Challenge Intermediate Math Challenge - May 2018

[QuantumQuest]

It's time for an intermediate math challenge! If you find the problems difficult to solve don't be disappointed! Just check our other basic level math challenge thread!

RULES:
1) In order for a solution to count, a full derivation or proof must be given. Answers with no proof will be ignored.
2) It is fine to use nontrivial results without proof as long as you cite them and as long as it is "common knowledge to all mathematicians". Whether the latter is satisfied will be decided on a case-by-case basis.
3) If you have seen the problem before and remember the solution, you cannot participate in the solution to that problem.
4) You are allowed to use google, wolframalpha or any other resource. However, you are not allowed to search the question directly. So if the question was to solve an integral, you are allowed to obtain numerical answers from software, you are allowed to search for useful integration techniques, but you cannot type in the integral in wolframalpha to see its solution.
5) Mentors, advisors and homework helpers are kindly requested not to post solutions, not even in spoiler tags, for the challenge problems, until 16th of each month. This gives the opportunity to other people including but not limited to students to feel more comfortable in dealing with / solving the challenge problems. In case of an inadvertent posting of a solution the post will be deleted by @fresh_42

$1.$ (solved by @julian ) Let $A$ be an $\text{n x n}$ matrix that is real skew symmetric, and with positive numbers $\{x_1, ..., x_k\}$
where $k \geq 2$. Prove that:

$\prod_{i = 1}^k \det\Big(A +x_i I\Big) \geq \det\Big(A +\big( \prod_{i = 1}^k x_i I \big)^\frac{1}{k}\Big)^k$ $\space$ (by @StoneTemplePython)

$2.$ (solved by @Biker ) Solve $\mathcal{I}=\int_{-1}^0 \,x \cdot \sqrt{x^2+x+1} \,dx$ . Hint: Use $\cosh^2x - \sinh^2 x=1$ . $\space$ (by @fresh_42)

$3.$ (solved by @nuuskur ) Show that there are infinitely many prime numbers of the form $4k + 3$, $k \in \mathbb{N} - \{0\}$ $\space$ (by @QuantumQuest)

$4.$ (solved by @lpetrich ) What's the 100th digit after the decimal point of $(1 + \sqrt{3})^{5000}$ $\space$ (by @StoneTemplePython)

$5.$ (solved by @lpetrich ) Given the differential operators $D_n := x^n \cdot \dfrac{d}{dx}\,\,(n \in \mathbb{Z})$ on smooth real valued functions $\mathcal{C}^\infty(\mathbb{R})$ .
Determine for which subsets $L \subseteq \mathbb{Z}$ the set $\{D_n \,\vert \,n \in L\}$ is a basis for a finite dimensional Lie algebra and which Lie algebra is it. $\space$ (by @fresh_42)

$6.$ (solved by @lpetrich, @julian ) Calculate the integral $I = \int_{0}^{\pi} \sin^{2n} x dx$ , $n \in \mathbb{N}$ $\space$ (by @QuantumQuest)

$7.$ (solved by @julian ) Solve $\sum_{k=1}^\infty \dfrac{1}{k \binom{2k}{k}}$ . $\space$ (by @fresh_42)

$8.$ (solved by @julian ) Find the coordinates of the center of gravity for the arc of the curve $y = a \cosh(\frac{x}{a})$ for $-a \leq x \leq a$ $\space$ (by @QuantumQuest)

$9.$ (resolved in post #62) For a given real Lie algebra $\mathfrak{g}$ , we define

$\mathfrak{A(g)} = \{\,\alpha \, : \,\mathfrak{g}\longrightarrow \mathfrak{g}\,\,: \,\,[\alpha(X),Y]=-[X,\alpha(Y)]\text{ for all }X,Y\in \mathfrak{g}\,\}$the set of antisymmetric transformations of $\mathfrak{g}$. Remember that a real Lie algebra is a real vector space equipped with a multiplication for which holds

• anti-commutativity: $[X,X]=0$
• Jacobi-identity: $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$

a) Show that $\mathfrak{A(g)}\subseteq \mathfrak{gl}(g)$ is a Lie subalgebra in the Lie algebra of all linear transformations of $\mathfrak{g}$ with the commutator as Lie product: $[\alpha, \beta]= \alpha \beta -\beta \alpha$ .
b) Show that $\mathfrak{g} \ltimes \mathfrak{A(g)}$ is a semidirect product ($\,\mathfrak{A(g)}$ is the ideal ) given by

$[X,\alpha]:=[\operatorname{ad}X,\alpha]=\operatorname{ad}X\,\alpha - \alpha\,\operatorname{ad}X$c) Show that for all $\alpha \in \mathfrak{A(g)}$ and $X,Y,Z \in \mathfrak{g}$

$[α(X),[Y,Z]]+[α(Y),[Z,X]]+[α(Z),[X,Y]]=0$ $\space$ (by @fresh_42)

$10.$ (solved by @julian ) Calculate the integral $\int_{c}^{} \frac{e^{kz}}{z}dz$ where $c$ is the circle $z = e^{i\theta}$ with $-\pi \leq \theta \leq \pi$ and then prove that $\int_{-\pi}^{\pi} e^{k\cos\theta} \cos(k\sin\theta)d\theta = 2\pi$ $\space$ (by @QuantumQuest)

 

[nuuskur]

I'll try problem 3. I have seen the original proof of there being infinitely many primes. I haven't done this particular exercise before, but a similar approach works. I employ only some introductory results about divisibility and the fundamental theorem of arithmetic.

Spoiler: Problem 3 solution

Denote by $\mathbb N$ the positive integers.
We assume towards a contradiction there are finitely many primes of the form $4k+3$ with $k\in\mathbb N$. Let these be $p_1,\ldots , p_t$. Consider
<br /> M := 4p_1\ldots p_t<br />
By hypothesis, $M+3$ is composite, therefore divisible by a prime. Note that $\neg (3\mid M+3)$, otherwise $3\mid M$, then, by Euclid's lemma, $3\mid p_j$ which is possible only if $k=0$. Since $M+3$ is odd, its prime factors must be of the form $4k+1$ or $4k+3$. However, if any $p_j\mid M+3$, then $p_j\mid 3$, which is impossible. Now, all the prime factors of $M+3$ are of the form $4k+1$. However
<br /> (4k+1)(4l+1) = 4(4kl + k + l) +1.\tag{1}<br />
Coupled with the fundamental theorem of arithmetic, (1) implies the number $M+3$ must be of the form $4k+1$, a contradiction. Therefore, there must be infinitely many primes of the form $4k+3$.

 

[mfb]

By hypothesis, $M+3$ is composite

Sure, it is divisible by 3.

 

[nuuskur]

Sure, it is divisible by 3.

If $3\mid M+3$, then $3\mid M$, that can't happen.

 

[mfb]

3 is one of your $p_i$. You constructed a number that is divisible by 3, and then claimed that it cannot be divisible by any prime of the form 4k+3. But it can, and we know an explicit example - it is divisible by 3.

 

[nuuskur]

3 is one of your $p_i$. You constructed a number that is divisible by 3, and then claimed that it cannot be divisible by any prime of the form 4k+3. But it can, and we know an explicit example - it is divisible by 3.

I have assumed $p_j = 4k+3, k\in\mathbb N$ (positive integers) as stated in problem 3. The number $M$ doesn't contain $3$.

 

[mfb]

Ah, I didn't see the "positive integers" part. Then it works (a comment that 3 is not a $p_i$ would have helped), although you should mention that M+3 is not a multiple of 3. Without that observation you don't get a contradiction.

 

[nuuskur]

Ah, I didn't see the "positive integers" part. Then it works (a comment that 3 is not a $p_i$ would have helped), although you should mention that M+3 is not a multiple of 3. Without that observation you don't get a contradiction.

I had assumed the observation is trivial, but you are right. Furthermore, I had invoked Euclid's lemma without mentioning it in my first draft.

 

[fresh_42]

I had assumed the observation is trivial, but you are right. Furthermore, I had invoked Euclid's lemma without mentioning it in my first draft.

I think you could have made it easier to read. I fought my way through, but with a little help from your side it would have been much easier to read ($\,p_i \neq 3$ per assumption, a few steps in the $3|M$ part and finally why $M+3=4N+1$ is impossible).

You reminded me of what I once told a student on how to write a thesis:

1. Write it down so that you can understand each step, even after a year.
2. Remove all but every third line (but keep a copy).
3. Undelete four consecutive lines every two pages, so that the professor has something to criticize.

 

[member 587159]

I think you could have made it easier to read. I fought my way through, but with a little help from your side it would have been much easier to read ($\,p_i \neq 3$ per assumption, a few steps in the $3|M$ part and finally why $M+3=4N+1$ is impossible).

You reminded me of what I once told a student on how to write a thesis:

1. Write it down so that you can understand each step, even after a year.
2. Remove all but every third line (but keep a copy).
3. Undelete four consecutive lines every two pages, so that the professor has something to criticize.

Can't see why the third one is necessary.

 

[fresh_42]

Can't see why the third one is necessary.

Because it referred to a Diplomarbeit which is corrected by the professor and I didn't know an English equivalent.

 

[QuantumQuest]

I'll try problem 3. I have seen the original proof of there being infinitely many primes. I haven't done this particular exercise before, but a similar approach works. I employ only some introductory results about divisibility and the fundamental theorem of arithmetic.

Checking your solution under spoiler I think that it is right but as already noted, it could be somewhat more well structured i.e. I needed quite a number of hops to understand some things. In all other respects it is good.

 

[Biker]

$2.$ Solve $\mathcal{I}=\int_{-1}^0 \,x \cdot \sqrt{x^2+x+1} \,dx$ . Hint: Use $\cosh^2x - \sinh^2 x=1$ . $\space$ (by @fresh_42)

You could use hyperbolic function and you could use normal trig function. I will go with trig functions instead. I hope it is correct

Spoiler: Solution

$\mathcal{I}=\int_{-1}^0 \,x \cdot \sqrt{(x+1/2)^2+3/4)} \,dx$
Use this substitution:
$\frac{\sqrt(3)}{2} tan(\theta) = x +1/2$
$\mathcal{I}=\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \, (\frac{ \sqrt{3}}{2} tan(\theta) - 1/2) \cdot \frac{3}{4} sec^3(\theta) \,d\theta$
$\mathcal{I}=\frac{3}{4} \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \, (\frac{ \sqrt{3}}{2} tan(\theta) sec^3(\theta) - 1/2 sec^3(\theta)) \,d\theta$
Notice that the $tan(\theta) sec^3(\theta)$ is an odd function, no need to calculate the integral and the other part is an even function
$\mathcal{I}=\frac{-3}{4} \int_{0}^{\frac{\pi}{6}} \, sec^3(\theta) \,d\theta$
Using integration by parts for $sec^3(\theta)$ which is easy. You get:
$\mathcal{I}=\frac{-3}{8} ( \left. tan(\theta) sec(\theta) + ln(sec(\theta) + tan(\theta)) \right|_0^{\frac{\pi}{6}}$
you get the answer -0.455989

 

[Biker]

@QuantumQuest Is the solution to 6 is

Spoiler

$\frac{(2n-1)^2 \cdot (2n-3)^2 ... 1^2}{(2n)!} \pi$ ?

 

[fresh_42]

You could use hyperbolic function and you could use normal trig function. I will go with trig functions instead. I hope it is correct

Spoiler: Solution

$\mathcal{I}=\int_{-1}^0 \,x \cdot \sqrt{(x+1/2)^2+3/4)} \,dx$
Use this substitution:
$\frac{\sqrt(3)}{2} tan(\theta) = x +1/2$
$\mathcal{I}=\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \, (\frac{ \sqrt{3}}{2} tan(\theta) - 1/2) \cdot \frac{3}{4} sec^3(\theta) \,d\theta$
$\mathcal{I}=\frac{3}{4} \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \, (\frac{ \sqrt{3}}{2} tan(\theta) sec^3(\theta) - 1/2 sec^3(\theta)) \,d\theta$
Notice that the $tan(\theta) sec^3(\theta)$ is an odd function, no need to calculate the integral and the other part is an even function
$\mathcal{I}=\frac{-3}{4} \int_{0}^{\frac{\pi}{6}} \, sec^3(\theta) \,d\theta$
Using integration by parts for $sec^3(\theta)$ which is easy. You get:
$\mathcal{I}=\frac{-3}{8} ( \left. tan(\theta) sec(\theta) + ln(sec(\theta) + tan(\theta)) \right|_0^{\frac{\pi}{6}}$
you get the answer -0.455989

That's correct, although you could have made life much easier for an old man. I don't know all the trig formulas in mind anymore. And to demonstrate the integration by parts would have been a nice service for the younger among us. However, the argument with the odd function was nice, so I'll not complain about the wrong round-off of the result, although an exact solution would have been better:
$$
\int_{-1}^0 \,x \cdot \sqrt{x^2+x+1} \,dx = -\frac{1}{4} - \frac{3}{16} \log 3 \approx -0.45598980\ldots \approx -0.456
$$

 

[Biker]

That's correct, although you could have made life much easier for an old man. I don't know all the trig formulas in mind anymore. And to demonstrate the integration by parts would have been a nice service for the younger among us. However, the argument with the odd function was nice, so I'll not complain about the wrong round-off of the result, although an exact solution would have been better:
$$
\int_{-1}^0 \,x \cdot \sqrt{x^2+x+1} \,dx = -\frac{1}{4} - \frac{3}{16} \log 3 \approx -0.45598980\ldots \approx -0.456
$$

Yeah, I am sorry. When I saw the question, I imagined having an exam with this question. So I tried the first thing that came to my mind, Trig subs. When I reached the answer. I tried the hint, Which is much simpler and you can still do the odd-even trick which will simplify it even more. So I just posted the answer that first came to my mind without the hint. Because that is a big hint.

 

[fresh_42]

So I just posted the answer that first came to my mind without the hint. Because that is a big hint.

No problem, keeps my mind busy . There are a few questions on this level which could have well been on "B" level, and it was only the machinery that made them "I". And the metric problem on the "B" list is actually the easiest of all, but still untouched. Seems we can't accurately predict the level, yet. As I said before, it's not as easy as we thought to find good problems ... and to assess their difficulty.

 

[QuantumQuest]

Is the solution to 6 is

No. I would really like to give the correct answer but I can't as this is a challenge. But what I can ask you is to think about the problem again. If you come up with a way which is in the right direction, I maybe able to give you some hint along the way.

 

[Biker]

No. I would really like to give the correct answer but I can't as this is a challenge. But what I can ask you is to think about the problem again. If you come up with a way which is in the right direction, I maybe able to give you some hint along the way.

Weird, I tried a few examples with my definite integral calculator and it seems to be correct. Isn't the integral (sin x)^2n?

Spoiler

I used integration by part to get a recursive formula for (sinx)^n then noticed that the integration is from 0 to pi which simplified the integral to the last term

 

[QuantumQuest]

@Biker

I see that you put good efforts so I'll ask can you change the limits of integration to a circle and maybe think outside real numbers?

 

[julian]

Is the solution to problem 8 this:

Spoiler: SOLUTION

The center of gravity is $(0,\overline{y})$ where

$\overline{y} = \lim_{\delta s \rightarrow 0} \frac{\sum y \; \delta s}{\sum \delta s} = \frac{1}{s} \int y ds$

where $\delta s$ is a small arc length. The $x$-component of the center of gravity is zero because $a \cosh (x/a)$ is symmetric in $x$. Now

$d s = \sqrt{(d x)^2 +(dy)^2} = \left( 1 + \left( \frac{dy}{dx} \right)^2 \right)^{1/2} dx$

So that

$s = \int_{-a}^a \left( 1 + \left( \frac{dy}{dx} \right)^2 \right)^{1/2} dx \quad \mathrm{and} \quad \overline{y} s = \int_{-a}^a y \left( 1 + \left( \frac{dy}{dx} \right)^2 \right)^{1/2} dx .$

We first calculate the arc length $s$:

\begin{align}
s & = \int_{-a}^a \left[ 1 + \left( \sinh \frac{x}{a} \right)^2 \right]^{1/2} dx
\nonumber \\
& = \int_{-a}^a \left[ \cosh^2 \frac{x}{a} \right]^{1/2} dx
\nonumber \\
& = \int_{-a}^a \cosh \frac{x}{a} dx
\nonumber \\
& = \left[ a \sinh \frac{x}{a} \right]_{-a}^a
\nonumber \\
& = 2a \sinh (1) .
\nonumber
\end{align}

And now

\begin{align}
\overline{y} s & = \int_{-a}^a a \cosh \frac{x}{a} \times \cosh \frac{x}{a} dx
\nonumber \\
& = \int_{-a}^a \frac{a}{2} (\cosh \frac{2x}{a} + 1) dx
\nonumber \\
& = \frac{a}{2} \left[ \frac{a}{2} \sinh \frac{2x}{a} + x \right]_{-a}^a
\nonumber \\
& = \frac{a^2}{4} 2 \sinh (2) +a^2= \frac{a^2}{2} \sinh (2) + a^2 .
\nonumber
\end{align}

So that

$\overline{y} = \frac{a^2}{2} (\sinh (2) +2) / 2a \sinh (1) = \frac{a}{4} \frac{\sinh (2) + 2}{\sinh (1)} = \frac{a}{4} \frac{e^2 - e^{-2} + 2}{e - e^{-1}} .$

 

[dRic2]

I'm a bit embarrassed to ask because I'm not skilled enough to solve any of this problems, but, out of curiosity,

if $F(z) = \int \frac {e^{kz}} z dz$

is it ok to assume:

$I = \int_c \frac {e^{kz}} z dz = F(z_a) - F(z_b) = F(e^{i\pi}) - F(e^{-i\pi})$ ?

or did I misunderstood the problem?

PS: Don't eat me alive if I wrote a blasphemy

 

[QuantumQuest]

@julian well done!

 

[nuuskur]

I'm a bit embarrassed to ask because I'm not skilled enough to solve any of this problems, but, out of curiosity,

if $F(z) = \int \frac {e^{kz}} z dz$

is it ok to assume:

$I = \int_c \frac {e^{kz}} z dz = F(z_a) - F(z_b) = F(e^{i\pi}) - F(e^{-i\pi})$ ?

or did I misunderstood the problem?

PS: Don't eat me alive if I wrote a blasphemy

Generally an integral is not determined by the end points of the path of integration.

 

[QuantumQuest]

I'm a bit embarrassed to ask because I'm not skilled enough to solve any of this problems, but, out of curiosity,

if $F(z) = \int \frac {e^{kz}} z dz$

is it ok to assume:

$I = \int_c \frac {e^{kz}} z dz = F(z_a) - F(z_b) = F(e^{i\pi}) - F(e^{-i\pi})$ ?

or did I misunderstood the problem?

PS: Don't eat me alive if I wrote a blasphemy

How did you get that? Note that there is a known formula that can make things a lot easier for this particular problem. Can you recognize which is it?

 

[lpetrich]

I will solve Problem 6.

Spoiler

First, set up a recursive solution, with $I(n) = \int_0^\pi \sin^{2n} x \, dx$ found in terms of I(n-1). With I(0), that gives the complete solution.

Change integration variables:
$$ I(n) = \int_0^\pi \sin^{2n} x \, dx = - \int_{x=0}^{x=\pi} \sin^{2n-1} x \, d(\cos x)$$
Integrate by parts:
$$ I(n) = - \sin^{2n-1} x \cos x |_{x=0}^{x=\pi} + \int_{x=0}^{x=\pi} \cos x \, d(\sin^{2n-1} x) = (2n-1) \int_0^\pi \sin^{2n-2} x \, \cos^2 x \, dx $$
Use a trigonometric identity:
$$ I(n) = (2n-1) \int_0^\pi \sin^{2n-2} x \, (1 - \sin^2 x) \, dx = (2n-1) I(n-1) - (2n-1) I(n)$$
giving us
$$ I(n) = \frac{2n-1}{2n} I(n-1) = \frac{(2n)(2n-1)}{4n^2} I(n-1) $$
This gives us
$$ I(n) = \frac{(2n)!}{2^{2n} n! n!} I(0) $$
The final step is to find I(0). That is easy; it is π. Thus,
$$ I(n) = \frac{(2n)!}{2^{2n} n! n!} \pi $$

 

[dRic2]

@QuantumQuest @nuuskur I read the problem carefully again. I apologize, I didn't study those kind of integrals. Sorry for my mistake

PS: it seems to me a line integral of a complex function, right? I googled it and I found out it is know as Contour integral, am I correct? If that is the case then I do not know how to solve it ( yet ;) )

 

[QuantumQuest]

it seems to me a line integral of a complex function, right? I googled it and I found out it is know as Contour integral, am I correct? If that is the case then I do not know how to solve it ( yet ;) )

It is a contour integral and one of the methods that are used for contour integration is the one I was talking about in post #25.

 

[dRic2]

@QuantumQuest Yes, but I don't know how to solve it. I didn't study it. anyway Thank you for the explanation! :)

 

[QuantumQuest]

I will solve Problem 6

Well done. There is also another very nice way to solve it but I won't say anything more right now, as there maybe people that want to tackle it another way;)

 

[lpetrich]

I will try Problem 4

Spoiler

I succeeded in solving it with brute force with Mathematica, but I will do an alternative that does not require bignums. That alternative was inspired by the Fibonacci series, and how member n of it can be expressed as a1*b1ⁿ + a2*b2ⁿ. So I seek a recurrence that has an exponential solution with a factor of $\sqrt{3}+1$ in it. That quantity solves equation $x^2 - 2x - 2$, and it gives us the recurrence $x(n) = 2x(n-1) + 2x(n-2)$. For x(0) = 2 and x(1) = 2, $x(n) = (\sqrt{3}+1)^n + (-\sqrt{3}+1)^n$. However, that recurrence guarantees that x(n) will be an integer for all nonnegative integer n.

So, $(\sqrt{3}+1)^{5000} = x(5000) - (-\sqrt{3}+1)^{5000}$. That final term is approximately $5.08 \cdot 10^{-678}$, and that makes the first 677 digits after the decimal point all 9's. Thus, the 100th digit after the decimal point is 9.

 

[StoneTemplePython]

I will try Problem 4

Spoiler

I succeeded in solving it with brute force with Mathematica, but I will do an alternative that does not require bignums. That alternative was inspired by the Fibonacci series, and how member n of it can be expressed as a1*b1ⁿ + a2*b2ⁿ. So I seek a recurrence that has an exponential solution with a factor of $\sqrt{3}+1$ in it. That quantity solves equation $x^2 - 2x - 2$, and it gives us the recurrence $x(n) = 2x(n-1) + 2x(n-2)$. For x(0) = 2 and x(1) = 2, $x(n) = (\sqrt{3}+1)^n + (-\sqrt{3}+1)^n$. However, that recurrence guarantees that x(n) will be an integer for all nonnegative integer n.

So, $(\sqrt{3}+1)^{5000} = x(5000) - (-\sqrt{3}+1)^{5000}$. That final term is approximately $5.08 \cdot 10^{-678}$, and that makes the first 677 digits after the decimal point all 9's. Thus, the 100th digit after the decimal point is 9.

I actually had Fibonacci in the back of my head when putting the question in. Well done.

A lot of problems can be solved if you find the right pairing.

- - - -
edit:
It may be worth pointing out that the magnitude of the second term is so small that the problem can actually be done without a calculator.

You know that that $1.5 ^2 = 2.25 \lt \big(\sqrt{3}\big)^2$ but you also know that
$\big(\sqrt{3}\big)^2 = 3 =\frac{48}{16} \lt \frac{49}{16} \to \sqrt{3} \lt \frac{7}{4}$, telling you that $\big \vert 1 - \sqrt{3}\big \vert \lt \frac{3}{4}$, and after raising to an even exponent of course the result is positive.

But the trick to the problem was finding the pairing.

 

[lpetrich]

I'll take a stab at solving problem 5.

Spoiler

The first thing I do is to find the commutators of the D_n's. For convenience, I will redefine them by doing D_n = old D_n+1 -- $D_n = x^{n+1} \frac{d}{dx}$.

Combining the operators,
$$ D_m(D_n(f(x))) = D_m( x^{n+1} f'(x) ) = x^{m+n+2} f''(x) + (n+1) x^{m+n+1} f'(x) $$
Thus, the commutator $[D_m,D_n] = (n-m)D_{n+n}$.
This I recognize as the Virasoro algebra.

I now consider what finite subalgebras are possible. From the commutators I deduce some constraints.

• The zero operator, D₀, does not affect the presence or absence of any other operators.
• If two operators are positive, D_m and D_n with m and n > 0, then their commutator contains an operator D_m+n with a higher arg value. Repeating the commutation with this new operator gives another one with even higher arg value. Thus, if more than one operator is positive, there are thus an infinite number of positive operators.
• The same argument shows that if two operators are negative, D_m and D_n with m and n < 0, then they generate an infinite number of negative operators.
• So there is at most one positive operator and at most one negative operator.
• Their commutator must be a multiple of the zero operator, or else they will make more than one positive or negative operator.

Thus, the possible finite subalgebras of this algebra are

• One element: D_n for any n.
• Two elements: D₀ and D_n for any nonzero n.
• Three elements: D_n, D_-n, and D₀ for any nonzero n.

The three-element one's commutators:
$[D_0, D_n] = n D_n$
$[D_0, D_{-n}] = - n D_{-n}$
$[D_{-n}, D_n] = 2n D_0$
These can be combined into
$[D_0, D_n+D_{-n}] = n (D_n - D_{-n})$
$[D_n-D_{-n},D_0] = - n (D_n + D_{-n})$
$[D_n+D_{-n}, D_n-D_{-n}] = 4n D_0$
The pattern of signs points to the algebra SO(2,1).

So the possible finite subalgebras are the one-element ones, the nontrivial two-element ones, and the three-element SO(2,1).

 

[julian]

I had realized that

$\frac{1}{2} (\sum_{m=0}^{5000} \frac{5000!}{m! (5000-m)!} 3^{m/2} + \sum_{m=0}^{5000} (-1)^m \frac{5000!}{m! (5000-m)!} 3^{m/2} ) = \frac{1}{2} ((1 + \sqrt{3})^{5000} + (1 - \sqrt{3})^{5000} )$

would give a part that was purely an integer, but I didn't make the next steps.

 

[Biker]

Well done. There is also another very nice way to solve it but I won't say anything more right now, as there maybe people that want to tackle it another way;)

That is literally the exact same solution I gave... No problem though. Cheers!

 

[mfb]

Well done. There is also another very nice way to solve it but I won't say anything more right now, as there maybe people that want to tackle it another way;)

As we have a solution already, a hint:

Spoiler

A few lines with complex numbers

 

[julian]

That is literally the exact same solution I gave... No problem though. Cheers!

Yep

$\frac{(2n-1)^2 (2n-3)^2 \dots 1^2}{(2n)!} = \frac{(2n-1)^2 (2n-3)^2 \dots 1^2}{(2n) (2n-1) (2n-2) (2n-3) \dots} = \frac{(2n-1) (2n-3) \dots}{2n (2n-2) \dots} = \frac{(2n-1) (2n-3) \dots}{2^n n!}$
$= \frac{2n (2n-1) (2n-2) (2n-3) \dots}{2^n 2n (2n-2) \dots n!} = \frac{(2n)!}{2^{2n} (n!)^2}$

So I think honours go to Biker.

 

[fresh_42]

I'll take a stab at solving problem 5.

Spoiler

The first thing I do is to find the commutators of the D_n's. For convenience, I will redefine them by doing D_n = old D_n+1 -- $D_n = x^{n+1} \frac{d}{dx}$.

Combining the operators,
$$ D_m(D_n(f(x))) = D_m( x^{n+1} f'(x) ) = x^{m+n+2} f''(x) + (n+1) x^{m+n+1} f'(x) $$
Thus, the commutator $[D_m,D_n] = (n-m)D_{n+n}$.
This I recognize as the Virasoro algebra.

I now consider what finite subalgebras are possible. From the commutators I deduce some constraints.

• The zero operator, D₀, does not affect the presence or absence of any other operators.
• If two operators are positive, D_m and D_n with m and n > 0, then their commutator contains an operator D_m+n with a higher arg value. Repeating the commutation with this new operator gives another one with even higher arg value. Thus, if more than one operator is positive, there are thus an infinite number of positive operators.
• The same argument shows that if two operators are negative, D_m and D_n with m and n < 0, then they generate an infinite number of negative operators.
• So there is at most one positive operator and at most one negative operator.
• Their commutator must be a multiple of the zero operator, or else they will make more than one positive or negative operator.

Thus, the possible finite subalgebras of this algebra are

• One element: D_n for any n.
• Two elements: D₀ and D_n for any nonzero n.
• Three elements: D_n, D_-n, and D₀ for any nonzero n.

The three-element one's commutators:
$[D_0, D_n] = n D_n$
$[D_0, D_{-n}] = - n D_{-n}$
$[D_{-n}, D_n] = 2n D_0$
These can be combined into
$[D_0, D_n+D_{-n}] = n (D_n - D_{-n})$
$[D_n-D_{-n},D_0] = - n (D_n + D_{-n})$
$[D_n+D_{-n}, D_n-D_{-n}] = 4n D_0$
The pattern of signs points to the algebra SO(2,1).

So the possible finite subalgebras are the one-element ones, the nontrivial two-element ones, and the three-element SO(2,1).

This is correct, well done.
However, the three-dimensional case is commonly noted as the simple Lie algebra of type $A_1$, i.e. $\mathfrak{sl}_2 \cong \mathfrak{su}_2$. The two-dimensional case is its Borel subalgebra, the maximal solvable subalgebra. It is the only non Abelian of dimension two. The one-dimensional is obviously Abelian. These are the only differential structures on the real line, if I remember it correctly. By the way, its the Witt algebra, Virasoro algebras are central extensions of Witt.

 

[fresh_42]

That is literally the exact same solution I gave... No problem though. Cheers!

Correct me, if I'm wrong, but as I could see, you gave the result without the way how to achieve it.

 

[Biker]

Correct me, if I'm wrong, but as I could see, you gave the result without the way how to achieve it.

I gave the way in a spoiler, But I don't really care. As long my answer was correct. The pleasure of solving it is enough.

 

[QuantumQuest]

That is literally the exact same solution I gave... No problem though. Cheers!

Sorry but where exactly is this solution? You just told me under spoiler in your post #19 just one headline of what you did. Also you had asked me for an answer which was not the final answer. Why didn't you post your full solution as lpetrich did?

 

[fresh_42]

I gave the way in a spoiler, But I don't really care. As long my answer was correct. The pleasure of solving it is enough.

Sorry, but I searched twice now for problem #6, and all I could find is post #14 with a result, not a way to the result. To be precise, it wasn't even a result, just a question. Anyway, there will be more questions. (And all of you are still invited to PM me good problems. Mine seem to be to easy for you .)

 

[Biker]

Sorry but where exactly is this solution? You just told me under spoiler in your post #19 just one headline of what you did. Also you had asked me for an answer which was not the final answer. Why didn't you post your full solution as lpetrich did?

The problems don't really need to show steps, If you know what to do then the rest is trivial.
I gave a solution which was the "final" solution for me, and a lot of alternative forms of the solution exist. I also said that the method is just recursive integration, and that the integration simplifies itself (last term) with the boundaries of the integral.

I really liked the problem though because it was the first time for me at least to make a recursive law in integration and i was quite happy when I noticed it, So thank you for the question.

 

[QuantumQuest]

The problems don't really need to show steps, If you know what to do then the rest is trivial.

If you solve a problem for yourself i.e. doing your own study you can do it any way you wish. But in the context of a challenge there are rules and for good reason. So, rules also apply for our challenges here. If you haven't already done so then please take a look at the rules.

I really liked the problem though because it was the first time for me at least to make a recursive law in integration and i was quite happy when I noticed it, So thank you for the question.

It is always good for all of us to learn so you're welcome. I think that it would be even more constructive to try to come up with a different way. A different solution - when we say different we mean it, deserves also credit no matter if the problem has already been solved in one way. So, if you want try a different approach but remember that here only full solutions (i.e. including all steps) are credited.

 

[lpetrich]

This is correct, well done.
However, the three-dimensional case is commonly noted as the simple Lie algebra of type $A_1$, i.e. $\mathfrak{sl}_2 \cong \mathfrak{su}_2$. The two-dimensional case is its Borel subalgebra, the maximal solvable subalgebra. It is the only non Abelian of dimension two. The one-dimensional is obviously Abelian. These are the only differential structures on the real line, if I remember it correctly.

The 3D Lie algebra A1 is something of a degenerate case. For real parameters,
so(3) ~ su(2) ~ sp(2) (or usp(2))
so(2,1) ~ su(1,1) ~ sp(2,R) ~ sl(2,R)
They are related by analytic continuation.

If you've ever worked with quantum-mechanical angular momentum, you've worked with this algebra.

 

[fresh_42]

What do you mean by degenerate? A term I would avoid by all means in the context of semisimple Lie algebras, as they can be defined by exactly the non-degeneracy of their Killing-forms. This adjective is highly confusing if used as you did.

The three dimensional $\mathfrak{sl}(2)$ is so to say the prototype of a simple Lie Algebra, one dimension of all what's needed. As there is only one, I've never really cared about the various realizations and $\mathfrak{sl}(2)$ is easiest to handle, although $\mathfrak{su}_\mathbb{R}(2,\mathbb{C})$ is the physics version of it.

 

[lpetrich]

What do you mean by degenerate?

"Degenerate" in the sense of "degeneracy" in quantum mechanics -- several algebra-family members looking alike.

 

[Fred Wright]

That's correct, although you could have made life much easier for an old man. I don't know all the trig formulas in mind anymore. And to demonstrate the integration by parts would have been a nice service for the younger among us. However, the argument with the odd function was nice, so I'll not complain about the wrong round-off of the result, although an exact solution would have been better:
$$
\int_{-1}^0 \,x \cdot \sqrt{x^2+x+1} \,dx = -\frac{1}{4} - \frac{3}{16} \log 3 \approx -0.45598980\ldots \approx -0.456
$$

I got this result using the substitution $u=x +1/2$ and the entries in a table of integrals for $$\int_\frac{-1}{2}^\frac{1}{2}u\sqrt{u^2 + \frac{3}{4}}du -\frac{1}{2}\int_\frac{-1}{2}^\frac{1}{2}\sqrt{u^2 + \frac{3}{4}}du$$

 

[julian]

Hello QuantumQuest. You said there is a nice way to do problem 6. Did you mean this:

First write

$I = \int_0^\pi \sin^{2n} \theta d \theta = \frac{1}{2} \int_{-\pi}^\pi \sin^{2n} \theta d \theta$

where we have used that $\sin^{2n} \theta$ is an even function in $\theta$. This can then be converted into a contour integral around the unit circle by the change of variables

$z = e^{i \theta} \qquad dz = i e^{i \theta} d \theta$

and writing

$\sin \theta = \frac{1}{2i} \Big( z - \frac{1}{z} \Big)$

so that

\begin{align}
I & = \frac{1}{2} \oint \Big[ \frac{1}{2i} \Big( z - \frac{1}{z} \Big) \Big]^{2n} \frac{dz}{iz}
\nonumber \\
& = \frac{1}{2i} \frac{(-1)^n}{2^{2n}} \oint \Big( z - \frac{1}{z} \Big)^{2n} \frac{dz}{z}
\nonumber
\end{align}

This countour integral is easily solved by finding the coefficient of $\frac{1}{z}$ using the binomial expansion:

\begin{align}
I & =
\frac{1}{2i} \frac{(-1)^n}{2^{2n}} \oint \Big( \cdots +
\begin{pmatrix}
2n \\ n
\end{pmatrix}
z^n \Big( - \frac{1}{z} \Big)^n + \cdots \Big) \frac{dz}{z}
\nonumber \\
& = \frac{1}{2i} \frac{(-1)^n}{2^{2n}} \oint \Big( \cdots + \frac{(-1)^n (2n)!}{(n!)^2} + \cdots \Big) \frac{dz}{z}
\nonumber \\
& = \frac{1}{2i} \frac{(-1)^n}{2^{2n}} \times 2 \pi i \frac{(-1)^n (2n)!}{(n!)^2}
\nonumber \\
& = \frac{(2n)!}{2^{2n} (n!)^2} \pi
\nonumber
\end{align}

 

[QuantumQuest]

@julian well done! You'll also get credit for this as you solved it in a different way.