Foundations of Data Science

Let
Substituting (12.2) into (12.1) yields
f(x) =
∞∑
f i x i . (12.2)
i=0
Thus, f(x) =
x
1−x−x 2
f(x) − f 0 − f 1 x = x (f(x) − f 0 ) + x 2 f(x)
f(x) − x = xf(x) + x 2 f(x)
f(x)(1 − x − x 2 ) = x
is the generating function for the Fibonacci sequence.
Note that generating functions are formal manipulations and do not necessarily converge
outside some region of convergence. Consider the generating function f (x) =
∞∑
∑
f i x i = for the Fibonacci sequence. Using ∞ f i x i ,
i=0
x
1−x−x 2
and using f(x) =
Asymptotic behavior
x
1−x−x 2 f(1) =
i=0
f(1) = f 0 + f 1 + f 2 + · · · = ∞
1
1 − 1 − 1 = −1.
To determine the asymptotic behavior of the Fibonacci sequence write
f (x) =
√
x
5
1 − x − x = 5
2 1 − φ 1 x + − √ 5
5
1 − φ 2 x
where φ 1 = 1+√ 5
and φ
2 1 = 1−√ 5
are the reciprocals of the two roots of the quadratic
2
1 − x − x 2 = 0.
Then
Thus,
f (x) =
√
5
5
(
1 + φ 1 x + (φ 1 x) 2 + · · · − ( 1 + φ 2 x + (φ 2 x) 2 + · · ·)) .
f n =
√
5
5 (φn 1 − φ n 2) .
√
Since φ 2 < 1 and φ 1 > 1, for large n, f n
∼ = 5
5 φn 1. In fact, since f n = √ 5
⌊
5 (φn 1 − φ n 2) is an
integer and φ 2 < 1, it must be the case that f n = f n + √ ⌋
⌊ √5 ⌋
5
2 φn 2 . Hence f n =
5 φn 1 for
all n.
Means and standard deviations of sequences
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