[seqfan] Re: Functional Equation -F(x)/F(-x) = exp(x)

Paul D Hanna pauldhanna at juno.com
Sat Jul 23 20:00:07 CEST 2011


SeqFans, 
     The question I posed is simpler than I thought at first; the functional equation 
(1)  -F(x)/F(-x) = exp(x) 
has an infinite number of solutions, simply equal to exp(x/2)*B(x), where B(x) = B(-x) is some odd function.   I think it is trivial after all. 
  
Yet, regarding the new sequence A193341, there is a non-trivial result in this regard. 
  
The unexpected result is that A(x) (which satisfies (3)) would be a product of exp(x/2) and an odd function B(x).  This B(x) begins:  
 
B(x) = x + 3*x^3/(3!*2^2) - 35*x^5/(5!*2^4) + 6111*x^7/(7!*2^6) - 3015207*x^9/(9!*2^8) + 3457389595*x^11/(11!*2^10) - 7910176435083*x^13/(13!*2^12) + 32652618744201015*x^15/(15!*2^14) +...
  
where A(x) = B(x)*exp(x) satisfies A(A(x)) = x*exp(A(x)). 
 
I will now submit the coefficients of this odd function B(x). 
 
Thanks anyway, 
    Paul 
 
---------- Original Message ----------
From: "Paul D Hanna" <pauldhanna at juno.com>
To: seqfan at list.seqfan.eu
Subject: [seqfan] Functional Equation -F(x)/F(-x) = exp(x)
Date: Sat, 23 Jul 2011 17:14:48 GMT

SeqFans, 
    Consider the functional equation: 
(1) -F(x)/F(-x) = exp(x), 
for which there are an infinite number of solutions. 
I would like to know the functions that satisfy (1)!  

Also consider the infinite family of functions that satisfy: 
(2) G(-G(-x)) = x. 

I was pleasantly surprised to find that both of these conditions are satisfied by the e.g.f. of a new sequence A193341, defined by: 
(3) A(A(x)) = x*exp(A(x)) 
where A(x) begins: 
A(x) = x + 2*x^2/(2!*2) + 6*x^3/(3!*4) + 16*x^4/(4!*8) - 144*x^6/(6!*32) + 5488*x^7/(7!*64) + 47104*x^8/(8!*128) - 2799360*x^9/(9!*256) - 29427200*x^10/(10!*512) +... 

Now, given any even function B(x) = B(-x), the product A(x)*B(x) will also satisfy (1) -F(x)/F(-x) = exp(x); this is trivial. 
 
Can anyone find another solution to (1) -F(x)/F(-x) = exp(x) that is not related to A(x) (defined by (3)) in a trivial manner? 

Thanks, 
     Paul  

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