Lucas Numbers and Generating Functions


by Johnny Numbers
Tags: functions, generating, lucas, numbers
Johnny Numbers
Johnny Numbers is offline
#1
Oct30-05, 01:51 PM
P: 9
Here is my problem and my attempt at the answer. Any help or advice is highly appreciated.
Problem
With the famous sequence of Lucas numbers 1, 3, 4, 7, 11, 18... (Where each number is the sum of the last two and the first two are defined as 1 and 3.) use generating functions to find an explicit formula for the Lucas function.
Attempted Solution
We have
[tex]\sum_{j=1}^{n}F_jx^j[/tex]
where Fj denotes the jth Fibonacci number and n is going to infinity. Then we add that to
[tex]\sum_{j=-1}^{n}F_jx^j^+^2[/tex]
Where F-1 = -1 and F0 = 0

And that should get us a function of Lucas numbers right?
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Tide
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#2
Oct30-05, 03:30 PM
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P: 3,149
The Lucas numbers satisfy the relation

[tex]L_{n+1} = L_n + L_{n-1}[/tex]

Just set [itex]L_n = a^n[/itex] and solve for a. Your generating function will be a linear combination of the two solutions. Apply your initial conditions ( [itex]L_1[/itex] and [itex]L_2[/itex]) to determine the two arbitrary constants and you're done! :)
Johnny Numbers
Johnny Numbers is offline
#3
Oct31-05, 12:01 AM
P: 9
I should've been more specific, but we have to use the Fibonacci numbers to generate the Lucas numbers in this manner.

EDIT: I changed it up a little bit as well.

Tide
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#4
Oct31-05, 12:40 AM
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P: 3,149

Lucas Numbers and Generating Functions


In that case, it should be apparent that [itex]L_n = F_{n+1} + F_{n-1}[/itex]
Johnny Numbers
Johnny Numbers is offline
#5
Nov1-05, 11:33 AM
P: 9
Ok so I believe that matches what I was intending on getting at. Thank you again.


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