sign in sign up
<<
Home , Fibonacci number

Seunghee Ye Ma 8: Week 10 Dec 1

Week 10 Summary

For our last week of Ma 8, we will talk about power .

Topics

Page

1 1 1.1 More Fun with Power Series ...... 2

1 Power Series

The last topic of the term is power series. We have seen an example of power series when discussing of smooth functions.

Definition (Power Series) A power series is an expression ∞ X n f(z) = anz n=0 where an ∈ C.

Before, we only considered power series over R but now, we will consider power series over C as well. To differentiate these two cases, a power series over the reals will be denoted f(x); and over the complex, f(z). One of the things we want to do is plug in various values of x and z. Once we do that, f(x) and f(z) will be series of real or complex . Of course, the natural thing to do is to determine whether these series converge. There are infinitely many real numbers and complex numbers but the following theorem will save us the trouble of checking them one by one (which, of course, would be impossible!).

Theorem 1.1. Let f(x) be a real valued power series. Suppose f(x) converges at some value a ∈ R. Then, f(x) converges at every value b ∈ (−a, a). Let f(z) be a complex valued power series. Suppose f(z) converges at some α ∈ C. Then, f(z) converges at every value of β such that |β| < |α|.

Theorem 1.1 says that once we know that the power series converges at some real/complex , we automatically know that it converges at all “smaller” numbers. Looking at the complex plane, we immediately see where the term radius of convergence comes from.

Definition (Radius of Convergence) Let f(z) be a power series. Then, the radius of convergence of f(z) is

sup{|α| | α ∈ C such that f(α) converges}

Example 1.1. Find the radius of convergence for

∞ X 1. f(z) = n! · xn n=0

∞ X 2. f(z) = xn n=0

Page 1 of 4 Seunghee Ye Ma 8: Week 10 Dec 1

∞ X xn 3. f(z) = n! n=0 We see that finding the radius of convergence for complex valued power series is exactly the same as for real valued power series.

1.1 More Fun with Power Series In this section, we will use the power series to understand the better. Given a sequence, we can define for the sequence. The generating function for a sequence {an} is the power series: ∞ X n f(x) = anx n=0 In other words, the generating function is the power series whose coefficients are the terms of the given sequence.

Example 1.2. Let f(x) be the generating function for the Fibonacci sequence. Show that x f(x) = 1 − x − x2

Proof. Let’s recall the definition of the Fibonacci sequence (starting from n = 0). If we denote by Fn the n-th Fibonacci number, we have:

F0 = 0,F1 = 1,Fn = Fn−1 + Fn−2 for all n ≥ 2

Then, the generating series for {Fn} is

∞ X n 2 3 4 5 f(x) = Fnx = x + x + 2x + 3x + 5x + ··· n=0

Now, using the recursive relation Fn = Fn−1 + Fn−2 for all n ≥ 2, we obtain the following:

∞ X n f(x) = Fnx (1) n=0 ∞ X n = 0 + x + Fnx (2) n=2 ∞ X n = x + (Fn−1 + Fn−2)x (3) n=2 ∞ ∞ X n X n = x + Fn−1x + Fn−2x (4) n=2 n=2 ∞ ∞ X n 2 X n = x + x Fnx + x Fnx (5) n=1 n=0 ∞ ! X n 2 = x + x 0 + Fnx + x f(x) (6) n=1 = x + xf(x) + x2f(x) (7)

Page 2 of 4 Seunghee Ye Ma 8: Week 10 Dec 1

Rearranging terms, we see that x f(x)(1 − x − x2) = x ⇒ f(x) = 1 − x − x2

Functions that can be written as the fraction of two polynomials is called rational functions. By expanding the denominator, we see that rational functions can always be written in their power series form. The converse, however, is not true: Given a general power series, it is not true that we can write it as the of two polynomials. The fact that the generating series for the Fibonacci sequence can be written as a rational function implies that the Fibonacci sequence has a nice structure (which we know is the recursive relation between the terms). Now, let’s try to use the power series to find the closed form for the Fibonacci sequence. If you are familiar with algorithms, the Fibonacci sequence is one of the examples that come up quite often. To compute the n-th Fibonacci number, you can use the straightforward recursive algorithm (using the recursive relation); or you can use iterative algorithm (the way). You might know that the recursive algorithm is “very bad” and the iterative algorithm is “good”: recursive algorithm has exponential time complexity, where as iterative algorithm has linear time complexity. With the closed form, we will have constant time complexity for computing the n-th Fibonacci number!

Example 1.3. Find the closed form for the Fibonacci sequence. In other words, express the Fibonacci sequence as Fn = F (n) where F (n) is a function of n.

Solution. We saw above that the generating series for Fn was a rational function. So let’s try to use the result we obtained above about f(x) to deduce the closed form for Fn. We showed that

∞ X x f(x) = F xn = n 1 − x − x2 n=0 x But this means that the coefficients of the power series expansion of are precisely the Fibonacci 1 − x − x2 x numbers. Matching terms, we see that the coefficient of xn of the power series expansion of is 1 − x − x2 Fn. First, we will try to split the right hand side using partial fractions. Note that 1 − x − x2 has two roots which are: √ √ −1 ± 1 + 4 −1 ± 5 = 2 2 Let’s call √ √ 1 + 5 1 − 5 ϕ := , ψ := 2 2 Then, we have 1 − x − x2 = −(x + ϕ)(x + ψ) Note that ϕ · ψ = −1. Hence, ψ = −ϕ−1. Using this, we get

1 − x − x2 = −(x + ϕ)(x + ψ) (8) = −(ϕ−1x + 1)ϕ(x + ψ) (9) = −(−ψx + 1)(ϕx + ϕ · ψ) (10) = (1 − ψx)(1 − ϕx) (11)

Page 3 of 4 Seunghee Ye Ma 8: Week 10 Dec 1

Let’s use this and partial fractions and see what we get x x = (12) 1 − x − x2 (1 − ϕx)(1 − ψx) √1 − √1 = 5 + 5 (13) 1 − ϕx 1 − ψx 1  1 1  = √ − (14) 5 1 − ϕx 1 − ψx

1 1 This is great! We know that 1−ϕx and 1−ψx are simply the ! In other words, their power series expansions are ∞ ∞ 1 X 1 X = ϕnxn and = ψnxn 1 − ϕx 1 − ψx n=0 n=0 Therefore, we find that

1  1 1  f(x) = √ − (15) 5 1 − ϕx 1 − ψx ∞ ∞ ! 1 X X = √ ϕnxn − ψnxn (16) 5 n=0 n=0 ∞ ! 1 X = √ (ϕn − ψn)xn (17) 5 n=0

n As we remarked earlier, the coefficient of x of this power series is precisely Fn. Hence, we conclude that

ϕn − ψn ϕn − (−ϕ)−n Fn = √ = √ 5 5

√ 1 + 5 A little aside: ϕ = is called the and comes up surprisingly often. 2

Page 4 of 4