[seqfan] Re: generating-function-ology

[Andrew N W Hone]

This is not the answer to the question that was asked. 

What is being asked for is the generating function for the sequence of reciprocals of the terms in a sequence, 
in terms of the generating function for the original sequence. 

This is completely different from the reciprocal of the generating function. 

To see this, consider Fibonacci again. 

The sequence of reciprocals must leave out F_0 = 0, so we start from index 1: 

1,1,1/2, 1/3, 1/5, 1/8, ... 

I do not know a simple formula for the generating function A(x) = x+x^2 +1/2 x^3+... 
of this sequence of rational numbers. 

However, it cannot be a rational function, since the explicit formula for the Fibonacci numbers has the form 

F_n = (u^n -v^n)(u-v) 

where u is the golden mean and v=-1/u, and the reciprocals are 

1/F_n = (u-v) / (u^n - v^n) 

which do not satisfy any linear recurrence relation.

All the best
Andy   
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Ed Jeffery [lejeffery7 at gmail.com]
Sent: 13 August 2012 05:12
To: seqfan at list.seqfan.eu
Subject: [seqfan] Re: generating-function-ology

Peter,

If I'm not mistaken again:

If A(x) = f(x) / g(x) is a rational function, then f(x) and g(x) are both
polynomials, so A(x) * g(x) / f(x) = 1, and your inverse is just the
expansion of b(x) = g(x) / f(x) to series. For example, for the inverse of
the Fibonacci series, you just get the inverse of the generating function
(up to an offset) for the Fibonacci sequence:

(1)     1 / (1 + x + 2*x*2 +3*x^3 + 5*x^4 + ...) = 1 - x - x^2.

Multiplying (1) through by 1 + x + 2*x^2 + ..., you can see (although this
is no proof) that all terms in x^k, for k > 0, on the right vanish except
for the coefficient of x^0 which is equal to 1.

Finally, if A(x) is a series and not a polynomial, then its inverse could
be another series or it could be a rational function.

LEJ

> if A(x) is the generating function for the sequence a(i), is there an
> equation for the generating function of b(i) = 1/a(i) ?

> thanks,
> Peter Lawrence.

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