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        riddles >> putnam exam (pure math) >> All n^th derivate at 0 are integers
        
(Message started by: Aryabhatta on Jul 20^th, 2009, 1:19am)

Title: All n^th derivate at 0 are integers
Post by Aryabhatta on Jul 20^th, 2009, 1:19am

Let f(x) = 2x/(1 + e^x).

Show that for all k,

f^k(0) is an integer.

f^k(x) is the k^th derivative of f(x).

Title: Re: All n^th derivate at 0 are integers
Post by Eigenray on Jul 20^th, 2009, 6:49am

Let A(t) = t/(e^t+1). Then
2A(t) - t = t(1-e^t)/(1+e^t) = -t tanh(t/2).

Similarly, if B(t) = t/(e^t-1), then
2B(t) + t = t coth(t/2).
So 2B(t) - 2A(t) + 2t = t [ coth(t/2) + tanh(t/2) ] = 2 t coth(t) = 2B(2t) + 2t,
so A(t) = B(t) - B(2t).
If A(t) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifa_ntⁿ/n!, B(t) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif b_ntⁿ/n!, then
a_n = b_n(1-2ⁿ)... and it is well known that these are half-integers ;)

Well, at least it is well known that the denominators of the Bernoulli numbers are square-free ::), so the denominator of 2a_n is odd. But we also have that
tan(x) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif A_2n+1 x²ⁿ⁺¹/(2n+1)!,
where the A_2n+1 are the [link=http://en.wikipedia.org/wiki/Tangent_numbers]tangent numbers[/link] (and therefore integers). Then

a_2n = (-1)ⁿ 2n/2²ⁿ A_2n-1,

which shows that the denominator of a_2n is a power of 2. So we are done (proof by Wikipedia ;D)

Title: Re: All n^th derivate at 0 are integers
Post by Aryabhatta on Jul 20^th, 2009, 8:13am

Any proof is acceptable, I was looking for a simpler proof though :)

Here is a sketch of a simpler proof.

Setting f(x)(1+e^x) = 2x and differentiating n times gives us

-2fⁿ⁺¹(0) = Sum_k=1ⁿ Choose(n+1,k) f^k(0)

Choose (n+1,k) = the n+1 choose k, the binomial coefficient.

We now prove the following for k>=1:

1) f^2k+1(0) = 0

2) f^2k(0) is an odd integer.

1) follows from the fact that f(x) - x is an even function.

2) To prove 2 , we use induction on k. Say it is true for 2k=n-1. To prove for n+1 we use 1) for 2k = n+2, couple with the fact that fⁿ⁺¹(0) is either an integer or odd integer/2.

Title: Re: All n^th derivate at 0 are integers
Post by Eigenray on Jul 20^th, 2009, 10:39am

So we have, with a_n = f⁽ⁿ⁾(0),
-2a_2n = 2n + C(2n,2) a₂ + C(2n,4) a₄ + ... + C(2n,2) a_2n-2
If all the a_2k are odd, then the RHS is even, so a_2n is an integer. But how do we know a_2n is odd?

Title: Re: All n^th derivate at 0 are integers
Post by Aryabhatta on Jul 20^th, 2009, 10:53am

on 07/20/09 at 10:39:22, Eigenray wrote:

Yes, that proves that it is an integer. To prove that it is odd, also consider

0 = -2a_2n+1 = (2n+1) + C(2n+1,2)a₂ + ... + C(2n+1,2n) a_2n

Title: Re: All n^th derivate at 0 are integers
Post by Eigenray on Jul 20^th, 2009, 1:06pm

Oh yes of course... I was completely ignoring half the equations :)