Math Challenge - December 2020

[fresh_42]

Summary: Circulation, Number Theory, Differential Geometry, Functional Equation, Group Theory, Infinite Series, Algorithmic Precision, Function Theory, Coin Flips, Combinatorics.1. (solved by @etotheipi ) Given a vector field
$$
F\, : \,\mathbb{R}^3 \longrightarrow \mathbb{R}^3\, , \,F(x,y,z)=\begin{pmatrix}x-y^2\\x^2\\z\end{pmatrix}
$$
Calculate the circulation of $F$ along the mathematically positive oriented unit circle in the $(x,y)$-plane
(a) by using the curl of a vector field.
(b) directly by an appropriate path.

2. An odd prime $p$ can be written as $p=x^2+y^2$ with integers $x,y\in \mathbb{Z}$ if and only is $p\equiv 1 \mod 4.$

3. Let $\mathbb{T}^n :=\mathbb{R}^n/\mathbb{Z}^n$ equipped with the quotient topology according to the projection
$$
\pi\, : \,\mathbb{R}^n \longrightarrow \mathbb{T}^n\, , \,\pi(a)=a+\mathbb{Z}^n.
$$
Show that $\mathbb{T}^n$ is a topological manifold.

4. Let $\alpha\in \mathbb{R}-\{0\}.$ Determine all functions $f:\mathbb{R}_{>0}\longrightarrow \mathbb{R}_{>0}$ which satisfy for all $x,y\in \mathbb{R}_{>0}$
$$
f(f(x)+y)=\alpha x+\dfrac{1}{f\left(\dfrac{1}{y}\right)}
$$

5. (solved by @fishturtle1 ) Show that the quaternion group $G=\{\pm 1,\pm i,\pm j,\pm k\}$ is a Hamilton group, and cannot be written as a semidirect product in a nontrivial way.

6. (solved by @LCSphysicist and @etotheipi )
(a) Calculate
$$
\dfrac{1}{2}-\dfrac{1}{12}+\dfrac{1}{40}-\dfrac{1}{112} \mp \ldots=\sum_{k=0}^{\infty}(-1)^k\dfrac{1}{(2k+1)2^{k+1}}\,.
$$
6. (b) Prove $\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\ldots=1\,.$

7. Prove
$$
\tan^{-1}(1/2)+\tan^{-1}(1/3)=\pi/4=4\tan^{-1}(1/5)-\tan^{-1}(1/239)
$$
and use the power series representation
$$
\tan^{-1}(x)=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^7}{7} \mp \ldots=\sum_{k=0}^{\infty}(-1)^k\dfrac{x^{2k+1}}{2k+1}
$$
to determine how many terms would it take to compute $\pi$ up to $100$ digits by $\pi=4\tan^{-1}(1)$ and by the formulas above.

8. (solved by @Gaussian97 ) Calculate the following integrals
(a) $\displaystyle{\int_0^{2\pi}e^{(e^{it})}\,dt}$
(b) $\displaystyle{\int_{|z|=1}\dfrac{\sin(z^2)}{(\sin z)^2}\,dz}$
(c) $\displaystyle{\int_{|z|=1}\sin \left(e^{1/z}\right)\,dz}$
(d) $\displaystyle{\int_{-\infty}^\infty \dfrac{1}{x^2-2x+2}\,dx}$

9. Suppose someone gives you a coin and claims that this coin is biased; that it lands on heads only $48\%$ of the time with an error margin of $2\%$. You decide to test the coin for yourself. If you want to be $95\%$ confident that this coin is indeed biased, how many times must you flip the coin? Compare the estimations by the weak law of large numbers and by the central limit theorem!

10. (solved by @Gaussian97 ) A hat-check boy at a congress held at Hilbert's hotel completely loses track of which of hats belong to which owners, and hands them back at random to their owners as the latter leave. What is the probability $P$ that nobody receives their own hat back?

High Schoolers only

11. (solved by @lekh2003 ) Prove that
$$
\dfrac{1}{1+x+\dfrac{1}{y}}+\dfrac{1}{1+y+\dfrac{1}{z}}+\dfrac{1}{1+z+\dfrac{1}{x}}\leq 1
$$
for all positive real numbers $x,y,z.$

12. Which is the smallest natural number greater than one such that the following statement holds:
In any set of $n$ natural numbers are at least two numbers, whose sum or difference is dividable by seven.

13. Determine
$$
\left[ \dfrac{1}{\sqrt{1}+\sqrt{2}} + \dfrac{1}{\sqrt{3}+\sqrt{4}} + \dfrac{1}{\sqrt{5}+\sqrt{6}}+ \ldots + \dfrac{1}{\sqrt{n^2-4}+\sqrt{n^2-3}} +\dfrac{1}{\sqrt{n^2-2}+\sqrt{n^2-1}} \right]
$$
for any odd natural number $n\geq 3$ where $[n]=\lfloor n\rfloor$ is the greatest integer smaller or equal $n.$

14. (solved by @YoungPhysicist ) Given a point $P$ inside an equilateral triangle $\triangle ABC$ with area $1.$ Show that for the lengths $x,y,z$ of the perpendiculars of $P$ onto the sides of the triangle holds $x+y+z=\sqrt[4]{3}.$

15. Prove: If for the edges of a tetrahedron $ABCD$ holds
$$
\overline{AD}=\overline{BD}=\overline{CD}=1 \text{ and } \overline{AB}=\overline{BC}=\overline{CA}\,,
$$
then its surface is smaller than $\dfrac{3\sqrt{3}}{2}.$

 

[etotheipi]

Spoiler: Problem 1

Firstly, just by doing the explicit line integral$$\begin{align*}

\oint_C \vec{F} \cdot \vec{r}'(\theta) d\theta &= \int_0^{2\pi} \vec{F} \cdot \hat{\theta}(\theta) d\theta \\&= \int_0^{2\pi} \left( -\sin{\theta} \left[ \cos{\theta} - \sin^2{\theta} \right] + \cos^3{\theta} \right) d\theta \\

&= \int_0^{2\pi} \left( \sin^{3}(\theta) + \cos^3{\theta} - \frac{1}{2} \sin{2\theta} \right) d \theta \\

& = \int_0^{2\pi} \left( \frac{3}{4} \left[ \sin{\theta} + \cos{\theta} \right] + \frac{1}{4} \left[ \sin{3\theta} + \cos{3\theta} \right] - \frac{1}{2} \sin{2\theta} \right) d\theta \\

&= \left[ \frac{3}{4} \left[\sin{\theta} - \cos{\theta} \right] + \frac{1}{12} \left[ \sin{3\theta} - \cos{3\theta} \right] + \frac{1}{4} \cos{2\theta} \right]_0^{2\pi} = 0\end{align*}$$Secondly, by using Stokes' theorem,$$\int_C \vec{F} \cdot d\vec{r} = \int_A (\nabla \times \vec{F}) \cdot d\vec{S} = \int_A 2(x+y) dS$$But since $g(x,y) \equiv 2(x+y)$ satisfies $g(x,y) = -g(-x,-y)$, and the region $A$ is the unit circle, by symmetry this must be zero.

 

[etotheipi]

Spoiler: Problem 7

For the first identity$$\tan{\left(\arctan{\left(\frac{1}{2} \right)} + \arctan{\left(\frac{1}{3} \right)} \right)} = \frac{1/2 + 1/3}{1-1/6} = 1$$and for the latter$$
\begin{align*}
\tan{\left( 4\arctan{\left( \frac{1}{5} \right)} - \arctan{\left( \frac{1}{239} \right)} \right)} &= \frac{ \tan{ \left( 4\arctan{\left( \frac{1}{5} \right)} \right) } }{1 + \frac{1}{239} \tan{ \left( 4\arctan{\frac{1}{239}} \right)}}

\\
\end{align*}
$$but we know that $$\tan{\left( 4\arctan{\frac{1}{5}} \right)} = \frac{2\tan{\left(2\arctan{\frac{1}{5}} \right)}}{1- [\tan{\left( 2\arctan{\frac{1}{5}} \right) }]}$$and also that$$\tan{ \left( 2\arctan{\left( \frac{1}{5} \right)} \right) = \frac{2\tan{\left(\arctan{\frac{1}{5}} \right)}}{1- [\tan{\left( \arctan{\frac{1}{5}} \right) }]}} = \frac{5}{12}$$which impies that $$\tan{ \left( 4\arctan{\left( \frac{1}{5} \right)} \right) } = \frac{120}{119}$$and we can finally substitute these back in,$$\tan{ \left( 4\arctan{\left( \frac{1}{5}\right)} - \arctan{\left( \frac{1}{239} \right)} \right)} = \frac{120/119 - 1/239}{1 + (1/139) \times (120/119)} = 1$$

Now for our estimations, for $\pi = 4\arctan{1}$, the general term is$$|u_k| = \frac{4}{2k-1}$$so we require $\frac{4}{2k-1} < 5 \times 10^{-101}$. For $\pi = 4\arctan{\frac{1}{2}} + 4\arctan{\frac{1}{3}}$, the general term is $$|v_k| = \frac{4\left [ \frac{1}{2}^{2k-1} + \frac{1}{3}^{2k-1}\right]}{2k-1}$$whilst for $\pi = 16\arctan{\frac{1}{5}} - 4\arctan{\frac{1}{239}}$, the general term is $$|w_k| = \frac{16 \left( \frac{1}{5} \right)^{2k-1} - 4\left( \frac{1}{239} \right)^{2k-1}}{2k-1}$$you can figure out $k$ by plugging some numbers in, so that $|v_k| < 5\times 10^{-101}$ and $|w_k| < 5 \times 10^{-101}$ respectively, but I'm too tired now.

 

[LCSphysicist]

Spoiler: Problem 6

a)
This expression is rather similar to the expansion of arc tan:

$$\frac{1/2}{1} - \frac{1/4}{3} + \frac{1/8}{5} + ...$$
but the terms in the denominator is really interesting, it follow this rule:
$$x^{2N+1} = 2^{-(N+1)}

Exp: N = 0 -> x = 1/2$$

So that this sum is he expansion of $tan^{-1}(\frac{(2^{-(N+1)})}{2N+1})$

Now we got a problem, it is not constant, but we can make the things be better, you can realize that if $x^{2N+1} = 2^{-(N+1/2)}$, we can cut off the term 2N+1, so to do that, we just need to multiply and divide all the series by some number, well... the square of two in fact, so that:
$$\frac{1/2}{1} - \frac{1/4}{3} + \frac{1/8}{5} + ...$$
become
$$1/\sqrt{2}(\frac{\sqrt{2}/2}{1} - \frac{\sqrt{2}/4}{3} + \frac{\sqrt{2}/8}{5} + ...)$$
Now our first expression, in fact, become
$$x^{2N+1} = 2^{-(N+1/2)} $$
That's the end, the series is now:
$$1/\sqrt{2}(\frac{\sqrt{2}/2}{1} - \frac{\sqrt{2}/4}{3} + \frac{\sqrt{2}/8}{5} + ... = \frac{tan^{-1}(\sqrt{2}^{-1})}{\sqrt{2}})$$

b)

B is easier, we are just summing $n/n! - 1/n! = 1/(n-1)! - 1/n!$

This is the trivial exponential series, so that
sum (from n = 2 to infty) $1/(n-1)! - 1/n! = (e - 1 ) - (e - 1 - 1) = 1$

 

[fresh_42]

Spoiler: Problem 1

Firstly, just by doing the explicit line integral$$\begin{align*}

\oint_C \vec{F} \cdot \vec{r}'(\theta) d\theta &= \int_0^{2\pi} \vec{F} \cdot \hat{\theta}(\theta) d\theta \\&= \int_0^{2\pi} \left( -\sin{\theta} \left[ \cos{\theta} - \sin^2{\theta} \right] + \cos^3{\theta} \right) d\theta \\

&= \int_0^{2\pi} \left( \sin^{3}(\theta) + \cos^3{\theta} - \frac{1}{2} \sin{2\theta} \right) d \theta \\

& = \int_0^{2\pi} \left( \frac{3}{4} \left[ \sin{\theta} + \cos{\theta} \right] + \frac{1}{4} \left[ \sin{3\theta} + \cos{3\theta} \right] - \frac{1}{2} \sin{2\theta} \right) d\theta \\

&= \left[ \frac{3}{4} \left[\sin{\theta} - \cos{\theta} \right] + \frac{1}{12} \left[ \sin{3\theta} - \cos{3\theta} \right] + \frac{1}{4} \cos{2\theta} \right]_0^{2\pi} = 0\end{align*}$$Secondly, by using Stokes' theorem,$$\int_C \vec{F} \cdot d\vec{r} = \int_A (\nabla \times \vec{F}) \cdot d\vec{S} = \int_A 2(x+y) dS$$But since $g(x,y) \equiv 2(x+y)$ satisfies $g(x,y) = -g(-x,-y)$, and the region $A$ is the unit circle, by symmetry this must be zero.

You could have elaborated the second part a bit for readers who are less familiar with the concept or are learning it, but o.k., the definition of the curl can be looked up and the cross product can be done in mind, but I missed the normal. You silently moved from a vector to a scalar without any explanations.

And you confused first and second part - just mentioning it for those who wonder.

 

[etotheipi]

I missed the normal. You silently moved from a vector to a scalar without any explanations.

I used area element $d\vec{S} = dS \hat{z}$, and then that $\hat{z}^2 = 1$

 

[fresh_42]

Spoiler: Problem 7

For the first identity$$\tan{\left(\arctan{\left(\frac{1}{2} \right)} + \arctan{\left(\frac{1}{3} \right)} \right)} = \frac{1/2 + 1/3}{1-1/6} = 1$$and for the latter$$
\begin{align*}
\tan{\left( 4\arctan{\left( \frac{1}{5} \right)} - \arctan{\left( \frac{1}{239} \right)} \right)} &= \frac{ \tan{ \left( 4\arctan{\left( \frac{1}{5} \right)} \right) } }{1 + \frac{1}{239} \tan{ \left( 4\arctan{\frac{1}{239}} \right)}}

\\
\end{align*}
$$but we know that $$\tan{\left( 4\arctan{\frac{1}{5}} \right)} = \frac{2\tan{\left(2\arctan{\frac{1}{5}} \right)}}{1- [\tan{\left( 2\arctan{\frac{1}{5}} \right) }]}$$and also that$$\tan{ \left( 2\arctan{\left( \frac{1}{5} \right)} \right) = \frac{2\tan{\left(\arctan{\frac{1}{5}} \right)}}{1- [\tan{\left( \arctan{\frac{1}{5}} \right) }]}} = \frac{5}{12}$$which impies that $$\tan{ \left( 4\arctan{\left( \frac{1}{5} \right)} \right) } = \frac{120}{119}$$and we can finally substitute these back in,$$\tan{ \left( 4\arctan{\left( \frac{1}{5}\right)} - \arctan{\left( \frac{1}{239} \right)} \right)} = \frac{120/119 - 1/239}{1 + (1/139) \times (120/119)} = 1$$

Now for our estimations, for $\pi = 4\arctan{1}$, the general term is$$|u_k| = \frac{4}{2k-1}$$so we require $\frac{4}{2k-1} < 5 \times 10^{-101}$. For $\pi = 4\arctan{\frac{1}{2}} + 4\arctan{\frac{1}{3}}$, the general term is $$|v_k| = \frac{4\left [ \frac{1}{2}^{2k-1} + \frac{1}{3}^{2k-1}\right]}{2k-1}$$whilst for $\pi = 16\arctan{\frac{1}{5}} - 4\arctan{\frac{1}{239}}$, the general term is $$|w_k| = \frac{16 \left( \frac{1}{5} \right)^{2k-1} - 4\left( \frac{1}{239} \right)^{2k-1}}{2k-1}$$you can figure out $k$ by plugging some numbers in, so that $|v_k| < 5\times 10^{-101}$ and $|w_k| < 5 \times 10^{-101}$ respectively, but I'm too tired now.

You are sloppy. O.k. I know why and I understand it, as well as I feel bad to request a bit more rigor: I did not ask for the tangent of the solution, and the tangent function isn't even injective!

I think it is easier and more transparent using complex numbers and polar coordinates which gets you automatically on the correct branch.

Anyway, this wasn't the important part. The important part was the algorithm accuracy given by the three different formulas and a rigorous estimation of the series to achieve the 100 digits in each case. But as you are sloppy and I do not want to argue about correct limits, I recommend to leave this problem for someone who appreciates strict estimations. The clue however, was the significantly different numbers of steps in order to demonstrate that the same algorithm (the Taylor series) has totally different run times depending on how it is used.

 

[fresh_42]

Spoiler: Problem 6

a)
This expression is rather similar to the expansion of arc tan:

$$\frac{1/2}{1} - \frac{1/4}{3} + \frac{1/8}{5} + ...$$
but the terms in the denominator is really interesting, it follow this rule:
$$x^{2N+1} = 2^{-(N+1)}

Exp: N = 0 -> x = 1/2$$

So that this sum is he expansion of $tan^{-1}(\frac{(2^{-(N+1)})}{2N+1})$

Now we got a problem, it is not constant, but we can make the things be better, you can realize that if $x^{2N+1} = 2^{-(N+1/2)}$, we can cut off the term 2N+1, so to do that, we just need to multiply and divide all the series by some number, well... the square of two in fact, so that:
$$\frac{1/2}{1} - \frac{1/4}{3} + \frac{1/8}{5} + ...$$
become
$$1/\sqrt{2}(\frac{\sqrt{2}/2}{1} - \frac{\sqrt{2}/4}{3} + \frac{\sqrt{2}/8}{5} + ...)$$
Now our first expression, in fact, become
$$x^{2N+1} = 2^{-(N+1/2)} $$
That's the end, the series is now:
$$1/\sqrt{2}(\frac{\sqrt{2}/2}{1} - \frac{\sqrt{2}/4}{3} + \frac{\sqrt{2}/8}{5} + ... = \frac{tan^{-1}(\sqrt{2}^{-1})}{\sqrt{2}})$$

b)

B is easier, we are just summing $n/n! - 1/n! = 1/(n-1)! - 1/n!$

This is the trivial exponential series, so that
sum (from n = 2 to infty) $1/(n-1)! - 1/n! = (e - 1 ) - (e - 1 - 1) = 1$

You have the correct solutions, but a horrible way to present them! Reading it is like having to solve it alone again in a different matter, which I did not do.

If anyone else wants to try: the calculation trick (for both) hasn't been found, yet, so a second solution is still possible,

 

[etotheipi]

@fresh_42 here's another solution for 6b. Define $$f(x) \equiv \frac{x^2}{2!} + \frac{2x^3}{3!} + \frac{3x^4}{4!} + \dots$$Then$$f'(x) = \frac{1}{0!}x + \frac{1}{1!}x^2 + \frac{1}{2!} x^3 + \dots$$and consequently$$\frac{1}{x} f'(x) = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!} x^3 + \dots = e^x$$To find $f(x)$, we need to solve$$f'(x) = xe^x \implies f(x) = \int xe^x dx = xe^x - \int e^x dx = (x-1)e^x + C$$Since $f(0) = 0$ (from the first line), we must have $C=1$. The series we're after is $f(1)$, which is$$f(1) = 0 \cdot e^0 + 1 = 1$$

 

[fresh_42]

@fresh_42 here's another solution for 6b. Define $$f(x) \equiv \frac{x^2}{2!} + \frac{2x^3}{3!} + \frac{3x^4}{4!} + \dots$$Then$$f'(x) = \frac{1}{0!}x + \frac{1}{1!}x^2 + \frac{1}{2!} x^3 + \dots$$and consequently$$\frac{1}{x} f'(x) = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!} x^3 + \dots = e^x$$To find $f(x)$, we need to solve$$f'(x) = xe^x \implies f(x) = \int xe^x dx = xe^x - \int e^x dx = (x-1)e^x + C$$Since $f(0) = 0$ (from the first line), we must have $C=1$. The series we're after is $f(1)$, which is$$f(1) = 0 \cdot e^0 + 1 = 1$$

That's the idea! It works for the series and the tangent in part (a), too. It's a nice useful trick to calculate series.

 

[fresh_42]

It's even a bit nicer in part (a) since you do not need to know the series for the tangent function, only an integral:
\begin{align*}
f(x)&:=\dfrac{x}{2}-\dfrac{x^3}{12}+\dfrac{x^5}{40}-\dfrac{x^7}{112} \mp \ldots=\sum_{k=0}^{\infty}(-1)^k\dfrac{x^{2k+1}}{(2k+1)2^{k+1}}\\
f'(x)&=\dfrac{1}{2}-\dfrac{x^2}{4}+\dfrac{x^4}{8}-\dfrac{x^6}{16} \mp \ldots=\sum_{k=0}^{\infty}(-1)^k\dfrac{x^{2k}}{2^{k+1}}=\dfrac{1}{2+x^2}
\end{align*}
As $f(0)=0$ we get
\begin{align*}
f(1)&=\sum_{k=0}^{\infty}(-1)^k\dfrac{1}{(2k+1)2^{k+1}}\\&=f(1)-f(0)=\int_0^1f'(x)\,dx\\&=\int_0^1 \dfrac{dx}{2+x^2}=\left[\dfrac{1}{\sqrt{2}}\cdot\tan^{-1}\left(\dfrac{x}{\sqrt{2}}\right)\right]_0^1\\
&=\dfrac{1}{\sqrt{2}}\cdot\tan^{-1}\left(\dfrac{1}{\sqrt{2}}\right)\approx 0.43521
\end{align*}

So one does not have to identify the series to apply the trick!

 

[benorin]

9. Suppose someone gives you a coin and claims that this coin is biased; that it lands on heads only 48% of the time with an error margin of 2%. You decide to test the coin for yourself. If you want to be 95% confident that this coin is indeed biased, how many times must you flip the coin? Compare the estimations by the weak law of large numbers and by the central limit theorem!

Spoiler: “CLT half of #9”

I think I've got the first half done ok: CLT.

From the Central Limit Theorem (CLT) we have that $Z=\tfrac{np - \mu _{p^{\prime}}}{\sigma _{p^{\prime}}}=1.645$ where the value $Z=1.645$ comes from the $95%$ confidence level from a Z-table, $\mu _{p^{\prime}} =np$ is the population mean, and $\sigma _{p^{\prime}} = \sqrt{\tfrac{p(1-p)}{n}}$ because it's CLT for a proportion and here $p=.48$ is the probability of heads for the sample of biased coin flips and we get

$$Z= \tfrac{0.48n-0.5n}{\sqrt{ n(0.48)(0.52) }} = 1.645 \Rightarrow n=\left( 1.645 \tfrac{\sqrt{ (0.48)(0.52)}}{0.02} \right) ^{\tfrac{2}{3}} = 12$$

where the ceiling function was applied because $n$ is integer.

Here's where it get fuzzy...

According to the weak law of large numbers, in particular the associated Chebyshev's inequality we have

$$P\left( \left| \bar{x}_n - \mu\right|\geq \epsilon\right)\leq \tfrac{\sigma ^2}{n\epsilon ^2}$$

From the margin of error we have $\epsilon =.02,$ and since it's binomial trials we have that $\sigma ^2 = np(1-p),$ here $p$ is the probability of heads for the sample being $p=0.48$. Hence
$$P\left( \left| p - \mu\right|\geq .02\right)\leq \tfrac{ n(0.48)(0.52)}{ n(0.02)^2}=0.05$$
since we are testing at the 95% confidence level--and freaking $n$ cancels instead of letting us solve for it: was I supposed to have used $\sigma _{p^{\prime}} ^2 = \tfrac{p(1-p)}{n}$ as in the CLT version? It might be wrong it's been a long while since my last stats course, and I'm kinda stoned so may have gremlins mucking about, speaking of which $n=12$ seems low.

 

[mathwonk]

hint for 2, try the only if direction. (the other direction is a theorem due to fermat; you might enjoy tackling it, but do not be discouraged if it is elusive.)

 

[etotheipi]

Yeah I think the "if $p = x^2 + y^2$ then $p \equiv 1 \, \, (\text{mod}\,\, 4)$" direction is fine, because $x^2, y^2 \equiv 0,1 \, \, (\text{mod}\,\, 4) \implies x^2 + y^2 \equiv 0,1,2 \, \, (\text{mod}\,\, 4)$ but then since $p$ is odd, $p = x^2 + y^2 \equiv 1 \, \, (\text{mod}\,\, 4)$, but I have no idea how to go the other way...

 

[fresh_42]

Yeah I think the "if $p = x^2 + y^2$ then $p \equiv 1 \, \, (\text{mod}\,\, 4)$" direction is fine, because $x^2, y^2 \equiv 0,1 \, \, (\text{mod}\,\, 4) \implies x^2 + y^2 \equiv 0,1,2 \, \, (\text{mod}\,\, 4)$ but then since $p$ is odd, $p = x^2 + y^2 \equiv 1 \, \, (\text{mod}\,\, 4)$, but I have no idea how to go the other way...

Consider involutions on $\{\,(x,y,z)\in \mathbb{N}^3\,|\,x^2+4yz=p\,\}$ and their fixed points, or alternatively Minkowski's lattice theorem.

 

[benorin]

@fresh_42 I assume you wrote #9, how was my attempt?

 

[fresh_42]

@fresh_42 I assume you wrote #9, how was my attempt?

I haven't checked since you said you did only half of it. I thought you would do the rest, too.

 

[Gaussian97]

10. A hat-check boy at a congress held at Hilbert's hotel completely loses track of which of hats belong to which owners, and hands them back at random to their owners as the latter leave. What is the probability $P$ that nobody receives their own hat back?

Well, let me try, I think I might overcomplicate it a little, but here comes my attempt:

Spoiler: Solution 10

To start, let's assume that $n$ persons go to the congress.
Then mathematically then, giving the hats at random is equivalent to do a permutation over a set of $n$ elements. So we must compute the number of permutations $\sigma \in S_n$ such that $$\sigma(i)\neq i, \quad \forall i$$
I think it's easier to compute the other way, so compute the number of permutations $\sigma \in S_n$ such that
$$\exists i: \sigma(i) = i$$
let's call it $N_n$.

We can start by computing the number of permutations for which $\exists i_1: \sigma(i_1)=i_1$.
Once fixed the value $i_1$ there are $(n-1)!$ such permutations, and since there are $n$ values to choose for $i_1$ we could think that $N_n = n (n-1)!$.

This, of course, is not true, because some of the permutations are counted more than once. In fact, a permutation such that $\sigma(i_1)=i_1,\ \sigma(i_2)=i_2$ is counted twice, so we need to subtract all such permutations. Now, for fixed values of $i_1, i_2$ there are $(n-2)!$ permutations, and we have $n \choose 2$ ways to choose $i_1, i_2$. So we could expect
$$N_n = n (n-1)! - {n \choose 2} (n-2)!$$
But now we have subtracted some permutations more than once, so we need to add all the permutations
$$\sigma(i_1)=i_1,\quad\sigma(i_2)=i_2,\quad\sigma(i_3)=i_3$$
which there are $n \choose 3$ options for the index and $(n-3)!$ permutations for each choice.
Repeating this process until we finish we obtain that
$$N_n = n (n-1)! - {n \choose 2} (n-2)! + {n \choose 3}(n-3)! - ... = \sum_{k=1}^n (-1)^{k+1}{n\choose k}(n-k)!$$
Since this argument may not be completely rigorous, I provided a more complete proof of this:

Spoiler: Proof

Let's define
$$n_{i_1\ldots i_k} = \text{Number of permutatios with } i_1\ldots i_k \text{ fixed} = (n-k)!$$
$$N_{k} = \text{Number of permutations with some } i\leq k \text{ fixed}$$
And, for $k < m$
$$N_{k|m} = \text{Number of permutations with some } i\leq k \text{ fixed with } m \text{ fixed}$$
To compute $N_1$ is easy, because by definition must be the number of permutations with $\sigma(1)=1$, i.e., $n_1$.
Then to compute $N_2$ we need to add all the permutations that fix 2, but without all the ones which fix 1 (which are already counted)
$$N_2 = N_1 + n_2 - N_{1|2} = n_1 + n_2 - n_{12}$$
Again, to compute $N_3$ we add all the permutations that fix 3, but without all the ones that fix 1 or 2:
$$N_3 = N_2 + n_3 - N_{2|3} = n_1 + n_2 - n_{12} + n_3 - n_{13} - n_{23} + n_{123}$$
In general, we can prove by induction that
$$N_{k} := N_{k-1} + n_{k} - N_{k-1|k} = -\sum_{j=1}^{k} (-1)^{j}{k\choose j} n_{12\ldots j}$$
In fact, the case $k=1$ is trivial, and
$$N_{k+1} = N_{k} + n_{k+1} - N_{k|k+1} = -\sum_{j=1}^{k} (-1)^{j}{k\choose j} n_{12\ldots j} + n_{k+1} +\sum_{j=1}^{k} (-1)^{j}{k\choose j} n_{12\ldots j(k+1)} $$
Using the fact that $n_{k+1}=n_1$ and $n_{12\ldots j(k+1)} = n_{1\ldots (j+1)}$ we can rewrite it as
$$N_{k+1} =k n_{1} -\sum_{j=2}^{k} (-1)^{j}{k\choose j} n_{12\ldots j}+ n_{1} -\sum_{j=1}^{k} (-1)^{j+1}{k\choose j} n_{1\ldots (j+1)}$$
$$N_{k+1} =(k+1) n_{1} -\sum_{j=2}^{k} (-1)^{j}{k\choose j} n_{12\ldots j} -\sum_{j=2}^{k+1} (-1)^{j}{k\choose j-1} n_{1\ldots j}$$
$$N_{k+1} =-(-1)^1{k+1\choose 1} n_{1} -\sum_{j=2}^{k} (-1)^{j}\left({k\choose j}+{k\choose j-1}\right) n_{12\ldots j} -(-1)^{k+1}{k\choose k} n_{1\ldots (k+1)}$$
which finally, using ${k\choose j}+{k\choose j-1} = {k+1\choose j}$ and ${k\choose k}={k+1\choose k+1}$ we obtain
$$N_{k+1} =-\sum_{j=1}^{k+1} (-1)^{j}{k+1\choose j}n_{12\ldots j}$$
As we wanted to prove.

Now we just need to modify some things, first remembering that
$${n\choose k}=\frac{n!}{k!(n-k)!} \Longrightarrow N_n = -n! \sum_{1}^n \frac{(-1)^k}{k!}$$
And because we want to compute the probability of a permutation having no fixed elements, such probability is
$$P_n = 1-\frac{N_n}{n!} = 1 + \sum_{1}^n \frac{(-1)^k}{k!} = \sum_{0}^n \frac{(-1)^k}{k!}$$
And finally, for the Hilber hotel we just need to take the limit $n \to \infty$
$$P = \lim P_n = e^{-1} \approx 36,79\%$$
is the probability nobody receives their own hat.

 

[fresh_42]

Well, let me try, I think I might overcomplicate it a little, but here comes my attempt:

Spoiler: Solution 10

To start, let's assume that $n$ persons go to the congress.
Then mathematically then, giving the hats at random is equivalent to do a permutation over a set of $n$ elements. So we must compute the number of permutations $\sigma \in S_n$ such that $$\sigma(i)\neq i, \quad \forall i$$
I think it's easier to compute the other way, so compute the number of permutations $\sigma \in S_n$ such that
$$\exists i: \sigma(i) = i$$
let's call it $N_n$.

We can start by computing the number of permutations for which $\exists i_1: \sigma(i_1)=i_1$.
Once fixed the value $i_1$ there are $(n-1)!$ such permutations, and since there are $n$ values to choose for $i_1$ we could think that $N_n = n (n-1)!$.

This, of course, is not true, because some of the permutations are counted more than once. In fact, a permutation such that $\sigma(i_1)=i_1,\ \sigma(i_2)=i_2$ is counted twice, so we need to subtract all such permutations. Now, for fixed values of $i_1, i_2$ there are $(n-2)!$ permutations, and we have $n \choose 2$ ways to choose $i_1, i_2$. So we could expect
$$N_n = n (n-1)! - {n \choose 2} (n-2)!$$
But now we have subtracted some permutations more than once, so we need to add all the permutations
$$\sigma(i_1)=i_1,\quad\sigma(i_2)=i_2,\quad\sigma(i_3)=i_3$$
which there are $n \choose 3$ options for the index and $(n-3)!$ permutations for each choice.
Repeating this process until we finish we obtain that
$$N_n = n (n-1)! - {n \choose 2} (n-2)! + {n \choose 3}(n-3)! - ... = \sum_{k=1}^n (-1)^{k+1}{n\choose k}(n-k)!$$
Since this argument may not be completely rigorous, I provided a more complete proof of this:

Spoiler: Proof

Let's define
$$n_{i_1\ldots i_k} = \text{Number of permutatios with } i_1\ldots i_k \text{ fixed} = (n-k)!$$
$$N_{k} = \text{Number of permutations with some } i\leq k \text{ fixed}$$
And, for $k < m$
$$N_{k|m} = \text{Number of permutations with some } i\leq k \text{ fixed with } m \text{ fixed}$$
To compute $N_1$ is easy, because by definition must be the number of permutations with $\sigma(1)=1$, i.e., $n_1$.
Then to compute $N_2$ we need to add all the permutations that fix 2, but without all the ones which fix 1 (which are already counted)
$$N_2 = N_1 + n_2 - N_{1|2} = n_1 + n_2 - n_{12}$$
Again, to compute $N_3$ we add all the permutations that fix 3, but without all the ones that fix 1 or 2:
$$N_3 = N_2 + n_3 - N_{2|3} = n_1 + n_2 - n_{12} + n_3 - n_{13} - n_{23} + n_{123}$$
In general, we can prove by induction that
$$N_{k} := N_{k-1} + n_{k} - N_{k-1|k} = -\sum_{j=1}^{k} (-1)^{j}{k\choose j} n_{12\ldots j}$$
In fact, the case $k=1$ is trivial, and
$$N_{k+1} = N_{k} + n_{k+1} - N_{k|k+1} = -\sum_{j=1}^{k} (-1)^{j}{k\choose j} n_{12\ldots j} + n_{k+1} +\sum_{j=1}^{k} (-1)^{j}{k\choose j} n_{12\ldots j(k+1)} $$
Using the fact that $n_{k+1}=n_1$ and $n_{12\ldots j(k+1)} = n_{1\ldots (j+1)}$ we can rewrite it as
$$N_{k+1} =k n_{1} -\sum_{j=2}^{k} (-1)^{j}{k\choose j} n_{12\ldots j}+ n_{1} -\sum_{j=1}^{k} (-1)^{j+1}{k\choose j} n_{1\ldots (j+1)}$$
$$N_{k+1} =(k+1) n_{1} -\sum_{j=2}^{k} (-1)^{j}{k\choose j} n_{12\ldots j} -\sum_{j=2}^{k+1} (-1)^{j}{k\choose j-1} n_{1\ldots j}$$
$$N_{k+1} =-(-1)^1{k+1\choose 1} n_{1} -\sum_{j=2}^{k} (-1)^{j}\left({k\choose j}+{k\choose j-1}\right) n_{12\ldots j} -(-1)^{k+1}{k\choose k} n_{1\ldots (k+1)}$$
which finally, using ${k\choose j}+{k\choose j-1} = {k+1\choose j}$ and ${k\choose k}={k+1\choose k+1}$ we obtain
$$N_{k+1} =-\sum_{j=1}^{k+1} (-1)^{j}{k+1\choose j}n_{12\ldots j}$$
As we wanted to prove.

Now we just need to modify some things, first remembering that
$${n\choose k}=\frac{n!}{k!(n-k)!} \Longrightarrow N_n = -n! \sum_{1}^n \frac{(-1)^k}{k!}$$
And because we want to compute the probability of a permutation having no fixed elements, such probability is
$$P_n = 1-\frac{N_n}{n!} = 1 + \sum_{1}^n \frac{(-1)^k}{k!} = \sum_{0}^n \frac{(-1)^k}{k!}$$
And finally, for the Hilber hotel we just need to take the limit $n \to \infty$
$$P = \lim P_n = e^{-1} \approx 36,79\%$$
is the probability nobody receives their own hat.

I couldn't avoid counting either.

 

[mathwonk]

In #2, obviously one must assume p > 0. (I make this absurdly obvious comment since x,y are allowed to be negative integers, and to some of us, the commutative algebra definition of a prime integer includes -3 as a prime congruent to 1, mod 4, for which the result fails. Actually I make it because I only just noticed it myself, and up until now I have been stating this result incorrectly.)

Detailed hint: Introduce the "Gaussian integers" Z[ i ], of form n +mi, with n,m integers, i^2 = -1.
0) show that an integer which is a sum of two squares, can be factored in the gaussian integers.
1) Show that if the prime integer p>0 is congruent to 1 mod 4, then p is not prime in the Gaussian integers, in the sense that there are Gaussian integers z,w such that p divides zw, but p does not divide either z or w. (by 0), it suffices to find an appropriate sum of two squares that is divisible by p, e.g. of form n^2+1. (see infrared's next hint.))
2) Show (or assume) that Z[ i ] is a "euclidean domain", hence enjoys unique factorization. (I.e. there is a division algorithm with remainder smaller than the divisor, as measured by the squared absolute value.)
3) Conclude that a (≠0) non - prime, non unit, in Z[ i ] must be factorable into two non - unit factors.
4) Deduce that a prime integer p >0 and congruent to 1 mod 4, is a sum of two positive integer squares.

 

[Infrared]

@mathwonk Try including spaces like [ i ]. Without spaces, it gets interpreted as the italics command.

I think even step (1) in the hint is pretty tricky. I'll give a hint to your hint: the standard trick is to count solutions to $x^2=-1$ in $\mathbb{Z}[ i ]/(p)$.

 

[Gaussian97]

8. Calculate the following integrals
(a) $\displaystyle{\int_0^{2\pi}e^{(e^{it})}\,dt}$
(b) $\displaystyle{\int_{|z|=1}\dfrac{\sin(z^2)}{(\sin z)^2}\,dz}$
(c) $\displaystyle{\int_{|z|=1}\sin \left(e^{1/z}\right)\,dz}$
(d) $\displaystyle{\int_{-\infty}^\infty \dfrac{1}{x^2-2x+2}\,dx}$

Spoiler: Solution (a)

We first see that this integral can be rewritten using complex variables as:
$$\int_{|z|=1} \dfrac{e^z}{iz}\,dz$$
This can be seen because using the parametrization of the path given by $z = e^{it}$ which implies $dz = i z dt$ we get the original integral. Now using the fact that $e^z$ is a holomorphic function we can use Cauchy's Integral Formula and write
$$\int_{|z|=1} \dfrac{e^z}{iz}\,dz = 2 \pi e^{0} = 2 \pi$$

Spoiler: Solution (b)

First, because $\sin{z}$ and $z^2$ are holomorphic functions and the composition of holomorphic functions is holomorphic both $\sin{z^2}$ and $\sin^2{z}$ are holomorphic, also, because the only $0$ of $\sin^2{z}$ in the region $|z| < \pi$ is $z=0$ the function $f(z)=\dfrac{\sin(z^2)}{(\sin z)^2}$ is holomorphic in some open set $U$ that includes the $|z|\leq 1$ region, except for the point $z=0$, in which the function is not defined, defining $f(0):=1$ we can see that $f(z)$ is differentiable at $z=0$ and therefore holomorphic in the open set $U$:

In fact, doing the limit by Taylor expanding the functions:
$$f'(0)=\lim_{z\to 0}\dfrac{f(z)-1}{z} = \lim_{z\to 0}\dfrac{\dfrac{z^2}{z^2-\frac{1}{3}z^4}-1}{z} = \lim_{z\to 0}\dfrac{1+\frac{1}{3}z^2-1}{z}=0$$

Therefore by Cauchy's integral theorem, the integral is zero
$$\displaystyle{\int_{|z|=1}\dfrac{\sin(z^2)}{(\sin z)^2}\,dz}=0$$

Spoiler: Solution (c)

Start by doing the change of variables $w = z^{-1}$, $dw = -w^2 dz$, also $|z|=1 \Longleftrightarrow |w|=1$ so the integral is
$$-\int_{|w|=1}\dfrac{\sin \left(e^{w}\right)}{w^2}\,dw$$
Notice that, if before the path was covered counterclockwise, now it is covered clockwise, i.e. the previous parametrization $z=e^{i\theta}$ is now translated to the parametrization $w = e^{-i\theta}$.
Using Cauchy's integral formula with $f(z)=\sin{\left(e^z\right)}$ we get
$$\int_{|w|=1}\dfrac{\sin \left(e^{w}\right)}{w^2}\,dw = - 2\pi i f'(0) = -2\pi i \cos{1}$$
where the - is to account for the clockwise orientations of the curve, and therefore
$$\int_{|z|=1}\sin \left(e^{1/z}\right)\,dz = 2\pi i \cos{1} \approx 3,395 i$$

Spoiler: Solution (d)

Finally, let's factorize the denomiator. Usign the quadratic formula is easy to see
$$x^2-2x+2 = (x-1-i)(x-1+i)$$
So, to compute this integral, let's compute first the same but with complex argument
$$\int_{\gamma} \frac{1}{(z-1-i)(z-1+i)}\, dz$$
where the path $\gamma$ is defined by the parametrization
$$\gamma(\theta)=\lim_{R\to \infty} \begin{cases}Re^{i\theta}&\theta \in [0,\pi] \\ R\cos{\theta} & \theta \in (\pi, 2\pi]\end{cases}$$
Now, inside the region delimited by the path the integrand has only a pole at $z=1+i$, so, using Cauchy's integral formula we get
$$\int_{\gamma} f(z)=\frac{2\pi i}{(1+i)-1+i}=\pi$$
Now, defining the paths $\gamma_1 = \gamma([0,\pi])$ and $\gamma_2 = \gamma([\pi, 2\pi])$, we have $\gamma = \gamma_1 + \gamma_2$ so the integral can be splitted into
$$\int_\gamma f(z)=\int_{\gamma_1} f(z) + \int_{\gamma_2} f(z)$$
The first integral is zero, because $f(z)$ decreases as $R^{-2}$ while the contour of the integral increases as $R$.
And the second integral is precisely the integral we want, so
$$\int_{-\infty}^\infty \dfrac{1}{x^2-2x+2}\,dx = \pi$$
Actually, I have computed the Principal Value of the integral, but it's not difficult to prove that the integral is convergent and therefore its value coincides with its principal value.

 

[mathwonk]

With help from infrared's suggestion, I'll try again on a detailed hint for #2. I'll try to make them roughly in increasing levels of difficulty:

0) Show an odd prime integer p>0 which is a sum of two integer squares, has form 4n+1, for some integer n.

1) If the prime integer p >0 is a sum of two integer squares, then p factors as a product of two Gaussian integers.

2) Conversely, if p>0 is a prime integer that factors as a product of two Gaussian integers, other than ±1,±i, then p is a sum of two integer squares. (use the fact that the absolute value of complex numbers is multiplicative.)

Hence the problem reduces to showing that an odd prime integer p>0, can be factored non trivially in Z[ i ], if p = 4n+1, for some integer n. Equivalently we want to shown that p = z.w, for Gaussian integers z,w, where p does not divide either z or w. As a first step, try to show that such a prime integer p divides some product z.w of non zero Gaussian integers, such that p does not divide either z or w, as follows:

3) Show that if p cannot divide a non zero product z.w of Gaussian integers without dividing one of the factors, then in the quotient ring Z[ i ]/(p), the product of two non zero elements is always again non zero, i.e. this ring is then an "integral domain". (The quotient ring Z[ i ]/(p) is defined analogously to the ring of modular integers, i.e. two Gaussian integers z,w become equal in the quotient ring if p divides z-w.)

4) Show that in an integral domain, a quadratic equation can have at most two distinct solutions.

5) Show: if an odd prime p>0 has form 4n+1, then X^2+1 = 0 has more than 2 solutions in Z[ i ]/(p).
(show that this equation already has solutions in the integers mod p. e.g. mod 5 we have the multiplicative group {1,2,3,4} = {1,2,-2,-1}, whose product, since only 1 and -1 are not paired inverses, is -1. thus the product 1.2.-2.-1 = 1.2.2.1 = 1^2.2^2 = (1.2)^2 = -1. thus there is a square root of -1.)

A non zero, non unit, element of a domain is called irreducible if it cannot be factored without using a unit factor.

6) If in an integral domain R, every non zero, non unit element factors into irreducible elements, uniquely up to order and multiplication by units, (i.e. "unique factorization" holds), then show that for every irreducible element x, the quotient ring R/(x) is again an integral domain.

7) Assuming Z[ i ] has unique factorization, prove an odd integer prime p >0 of form 4n+1 can be factored in the Gaussian integers, hence can be written as a sum of two integer squares.

8) Using the squared absolute value as a size measure, show that there is a division algorithm in Z[ i ] where the remainder is smaller than the divisor. this is hard, but if you use thev fact that there is such a division algorithm in Q[ i ], you can deduce this by estimating the distances between complex numbers in the plane, at least as i recall.

9) Deduce that unique factorization holds in Z[ i ], completing problem #2.

Boy, there is a lot more to this approach than I thought. but i believe it is a valuable exercise to attempt these steps with the hints.

 

[fresh_42]

Boy, there is a lot more to this approach than I thought.

Both of my solutions (see post #15: fix points of involutions or Minkowski) are shorter than your entire hint. And the involution method is shorter than Minkowski. It's a bit tricky but I think nice. Minkowski is a bit brute force.

 

[julian]

Problem #8

I did (a), (c), and (d) the same way @Gaussian97. I was moving on to (b).

Here is a different way of performing the integral (d):

Write

\begin{align*}
I & = \int_{- \infty}^\infty \frac{1}{x^2 -2x + 2} dx \\
& = \int_{- \infty}^\infty \left( \int_0^\infty e^{- \alpha (x^2 -2x + 2)} d \alpha \right) dx \\
& = \int_0^\infty \left( \int_{- \infty}^\infty e^{- \alpha (x^2 -2x + 2)} dx \right) d \alpha \\
& = \int_0^\infty \left( \int_{- \infty}^\infty e^{- \alpha ((x - 1)^2 + 1)} dx \right) d \alpha \\
& = \int_0^\infty e^{- \alpha} \left( \int_{- \infty}^\infty e^{- \alpha x^2} dx \right) d \alpha \\
& = \int_0^\infty e^{- \alpha} \sqrt{\frac{\pi}{\alpha}} d \alpha
\end{align*}

Make the substitution $u^2 = \alpha$ and then we get

\begin{align*}
I & = \sqrt{\pi} \int_0^\infty \frac{e^{- u^2}}{u} 2 u d u \\
& = 2 \sqrt{\pi} \int_0^\infty e^{- u^2} d u \\
& = \sqrt{\pi} \int_{- \infty}^\infty e^{- u^2} d u \\
& = \pi .
\end{align*}

 

[mathwonk]

Hint for #3: Since an n - manifold is a space locally homeomorphic to R^n, it suffices to show that every point a has a neighborhood on which π is a homeomorphism to a nbhd of π(a).

 

[Infrared]

@mathwonk Don't forget second-countability and Hausdorff!

They're straightforward to check, but so is the locally Euclidean property.

 

[mathwonk]

Hint for #5: my notes linked below cover semi direct products on pages 43-50.

https://www.math.uga.edu/sites/default/files/inline-files/843-1.pdf@Infrared: I thought those were called "nice manifolds".

 

[fresh_42]

@mathwonk Don't forget second-countability and Hausdorff!

$\langle \rm{irony}\rangle$
Proof by contradiction: If the tori weren't then topologist would have a massive problem.
$\langle /\rm{irony}\rangle$

 

[Infrared]

$\langle \rm{irony}\rangle$
Proof by contradiction: If the tori weren't then topologist would have a massive problem.
$\langle /\rm{irony}\rangle$

Fair enough! But I would go further and apply that logic to the problem statement (if tori weren't manifolds...). I can forgive @mathwonk for not checking Hausdorff if he's more used to varieties and the Zariski topology.

 

[fresh_42]

Fair enough! But I would go further and apply that logic to the problem statement (if tori weren't manifolds...).

Yeah, that's true. It is not so difficult, but I still want to offer problems from which students can learn something, and if it is to look up the definitions and check them. And I am really bored by the integral questions ...

I can forgive @mathwonk for not checking Hausdorff if he's more used to varieties and the Zariski topology.

You mean the only reasonable and intuitive topology!

 

[mathwonk]

thanks for the reminder many people make these extra assumptions. Among my books, this is not assumed in Lang, Loomis-Sternberg, or Hicks, but is assumed in Spivak and Warner. Bott-Tu and Hirsch also do not assume this in their definition, but state explicitly that all manifolds they consider will satisfy these conditions. One should keep in mind however that a very interesting example of a locally euclidean but non - hausdorff space is the line with a double origin, obtained by gluing R to R along R-{0}. Note too that this example makes it clear that the classification of 1-manifolds is more complicated without those assumptions, in particular there are infinitely many compact, connected 1-manifolds in the more general case.

Algebraic varieties in projective space, with the zariski topology, are of course separated, when one gives the correct definition of separated, namely that X is separated if the diagonal is closed in the product variety XxX. (The paradox is resolved by the fact that the zariski topology on this product is not the usual product topology!)

 

[Infrared]

Note too that this example makes it clear that the classification of 1-manifolds is more complicated without those assumptions, in particular there are infinitely many compact 1-manifolds in the more general case.

I think this is actually a rather large loss. For example, intersection theory on smooth manifolds relies on the fact that there are zero boundary points on a compact $1$-manifold, counted with orientation. So theorems like Poincare-Hopf, etc. would be in danger.

Another obvious loss is the Whitney embedding theorem.

Algebraic varieties in projective space, with the zariski topology, are of course separated, when one gives the correct definition of separated, namely that X is separated if the diagonal is closed in the product variety XxX. (The paradox is resolved by the fact that the zariski topology on this product is not the usual product topology!)

Okay, but considering projective varieties would be roughly analogous to studying submanifolds of $\mathbb{R}^n$ (okay I guess these are affine varieties, but the essential point is just that they live in a "nice" space you already know) and those are automatically Hausdorff. Removing the Hausdorff condition only affects abstract manifolds, just like abstract varieties aren't necessarily separated.

Edit: Maybe "abstract variety" is vague, and there can be ambiguity as to whether it should be separated (just like there is ambiguity as to whether manifolds should be Hausdorff).

 

[mathwonk]

That is true of course about projective varieties being automatically separated. I gave that example to simplify my statement about the hausdorff condition in algebraic geometry. The point is that one does define general schemes without the separated condition, and also abstract varieties as you say, and then one adds on the condition of separatedness when desired.

I don't consider those examples you mention as losses, since the theorems are still true, but one must simply make clear the hypotheses that are needed to make the theorems hold. There is also a gain in remembering non - separated varieties. I once heard a conference presentation where the speaker said all he had to do to prove his theorem was exhibit a certain type of mapping (a finite birational map of normal varieties that is not an isomorphism). (Must an injective continuous map of compact connected 1-manifolds be surjective?) Someone in the audience pointed out that the desired example would violate Zariski's "main theorem". The speaker said "that bothered me too, at first" to general laughter. then he pointed out that relaxing the condition of separation would solve the contradiction, but would still suffice to prove his result. In other words, consideration of a non - separated example allowed a proof of a result which he wanted mainly to apply to separated ones!

I guess the main issue is to remember that if one does not give ones definitions, we cannot assume everyone will use the same ones.

 

[benorin]

I haven't checked since you said you did only half of it. I thought you would do the rest, too.

There's a failed attempt at solving the other side in that post, I'm not knowing how to use the weak law of large numbers here, since a statement of it is $\lim_{n\to\infty}P\left( \left|\bar{x}_n-\mu \right|\geq \epsilon\right) =0$, not really knowing how to approximate the required $n$ value from that, the associated Chebychev's inequality might be useful though, namely $P\left( \left|\bar{x}_n-\mu \right|\geq \epsilon\right)\leq \tfrac{\sigma ^2}{n\epsilon ^2}$? Anyhow, could you please take a look and my original post? Thanks @fresh_42

The rest will hopefully follow soon...