f(x) = g(x) + g(x+1) ------ f(x) is known, g(x)=?
From: denoxis (google_at_deniznet.com)
Date: 02/28/05
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Date: 28 Feb 2005 10:42:48 -0800
Hi,
I was working on a problem and I finally came across this:
f(x) = g(x) + g(x+1)
I know f(x) = (x(x+1)(x+2))/6
Is there any method to find what g(x) could be?
Thanks.
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- Reply: Hero: "Re: f(x) = g(x) + g(x+1) ------ f(x) is known, g(x)=?"
- Reply: Ignacio Larrosa Caņestro: "Re: f(x) = g(x) + g(x+1) ------ f(x) is known, g(x)=?"
- Reply: Will Twentyman: "Re: f(x) = g(x) + g(x+1) ------ f(x) is known, g(x)=?"
- Reply: Abraham Buckingham: "Re: f(x) = g(x) + g(x+1) ------ f(x) is known, g(x)=?"
- Reply: Zdislav V. Kovarik: "Re: f(x) = g(x) + g(x+1) ------ f(x) is known, g(x)=?"
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