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Topic: All n^th derivate at 0 are integers (Read 2303 times) |
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Aryabhatta
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All n^th derivate at 0 are integers
« on: Jul 20th, 2009, 1:19am » |
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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 » |
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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 » |
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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.
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| « Last Edit: Jul 20th, 2009, 8:14am by Aryabhatta » |
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Eigenray
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Re: All n^th derivate at 0 are integers
« Reply #3 on: Jul 20th, 2009, 10:39am » |
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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 » |
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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? |
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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 » |
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Oh yes of course... I was completely ignoring half the equations
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