Ex10 - Cs.ioc.ee

Concrete Mathematics, Lesson 10: Exercises
Exercise
Silvio Capobianco
Revised: November 20, 2012
Define the Fibonacci number fn for negative integer n.
Solution. What we really want, is that the defining relation fn+1 = fn + fn−1
with the conditions f0 = 0, f1 = 1 still holds when n is an arbitrary integer.
So: what can f−1 be? From the defining relation, we must have f1 = f0+f−1,
that is, 1 = 0 + f−1: which yields f−1 = 1 = f1.
Could it be that f−n = fn for arbitrary n > 0? Well, it must be f0 =
f−1 + f−2, that is, 0 = 1 + f−2, which yields f−2 = −1 = −f2: so our
hypothesis is disproved.
But could it be that f−n = (−1) n−1 fn for arbitrary n > 0? Well, we have
two consecutive cases where this is true, and the defining relation is of the
second order, so we can just go by induction: suppose f−n = (−1) n−1 fn and
f−(n−1) = (−1) n−2 fn−1. Then, as f−n+1 = f−n + f−n−1, we get
f−(n+1) = f−(n−1) − f−n
= (−1) n−2 fn−1 − (−1) n−1 fn
= (−1) n−2 fn−1 + (−1) n fn
= (−1) n fn+1 ,
because n − 2 and n are either both even or both odd.
Exercise 6.7
Show that Cassini’s identity
fn+1fn−1 − f 2 n = (−1) n
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(1)

is in fact an instance of the more general equation
fn+k = fkfn+1 + fk−1fn , (2)
and prove (2).
Solution. We have just seen that Fibonacci numbers may be defined for
negative index as
f−n = (−1) n−1 fn ,
which preserves the fundamental identity fn = fn−1+fn−2. If we set k = 1−n
in (2), we get
f1 = (−1) n−2 fn−1fn+1 + (−1) n fnfn ,
which is just (1) multiplied by (−1) n .
To prove (2), define the proposition P (n) as
Then P (0) is
∀k ∈ Z : fn+k = fkfn+1 + fk−1fn .
∀k ∈ Z : fk = fkf1 + fk−1f0 ,
which is true because f1 = 1 and f0 = 0, while P (1) is
∀k ∈ Z : fk+1 = fkf2 + fk−1f1 ,
which is true because f1 = f2 = 1. Suppose now that n ≥ 1, and that P (n)
and P (n − 1) are true: then for every k ∈ Z,
fkfn+2 + fk−1fn+1 = fkfn+1 + fkfn + fk−1fn + fk−1fn−1
= (fkfn+1 + fk−1fn) + (fkf(n−1)+1 + fk−1fn−1)
= fn+k + fn−1+k by inductive hypothesis
= fn+1+k
which means that P (n + 1) is true as well: this proves P (n) for every n ≥ 0.
For n < 0, suppose that P (n + 1) and P (n + 2) are true: then for every
k ∈ Z,
fn+k = fn+2+k − fn+1+k
= fkfn+2+1 + fk−1fn+2 − (fkfn+1+1 + fk−1fn+1)
= fk(fn+3 − fn+2) + fk−1(fn+2 − fn+1)
= fkfn+1 − fk−1fn ,
that is, P (n) is true as well.
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Exercise 6.27
Prove the law of greatest common divisor (6.111) for Fibonacci numbers.
Solution. We are required to prove that, for positive m and n,
gcd(fm, fn) = fgcd(m,n) . (3)
We observe that this is true for n = m + 1, because of Cassini’s identity: in
fact, every common factor of fm and fm+1 is also a factor of fm+1fm−1 −f 2 m =
(−1) m , so indeed gcd(fm, fm+1) = 1 = f1 = fgcd(m,m+1).
Let us now prove the general case. Suppose for convenience n > m. By
taking (2) and replace n with n − m and k with m, we get fn = fmfn−m+1 +
fm−1fn−m: then
gcd(fm, fn) = gcd(fn mod fm, fm) = gcd(fm−1fn−m, fm) = gcd(fn−m, fm) ,
because consecutive Fibonacci numbers are relatively prime. But we can
continue subtracting m until n − km becomes smaller than m, that is, until
k = ⌊n/m⌋ and n − km = n mod m: then
gcd(fm, fn) = gcd(fn mod m, fn) .
But the equality above means that we can run the Euclidean algorithm on the
indices of the Fibonacci numbers, instead of the Fibonacci numbers themselves!
The thesis clearly follows.
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