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Topic: Summation (Read 1603 times) |
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ThudnBlunder
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 (3/5)n*F(n) where F(n) is the nth Fibonacci number.
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towr
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Re: Summation
« Reply #1 on: Mar 20th, 2008, 1:15pm » |
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Seems like we could just use the closed form for fibonacci numbers
I'll see what I can do after House
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towr
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Re: Summation
« Reply #2 on: Mar 20th, 2008, 1:37pm » |
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(3/5)n F(n)
[(3/5)n (phin - (-1/phi)n)/ 5]
1/5 [(3/5 phi)n - (-3/5 1/phi)n]
1/5 [ (3/5 phi)n - (-3/5 1/phi)n ]
1/5 [ (3/5 phi)n - (-3/5 1/phi)n ]
1/5 [ 1/(1-3/5 phi) - 1/(1+ 3/5 1/phi)]
15
Seems like something which might be the answer..
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| « Last Edit: Mar 20th, 2008, 1:38pm by towr » |
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ThudnBlunder
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Re: Summation
« Reply #3 on: May 8th, 2008, 9:33pm » |
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How about
F(n)/10n+1
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| « Last Edit: May 31st, 2008, 9:52pm by ThudnBlunder » |
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towr
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Re: Summation
« Reply #4 on: May 9th, 2008, 12:37am » |
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You can use the same approach, just replace 3/5 by 1/10 and divide the whole by ten.
1/89
In general an F(n) = a/(1 - a - a2) if it converges.
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| « Last Edit: May 9th, 2008, 12:53am by towr » |
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ThudnBlunder
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Re: Summation
« Reply #5 on: May 9th, 2008, 10:15am » |
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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
F(n)/10n+1 = (0.1)2/(1 - 0.1 - 0.01) = 1/89
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| « Last Edit: May 10th, 2008, 11:57am by ThudnBlunder » |
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towr
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Re: Summation
« Reply #6 on: May 10th, 2008, 1:42am » |
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on May 9th, 2008, 10:15am, ThudanBlunder wrote:| Ah, I wasn't aware of that formula. |
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Neither was I, until I worked it out. I had a bit of trouble getting quickmath to simplify it, too.
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.)
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Eigenray
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Re: Summation
« Reply #7 on: May 10th, 2008, 11:11am » |
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on May 10th, 2008, 1:42am, towr wrote:
Neither was I, until I worked it out. I had a bit of trouble getting quickmath to simplify it, too.
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.) |
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5? But suppose An+2 = a An+1 + b An, and let A(x) = Anxn. What is A(x)*(1 - a x - b x2)? Of course this works for linear recurrences of any order.
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towr
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Re: Summation
« Reply #8 on: May 11th, 2008, 7:19am » |
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on May 10th, 2008, 11:11am, Eigenray wrote:Yes, the two starting values, the two factors in the recurrence, and the geometric ratio.
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Eigenray
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Re: Summation
« Reply #9 on: May 11th, 2008, 9:38am » |
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on May 11th, 2008, 7:19am, towr wrote:| Yes, the two starting values, the two factors in the recurrence, and the geometric ratio. |
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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.
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| « Last Edit: May 11th, 2008, 9:41am by Eigenray » |
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towr
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Re: Summation
« Reply #10 on: May 11th, 2008, 10:04am » |
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on May 11th, 2008, 9:38am, Eigenray 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. |
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Yes; after all, you need x as well before you get a value. I suppose one could argue about terminology, but I just deemed everything you need to fill in to get an answer a parameter.
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