

Title: Summation Post by ThudanBlunder on Mar 20^{th}, 2008, 1:00pm http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif(3/5)^{n}*F(n) where F(n) is the nth Fibonacci number. 0 

Title: Re: Summation Post by towr on Mar 20^{th}, 2008, 1:15pm [hide]Seems like we could just use the closed form for fibonacci numbers I'll see what I can do after House[/hide] 

Title: Re: Summation Post by towr on Mar 20^{th}, 2008, 1:37pm [hide]http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif(3/5)^{n} F(n) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif[(3/5)^{n} (phi^{n}  (1/phi)^{n})/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5] 1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif[(3/5 phi)^{n}  (3/5 1/phi)^{n}] 1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5 [ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif(3/5 phi)^{n}  http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif(3/5 1/phi)^{n} ] 1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5 [ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif(3/5 phi)^{n}  http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif(3/5 1/phi)^{n} ] 1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5 [ 1/(13/5 phi)  1/(1+ 3/5 1/phi)] 15[/hide] Seems like something which might be the answer.. 

Title: Re: Summation Post by ThudanBlunder on May 8^{th}, 2008, 9:33pm How about http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifF(n)/10^{n+1} 1 

Title: Re: Summation Post by towr on May 9^{th}, 2008, 12:37am You can use the same approach, just replace 3/5 by 1/10 and divide the whole by ten. [hide] 1/89 [/hide] In general http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifa^{n} F(n) = a/(1  a  a^{2}) if it converges. 

Title: Re: Summation Post by ThudanBlunder on May 9^{th}, 2008, 10:15am Ah, I wasn't aware of that formula. The reason I added the second summation is that we have n 1 > 0.01 2 > 0.001 3 > 0.0002 4 > 0.00003 5 > 0.000005 6 > 0.0000008 7 > 0.00000013 etc Adding we get 0.011235955056179775280898876404494...... which is 1/89, as you say. I was wondering why 89, but from the formula it is now clear, as we get http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifF(n)/10^{n+1} = (0.1)^{2}/(1  0.1  0.01) = 1/89 1 

Title: Re: Summation Post by towr on May 10^{th}, 2008, 1:42am on 05/09/08 at 10:15:37, ThudanBlunder wrote:
It's neat though. And you can probably generalize it further for other second order recurrences. (Although I'm not sure how pretty the result will be; because it has 5 parameters.) 

Title: Re: Summation Post by Eigenray on May 10^{th}, 2008, 11:11am on 05/10/08 at 01:42:16, towr wrote:
5? But suppose A_{n+2} = a A_{n+1} + b A_{n}, and let A(x) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif A_{n}x^{n}. What is A(x)*(1  a x  b x^{2})? Of course this works for linear recurrences of any order. 

Title: Re: Summation Post by towr on May 11^{th}, 2008, 7:19am on 05/10/08 at 11:11:13, Eigenray wrote:


Title: Re: Summation Post by Eigenray on May 11^{th}, 2008, 9:38am on 05/11/08 at 07:19:42, towr wrote:
Can you give an example? [edit] Oh wait, do you mean x? I was thinking of the generating function itself, which only has 4 parameters. 

Title: Re: Summation Post by towr on May 11^{th}, 2008, 10:04am on 05/11/08 at 09:38:56, Eigenray wrote:


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