Susanne217
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1 If I am given the function \frac{1}{-x^2-x+1} = \sum_{j=0}^{\infty} F_{j} x^{j}
Which discribes a sequence of fibunacci numbers 1,1,2,3,5,8,13,21...
Find the formula for the fibunacci sequence for n>=2 and where F_j = F_{j-1}+F_{j-2}
I know that the recusive relation can be written as F_j = \alpha_1(r_1)^j + \alpha_2 (r_2)^j
With the inital conditions F_0 = F_1 = 1
Since the poles of the function are r_1 = \frac{\sqrt{5}-1}{2} and r_2 = \frac{-(\sqrt{5}+1)}{2}
which gives me the expression F_j = \alpha_1(\frac{\sqrt{5}-1}{2})^n + \alpha_2 (\frac{-(\sqrt{5}+1)}{2}})^n
so this gives me F_0 = \alpha_1 + \alpha_2 = 1
and F_1 = \alpha_1(\frac{\sqrt{5}-1}{2}) + \alpha_2 (\frac{-(\sqrt{5}+1)}{2}}) = 1 and I end up with
\alpha_1, \alpha_2 = \pm \frac{1}{\sqrt{5}}
and this j >= 2 then the formula for the jth fibunacci number must
F_j = \frac{1}{\sqrt{5}}(\frac{\sqrt{5}+1}{2})^{j+1} -\frac{1}{\sqrt{5}} (\frac{-(\sqrt{5}+1)}{2}})^{j+1}
How is that HallsoftIvy?
Susanne
Which discribes a sequence of fibunacci numbers 1,1,2,3,5,8,13,21...
Find the formula for the fibunacci sequence for n>=2 and where F_j = F_{j-1}+F_{j-2}
Homework Equations
The Attempt at a Solution
I know that the recusive relation can be written as F_j = \alpha_1(r_1)^j + \alpha_2 (r_2)^j
With the inital conditions F_0 = F_1 = 1
Since the poles of the function are r_1 = \frac{\sqrt{5}-1}{2} and r_2 = \frac{-(\sqrt{5}+1)}{2}
which gives me the expression F_j = \alpha_1(\frac{\sqrt{5}-1}{2})^n + \alpha_2 (\frac{-(\sqrt{5}+1)}{2}})^n
so this gives me F_0 = \alpha_1 + \alpha_2 = 1
and F_1 = \alpha_1(\frac{\sqrt{5}-1}{2}) + \alpha_2 (\frac{-(\sqrt{5}+1)}{2}}) = 1 and I end up with
\alpha_1, \alpha_2 = \pm \frac{1}{\sqrt{5}}
and this j >= 2 then the formula for the jth fibunacci number must
F_j = \frac{1}{\sqrt{5}}(\frac{\sqrt{5}+1}{2})^{j+1} -\frac{1}{\sqrt{5}} (\frac{-(\sqrt{5}+1)}{2}})^{j+1}
How is that HallsoftIvy?
Susanne
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