Challenge Intermediate Math Challenge - May 2018

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I will solve Problem 6.
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

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@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

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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

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

QuantumQuest

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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;)
 
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I will try Problem 4
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

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I will try Problem 4
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.
 
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I'll take a stab at solving problem 5.
The first thing I do is to find the commutators of the Dn's. For convenience, I will redefine them by doing Dn = old Dn+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, Dm and Dn with m and n > 0, then their commutator contains an operator Dm+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, Dm and Dn 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: Dn for any n.
  • Two elements: D0 and Dn for any nonzero n.
  • Three elements: Dn, 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

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I had realised 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.
 
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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!
 
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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:
A few lines with complex numbers
 

julian

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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

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I'll take a stab at solving problem 5.
The first thing I do is to find the commutators of the Dn's. For convenience, I will redefine them by doing Dn = old Dn+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, Dm and Dn with m and n > 0, then their commutator contains an operator Dm+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, Dm and Dn 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: Dn for any n.
  • Two elements: D0 and Dn for any nonzero n.
  • Three elements: Dn, 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.
 
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fresh_42

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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.
 
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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

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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

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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 :wink:.)
 
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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

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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.
 
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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

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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.
 
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What do you mean by degenerate?
"Degenerate" in the sense of "degeneracy" in quantum mechanics -- several algebra-family members looking alike.
 
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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

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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}
 
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QuantumQuest

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

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