\rangle
$$
for all $\mathbf v$ and $\mathbf w$ in $\mathbf C^n$, so $M^* = \overline{M}^\top$. This motivates the definition of $T^*$ in the abstract case, but you can't say in some abstract setting that $T^*$ is a "transpose" since abstract $T$'s are not matrices.

Proof of (i).  Express the norm on $V$ in terms of the inner product by squaring:
$$
||Tv||^2 = \langle Tv,Tv\rangle = \langle v,T^*(Tv)\rangle =
\langle v,T(T^*v)\rangle = \langle T^*v,T^*v\rangle = ||T^*v||^2.
$$
Thus $||Tv|| = ||T^*v||$.

Proof of (ii). First I'll show a property of all linear operators $T : V \to V$, not just normal ones: if $\lambda$ is an eigenvalue of $T$ then $\overline{\lambda}$ is an eigenvalue of $T^*$.  From $Tv = \lambda v$, where $v \not= 0$ in $V$, we get for all $w \in V$ that
$$
\langle Tv,w\rangle = \langle v,T^*w\rangle \Longrightarrow
\langle \lambda v,w\rangle = \langle v,T^*w\rangle \Longrightarrow \lambda\langle v,w\rangle = \langle v,T^*w\rangle.
$$
We can put $\lambda$ back into the inner product on the right side using its complex conjugate:
$$
\langle v,\overline{\lambda}w\rangle = \langle v,T^*w\rangle \Longrightarrow \langle v,(T^* - \overline{\lambda}I)w\rangle = 0.
$$
Here $v$ is a nonzero element of $V$ and that vanishing of the inner product is true for all $w \in V$. That means the subspace
${\rm im}(T^* - \overline{\lambda}I) = \{(T^* - \overline{\lambda} I)w : w \in V\}$ can't be all of $V$ (we certainly can't have $v$ in the image, since $v$ isn't $0$!), so $T^* - \overline{\lambda}I$ is not surjective. Therefore (since $V$ is finite-dimensional) $T^* - \overline{\lambda} I$ is not injective, so there is a nonzero $w \in V$ such that $(T^* - \overline{\lambda} I)(w) = 0$, which makes $T^*w = \overline{\lambda}w$.  We have shown that if $\lambda$ is an eigenvalue of an arbitrary linear operator $T \colon V \to V$, then $\overline{\lambda}$ is an eigenvalue of $T^* \colon V \to V$. This is not going to be used in the proof of (ii), but it may be helpful to compare such reasoning to what I now do to prove (ii).

Suppose $T$ is normal and $Tv = \lambda v$. We want to show $T^*v = \overline{v}$. Since $Tv - \lambda v = 0$,
$$
0 = \langle Tv - \lambda v,Tv - \lambda v\rangle =
\langle (T - \lambda I)v,(T - \lambda I)v\rangle =
\langle v,(T-\lambda I)^*(T-\lambda I)v\rangle.
$$
Since $(T - \lambda I)^* = T^* - (\lambda I)^* = T^* - \overline{\lambda}I$,
$$
0 = \langle v,(T^*- \overline{\lambda} I)(T-\lambda I)v\rangle.
$$
Since $T$ and $T^*$ commute, while all linear operators commute with scaling, $T^* - \overline{\lambda}I$ and $T - \lambda I$ commute. Thus
$$
0 = \langle v,(T - \lambda I)(T^*- \overline{\lambda} I)v\rangle =
\langle (T - \lambda I)^*v,(T^* - \overline{\lambda}I)v\rangle =
\langle (T^* - \overline{\lambda} I)v,(T^* - \overline{\lambda}I)v\rangle.
$$
Therefore $(T^* - \overline{\lambda})v = 0$, so $T^*v = \overline{\lambda}v$.  If $v \not= 0$ then $v$ is an eigenvector of $T$ with eigenvalue $\lambda$ and we showed it is also an eigenvector of $T^*$ with eigenvalue $\overline{\lambda}$>

Proof of (iii).  Suppose $Tv = \lambda v$ and $Tw = \mu w$. Then
$$
\lambda \langle v,w\rangle = \langle \lambda v,w\rangle =
\langle Tv,w\rangle = \langle v,T^*w\rangle.
$$
Since $T$ is normal, by (ii) we have $T^*w = \overline{\mu} w$. Thus
$$
\lambda \langle v,w\rangle = \langle v,\overline{\mu}w\rangle =
\mu\langle v,w\rangle.
$$
Therefore the complex number $\langle v,w\rangle$ multiplied by $\lambda$ and $\mu$ has the same effect. Since $\lambda \not= \mu$, $\langle v,w\rangle$ has to be $0$.

Body:
I did an undergrad in mathematics but I've been working in the financial sector for a few years and it seems my math skills have all but deteriorated... I'm going problems in my old linear algebra textbook for fun and I've been trying to figure out these three factoids for awhile now... Would appreciate some help, thanks

Let $T: V \rightarrow V$ be a normal linear operator on a finite dimensional inner product space. I'm trying to understand why the following things are true:

$(i):$ $||T(x)|| = ||T^*(x)||$ for all $x \in V$

$(ii):$ If $\lambda$ is an eigenvalue for $T$ with corresponding eigenvector $x$, then $T^*(x) = \bar{\lambda}x$, where $\bar{\lambda}$ is the complex conjugate of $\lambda$.

$(iii):$ If $\lambda,\mu$ are distinct eigenvalues, with corresponding eigenvectors $x,y$, then $\langle x,y \rangle = 0$

My thoughts:

$(i)$: I was thinking that this should be true, since the $^*$ just takes the transpose and the conjugate, so the norm should remain the same... Would obviously like to understand this deeper though.

$(ii):$ Seems obvious but I don't know how to show it...

$(iii)$ I have no clue

If somebody could help me out with this and provide some details that'd be great, my math skills arn't what they used to be...


问:
What do cones have to do with quadratics? Why is $2$ special?
I've always been nagged about the two extremely non-obviously related definitions of conic sections (i.e. it seems so mysterious/magical that somehow slices of a cone are related to degree 2 equations in 2 variables). Recently I came across the following pages/videos:

This 3B1B video about ellipses, which reignited my desire to understand conics

Why are quadratic equations the same as right circular conic sections?, which offers a very computational approach to trying to resolve this question

Another 3B1B video on visualizing Pythagorean triples (i.e. finding the rational points of a circle)

and Manjul Bhargava's lecture on the Birch-Swinnerton-Dyer Conjecture, where minutes ~10-15 discuss the complete solution to the problems of rational points on conics.

While 3B1B's video makes a lot of sense and is very beautiful from a geometric standpoint, it does not talk about any of the other conics, or discuss the relationship with "degree 2". Moreover, the 2nd 3B1B video I linked and then Bhargava's lecture highlights "degree 2" as something we understand well, compared to higher degrees (reminds me a little bit of Fermat's last theorem and the non-existence of solutions for $n>2$).

So, I suppose my questions are as follows:

Why, from an intuitive standpoint, should we expect cones to be deeply related to zero-sets of degree 2 algebraic equations?

and more generally:

Is there some deep reason why "$2$" is so special? I've often heard the quip that "mathematics is about turning confusing things into linear algebra" because linear algebra is "the only subject mathematicians completely understand"; but it seems we also understand a lot of nice things about quadratics as well  -- we have the aforementioned relationship with cones, a complete understanding of rational points, and the Pythagorean theorem (oh! and I just thought of quadratic reciprocity). 2 is also special in all sorts of algebraic contexts, as well as being the only possible finite degree extension of $\mathbb R$, leading to in particular $\mathbb C$ being 2-dimensional.

Also interesting to note that many equations in physics are related to $2$ (the second derivative, or inverse square laws), though that may be a stretch. I appreciate any ideas you share!

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EDIT 3/12/21: was just thinking about variances, and least squares regression. "$2$" is extremely special in these areas: Why square the difference instead of taking the absolute value in standard deviation?, Why is it so cool to square numbers (in terms of finding the standard deviation)?, and the absolutely mindblowing animation of the physical realization of PCA with Hooke's law: Making sense of principal component analysis, eigenvectors & eigenvalues.

In these links I just listed, seems like the most popular (but still not very satisfying to me) answer is that it's convenient (smooth, easy to minimize, variances sum for independent r.v.'s, etc), a fact that may be a symptom of a deeper connection with the Hilbert-space-iness of $L^2$. Also maybe something about how dealing with squares, Pythagoras gives us that minimizing reconstruction error is the same as maximizing projection variance in PCA. Honorable mentions to Qiaochu Yuan's answer about rotation invariance, and Aaron Meyerowitz's answer about the arithmetic mean being the unique minimizer of sum of squared distances from a given point. As for the incredible alignment with our intuition in the form of the animation with springs and Hooke's law that I linked, I suppose I'll chalk that one up to coincidence, or some sort of SF ;)

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EDIT 2/11/22:
I was thinking about Hilbert spaces, and then wondering again why they behave so nice, namely they have the closest point lemma (leading to orthogonal decomposition $\mathcal H = \mathcal M \oplus \mathcal M^\perp$ for closed subspaces $\cal M$), or orthonormal bases (leading to Parseval's identity, convergence of a series of orthogonal elements if and only if the sum of the squared lengths converge), and I came to the conclusion that the key result each time seemed to be the Pythagorean theorem (e.g. the parallelogram law is an easy corollary of Pythag). So that begs the questions, why is the Pythagorean theorem so special? The linked article in the accepted answer of this question:  What does the Pythagorean Theorem really prove? tells us essentially the Pythagorean theorem boils down to the fact that right triangles can be subdivided into two triangles both similar to the original.

The fact that this subdivision is reached by projecting the vertex onto the hypotenuse (projection deeply related to inner products) is likely also significant... ahh, indeed by the "commutativity of projection", projecting a leg onto the hypotenuse is the same as projecting the hypotenuse onto the leg, but by orthogonality of the legs, the projection of the hypotenuse onto the leg is simply the leg itself! The square comes from the fact that projection scales proportionally to the scaling of each vector, and there are two vectors involved in the operation of projection.

I suppose this sort of "algebraic understanding" of the projection explains the importance of "2" more than the geometry, since just knowing about the "self-similarity of the subdivisions" of the right triangle, one then has to wonder why say tetrahedrons or other shapes in other dimensions don't have this "self-similarity of the subdivisions" property. However it is still not clear to me why projection seems to be so fundamentally "2-dimensional". Perhaps 1-dimensionally, there is the "objective" perception of the vector, and 2-dimensionally there is the "subjective" perception of one vector in the eyes of another, and there's just no good 3-dimensional perception for 3 vectors?

There might also be some connection between the importance of projection and the importance of the Riesz representation theorem (all linear "projections" onto a 1-dimensional subspace, i.e. linear functionals, are actually literal projections against a vector in the space).

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EDIT 2/18/22: again touching on the degree 2 Diophantine equations I mentioned above, a classical example is the number of ways to write $k$ as the sum of $n$ squares $r_n(k)$. There are a number of nice results for this, the most famous being Fermat's 2-square theorem, and Jacobi's 4-square theorem. A key part of this proof was the use of the Poisson summation formula for the Euler/Jacobi theta function $\theta(\tau) := \sum_{n=-\infty}^\infty e^{i \pi n^2 \tau}$, which depends on/is heavily related to the fact that Gaussians are stable under the Fourier transform. I still don't understand intuitively why this is the case (see Intuitively, why is the Gaussian the Fourier transform of itself?), but there seems to be some relation to Holder conjugates and $L^p$ spaces (or in the Gaussian case, connections to $L^2$), since those show up in generalizations to the Hardy uncertainty principle (“completing the square”, again an algebraic nicety of squares, was used in the proof of Hardy, and the Holder conjugates may have to do with the inequality $-x^p + xu \leq u^q$ -— Problem 4.1 in Stein and Shakarchi’s Complex analysis, where the LHS basically comes from computing the Fourier transform of $e^{-x^p}$) Of course why the Gaussian itself appears everywhere is another question altogether: https://mathoverflow.net/questions/40268/why-is-the-gaussian-so-pervasive-in-mathematics.

This (squares leading to decent theory of $r_n(k)$, and squares leading to nice properties of the Gaussian) is probably also connected to the fact that $\int_{\mathbb R} e^{-x^2} d x$ has a nice explicit value, namely $\sqrt \pi$. I tried seeing if there was a connection between this value of $\pi$ and the value of $\pi$ one gets from calculating the area of a circle "shell-by-shell" $\frac 1{N^2} \sum_{k=0}^N r_2(k) \to \pi$, but I couldn't find anything: Gaussian integral using Euler/Jacobi theta function and $r_2(k)$ (number of representations as sum of 2 squares).

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EDIT 10/13/22: I was recently learning about Riemannian geometry, and indeed the metric tensor is a bilinear form (cf. above discussion on inner products), and the Riemann curvature tensor (or curvature in general) is all about the second (covariant) derivative. Taking traces we arrive at the Ricci curvature tensor (used in no less important things than Einstein's general relativity, whose "classical approximation" Newtonian gravity follows an inverse square law; and Perelman's proof of the Poincare conjecture!) or scalar curvature, which can be interpreted geometrically as the second-order change in volume of balls/base of cones (see https://math.stackexchange.com/a/469005/405572 or Equation (10) of https://arxiv.org/pdf/2201.04923v1.pdf).

And of course like Ricci flow we also have the heat/diffusion differential equations (and the Schrodinger equation, quoting James Gleck, is "diffusion through imaginary time"), and endless equations involving the Laplacian, all second-order differential equations (how AMAZING that the two great pillars of 20th century physics, general relativity and quantum mechanics both have at their hearts second-order differential equations! And the fact that observables in quantum mechanics are modeled as operators on Hilbert spaces). Relating to the Laplacian, we have the important concept of harmonicity, with beautiful manifestations/consequences in complex analysis, or PDEs, in the form of elliptic regularity. Besides tensors of 2nd derivatives, we have the ever-present Hessian matrix of second derivatives. I'll end with some quotes from a MO answer I linked here in a comment several months ago (https://mathoverflow.net/a/171743/112504) which tries to argue that "Polynomials are useful because quadratic polynomials are useful.":

So the question becomes: why are quadratic polynomials useful? There seem to be two different but interacting reasons. The first is that quadratic functions of a real variable are always either convex or concave and therefore have a unique maximum or minimum. The second is that quadratic functions are intimately related to bilinear forms and therefore can be accessed using linear algebra. The combination of these two reasons seems to explain the success of quadratic algebra in analysis and geometry (e.g. Hilbert spaces, Riemannian manifolds)

答:
There is a paper by David Mumford which may be hard to read depending upon your level of preparation.

The gist of that paper is to say that any system of polynomial equations can be replaced (by adding more variables and more equations) to a system of quadratic and linear equations.

One can probably generalise this further to show that if the polynomial system has parameters, then one can ensure that these parameters only appear in the linear equations.

The very special early case of this is the one you have mentioned.


问:
Deriving AM-GM in a way that reflects my intuition.
One of my core intuitions with inequalities (especially the olympiad kind in which this is very explicit, but this has more profound influences on how I understand the relationship between inequalities) is that in a homogenous inequality, on average, terms that are more mixed ($x^2yz$) are smaller than their less-mixed counterparts ($x^4$).

With this bit of intuition, many inequalities seem trivially true, like am-gm:

$$x^3+y^3+z^3\geq xyz + xyz +xyz$$

Or this one:

$$x^2y + y^2z + z^2x\geq xyz+xyz+xyz$$

And a generalisation of part of this idea: Murihead's Inequality.

I want to find a proof (by which I mean not particularly formal derivation) of AM-GM, that adequately captures 'mixed terms being smaller on average' as being the reason for the inequality. I think this would satisfy some innate and inexplicable urge.

There may not exist a satisfying answer to this question (though I would certainly no one if I saw it), but hopefully, I have communicated the kind of thing I'm looking for.

答:
I think there is a proof satisfying your intuition. I will talk just about the common AM-GM for three numbers $x^3+y^3+z^3\geq 3xyz$ but you can straightforwardly generalize the idea for more variables.

Very intuitive rearrangement inequality can be generalized for three ordered sequences of nonnegative numbers. For $a_1 \geq a_2 \geq a_3\geq 0$ and $b_1 \geq b_2 \geq b_3\geq 0$ and $c_1 \geq c_2 \geq c_3\geq 0$ the generalized rearrangement inequality states that
$$a_1b_1c_1 + a_2b_2c_2 + a_3b_3c_3 \geq a_{\sigma(1)}b_{\tau(1)}c_1 + a_{\sigma(2)}b_{\tau(2)}c_2 + a_{\sigma(3)}b_{\tau(3)}c_3$$
for any permatutions $\sigma, \tau:\{1,2,3\}\rightarrow\{1,2,3\}$.

Since the AM-GM inequality is symmetric, we may assume that $x\geq y\geq z$. Then AM-GM is an immediate consequence of the generalized rearrangement inequality for three ordered sequences $x\geq y\geq z\geq 0$, $x\geq y\geq z\geq 0$, $x\geq y\geq z\geq 0$ since on the LHS of AM-GM we have the greatest possible sum of products and on the RHS we have some "mixture" of products.

You may wonder how to (intuitively) prove rearrangement inequality and its generalization. If you read the paragraph called Intuition on Wikipedia, it should be obvious why rearrangement inequality holds for two sequences. Nevertheless I will present slightly unusual proof with something I call adding level by level (someone told me it is actually Abel's summation but I admit I was lazy to check it).

I will prove the following inequality (all numbers are nonnegative)
$$\color{red}{a_1b_1} + \color{fuchsia}{a_2b_2} + \color{maroon}{a_3b_3} \geq \color{blue}{a_1b_2} + \color{lime}{a_2b_3} + \color{green}{a_3b_1}$$
and I believe you will understand how it works for any other permutation. I start with $b_3$-level (the first line which is actually equality), the second line is $b_2$-level and the third is $b_1$-level.
\begin{align}
\color{red}{a_1b_3}\phantom{-x_y)} + \color{fuchsia}{a_2b_3}\phantom{-x_y)} + \color{maroon}{a_3b_3} &\geq \color{blue}{a_1b_3}\phantom{-x_y)} +\color{lime}{a_2b_3}  +\color{green}{a_3b_3} \\
\color{red}{a_1(b_2-b_3)} + \color{fuchsia}{a_2(b_2-b_3)} \phantom{+x_yx_y}  &\geq \color{blue}{a_1(b_2-b_3)}\phantom{+x_yx_y} +\color{green}{a_3(b_2-b_3)} \\
\color{red}{a_1(b_1-b_2)}\phantom{+ a_2(b_2-b_3) +x_yx_y}   &\geq \phantom{a_1(b_2-b_3) + x_yx_y+} \color{green}{a_3(b_1-b_2)} \\
\end{align}
Just observe that we get the desired inequality after summing all these three "levels".

The idea for three sequences is exactly the same! Assume we want to prove this inequality
$$\color{red}{a_1b_1c_1} + \color{fuchsia}{a_2b_2c_2} + \color{maroon}{a_3b_3c_3} \geq \color{blue}{a_1b_2c_3} + \color{lime}{a_2b_3c_1} + \color{green}{a_3b_1c_2}.$$
Again observe that it is just a sum of these three inequalities that follow from the rearrangement inequality for two sequences:
\begin{align}
\color{red}{a_1b_1c_3}\phantom{-x_y)} + \color{fuchsia}{a_2b_2c_3}\phantom{-x_y)} + \color{maroon}{a_3b_3c_3} &\geq \color{blue}{a_1b_2c_3}\phantom{-x_y)} +\color{lime}{a_2b_3c_3}\phantom{x}  +\color{green}{a_3b_1c_3} \\
\color{red}{a_1b_1(c_2-c_3)} + \color{fuchsia}{a_2b_2(c_2-c_3)} \phantom{+x_yx_yx_y}  &\geq \phantom{x_yx_y++}  \color{lime}{a_2b_3(c_2-c_3)} + \color{green}{a_3b_1(c_2-c_3)} \\
\color{red}{a_1b_1(c_1-c_2)}\phantom{+ a_2b_2(c_2-c_3) +x_yx_yx_y}   &\geq \phantom{x_yx_y++} \color{lime}{a_2b_3(c_1-c_2)}\\
\end{align}


问:
proving $n!>2^n\;\;\forall \;n≥4\;$ by mathematical induction
My teacher proved the following $n!>2^n\;\;\;\forall \;n≥4\;$ in the following way

Basis step: $\;\;4!=24>16$ ok

Induction hypothesis: $\;\;k!>2^k$

Induction step: $\qquad\qquad(k+1)!=k!(k+1)>(k+1)2^k>2^k\cdot 2=2^{k+1}$

I wonder how did he assume that $2^k(k+1)>2^{k}\cdot 2\quad\forall k≥4$?

Don't we have to show it by induction too?

答: You assume by induction hypothesis that $k!>2^k$. Also, you have $k\ge 4$, hence surely $k+1>2$. Multiplying $k+1> 2$ with the positive number $2^k$, you find $(k+1)2^k>2^{k+1}$. And multiplying $k!>2^k$ with the positive number $k+1$ you find $(k+1)!>(k+1)2^k$, whence together the claim $(k+1)!>2^{k+1}$.</p>


问:
What does a generated ring look like?
The motivation for this question comes from one of my previous posts: Non-Noetherian subring of $\mathbb{C}[X,Y]$. There, I encounter the following statement,

Consider the subring $R$ of $\mathbb{C}[X, Y]$ generated (over $\mathbb{C})$ by $X, X Y, X Y^{2}, X Y^{3}, \ldots$.

My question is, what does it mean to 'generate a ring' in that context? Usually, when I was given a set $S \subset K$ where $K$ is a ring, and I needed to generate a ring with $S$, I would do: Generated Ring $= \{\sum_{i=1}^n s_i k_i:$ $n$ is finite, $s_i \in S$ and $k_i \in K \}$.

But in the above statement, the context in which it appears makes it clear that $\mathbb{C} \subset R$. So the previous notion I was taught doesn't work, as $X$ does not divide every element of $\mathbb{C}$. What does the ring $R$ of $\mathbb{C}[X, Y]$ generated (over $\mathbb{C})$ by $X, X Y, X Y^{2}, X Y^{3}, \ldots$ even look like? What does it mean to generate such a ring?

I've read somewhere that this has to do with the notion of an 'Algebra'. But I've never heard of an Algebra, not in this context anyway, and googling shows that Algebra is a concept that shows up in Commutative Algebra, not in an elementary UG Algebra course like the one I'm taking now. So I'm fairly certain that this can be understood without referring to the notion of an Algebra.

答:
Your description of what it means to generate a subring is incorrect; you've described generating an ideal. If $S$ is a subset of a commutative ring $R$, the subring of $R$ generated by $S$ consists of polynomials in the elements of $S$ with coefficients in $\mathbb{Z}$. Explicitly, these consist of sums (with integer coefficients) of monomials $s_1^{e_1} s_2^{e_2} \dots s_k^{e_k}$ where $s_i \in S$ and $e_i \in \mathbb{Z}_{\ge 0}$. This is the intersection of all subrings of $R$ containing $S$.

"Generated over $\mathbb{C}$" means that $R$ is a $\mathbb{C}$-algebra (a ring which is also a vector space over $\mathbb{C}$ such that multiplication is $\mathbb{C}$-bilinear) and we consider sums of monomials with coefficients in $\mathbb{C}$. This is the intersection of all $\mathbb{C}$-subalgebras of $R$ containing $S$ (a $\mathbb{C}$-subalgebra being a subring closed under scalar multiplication by $\mathbb{C}$). So the $\mathbb{C}$-subalgebra of $\mathbb{C}[X, Y]$ generated by $X, XY, XY^2, XY^3$ consists of $\mathbb{C}$-linear combinations of monomials in $X, XY, XY^2, \dots$, of the form

$$\prod_{i=0}^{k-1} (XY^i)^{e_i}.$$

It's then a straightforward exercise to show that the monomials $X^i Y^j$ which occur in this way are exactly the monomials such that $i \neq 0$. In other words, $R$ can equivalently be described as the $\mathbb{C}$-linear span of the ideal $(X)$ and the identity $1$.

Title: What does a generated ring look like?


问:
Proving $\left\lVert f\right\rVert_{\infty}:=\sup_{x \in X}\left\lVert f(x)\right\rVert_2$ is finite and the norm of a continuous subset
I'm trying to answer this Real Analysis question below,

Let $K$ be a compact subset of $R^2$, and let $C(K,\mathbb R^m)$ denote the vector space of all continuous functions from $K$ into $\mathbb R^m$. Show that for $f$ in $C(K,\mathbb R^m)$ , the quantity $\left\lVert f\right\rVert_{\infty}:=\sup_{x \in X}\left\lVert f(x)\right\rVert_2$ is finite and $\left\lVert
 \cdot\right\rVert_\infty$ is the norm on $C(K,\mathbb R^m)$.

I have absolutely no idea how to answer this question. I have a very basic understand for the Uniform Norm and a very basic understanding of compactness (I understand the definitions and nothing else). All I can figure out about this question is how to state the expansion of a two-norm,

$\left\lVert f(x)\right\rVert_2=(\sum_{n=1}^{m}|f_n(x)|^2)^{1/2}$

Could I please receive some hints on how I'm supposed to solve this question? I'm extremely incapable with abstract mathematics so keeping the hints very simple would help a lot too.

答: What kind of a function is $\left\|f(x)\right\|_2$? What do you know about such functions on compact domains?


问:
Pictorial representation of a tangent plane to a right circular cone
Can anyone pictorially show me of a tangent plane to a right circular double cone? Since I think it doesn't have a tangent plane through the origin (correct me if I am wrong), I am not able to visualize it.</p>

答:
A right circular cone can be constructed by choosing a line to serve as the axis of the cone, choosing a point on the axis to serve as the vertex of the cone, taking a plane perpendicular to the axis, and constructing a circle in that plane whose center is the point where the axis intersects the plane.

From the vertex you can construct a line segment to any point on the circle. If you take the set of all such line segments you get a finite cone. If you extend each segment from the vertex through the circle and continuing indefinitely beyond the circle, you get a single infinite cone.
And if you extend each segment indefinitely in both directions, you get a double infinite cone.

In order for a plane to be tangent to the cone (in either the finite, single infinite, or double infinite case), the plane must contain one of the lines through the vertex and the circle, and the tangent plane must intersect the plane of the circle in a line that is tangent to the circle.

Any other plane will cut through the cone forming a conic section, will cut the cone along two intersecting lines, or will pass through the vertex only and no other point of the cone; none of those cases produces a tangent plane.

It follows that the only place the tangent plane touches the cone is along one single line through the vertex and one single point on the given circle.
You can choose any of infinitely many such lines, but once you have chosen one, it is the only one that will lie in the tangent plane.
It is not possible for a tangent plane to contain two lines that lie on the cone's surface, let alone three.

If a plane is tangent to a finite right circular cone, however, it is also tangent to the infinite single cone extended from that finite cone, because as the tangent line is extended along the surface of the cone it is still in the plane. Likewise the same plane is also tangent to the double cone, because when the line is extended in the opposite direction it again remains in the same plane.


问:
Four dice are tossed let $N$ be the number of distinct outcomes. Find $E[N]$
Four dice are tossed let $N$ be the number of distinct outcomes. Find $E[N]$

If we let $X$ be the number of distinct outcomes in a given toss.

E.g. if we get the toss $1,2,2,4$ then $X=3$

Calculating some probabilities:

$P(X=1) = 6 \cdot 6^{-4}$ since we have six ways we can end up with all four dice having the same value face-up.

Now according to a solution I have for this:

$\displaystyle P(X = 2) = 6^{-4}\cdot 6 \cdot 5 \cdot \binom{4}{2}$

This is where I am not so sure; why $4$ choose $2$ as opposed to $6 $ choose $2$ ?

$P(X= 4) = 6^{-4}\cdot 6 \cdot 5 \cdot 4 \cdot 3$

and finally $P(X = 3) $ I haven't calculated as I am not clear on $P(X=2)$.

The issue for me seems to be the combinatorial part of the argument. i.e. The number of ways each event occurs.

答:
You're counting the number of distinct ways to:

Select two numbers to show up ($6\times5$) with order

Then, you're selecting which two dice to show the first (vs the second) number, hence $\binom{4}{2}$

Which for example, double-counts 3,3,4,4. Select 3, 4, select the first two dice to show 3. Select 4, 3, select the second two dice to show 4.

Additionally, some ways to have $X=2$ have not been counted.


问:
Prove absolute convergence of $\sum_{n=0}^{\infty} a_{f(n)}$ if $f$ is increasing and $\sum_{n=0}^{\infty} a_n$ is absolutely convergent
Synopsis

In Terence Tao's, Analysis 1, we are given the following exercise:

Let $\displaystyle\sum_{n=0}^{\infty} a_n$ be an  absolutely convergent series of real numbers. Let $f: \mathbb{N} \to \mathbb{N}$ be an increasing function. Show that $\displaystyle\sum_{n=0}^{\infty} a_{f(n)}$ is also an absolutely convergent series.

I would like someone to help verify my proof.

Proof

Let $\epsilon > 0$. Since $\displaystyle\sum_{n=0}^{\infty} a_n$ is absolutely convergent, we know by the Cauchy Convergence Criterion that there exists some $M \geq 0$ such that for all $p,q \geq M$ we have $$ \sum_{n=p}^q |a_n|  \leq \epsilon.$$ Note, however, that $\displaystyle\sum_p^q |a_{f(n)}| \leq \sum_{f(p)}^{f(q)} |a_n|$ for all $p,q \in \mathbb{N}$ (since each term is positive and for all $n \in \mathbb{N}$, we have $n \leq f(n)$ since $f$ is increasing), so if $p,q \geq M$, we obtain $$\displaystyle\sum_p^q |a_{f(n)}| \leq \displaystyle\sum_{f(p)}^{f(q)} |a_n| \leq \epsilon$$ since $f(p),f(q) \geq M$. Hence, $\displaystyle\sum_{n=0}^{\infty} a_{f(n)}$ is also absolutely convergent by the Cauchy Convergence Criterion.

答:
This means you have to prove $$\lim_{n \to \infty} \sum_{k=1}^n a_{f(k)}$$ $$=\lim_{n \to \infty}(a_{f(1)}+...+a_{f(n)})$$ is absolutely convergent where $f:\mathbb N \mapsto \mathbb N$ is strictly increasing function.
I am giving my argument through an example.

Let $f(n)=n^2$. Then $\sum a_{n^2}=a_1+a_4+a_9+...=a_1+0+0+a_4+0+0+0+0+a_9+...=\sum b_n$ where $a_n=b_n$ if $n$ is a perfect square number and $b_n=0$ otherwise.

$\sum |a_n|$ is convergent.
Since $0\le |b_n|\le |a_n|  \forall n \in \mathbb N$, by Comparison Test $\sum b_n=\sum a_{n^2}$ is absolutely convergent.


问:
Conductor of product of characters is the product of the conductors
First of all, I know that there are other questions similar to this, but these use other definitions or don't have a complete answer.

I'm trying to prove that if $\chi_{1}$ and $\chi_{2}$ are two Dirichlet characters mod $a$ and $b$ respectively, with $(a,b)=1$, then $$c(\chi_{1}\cdot\chi_{2})=c(\chi_{1})\cdot c(\chi_{2})$$ where $c(\chi)$ is the conductor of the character $\chi$.

The definitions that I'm using are the following:

A Dirichlet character $\chi$ mod $q$ is a function $\chi:\mathbb{N}\to\mathbb{C}$ such that $\chi(mn)=\chi(m)\chi(n)$ for
all $m,n\in\mathbb{N}$, $\chi$ is periodic of period $q$, and
$\chi(n)=0$ if $(n,q)>1$.

The conductor $c(\chi)$ of a Dirichlet character $\chi$ is the least period of $\chi$ restricted to values of $n$ such that
$(n,q)=1$. In particular $c(\chi)\mid q$. I will use the notation
$\chi^{*}$ for the function ''$\chi$ restricted for the corresponding
values'', so the conductor of $\chi$ is simply the least period of
$\chi^{*}$.

I know that if $(n,ab)=1$ then $(n,a)=1$ and $(n,b)=1$, so $(\chi_{1}\cdot\chi_{2})^{*}=\chi_{1}^{*}\cdot\chi_{2}^{*}$. Now $\chi_{1}^{*}$ has period $c(\chi_{1})$ and $\chi_{2}^{*}$ has period $c(\chi_{2})$, so $(\chi_{1}\cdot\chi_{2})^{*}$ has period $\text{lcm}(c(\chi_{1}),c(\chi_{2}))=c(\chi_{1})\cdot c(\chi_{2})$, therefore $c(\chi_{1}\cdot\chi_{2})\mid c(\chi_{1})\cdot c(\chi_{2})$.

My problem is that I'm unable to prove that $c(\chi_{1})\cdot c(\chi_{2})\mid c(\chi_{1}\cdot\chi_{2})$, and therefore these two numbers are equal.

I tried this:

For given $n$ such that $(n,ab)=1$ we can find $m$ such that $m\equiv 1\mod{a}$ and $m\equiv n\mod{b}$, so $$\chi_{1}(n)\cdot\chi_{2}(n)=\chi_{1}(1)\cdot\chi_{2}(m)=\chi_{2}(m)$$ therefore $$\{\chi_{1}\cdot\chi_{2}(n)\mid (n,ab)=1\}\subseteq\{\chi_{2}(m)\mid (m,b)=1\text{ and }m\equiv 1\mod{a}\}\subseteq\{\chi_{2}(m)\mid (m,b)=1\}$$
from where (I think) we conclude that $(\chi_{1}\cdot\chi_{2})^{*}$ has period $c(\chi_{2})$, and similarly one proves that it has period $c(\chi_{1})$, but this would imply that it has period $(c(\chi_{1}),c(\chi_{2}))=1$ which doesn't have any sense at all.

Any help would be appreciated.

答:
Use that
$$\chi_1(n) = \chi_1\chi_2(un+(1-u))\qquad where\quad  u=1\bmod a,u= 0\bmod b$$
If $\chi_1\chi_2$ is $r$-periodic then so is $\chi_1$.

Thus the least period of $\chi_1$ divides that of $\chi_1\chi_2$.

The idea is the same when restricting to the $n$ coprime with $a,b$.


问:
Sufficient Condition for an Affine Map
Let $f$ be a map on an affine space $A$ (i.e. any map taking points to points, not necessarily an affine map). Suppose that for any two pairs of points $(p_1,q_1)$ and $(p_2,q_2)$ in $A$ with $\overrightarrow{p_1q_1} = \overrightarrow{p_2q_2}$ we have $\overrightarrow{f(p_1)f(q_1)} = \overrightarrow{f(p_2)f(q_2)}$. Then $f$ induces a well defined map on the vector space associated with $A$, call it $f'$, via $f'(\overrightarrow{pq}) = \overrightarrow{f(p)f(q)}$. Is $f'$ necessarily linear, or in other words, is $f$ necessarily an affine map? If so, how can one prove this? If not, what is a counterexample?

答:
By picking an origin we can think of $A$ as a vector space $V$ WLOG, and then by replacing $f$ with $f - f(0)$ (which preserves the desired condition) we can assume that $f(0) = 0$ WLOG. Then the condition is that if $p_1 - q_1 = p_2 - q_2$ then $f(p_1) - f(q_1) = f(p_2) - f(q_2)$. Rearranging and setting $q_2 = 0$, the condition is that

$$f(a + b) = f(a) + f(b)$$

(where $a = p_2, b = q_1, a + b = p_1$). In other words, $f$ is additive; this condition is necessary and sufficient, since if $f$ is additive then $f(p_i) - f(q_i) = f(p_i - q_i)$.

So the question is equivalent to asking whether an additive function $f : V \to V$ is linear. This is false over every field except the prime vector spaces $\mathbb{Q}, \mathbb{F}_p$, since every other field $F$ is a vector space of dimension $\ge 2$ over its prime subfield and hence admits nontrivial additive maps $F \to F$ which are not $F$-linear.

When $F = V = \mathbb{R}$ the equation $f(a + b) = f(a) + f(b)$ is called Cauchy's functional equation and nonlinear solutions to it (which are necessarily everywhere non-measurable) can be "constructed" using the axiom of choice to find a (Hamel) basis of $\mathbb{R}$ over $\mathbb{Q}$.  However, in this case $f$ is $\mathbb{R}$-linear as soon as one imposes any of a number of very mild additional hypotheses, e.g. $f$ is continuous at any point, measurable at any point, bounded on any interval...

Title: Sufficient Condition for an Affine Map


问:
Derivation of equation. Mackay pg. 17
Can someone explain how the final equation is obtained? I cannot understand how the exponential term occurs. from Mackay Pg. 17

$1=\sum_{K}\left(\begin{array}{l}N \\ K\end{array}\right) 2^{-N} \simeq 2^{-N}\left(\begin{array}{c}N \\ N / 2\end{array}\right) \sum_{r=-N / 2}^{N / 2} e^{-r^{2} / 2 \sigma^{2}} \simeq 2^{-N}\left(\begin{array}{c}N \\ N / 2\end{array}\right) \sqrt{2 \pi} \sigma$

where $\sigma$ = $\sqrt{N/4}$ from
$\left(\begin{array}{c}N \\ N / 2\end{array}\right) \simeq 2^{N} \frac{1}{\sqrt{2 \pi N / 4}}$

答:
It's a rather awkward derivation of the asymptotic value of the central binomial coefficient.

I would put in a slightly simpler form: We know that $p_k = \binom{N}{k} 2^{-N}$ corresponds to a Binomial $(N,\frac12)$ probability distribution, which is the sum of $N$ fair Bernoulli variables. For large $N$, the CLT applies, and $p_k$ can be approximated by a gaussian $g_k =\frac{1}{\sqrt{2 \pi \sigma^2}} \exp( -\frac12(k- \mu)^2)$ with mean $\mu=N/2$ and variance $ \sigma^2=N/4$.

Now, we evaluate that approximation for $k=N/2=\mu$:

$$ \binom{N}{N/2} 2^{-N} \approx \frac{1}{\sqrt{2 \pi \sigma^2}} = \frac{1}{\sqrt{2 \pi N / 4}} $$

Title: Derivation of equation. Mackay pg. 17


答:
Alternatively, split into indicator variables and use linearity of expectation.

For $1\leq i\leq 6$, let $X_i$ be $1$ if we roll $i$ at least once, and $0$ otherwise. Then $$\mathbb E[X]=\sum_{i=1}^6\mathbb E[X_i]=6\mathbb E[X_1].$$
But $\mathbb E[X_1]=\mathbb P(\text{roll a }1)=1-\left(\frac56\right)^4$, so we get $\mathbb E[X]=\frac{671}{216}$.

This method has the benefit of generalising easily. For a $d$ sided die rolled $n$ times, the expected number of distinct outcomes is
$$\mathbb E[X]=d\left[1-\left(\frac{d-1}{d}\right)^n\right].$$

Body:
Four dice are tossed let $N$ be the number of distinct outcomes. Find $E[N]$

If we let $X$ be the number of distinct outcomes in a given toss.

E.g. if we get the toss $1,2,2,4$ then $X=3$

Calculating some probabilities:

$P(X=1) = 6 \cdot 6^{-4}$ since we have six ways we can end up with all four dice having the same value face-up.

Now according to a solution I have for this:

$\displaystyle P(X = 2) = 6^{-4}\cdot 6 \cdot 5 \cdot \binom{4}{2}$

This is where I am not so sure; why $4$ choose $2$ as opposed to $6 $ choose $2$ ?

$P(X= 4) = 6^{-4}\cdot 6 \cdot 5 \cdot 4 \cdot 3$

and finally $P(X = 3) $ I haven't calculated as I am not clear on $P(X=2)$.

The issue for me seems to be the combinatorial part of the argument. i.e. The number of ways each event occurs.


问:
Simple $R$-Modules are isomorphic to quotient of maximal ideal (proof explanation)
I'm trying to understand the proof of the following proposition.

An $R$ -module is $M$ said to be cyclic if $M$ it is generated by one element, and simple if $M$ has no $R$ -submodules other than $0$ and $M$. Show that $M$ is a simple $R$ -module if, and only if, $M$ is isomorphic to $R / \mathfrak{m}$ for some maximal ideal $\mathfrak{m}$ of $R$.

Proof: Due to previous results, we know that:
(a) Any cyclic $R$ module is isomorphic to $R / I$ for some ideal $I$ of $R$, (b) Any simple $R$ -module is cyclic. Therefore, we must show that $R / I$ is simple if, and only if, $I$ is maximal. Let $f: R \rightarrow R / I$ be the natural quotient map. Then given any submodule $J$ of $R / I,$ its preimage $f^{-1}(J)$ is an ideal of $R$ containing $I .$ This gives a bijection between the ideals of $R$ containing $I$ and the submodules of $R / I .$ In particular we see that $R / I$ is simple if, and only if, the only ideals containing $I$ are $I$ itself and the unit ideal; that is, if and only if $I$ is maximal.

Here are my confusions.

If $M$ is a simple $R$-module, then (b) implies it is cyclic, and (a) implies it is isomorphic to $R/I$ for some ideal $I \subset R$. How can we conclude from this that we must show $R / I$ is simple $\iff$ $I$ is maximal? I get that we want to show $M$ is a simple $R$-module $\implies$ $R / I$ is simple $\implies$ $I$ is maximal. But what about the converse direction?

For such $f$ as above, I have managed to prove that for any $R$-submodule(!) $J$ of $R / I,$ its preimage $f^{-1}(J)$ is an ideal of $R$ containing $I .$ I have also proven that if $K \subset R$ is an ideal of $R$ and $I \subset K$, then $f(K)$ is an $R$-submodule of $R/I$ (i.e. $f(K) \subset R/I$). Is this enough to imply that $f$ is a bijection? I have seen this picture many times before, when $\phi: \text{set 1} \to \text{set 2}$ is a function, and $\phi^{-1}:\text{set 2} \to \text{set 1}$ is a function. But I don't know how to justify $\phi$ is a bijection from this.

Partial answers (answering only some of the questions) are very welcome! I will upvote them too.

Edit: I've moved some of my proofs in (2) to a different question.

答:
Remember the original statement we want to prove:

$M$ is simple if and only if $M \cong R/\mathfrak{m}$ for some maximal ideal $\mathfrak{m}$.

We've already shown:

$M$ is simple if and only if $M \cong R/I$ for some ideal $I$ such that $R/I$ is simple.

Thus, we have left to prove:

$\color{green}{M \cong R/I ~\text{for some ideal}~ I ~\text{such that}~ R/I ~\text{is simple}}$ if and only if $\color{blue}{M \cong R/\mathfrak{m} ~\text{for some maximal ideal}~ \mathfrak{m}}$.

If we knew
$R/I$ simple $\iff$ $I$ maximal, the first (green) statement would be
equivalent to "$M \cong R/I$ for some ideal $I$ such that $I$ is
maximal", which is exactly the second (blue) statement. Thus, it suffices
to prove $R/I$ simple $\iff$ $I$ maximal.

$f$ is never a bijection unless $I = 0$! I think you want to say that $J \mapsto f^{-1}(J)$ is a bijection from the set of submodules of $R/I$ to the set of ideals of $R$ which contain $I$. Anyway, your argument is not enough. You have correctly constructed the inverse map, but you never proved it's an inverse! You then need to show that $f^{-1}(f(K)) = K$ and $f(f^{-1}(J)) = J$ for all submodules $J$ of $R/I$ and all ideals $K$ of $R$ such that $I \subseteq K$. Then the two functions (which you correctly showed are well-defined) will be inverses, hence both bijective.


答: If $k \geq 4$, it follows that $k + 1 \geq 5 > 2$. Hence it is safe to say that $2^k \cdot (k + 1) > 2^k \cdot 2$.</p>

Title: proving $n!>2^n\;\;\forall \;n≥4\;$ by mathematical induction

Body:
My teacher proved the following $n!>2^n\;\;\;\forall \;n≥4\;$ in the following way

Basis step: $\;\;4!=24>16$ ok

Induction hypothesis: $\;\;k!>2^k$

Induction step: $\qquad\qquad(k+1)!=k!(k+1)>(k+1)2^k>2^k\cdot 2=2^{k+1}$

I wonder how did he assume that $2^k(k+1)>2^{k}\cdot 2\quad\forall k≥4$?

Don't we have to show it by induction too?


问:
How to prove that a sum: $\sum_{n=1}^\infty \cos(n\sqrt 2)$ is uniformly bounded?
How to prove that a sum:

$$\sum_{n=1}^\infty \cos(n\sqrt 2)$$
is uniformly bounded?

That problem is connected to an answer to other question that I have asked before:

Convergence of $\sum_{n=0}^{\infty}\frac{cos(n \sqrt{2})}{\sqrt{n}}$. Is my thinking correct?

答:
I'll provide two methods, one using complex numbers and one using sum-of-angle formulas. Let $b=\sqrt 2$.

Note that by Euler's formula:
\begin{align*}
\sum_{n=1}^N\cos\left(bn\right)&=\Re\sum_{n=1}^N\left(\cos\left(bn\right)+i\sin\left(bn\right)\right)\\
&=\Re\sum_{n=1}^Ne^{ibn}\\
&=\Re\frac{e^{ib(N+1)}-e^{ib}}{e^{ib}-1}
\end{align*}

Here $e^{ib}-1\ne 0$ and $e^{ib(N+1)}$ is uniformly bounded, so the partial sum is also uniformly bounded.

To use trigonometry, you want to recall formulas for $f\left((n\pm 1)b\right)$ where $f$ is $\sin$ or $\cos$. In this case $\sin$ works:

\begin{align*}
\Delta_n&=\sin\left((n+1)b\right)-\sin\left((n-1)b\right)\\
&=\sin(nb)\cos(b)+\cos(nb)\sin(b)-\left\{\sin(nb)\cos(b)-\cos(nb)\sin(b)\right\}\\
&=2\cos(nb)\sin(b)
\end{align*}

Therefore we have a telescoping sum:

\begin{align*}
\sum_{n=1}^N\cos\left(bn\right)&=\frac{1}{2\sin\left(b\right)}\sum_{n=1}^N\left(\sin\left((n+1)b\right)-\sin\left((n-1)\right)b\right)\\
&=\frac{\sin\left((N+1)b\right)+\sin\left(Nb\right)-\sin\left(b\right)-\sin\left(0\right)}{2\sin(b)}
\end{align*}

Now again $\sin(b)\ne 0$ and $\sin((N+1)b)+\sin(Nb)$ is uniformly bounded, so the partial sum is uniformly bounded.


问:
Why does the multiplicative order divide $|\mathbb{F}|-1$?
This source describes Lagrange's theorem for finite fields (page 6). It generalizes Fermat's little theorem. In particular, it states that if $\mathbb{F}$ is a field with finitely many elements, with $|\mathbb{F}|=m$, then
$$a^{m-1}=1$$
for every $a\in\mathbb{F}\setminus\{0\}$.

To prove this, they note that $\mathbb{F}\setminus\{0\}$ has $m-1$ elements. But then, they suddenly conclude that this implies $\text{ord}_{\mathbb{F}}(a)$ divides $m-1$. The result then follows.

How is the divisibility claim reached?

答: $\mathrm{ord}_{\Bbb F}(a)=|\langle a\rangle|\mid|\Bbb F\setminus\{0\}|$ for any $a\in\Bbb F\setminus\{0\}$. This is Lagrange's theorem for groups


问:
Proof that the orthogonal projection onto a subspace satisfies the Cauchy Schwarz inequality
I am trying to prove

$\left\|\hat{P}_{s}x\right\|^{2}_{2} \leqslant \left\| x \right\|_{2}\left\|\hat{P}_{s}x \right\|_{2} $ for all $x_1, x_2 \in X$

I am aware that I will need to use the Cauchy Schwarz inequality to prove this, however am not able to justify why $|<x, \hat{P}_{s}x>|_{2} =  \left\|\hat{P}_{s}x\right\|^{2}_{2}$ if I do use the Cauchy Schwarz inequality.

答:
If $P$ is an orthogonal projection, then $P^\top P =P$, so
$$x^\top P^\top P x = x^\top P x.$$


问:
Induced map on Eilenberg-MacLane space
Let $X$ be an $n-1$ connected space. Why does a map $X\rightarrow K(\pi_n(X),n)$ that induces an isomorphism on $\pi_n$ exist and what is this map?</p>

答:
Let me assume $n>1$ and —so that your claim is true— that $X$ is a CW complex.

The Hurewicz theorem, in view of the hypothesis, gives us an isomorphism $$\phi:H_n(X,\mathbb Z)\to\pi_n(X).$$
On the other hand, in view of the hypothesis made on $X$, the universal coefficient theorem for cohomology gives us an isomorphism
  $$\alpha:H^n(X,\pi_n(X))\to \hom(H_n(X,\mathbb Z), \pi_n(X)) $$
Moreover, there is a canonical bijection
  $$\beta:[X,K(\pi_n(X), n)]\to H^n(X,\pi_n(X))$$
with $[X,K(\pi_n(X), n)]$ the set of homotopy classes of maps $X\to K(\pi_n(X),n)$.

The map you are looking for is any $f:X\to K(\pi_n(X),n)]$ whose homotopy class $[f]$ is such that $\alpha(\beta([f]))=\phi$.

(This works for any $K(\pi_n(X),n)$ that you pick)

Title: Induced map on Eilenberg-MacLane space


问:
Summation of a finite series
Let

$$f(n) = \frac{1}{1} + \frac{1}{2} + \frac{1}{3}+ \frac{1}{4}+...+ \frac{1}{2^n-1} \forall \ n \ \epsilon \ \mathbb{N} $$

If it cannot be summed , are there any approximations to the series ?

答:
$$1+(\frac12+\frac12)+ (\frac14+\frac14+\frac14+\frac14)+ \dots (\dots \frac1{2^{n-1}})\gt$$

$$\frac{1}{1} + \frac{1}{2} + \frac{1}{3}+ \frac{1}{4}+...+ \frac{1}{2^n-1}\gt$$  $$1+\frac12+(\frac14+\frac14)+(\frac18+\frac18+\frac18+\frac18)+\dots +( \dots \frac1{2^n})$$ (compare equivalent terms, and count the elements in the last bracket carefully) gives some bounds (the summation is straightforward). Since $\frac 1n$ is monotone decreasing, it is easy to use the integral to give tighter bounds than this.

The question is tagged as a contest, so integration may not be appropriate (depends on contest parameters). On the other hand the sums above, though rather crude, can be computed in a contest, and make sense of the relationship of the sum with a power of $2$.

I'd be interested in a better elementary estimate not involving integration.

Title: Summation of a finite series


答:
By defintion of $\hat P_S$ being an orthogonal projection,
$$\langle \hat P_Sx, x-\hat P_S x\rangle = 0$$
This can be rewritten,  $$\langle \hat P_Sx,  x\rangle =\|\hat P_S x\|^2\tag{1}$$
Now, Cauchy-Schwarz means $$\langle \hat P_Sx,  x\rangle  \leq \|\hat  P_S x\|\|x\|\tag{2}$$
Injecting $(2)$ into $(1)$ yields the result.

Title: Proof that the orthogonal projection onto a subspace satisfies the Cauchy Schwarz inequality

Body:
I am trying to prove

$\left\|\hat{P}_{s}x\right\|^{2}_{2} \leqslant \left\| x \right\|_{2}\left\|\hat{P}_{s}x \right\|_{2} $ for all $x_1, x_2 \in X$

I am aware that I will need to use the Cauchy Schwarz inequality to prove this, however am not able to justify why $|<x, \hat{P}_{s}x>|_{2} =  \left\|\hat{P}_{s}x\right\|^{2}_{2}$ if I do use the Cauchy Schwarz inequality.


问:
Non-affine transform for visual detail
I'm designing a transformation that I think needs to be non-affine by definition. Technically it's a chain of two transformations:

The first transforms from the frequency space to a logarithmic chromatic music note space, in units of cents,  via

$$ x = 100 \cdot 12 \log_2 \frac f {55} - t $$

where $t$ is some known, desired tuning note.

The second is an antisymmetric non-affine transformation $y(x)$ for the purposes of graphing with the following requirements:

$$y(0) = 0$$
$$y(600) = 600$$
$$y(x) = -y(-x)$$
$$ \text{For all } 0 \le x \le 600, \frac {\partial y}{\partial x} > 0$$
$$ \text{For all } 0 < x < 600, y(x) < x $$

The purpose is to zoom into $x=0$ while still showing $x = -600, x=600$. The function must be invertible and the inverse will be used about as frequently as the function itself, due to the way that matplotlib is written.

The two candidates I thought of are a cubic and a tangent. The cubic implementation is

$$ y = a x^3 + b x , b > 0$$
$$ a = \frac {s - 100} {s(600^2 - s^2)} $$
$$ b = 1 - 600^2 a $$

where $s$ is a scaling parameter, interpreted as the value of $x$ where $y = 100$.

This works well enough; here is a $1 - \left| \frac x {600} \right|$ triangle shown with this horizontal zoom for $s = 320$:

This formulation has two big disadvantages:

Values of $s > 330$ are not possible without making $b < 0$, in turn giving the curve undesired inflection points

The inverse $x(y)$ is truly monstrous and I won't include it here unless requested

Is there a different function that will meet these requirements but is more easily invertible?

答:
Well, the best I've found so far is neither $\tan$ nor cubic, but a piecewise hyperbola; with a scaling parameter $a$ the following applies to the right half of the domain, with the left half mirrored:

$$a = 1 \times 10^{-5} $$
$$b = \frac {600a + \sqrt{600^2 a^2 + 4a}} 2 $$
$$y = \frac 1 {b - ax} - \frac 1 b $$

This is trivially invertible and does not have the same scaling parameter limit as the cubic.

Title: Non-affine transform for visual detail


答:
Expected value

$$E[X] = \sum_{k=1}^4 [k \times p(k)]$$

where, for each $k, ~p(k)$ will represent the probability of having exactly $k$ distinct numbers, and will equal

$$\frac{N\text{(umerator)}_k}{D\text{(enominator)}}.$$

For all values of $k$,

$$D = 6^4.$$

As the OP indicated,

$N_1 = 6 \times \binom{4}{4} = 6.$

To compute $N_2$ note that you can have either [3 of one number, 1 of a different number] or [2 of one number, 2 of another number].

In the first case, there are $(6 \times 5)$ ways of choosing the two numbers that show and $\binom{4}{1}$ ways of choosing the odd die out.  In the second case, superficially, one would imagine that there are [again] $(6 \times 5)$ ways of choosing the two numbers and $\binom{4}{2}$ ways of grouping the 4 dice in pairs.  This is wrong, as has already been indicated, because of overcounting.  You have to divide the second case by $2$.

A safer way of enumerating the second case is to pretend that one of the dice is green, and the others are red.  There are 3 potential partner dice that can be joined with the green die.  Once the partnership is made, there are $6$ choices for the [green die + partner] and then $5$ choices for the other partnership.

Therefore

$N_2 = [6 \times 5 \times 4] + [6 \times 5 \times 3] = 210.$

To compute $N_3$, there must be 2 dice that roll the same number and two other dice that roll different numbers.  There are $\binom{4}{2}$ ways of deciding which two dice get partnered.  Then, there are $6$ choices for which number this partnership will roll.

For the other two dice, although it (wrongly) appears that an overcounting trap is present, similar to the second case for $N_2$, that is false.  The easiest way to see this is to pretend that the two single dice are two colors, blue and green.
After one of the 6 numbers is assigned to the partnership, there are then $5$ choices for the blue die and then $4$ [remaining] choices for the green die.

Therefore

$N_3 = \binom{4}{2} \times 6 \times 5 \times 4 = 720.$

To compute $N_4$, simply assume that each die has a different color (e.g red, green, blue, yellow) and that there are 6 choices for the red die, then 5 choices for the green die, then 4 choices for the blue die, and then 3 choices for the yellow die.

Therefore

$N_4 = 6 \times 5 \times 4 \times 3 = 360.$

Before proceeding further, there is a sanity check.  There are $6^4$ ways that dice of 4 different colors may be rolled.  Therefore, $N_1 + N_2 + N_3 + N_4$ must equal $6^4 = 1296.$

Final answer:

$$\frac{1}{D} \times \sum_{k=1}^4 [k \times N_k] ~=~
\frac{(1 \times 6) + (2 \times 210) + (3 \times 720) + (4 \times 360)}{1296}$$

$$=~ \frac{4026}{1296} ~=~ \frac{671}{216}.$$

Title: Four dice are tossed let $N$ be the number of distinct outcomes. Find $E[N]$


问:
A question regarding algebra
A book costs $1$ plus half its price. How much does it cost?

The answer to the question as suggested by some is $2$.

They equated this by taking a variable for price and forming a linear equation:

$$p=1+p/2\\p=2$$

I think there is some problem in this answer but I am not getting what it actually is.

答:
If p is its price then

p/2 is "half-its-price".

So 1+p/2 is 1+"half-its-price", which is assumed to be  its cost,

which is its price, which is p.

So 1+p/2=p.

Unless you are an accountant, who would object to using "price" and "cost" synonymously.

Title: A question regarding algebra