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Topic: All n^th derivate at 0 are integers (Read 2283 times) 

Aryabhatta
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All n^th derivate at 0 are integers
« on: Jul 20^{th}, 2009, 1:19am » 
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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).


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Eigenray
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Re: All n^th derivate at 0 are integers
« Reply #1 on: Jul 20^{th}, 2009, 6:49am » 
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Let A(t) = t/(e^{t}+1). Then 2A(t)  t = t(1e^{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) = a_{n}t^{n}/n!, B(t) = b_{n}t^{n}/n!, then a_{n} = b_{n}(12^{n})... and it is well known that these are halfintegers Well, at least it is well known that the denominators of the Bernoulli numbers are squarefree , so the denominator of 2a_{n} is odd. But we also have that tan(x) = A_{2n+1} x^{2n+1}/(2n+1)!, where the A_{2n+1} are the tangent numbers (and therefore integers). Then a_{2n} = (1)^{n} 2n/2^{2n} A_{2n1}, which shows that the denominator of a_{2n} 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 20^{th}, 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 2f^{n+1}(0) = Sum_{k=1}^{n} 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=n1. To prove for n+1 we use 1) for 2k = n+2, couple with the fact that f^{n+1}(0) is either an integer or odd integer/2.

« Last Edit: Jul 20^{th}, 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 20^{th}, 2009, 10:39am » 
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So we have, with a_{n} = f^{(n)}(0), 2a_{2n} = 2n + C(2n,2) a_{2} + C(2n,4) a_{4} + ... + C(2n,2) a_{2n2} 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?


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Aryabhatta
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Re: All n^th derivate at 0 are integers
« Reply #4 on: Jul 20^{th}, 2009, 10:53am » 
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on Jul 20^{th}, 2009, 10:39am, Eigenray wrote:So we have, with a_{n} = f^{(n)}(0), 2a_{2n} = 2n + C(2n,2) a_{2} + C(2n,4) a_{4} + ... + C(2n,2) a_{2n2} 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? 
 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)a_{2} + ... + C(2n+1,2n) a_{2n}


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Eigenray
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Re: All n^th derivate at 0 are integers
« Reply #5 on: Jul 20^{th}, 2009, 1:06pm » 
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Oh yes of course... I was completely ignoring half the equations


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