Department of Mathematics Nebraska Wesleyan University June 14, 01
Fibonacci Sequence Fun Fact: November 3rd is Fibonacci Day! (1, 1,, 3) Definition The Fibonacci sequence is defined by the recurrence relation F n+1 = F n + F n 1 for all n with F 0 = 0 and F 1 = 1. So, the sequence is 1, 1,, 3,, 8, 13...
Fibonacci Sequence What if? We want to know the 10th number of the sequence? We want to know the 0th or 0th term?
Fibonacci Sequence What if? We want to know the 10th number of the sequence? We want to know the 0th or 0th term?
Generating Functions Definition Let {f n } be a sequence of real numbers. Then, the formal power series F (x) = f n x n is called the ordinary generating function of the sequence {f n }.
Example We have invested 1000 dollars into a savings account that pays five percent interest at the end of each year. At the beginning of each year, we deposit another 00 dollars into this account. How much money will be in this account after n years? The recurrence relation is a n+1 = 1.0a n + 00 with a 0 = 1000.
Example We have invested 1000 dollars into a savings account that pays five percent interest at the end of each year. At the beginning of each year, we deposit another 00 dollars into this account. How much money will be in this account after n years? The recurrence relation is a n+1 = 1.0a n + 00 with a 0 = 1000.
Example The recurrence relation is a n+1 = 1.0a n + 00 with a 0 = 1000. Let G(x) = a n x n be the generating function of the sequence {a n }. Now, multiply both sides of the recurrence relation by x n+1 and sum over all non-negative integers n. We get 00x n+1. Note: a n+1 x n+1 = 1.0a n x n+1 + a n+1 x n+1 = G(x) a 0 and 1.0a n x n+1 = 1.0xG(x).
Example The recurrence relation is a n+1 = 1.0a n + 00 with a 0 = 1000. Let G(x) = a n x n be the generating function of the sequence {a n }. Now, multiply both sides of the recurrence relation by x n+1 and sum over all non-negative integers n. We get 00x n+1. Note: a n+1 x n+1 = 1.0a n x n+1 + a n+1 x n+1 = G(x) a 0 and 1.0a n x n+1 = 1.0xG(x).
Example The recurrence relation is a n+1 = 1.0a n + 00 with a 0 = 1000. Let G(x) = a n x n be the generating function of the sequence {a n }. Now, multiply both sides of the recurrence relation by x n+1 and sum over all non-negative integers n. We get 00x n+1. Note: a n+1 x n+1 = 1.0a n x n+1 + a n+1 x n+1 = G(x) a 0 and 1.0a n x n+1 = 1.0xG(x).
Example Reminder: The Taylor Series for 1 1 x is x n. So, By substitution, we have Solving for G(x), we obtain By partial fractions, 00x n+1 = 00x 1 x. G(x) a 0 = 1.0xG(x) + 00x 1 x. G(x) = 1000 1 1.0x + 00x (1 x)(1.10x). 00x (1 x)(1.10x) = 10000 ( 1 1 1.0x 1 1 x ).
Example Reminder: The Taylor Series for 1 1 x is x n. So, By substitution, we have Solving for G(x), we obtain By partial fractions, 00x n+1 = 00x 1 x. G(x) a 0 = 1.0xG(x) + 00x 1 x. G(x) = 1000 1 1.0x + 00x (1 x)(1.10x). 00x (1 x)(1.10x) = 10000 ( 1 1 1.0x 1 1 x ).
Example Reminder: The Taylor Series for 1 1 x is x n. So, By substitution, we have Solving for G(x), we obtain By partial fractions, 00x n+1 = 00x 1 x. G(x) a 0 = 1.0xG(x) + 00x 1 x. G(x) = 1000 1 1.0x + 00x (1 x)(1.10x). 00x (1 x)(1.10x) = 10000 ( 1 1 1.0x 1 1 x ).
Example Reminder: The Taylor Series for 1 1 x is x n. So, By substitution, we have Solving for G(x), we obtain By partial fractions, 00x n+1 = 00x 1 x. G(x) a 0 = 1.0xG(x) + 00x 1 x. G(x) = 1000 1 1.0x + 00x (1 x)(1.10x). 00x (1 x)(1.10x) = 10000 ( 1 1 1.0x 1 1 x ).
Example Replacing with Taylor series, we obtain G(x) = 1000 (1.0x) n + 10000 (1.0 n 1)x n. This means, a n = 1000 1.0 n + 10000(1.0 n 1) = 11000 1.0 n 10000. This how much money will be in the account after n years!
Example Replacing with Taylor series, we obtain G(x) = 1000 (1.0x) n + 10000 (1.0 n 1)x n. This means, a n = 1000 1.0 n + 10000(1.0 n 1) = 11000 1.0 n 10000. This how much money will be in the account after n years!
Example Replacing with Taylor series, we obtain G(x) = 1000 (1.0x) n + 10000 (1.0 n 1)x n. This means, a n = 1000 1.0 n + 10000(1.0 n 1) = 11000 1.0 n 10000. This how much money will be in the account after n years!
Fibonacci Example The recurrence relation F n+1 = F n + F n 1 with F 0 = 0 and F 1 = 1. Let F (x) = f (n)x n be the generating function. Multiply both sides of the recurrence relation by x n+ and sum it over all non-negative integers n. We get f (n + )x n+ = f (n + 1)x n+ + f (n)x n+. Using the defined generating function, we obtain F (x) x 1 = x(f (x) 1) + x F (x).
Fibonacci Example The recurrence relation F n+1 = F n + F n 1 with F 0 = 0 and F 1 = 1. Let F (x) = f (n)x n be the generating function. Multiply both sides of the recurrence relation by x n+ and sum it over all non-negative integers n. We get f (n + )x n+ = f (n + 1)x n+ + f (n)x n+. Using the defined generating function, we obtain F (x) x 1 = x(f (x) 1) + x F (x).
Fibonacci Example The recurrence relation F n+1 = F n + F n 1 with F 0 = 0 and F 1 = 1. Let F (x) = f (n)x n be the generating function. Multiply both sides of the recurrence relation by x n+ and sum it over all non-negative integers n. We get f (n + )x n+ = f (n + 1)x n+ + f (n)x n+. Using the defined generating function, we obtain F (x) x 1 = x(f (x) 1) + x F (x).
Fibonacci Example Solving for F(x), we obtain F (x) = 1 1 x x. Note that the two roots of 1 x x are r 1 = 1 + and r = 1. By partial fractions, we get F (x) = 1 ( 1 1 ). x r x r 1
Fibonacci Example Solving for F(x), we obtain F (x) = 1 1 x x. Note that the two roots of 1 x x are r 1 = 1 + and r = 1. By partial fractions, we get F (x) = 1 ( 1 1 ). x r x r 1
Fibonacci Example Now, F (x) = 1 ( 1 1 ) x r x r 1 = 1 ( r1 1 r ) 1 r 1 x r r x r 1 = 1 ( r1 r 1 x 1 r ) r x 1 = 1 ( r 1 r x r ) 1 1 r 1 x = 1 r r n x n r 1 r1 n x n.
Fibonacci Example Now, F (x) = 1 ( 1 1 ) x r x r 1 = 1 ( r1 1 r ) 1 r 1 x r r x r 1 = 1 ( r1 r 1 x 1 r ) r x 1 = 1 ( r 1 r x r ) 1 1 r 1 x = 1 r r n x n r 1 r1 n x n.
Fibonacci Example Now, F (x) = 1 ( 1 1 ) x r x r 1 = 1 ( r1 1 r ) 1 r 1 x r r x r 1 = 1 ( r1 r 1 x 1 r ) r x 1 = 1 ( r 1 r x r ) 1 1 r 1 x = 1 r r n x n r 1 r1 n x n.
Fibonacci Example Now, F (x) = 1 ( 1 1 ) x r x r 1 = 1 ( r1 1 r ) 1 r 1 x r r x r 1 = 1 ( r1 r 1 x 1 r ) r x 1 = 1 ( r 1 r x r ) 1 1 r 1 x = 1 r r n x n r 1 r1 n x n.
Fibonacci Example So, Thus, F (x) = 1 r r n x n r 1 r1 n x n. f n = 1 ( r n+1 r n+1 ) 1. By substituting r 1 and r into the equation, we get ( f n = 1 1 + ) n+1 ( 1 1 ) n+1. Note: this equation is set up so f n = F n+1 since the series indices begin at 0, but the Fibonacci indices begin at 1.
Fibonacci Example So, Thus, F (x) = 1 r r n x n r 1 r1 n x n. f n = 1 ( r n+1 r n+1 ) 1. By substituting r 1 and r into the equation, we get ( f n = 1 1 + ) n+1 ( 1 1 ) n+1. Note: this equation is set up so f n = F n+1 since the series indices begin at 0, but the Fibonacci indices begin at 1.
Fibonacci Example So, Thus, F (x) = 1 r r n x n r 1 r1 n x n. f n = 1 ( r n+1 r n+1 ) 1. By substituting r 1 and r into the equation, we get ( f n = 1 1 + ) n+1 ( 1 1 ) n+1. Note: this equation is set up so f n = F n+1 since the series indices begin at 0, but the Fibonacci indices begin at 1.
Does it work? ( F 10 = f 9 = 1 1 + ) 10 ( 1 1 ) 10 = If we go through the sequence, 1, 1,, 3,, 8, 13, 1, 34, is the 10th term! It works!
It works! ( F 0 = f 19 = 1 1 + ) 0 ( 1 1 ) 0 = 6, 76 ( F 0 = f 49 = 1 1 + ) 0 ( 1 1 ) 0 = 1, 86, 69, 0
Take a Closer Look With our equation, ( f n = 1 1 + ) n+1 ( 1 1 ) n+1 the second term goes to zero quickly. So, we can remove that term ( f n = 1 1 + ) n+1 and still get fairly accurate results.
Take a Closer Look With our equation, ( f n = 1 1 + ) n+1 ( 1 1 ) n+1 the second term goes to zero quickly. So, we can remove that term ( f n = 1 1 + ) n+1 and still get fairly accurate results.
Take a Closer Look ( F 10 = f 9 = 1 1 + ) 10 =.00368 ( F 0 = f 19 = 1 1 + ) 0 = 6, 76.00003 ( F 0 = f 49 = 1 1 + ) 0 = 1, 86, 69, 0.00000
Thanks! Thanks for listening! Search the web for Generating Functions to learn more.