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Aryabhatta
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 All n^th derivate at 0 are integers   « on: Jul 20th, 2009, 1:19am » Quote Modify

Let f(x) = 2x/(1 + ex).

Show that for all k,

fk(0) is an integer.

fk(x) is the kth derivative of f(x).
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Eigenray
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 Re: All n^th derivate at 0 are integers   « Reply #1 on: Jul 20th, 2009, 6:49am » Quote Modify

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

Similarly, if B(t) = t/(et-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) = antn/n!, B(t) = bntn/n!, then
an = bn(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 2an is odd.  But we also have that
tan(x) = A2n+1 x2n+1/(2n+1)!,
where the A2n+1 are the tangent numbers (and therefore integers).  Then

a2n = (-1)n 2n/22n A2n-1,

which shows that the denominator of a2n is a power of 2.  So we are done (proof by Wikipedia )
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Aryabhatta
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 Re: All n^th derivate at 0 are integers   « Reply #2 on: Jul 20th, 2009, 8:13am » Quote Modify

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) = Sumk=1n Choose(n+1,k) fk(0)

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

We now prove the following for k>=1:

1) f2k+1(0)  = 0

2) f2k(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.
 « Last Edit: Jul 20th, 2009, 8:14am by Aryabhatta » IP Logged
Eigenray
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 Re: All n^th derivate at 0 are integers   « Reply #3 on: Jul 20th, 2009, 10:39am » Quote Modify

So we have, with an = f(n)(0),
-2a2n = 2n + C(2n,2) a2 + C(2n,4) a4 + ... + C(2n,2) a2n-2
If all the a2k are odd, then the RHS is even, so a2n is an integer.  But how do we know a2n is odd?
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Aryabhatta
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 Re: All n^th derivate at 0 are integers   « Reply #4 on: Jul 20th, 2009, 10:53am » Quote Modify

on Jul 20th, 2009, 10:39am, Eigenray wrote:
 So we have, with an = f(n)(0), -2a2n = 2n + C(2n,2) a2 + C(2n,4) a4 + ... + C(2n,2) a2n-2 If all the a2k are odd, then the RHS is even, so a2n is an integer.  But how do we know a2n is odd?

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

0 = -2a2n+1 = (2n+1) + C(2n+1,2)a2 + ... + C(2n+1,2n) a2n
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Eigenray
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 Re: All n^th derivate at 0 are integers   « Reply #5 on: Jul 20th, 2009, 1:06pm » Quote Modify

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