Download Mathematical induction Elad Aigner-Horev

Mathematical induction
Elad Aigner-Horev
§1. Weak induction
We have defined the natural numbers to be the least set N that can be generated using
(N.1) 0 ∈ N; and
(N.2) if n ∈ N then n + 1 ∈ N.
This set has the following property
Well-ordering principle (WOP). Every non-empty set S ⊆ N contains a least element.
§1.1 The definition of weak induction. The definition of N and the WOP give rise to a
powerful proof technique called mathematical induction.
Theorem 1.1. (First principle of finite induction) Let S ⊆ Z+ be a set satisfying:
(I.1) 1 ∈ S; and
(I.2) if k ∈ S then k + 1 ∈ S.
Then S = Z+ .
Proof. Assume towards a contradiction that the claim is false. Then Z+ \ S is non-empty. By the WOP
Z+ \ S contains a least element, namely a. As 1 ∈ S, it follows that a > 1 (recall that 0 ∈
/ Z+ ). By the
minimality of a the number a − 1 ∈
/ T and thus a − 1 ∈ S. By (I.2), we have that a ∈ S contradicting the
assumption that a ∈
/ S.
A more familiar form of Theorem 1.1 is the following.
Theorem 1.2. (Weak mathematical induction) Let S(n) denote a mathematical statement that
depends on n ∈ Z+ .
(Base) If S(1) is true; and
(Step) if whenever S(k) is true then S(k + 1) is true as well,
then S(n) is true for all n ∈ Z+ .
§1.1.1 Jokes. We start with a couple of jokes. For our first joke let us use induction in order to prove
that all positive integers are small.
1. Basis. Surely 1 is a small number.
2. Step. If n is small surely n + 1 is small.
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It follows by induction that all numbers in Z+ are small. Do you agree?
Here is another joke. Let us use induction to prove that all horses in the world have the same colour.
Surely in a set consisting of one horse the claim is true. Suppose then that the statement is true for sets of
horses of size n. We show it holds for n + 1. Let our horses be labelled {1, . . . , n, n + 1}. By the induction
hypothesis the n horses labelled {1, . . . , n} are coloured with the same colour. Also, by the induction
hypothesis, the n horses labelled {2, . . . , n + 1} are coloured with the same colour. It follows that all n + 1
horses are coloured the same. Do you agree?
§1.2 Examples. Let us now demonstrate the use of Theorems 1.1 and 1.2.
Lemma 1.3. For every n ∈ Z+ :
n
X
i=
i=1
n(n + 1)
.
2
Proof. In the practical session.
Lemma 1.4. For every n ∈ Z+ :
n
X
(2i − 1) = n2 .
(1.5)
i=1
Proof. Let S ⊆ Z+ denote the set of positive integers satisfying (1.5). We verify that S satisfied both (I.1)
and (I.2).
Verifying (I.1) As 2 · 1 − 1 = 1 we have that 1 ∈ S. That is we have verified that S satisfies (I.1).
Verifying (I.2) We seek to prove that S satisfies (I.2). To that end, we have to show that whenever
1 ≤ k ∈ S then k + 1 ∈ S as well. The assumption that k ∈ S means
k
X
(2i − 1) = k 2 .
i=1
This assumption is then our induction hypothesis.
k+1
X
k
X
(2i − 1) =
(2i − 1) +(2(k + 1) − 1)
i=1
i=1
|
{z
}
=k2 by I.H.
2
= k + 2k + 1
= (k + 1)2 .
As S satisfies both (I.1) and (I.2) we have that S = Z+ , by Theorem 1.1.
Lemma 1.6. For every n ∈ Z+ :
n
X
i=1
i2 =
n(2n + 1)(n + 1)
.
6
(1.7)
Proof. Let S ⊆ Z+ be the set of positive integers satisfying (1.7). As 12 = 1·(2·1+1)(1+1)
= 66 we have
6
that 1 ∈ S so that S satisfies (I.1). It remains to verify (I.2). Suppose then that 1 ≤ k ∈ S, we show that
k+1
X
i=1
i2 =
k
X
i=1
i2 + (k + 1)2 =
(k + 1)(2(k + 1) + 1)((k + 1) + 1)
(k + 1)(2k + 3)(k + 2)
=
.
6
6
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By the assumption that k ∈ S this reduces to showing that
k(2k + 1)(k + 1)
(k + 1)(2(k + 1) + 1)((k + 1) + 1)
+ (k + 1)2 =
.
6
6
This is indeed true:
k(2k + 1)(k + 1)
+ (k + 1)2 = (k + 1)
6
k(2k + 1)
+ (k + 1)
6
k(2k + 1) + 6(k + 1)
= (k + 1)
6
2
2k + 7k + 6
= (k + 1)
6
(2k + 3)(k + 2)
= (k + 1)
6
The proof of (1.7) was a bit tedious. Indeed, we had to resort to some annoying algebraic manipulations.
A weaker statement that would capture the spirit of (1.7) would be that
there exists a C ∈ R such that
n
X
i2 ≥ Cn3 for any sufficiently large n ∈ Z+ .
(1.8)
i=1
3
2
To see (1.8), let us consider the r.h.s. of (1.7); this is equal to n3 + n2 + n6 . As n → ∞ we see that n3
3
2
becomes the dominant term in n3 + n2 + n6 allowing us to "neglect" the other terms. We now make this
precise.
Claim 1.9. For every ε > 0 there exists an n0 = n0 (ε) such that n2 ≤ εn3 whenever n ≥ n0 .
Proof. Given ε we set
1
n0 =
.
ε
We prove by induction on n that n2 ≤ εn3 whenever n ≥ n0 . For the induction basis set n = n0 and note
that n20 < εn30 . Assuming that the claim holds for n ≥ n0 we prove that it holds for n + 1. That is we
prove that (n + 1)2 ≤ ε(n + 1)3 whenever n ≥ n0 .
ε(n + 1)3 = εn3 + 3εn2 + 3εn + ε
≥ n2 + 3εn2 + 3εn + ε,
by the induction hypothesis
≥ n2 + 3εn2
3
≥ n2 + n2 ,
n
as n ≥ n0 ≥
1
ε
so that ε ≥ n1 ;
= n2 + 3n
≥ n2 + 2n + 1
= (n + 1)2 ,
and the claim follows.
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In a similar manner we have the following claim proof of which is left to the reader.
Claim 1.10. For every ε > 0 there exists an n0 = n0 (ε) such that n ≤ εn3 whenever n ≥ n0 .
By Claims 1.9 and 1.10 we have that there exists an n0 such that
we have that in the sum
n3 n2 n
+
+
3
2
6
n3
1010
≥
n2
2
+ n6 for all n ≥ n0 . As a result
3
the term n3 is the dominant one as stated above.
We have verified (1.8) using (1.7). The analysis above serves as a justification as to why C as in (1.8)
exists. Suppose now that we had no knowledge of (1.7); how can we come up with C in this case?
Let us try to prove (1.8) directly
P using induction. Let us start with the induction step. Suppose that
the inductive hypothesis is that ni=1 i2 ≥ Cn3 where C is yet to be determined whenever n ≥ n0 where
n0 is also yet to be determined. The burden of proof in the induction step would then be to show that
n
X
i2 +(n + 1)2 ≥ C(n + 1)3
i=1
| {z }
≥Cn3
so that
Cn3 + (n + 1)2 ≥ Cn3 + 3Cn2 + 3Cn + C.
We see that the term Cn3 cancels. We are left with
n2 + 2n + 1 ≥ 3Cn2 + 3Cn + C
which holds providing that the following three conditions hold for C
1
(by the n2 term)
3
2
C≤
(by the n term)
3
C ≤ 1 (by the constant term).
C≤
The choice C = 1/3 makes sense then in the induction step. It remains to find n0 . But this turns out to
3
be quite easy as 12 ≥ 13 so we can take n0 = 1 as usual. Indeed, this is not surprising in view of (1.7).
We have just proved the following lemma.
Lemma 1.11. There exists a C ∈ R such that for every n ∈ Z+
n
X
i2 ≥ Cn3
i=1
holds.
The quantification in Lemma 1.11 is ∃C ∈ R such that ∀n ∈ Z+ . Suppose we were to change this
quantification to ∀n ∈ Z+ ∃C ∈ R would that have any effect?
Lemma 1.12. For every n ∈ Z+
n
X
i=1
i3 =
n2 (n + 1)2
4
holds.
Proof. In the practical session
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§1.2.1 The n! function. Mathematical induction is often used in definitions. For instance, the definition
of n! reads as follows.
Definition 1.13. Let n ∈ Z+ . Then n! is given by the following.
(a) 1! = 1;
(b) n! = n · (n − 1)!.
We call such definitions recursive definition. A trivial upper bound on n! is the following.
Lemma 1.14. n! ≤ nn for every n ∈ Z+ .
Proof. The proof is by induction on n.
Basis. For n = 1 the assertion is true.
Step: Suppose the claim is true for 1 ≤ n ∈ Z+ . We show that the assertion also holds for n + 1.
(n + 1)! = (n + 1) · n!
≤ (n + 1)nn (by the induction hypothesis)
< (n + 1)(n + 1)n (as n ≥ 1)
= (n + 1)n+1 .
To get a sense for the rate of growth of n! let us compare it with the function 2n . For small values of
n we see that 2n prevails over n!:
1! = 1 < 21 = 2
2! = 2 < 22 = 4
3! = 6 < 23 = 8.
But at n = 4 we have that 4! = 24 > 24 = 16. This in fact holds for all n ≥ 4.
Lemma 1.15. n! ≥ 2n for all n ≥ 4
Proof. We prove the assertion by induction on n. The basis of the induction we have already seen above.
We proceed to the induction step. Suppose then that the claim is true for n ≥ 4. We show that it holds
for n + 1.
(n + 1)! = n!(n + 1)
≥ 2n (n + 1) (by I.H.)
≥ 2n · 2 (n ≥ 4 =⇒ n + 1 ≥ 2)
≥ 2n+1 .
§1.2.2 Powers of 2. The function 2n is a fast growing function. Can we quantify "fast"? We will now
show that this function grows so fast that 2n is larger than the sum of all the powers of 2 preceding it.
Lemma 1.16. For all n ∈ Z+
n
X
2i ≤ 2n+1
i=1
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Elad Aigner-Horev
holds.
Proof. The proof is by induction on n. For n = 1 the assertion holds.
then that the claim holds
Pn+1Assume
i
n+2
. This is indeed true as
for n ≥ 1. We show that it holds for n + 1. That is, we show that i=1 2 ≤ 2
n+1
X
i
2 =
i=1
n
X
2i +2n+1 ≤ 2 · 2n+1 = 2n+2 .
i=1
| {z }
≤2n+1
§1.2.3 Stronger assertions. The following lemma demonstrates an important principle. This being
the fact that sometimes one has to prove a stronger assertion
Pn 1 than the one intended. Let us consider the
sum of the reciprocals of the first n squares, namely i=1 i2 and let us try to prove that
n
X
1
≤ 2.
i2
i=1
A naive attempt would be to have the latter as our induction hypothesis. However this will clearly end in
failure. Indeed, in the induction step we would have to prove that
n
X
1
1
+
≤ 2.
2
i
(n + 1)2
i=1
| {z }
≤2
We see that the inductive hypothesis is simply too weak. The problem is that we did not make room for
1
in the inductive hypothesis.
(n+1)2
Lemma 1.17. For every n ∈ Z+
n
X
1
≤2
i2
i=1
holds.
Proof. We will prove a stronger assertion. Namely that for every n ∈ Z+
n
X
1
1
≤2−
i2
n
(1.18)
i=1
holds.
We prove (1.18) by induction on n. For n = 1 note that 11 = 1 ≤ 2 − 11 = 2 − 1 = 1. Assume then
that (1.18) holds for n, we show it holds for n + 1. By the induction hypothesis we have
n
X
1
1
1
1
+
≤2− +
.
i2 (n + 1)2
n (n + 1)2
i=1
1
1
1
1
To conclude it remains to show that 2 − n1 + (n+1)
2 ≤ 2 − n+1 . That is, (n+1)2 + n+1 =
now that n(n + 2) = n2 + 2n while (n + 1)2 = n2 + 2n + 1. The claim then follows.
n+2
(n+1)2
≤ n1 . Note
How did we come up with 2 in Lemma 1.17? Can we replace it with a smaller number? What about
1? Let us try to prove that
n
X
1
1
≤1− .
2
i
n
i=1
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For the induction step we have
n
X
1
1
1
1
1
+
≤1− +
≤1−
.
2
2
2
i
(n + 1)
n (n + 1)
(n + 1)2
i=1
So at this point we might be inclined to believe that replacing 2 with 1 is possible. However, the fact that
we can prove the induction step is meaningless unless we can find
P some n from which we can start. Here,
there cannot be such an n. Indeed, for every n ∈ Z+ we have ni=1 i12 ≥ 1 as the first element in the sum
is 1. On the other hand 1 − n1 < 1 for every n ∈ Z+
§1.2.4 Versatility of induction. So far we we used induction to prove inequalities. Inductions are
much more versatile than this.
Lemma 1.19. For all n ∈ Z+ : 3 n3 + 2n.
Proof. The proof is by induction on n. For the induction basis let us note that for n = 1 we have that
13 + 2 · 1 = 3 and so the claim holds for
n = 1. We proceed to the induction step. Suppose then that
3 n3 + 2n and we seek to show that 3 (n + 1)3 + 2(n + 1). Expand (n + 1)3 to n3 + 3n2 + 3n + 1 and
obtain
(n + 1)3 + 2(n + 1) = n3 + 3n2 + 5n + 3 = (n3 + 2n) + (3n2 + 3n + 3).
Surely 3 3(n2 + n + 1). Furthermore, 3 (n3 + 2n) by the induction hypothesis. The claim follows. §1.2.5 Multiple invocations of the inductive hypothesis. In all of the examples so far we always
invoked the inductive hypothesis only once. In fact, we are allowed to invoke it more than once. Here is
an example of that. For a set X let P(X) denote the set comprised of all subsets of X, that is
P(X) = {Y : Y ⊆ X}.
We refer to P(X) as the power set of X. For instance, if X = {1, 2} then P(X) = {∅, {1}, {2}, {1, 2}}.
Lemma 1.20. Let X be a finite set of size n. Then |P(X)| = 2n .
Proof. The proof is by induction on n. For any set of size 1, say {1} we have that P({1}) = {∅, {1}}.
Hence |P({1})| = 21 . Assume then that |P({1, . . . , n})| = 2n . We show that the claim holds for n + 1.
The subsets of {1, . . . , n, n + 1} can be partitioned into two types. Those that contain n + 1 and those that
do not. Any subset not containing n + 1 is a subset of {1, . . . , n}. Hence, there are 2n sets of this type,
by the induction hypothesis. Any subset containing n + 1 has the form A ∪ {n + 1} where A ⊆ {1, . . . , n}.
The number of choices for A then is 2n , by the inductive hypothesis. Altogether the number of subsets
we have accounted for is 2n + 2n = 2n+1 .
§2. Strong/complete induction
The step of a weak induction has the form S(n) =⇒ S(n + 1) where S(n) is assumed to be true (see
Theorems 1.1 and 1.2). There is an additional type of induction which allows us to assume S(1), . . . , S(n)
and using all of these (or perhaps just a part of them) prove that S(n + 1) is true. This form of induction
is called strong or complete induction. We have the following two analogous versions of Theorems 1.1
and 1.2.
Theorem 2.1. (The second principle of finite induction)
Let S ⊆ N be a set satisfying:
(S.1) 0 ∈ S; and
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(S.2) if {0, 1, . . . , n} ⊆ S then n + 1 ∈ S.
Then S = N.
Theorem 2.2. (Strong/Complete mathematical induction) Let S(n) denote a mathematical statement that depends on n ∈ Z+ . In addition, let n0 , n1 ∈ Z+ satisfy n0 ≤ n1 .
(Base) If S(n0 ), S(n0 + 1), . . . , S(n1 ) are all true; and
(Step) if whenever S(n0 ), S(n0 + 1), . . . , S(k − 1), S(k) are true then S(k + 1) is true as well,
then S(n) is true for all n ≥ n0 .
(Strong Base) It is a bit difficult to motivate at this stage why the basis of this induction has this specific
form. Below we shall consider examples that will clarify this point.
It is clear that anything that can be proved using weak induction can be proved using strong induction.
The converse holds as well. In fact, these two techniques are equivalent.
Theorem 2.3. The principles of weak and strong inductions are equivalent.
We omit the proof of this result here. Let us however mention the following.
Theorem 2.4. The WOP and the principles of weak induction and strong induction are all equivalent.
So far we have seen that WOP implies the principle of weak induction (see the proof Theorem 1.1). In
the practical session, it will be shown that strong induction implies the WOP.
§2.1 Examples. Let us demonstrate how Theorem 2.2 can be useful.
§2.1.1 Cauchy’s inequality. One of the most heavily used inequalities in mathematics is the CauchySchwartz inequality. This inequality for real sequences reads as follows.
Lemma 2.5. Let a1 , . . . , an and b1 , . . . , bn be two sequences of real numbers. Then
n
X
!2
ai bi
≤
i=1
n
X
i=1
a2i ·
n
X
b2i
i=1
Proof. The proof is by induction on n. For n = 1 the assertion is trivial formally this is enough. However,
for pedagogical reasons we consider n = 2. Here we need to show that
(a1 b1 + a2 b2 )2 ≤ (a21 + a22 )(b21 + b22 ).
(2.6)
Expand both sides to obtain
(a1 b1 )2 + 2a1 a2 b1 b2 + (a2 b2 )2 ≤ (a1 b1 )2 + (a1 b2 )2 + (a2 b1 )2 + (a2 b2 )2
which is equivalent to
0 ≤ (a1 b2 )2 − 2a1 a2 b1 b2 + (a2 b1 )2 = (a1 b2 )2 − 2(a1 b2 )(a2 b1 ) + (a2 b2 )2 = (a1 b2 − a2 b1 )2
The term (a1 b2 −a2 b1 )2 is clearly non-negative due to the square. This completes the basis of the induction.
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We proceed to the induction step. For n ∈ Z+ let H(n) denote the assertion that Cauchy’s inequality
is true for sequences of length n. Assuming that H(1), . . . , H(n) hold, we show that H(n + 1) follows. In
fact, we will show that H(2) and H(n) imply H(n + 1). By H(n) we have that
v
v
u n
u n
n
X
uX uX
2
t
ai bi + an+1 bn+1 ≤
ai t
b2i + an+1 bn+1
(2.7)
i=1
Put α :=
have
qP
n
2
i=1 ai and β :=
i=1
qP
n
2
i=1 bi
i=1
so that the r.h.s. of (2.7) becomes αβ + an+1 bn+1 . By (2.6) we
αβ + an+1 bn+1 ≤ (α2 + a2n+1 )1/2 (β 2 + b2n+1 )1/2
and the claim follows.
§2.1.2 Lucas numbers. The Lucas numbers are given by the sequence
1, 3, 4, 7, 11, 18, 29, 47, 76, . . . .
Alternatively, these can be defined recursively as follows.
Definition 2.8. The Lucas numbers `n are given by
`1 = 1
`2 = 3
`n = `n−1 + `n−2 ,
Lemma 2.9. For all n ∈ Z+
∀n ≥ 3.
n
7
`n <
4
holds.
Proof. The proof is by induction on n. For n = 1, 2 the assertion holds. (In fact we must insist on
verifying the induction basis for both n = 1 and n = 2). This is because in the induction step we shall
invoke the hypothesis once for `n−1 and once for `n−2 . This is the point that we raised in (Strong Base).
n−1
n−2
Assume then that for n ≥ 3 we have that `n−1 < 74
and `n−2 < 74
. Then
`n = `n−1 + `n−2
Note now that
11
4
<
7 2
4
n−1 n−2
7
7
<
+
4
4
n−2 7
7
=
1+
4
4
n−2 7
11
=
.
4
4
(indeed, 16 · 11 < 49 × 4) and so we have
n
7
<
4
and the claim follows.
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It the sequel it will be convenient to have `0 also defined.
Definition 2.10. The extended Lucas numbers Ln are given by
L0 = 2
L1 = 1
Ln = Ln−1 + Ln−2 ,
∀n ≥ 2.
§3. Fibonacci numbers
Perhaps the most famous recursively defined sequence is that of Fibonacci that first appeared in
Fibonacci’s book Liber Abaci in 1202. Fibonacci was faced with the following problem involving rabbits.
Suppose one places a pair of rabbits one of each sex on an island. Assuming that rabbits do not breed
until these are two months old, and assuming that each pair produces another pair each month, how many
pairs of rabbits does one have after n months?
Let fn denote the number of pairs of rabbits on the island after n months. Then f1 = 1 and also
f2 = 1 as we have to wait two months for the first breeding. On the third month we have two pairs so
f3 = 2. More generally, fn = fn−1 + fn−2 because for each newly born pair we must wait two months.
Definition 3.1. The Fibonacci sequence is defined as follows.
f1 = 1
f2 = 1
fn = fn−1 + fn−2 ,
∀n ≥ 3.
With this definition we have
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144
as the first twelve numbers of the Fibonacci sequence.
We notice that apart from their initialisations the Lucas sequence (see § 2.1.2) and the Fibonacci
sequence have the same definition. Indeed, it was Lucas who named the Fibonacci sequence after Fibonacci
and discovered many of its properties. To clarify the connection between these sequences let us first extend
the Fibonacci sequence to have f0 defined.
Definition 3.2. The extended Fibonacci sequence is defined as follows.
F0 = 0
F1 = 1
Fn = Fn−1 + Fn−2 ,
∀n ≥ 2.
Lemma 3.3. For all n ∈ Z+
Ln = Fn−1 + Fn+1
holds.
Proof. The proof is by induction on n. For the induction base note that
L1 = 1 = 0 + 1 = F0 + F2 = F1−1 + F1+1
L2 = 3 = 1 + 2 = F1 + F3 = F2−1 + F2+1
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so the claim holds for both n = 1, 2 (recall (Strong Base)).
Suppose then that Ln = Ln−1 + Ln+1 for n = 1, . . . , k − 1, k, k ≥ 2, and let us consider Lk+1 .
Lk+1 = Lk + Lk−1
so by the induction hypothesis
= (Fk−1 + Fk+1 ) + (Fk−2 + Fk )
rearranging we get
= (Fk−2 + Fk−1 ) + (Fk + Fk+1 )
= Fk + Fk+2
and the claim follows.
§3.1 Growth of Fibonacci numbers. How fast P
does the Fibonacci sequence grow? In § 1.2.2
n
we have seen that the function 2 grows in the sense that ni=1 2i ≤ 2n+1 . For the Fibonacci sequence a
similar phenomenon occurs.
Lemma 3.4. For all n ∈ Z+
n
X
fi = fn+2 − 1
i=1
holds.
Proof. The proof is by induction on n (a non-inductive proof will be presented in the practical session).
For n = 1 note that f1 = 1 and that f1+2 − 1 = f3 − 1 = 2 − 1 = 1.
Assume then that the claim holds for n ≥ 1, that is
n
X
fi = fn+2 − 1;
i=1
we show that it holds for n + 1, that is
n+1
X
fi = fn+3 − 1.
i=1
Indeed, we have
n
X
fi +fn+1 = fn+2 − 1 + fn+1 = fn+3 − 1
i=1
| {z }
I.H.
and the claim follows.
With Lemma 3.4 we might begin to suspect that fn and 2n are rather close. This is misleading.
Lemma 3.4 is somewhat superficial and with it we cannot really estimate the rate of growth of fn . The
next lemma does.
√
Lemma 3.5. Let α = (1 + 5)/2 ≈ 1.61803398875.
fn > αn−2
for all n ≥ 3.
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Proof. The proof is by induction on n. For the induction basis we verify the claim for n = 3 and n = 4.
Indeed, here we see that α3−2 = α < 2 = f3 and that α4−2 = α2 ≈ 2.61803398875 < 3 = f4 .
Let n > 4 and assume that fk > αk−2 for all 3 ≤ k < n. We consider fn , and by the induction
hypothesis we have
fn = fn−1 + fn−2 > αn−3 + αn−4 .
We seek to show that αn−3 + αn−4 ≥ αn−2 . To that end note that
αn−3 + αn−4 = (α + 1)αn−4
Next, we note that α + 1 = α2 . Indeed, α is a solution to x2 − x − 1 = 0 (or put another way, x2 = x + 1)
and so we have that
(α + 1)αn−4 = α2 αn−4 = αn−2
and the claim follows.
To distinguish fn from 2n we prove the following.
Lemma 3.6. For all n ≥ 11
fn < 2n−4
holds
Proof. The proof is by induction on n. For the induction basis we consider n = 11 and n = 12. Indeed
here we see that 211−4 = 128 > 89 = f11 and that 212−4 = 256 > 144 = f12 .
Let n ≥ 13 and assume that fk < 2k−4 for all 11 ≤ k ≤ n − 1. For fn we have
fn = fn−1 + fn−2
< 2n−5 + 2n−6
= 2n−6 · (2 + 1)
< 2n−6 · 22
= 2n−4 .
The claim follows.
So far we were able to completely distinguish fn from 2n as follows.
Corollary 3.7. For all n ≥ 11
holds.
In fact much more is known. Let β = (1 −
αn−2 < fn < 2n−4
√
5)/2.
Theorem 3.8. For all n ∈ Z+
fn =
αn − β n
√
5
holds.
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