Lucas Numbers and Generating Functionsby Johnny Numbers Tags: functions, generating, lucas, numbers 

#1
Oct3005, 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 F_{j} denotes the j^{th} 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 F_{0} = 0 And that should get us a function of Lucas numbers right? 



#2
Oct3005, 03:30 PM

Sci Advisor
HW Helper
P: 3,149

The Lucas numbers satisfy the relation
[tex]L_{n+1} = L_n + L_{n1}[/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! :) 



#3
Oct3105, 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. 



#4
Oct3105, 12:40 AM

Sci Advisor
HW Helper
P: 3,149

Lucas Numbers and Generating Functions
In that case, it should be apparent that [itex]L_n = F_{n+1} + F_{n1}[/itex]




#5
Nov105, 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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