Download Induction 4 Solutions

Fun with Fibonaccis!
Mathematics studies many sequences of numbers—the evens, odds, primes, squares, powers of
two, and so on. Some of the sequences studied are defined by referring back to previous
numbers. This is called a recursive definition. One famous sequence that is defined like this is
the Fibonacci numbers. They start with F0 = 0, and continue with F1 = 1, F2 = 1, F3 = 2, then 3,
5, 8, 13, 21, 34, …. The pattern is that the next number is generated by adding the previous 2.
Formally, F0 = 0, F1 = 0, and then Fk = Fk–1 + Fk–2 for k ≥ 2. For instance, when k = 2, we have
k – 1 = 1 and k – 2 = 0, so that F2 = F1 + F0 = 1 + 0 = 1. Then F3 is found by setting k = 3, to
find that F3 is F2 + F1 = 1 + 1 = 2, and so on.
Since this kind of recursive definition, just like induction, depends on previous values of k,
induction is the tool of choice when proving things about recursively defined sequences of
numbers. For example:
Theorem: Fn < 2n for all n.
Proof: let’s make the base case the lowest number, k = 0, where F0 = 0 < 1 = 20, so the theorem
is true in this case.
Inductive step: If, in fact, we really do know that Fk–1 < 2k–1, well… hmm… we actually need
both Fk–1 and Fk–2 to get at Fk. OK, so let’s say we really do know that Fk–1 < 2k–1 and Fk–2 < 2k–2.
Now we’re in good shape, because Fk = Fk–1 + Fk–2 < 2k–1 + 2k–2 < 2k–1 + 2k–1 = 2⋅2k–1 = 2k. So it
worked!
That worked just fine, except for one thing. Can you spot the fly in the ointment?
What happens when k = 1? The base case covered k = 0. The k = 1 case isn’t covered by the
basis. But it’s not covered by the induction either! This is because F1 is not defined
recursively—the recursive part of the definition starts with k = 2. We would have F1 = F0 + F–1
and there isn’t F–1. Now fortunately F1 = 1 < 2 = 21, so the theorem works here, too. The
important thing is that we checked to make sure!
If your induction needs to go back multiple steps, you need to check multiple base cases—
the same number as the farthest back your induction goes.
Since our Fibonacci induction needs to go back to k – 1 and k – 2, one and two steps back, we
need to check the lowest two cases when we do the base.
Notice something that could have happened. What if we had said G0 = 1, G1 = 6 for our starting
values instead of 0 and 1? Then the sequence would have begun 1, 6, 7, 13, 20, 33, 53, 86, 139.
Notice that our inductive step is still correct! If we really knew that a pair of G’s were less than
2n, then the next one would also be. But this doesn’t happen in the modified sequence until we
get to G6 and G7. Every G after that point is < 2n, but if you don’t make sure to check the base
case(s) you might end up “proving” something that isn’t always true!
Try a few for yourself:
1.
Prove the slightly tougher inequality Fn < (5/3)n.
If we look ahead to the induction, we will be going both one and two steps back, so we need two base cases. These two bases, for k = 0 and 1 are as before: F0 = 0 < 1 = (5/3)0 and F1 = 1 < 5/3 = (5/3)1. Now, our inductive hypothesis will be that both Fk-­‐1 < (5/3)k-­‐1 and Fk-­‐2 < (5/3)k-­‐2. Then we simply add these: Fk = Fk-­‐1 + Fk-­‐2 < (5/3)k-­‐1 + (5/3)k-­‐2. Factor this to get (5/3)k-­‐2⋅(5/3 + 1). But (5/3)k-­‐2⋅(8/3) < (5/3)k-­‐2⋅(25/9) = (5/3)k-­‐2⋅(5/3)2 = (5/3)k. So we find Fk < (5/3)k. That is, if we know that the theorem is true for k – 1 and k – 2, then it is also true for k. Since we started with two base cases, we are now done by induction. 2.
Prove that for any n, Fn + Fn + 3 = 2⋅Fn + 2.
By now, we know that working with Fibonaccis we need two base cases. So for k = 0, we see that F0 + F3 = 0 + 2 = 2 = 2⋅1 = 2⋅F2, and F1 + F4 = 1 + 3 = 4 = 2⋅2 = 2⋅F3. Now if the statement is true when n = k – 1 (i.e. Fk – 1 + Fk + 2 = 2⋅Fk + 1) and when n = k – 2 (Fk – 2 + Fk + 1 = 2⋅Fk), then we use the definition of Fibonacci numbers: Fk + Fk + 3 = (Fk – 1 + Fk – 2) + (Fk + 2 + Fk + 1), which we can rearrange as (Fk – 1 + Fk + 2) + (Fk – 2 + Fk – 1) = 2⋅Fk + 1 + 2⋅Fk = 2⋅(Fk + 1 + Fk) = 2⋅Fk + 2 which is what we wanted! 3.
Prove that for any positive n, Fn+1Fn-1 – Fn2 = (–1)n. (This is called Cassini’s identity.
Hint: in the induction step, you will have to use the defining recursion more than once.)
As usual, we will check two base cases. For k = 1, we have F2F0 – F12 = 1⋅0 – 12 = –1 = (–
1)1. When k = 2, we check F3F1 – F22 = 2⋅1 – 12 = 1 = (–1)2. Note that the formula doesn’t make sense when k = 0 because there is no F–1. Now for the inductive step. Assume it works when n = k – 1, i.e. FkFk-­‐2 – Fk-­‐12 = (–1)k-­‐1. Then Fk+1Fk-­‐1 – Fk2 = (Fk + Fk-­‐1)Fk-­‐1 – Fk (Fk-­‐1 + Fk-­‐2). When these are multiplied out, there are two copies of FkFk-­‐1 that cancel each other out, leaving Fk-­‐12 – FkFk-­‐2 = –(FkFk-­‐2 – Fk-­‐12) which is –(–1)k-­‐1 or (–1)k, which is what we needed. Notice something interesting—we didn’t actually go back two steps! One base case would have sufficed this time. 4.
Prove that F1 + F3 + F5 +  + F2n– 1 = F2n. (There is a similar formula for the sum of
even-index Fibonacci numbers—can you find and prove it?)
Maybe we only need one base this time? Let’s do just one, and if the induction part goes back multiple steps, then we’ll come back and do additional base cases as needed. The base case is when n = 1. Then the sum on the left goes all the way up to F1. So we need to know if F1 = F2? Yes—both are equal to one. Now for the inductive step. Assume it works for k – 1, i.e. F1 + F3 +  F2(k – 1) – 1 = F2(k – 1). Simplifying, F1 + F3 +  + F2k – 3 = F2k – 2. To get to step k, we need F1 + F3 +  F2k – 1 which is just the old left-­‐hand side with F2k – 1 added on. So we add it to the right side as well, to obtain F2k – 2 + F2k – 1 which the definition of Fibonacci numbers tells us is F2k. So we’re done! 5.
Consider the “tribonacci” numbers Tn, where you start the same way (0, 1, 1, …) but
continue by adding three in a row: 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, …. Prove that Tn < (1.9)n.
It’s getting confusing how many base cases we need—so we’ll do the induction part first, see how many steps backward we go, and then do the base cases second, once we know how many we need! So we need to know about Tk which is the sum of three terms: Tk = Tk-­‐1 + Tk-­‐2 + Tk-­‐3. We assume, by induction, that each of these is less than 1.9 to the appropriate power. So we find that Tk = Tk-­‐1 + Tk-­‐2 + Tk-­‐3 < (1.9)k-­‐1 + (1.9)k-­‐2 + (1.9)k-­‐3. Factor to get Tk < (1.9)k-­‐
3
(1.92 + 1.9 + 1). The last term equals 6.51. But 1.93 = 6.859, so Tk < (1.9)k-­‐3(1.9)3 = (1.9)k which is what we needed. Notice we went back three steps. So our three base cases are 0, 1, and 2. But T0 = 0 < 1 = 1.90, T1 = 1 < 1.9 = 1.91, and T2 = 1 < 3.61 = 1.92. 6.
a.
1+ 5
1− 5
and ψ =
.
2
2
Check that each of these solves the equation x2 = x + 1.
Let φ be the crazy number
If you use the quadratic formula on the rearranged equation x2 – x – 1 = 0, it is easy to see that these two numbers are the solutions. φ n −ψ n
. Yes, all those halves and square roots work
5
themselves out to give you an integer! This is called Binet’s formula.
b.
Now prove by induction that Fn =
The base cases are: φ 0 − ψ 0 1− 1
φ −ψ
5
=
= 0 = F0 and =
= 1 = F1 . 5
5
5
5
The induction step looks like: φ k−1 − ψ k−1 φ k−2 − ψ k−2 (φ k−1 + φ k−2 ) − (ψ k−1 + ψ k−2 )
+
=
Fk−1 + Fk−2 =
Now when we factor 5
5
5
we get (φk-­‐1 + φk-­‐2) = φk-­‐2(φ + 1) = φk-­‐2φ2 = φk, and similarly for the ψ term. Substituting we get exactly the formula we want.