9. Aplicacions dels làsers
9.5. Aplicacions Quotidianes
The reader familiar with induction can either omit this section or skim the examples. In this section we highlight how to use induction to complement our combinatorial proof techniques. Sometimes you might first discover the truth of an identity using induc- tion, and then later realize a combinatorial proof. Other times, induction ends up being the only thing that works.
3.2. Mathematical induction 95
The principle of mathematical induction
The principle of mathematical induction provides a sufficient condition to guarantee the truth of a statement that depends on an integer.
Theorem 3.2.1 (mathematical induction) Let n0 be an integer, and supposeS.n/ is a statement involving an integern. If the following two conditions are true, then S.n/ is true for alln > n0.
Base case: S.n0/ is true.
Inductive step: If k is an integer, k > n0, andS.k/ is true, then S.kC 1/ is true. The proof appears at the end of this section.
To use mathematical induction you must verify the base case and the inductive step of the theorem. The base case must involve a demonstration that S.n0/is true. The inductive step is an if-then proof in itself. You must (1) assume k is an integer, k > n0, (2) assume that S.k/ is true, and then (3) prove that S.kC 1/ is true.
It is step (3) where all the work and creativity occur. The truth of S.k/ in step (2) is known as the inductive hypothesis. Every proof by induction must use the inductive hypothesis (abbreviated IHYP in the examples to follow) at some point. If it doesn’t then it’s mostly likely an incorrect proof.
Example #1: partial geometric series
In the sections on generating functions soon to follow we make good use of the identity
1C x C x2C C xnD 1 x nC1
1 x
which holds for any real number x 6D 1 and for any integer n > 0. You have used this formula in calculus to find the partial sum of a geometric series.
Question 101 Compute1 2C 4 8C 16 32C 64 128 using the formula. We’ll prove the theorem by induction on n.
Theorem 3.2.2 Ifx is a real number, x6D 1, then for all n > 0, n X jD0 xj D 1 x nC1 1 x :
Proof by induction onn:Assume that x is a real number, x6D 1. For n > 0, define S.n/ to be the statement S.n/: n X jD0 xj D 1 x nC1 1 x :
When n D 0, the left-hand side of the equation S.0/ isPj0D0xj D x0 D 1 since x0 is defined to be 1 for all real numbers x (including 0). The right-hand side of S.0/ is
1 x0C1
1 x D
1 x
Now assume k is an integer, k > 0, and that S.k/ is true, namely IHYP: k X jD0 xj D 1 x kC1 1 x :
We must prove that S.kC 1/ is true, namely kC1 X jD0 xj D 1 x kC2 1 x :
To do so, start with the left-hand side: kC1 X jD0 xj D 0 @ k X jD0 xj 1
AC xkC1 peel off last term
D 1 x kC1 1 x C x kC1 use IHYP D 1 x kC1C xkC1.1 x/ 1 x common denominator D 1 x kC1C xkC1 xkC2 1 x D 1 x kC2 1 x :
Therefore S.kC 1/ is true. Therefore S.n/ is true for n > 0.
Example #2: proving an inequality
It appears that the inequality
n X jD1
j Š < .nC 1/Š
might be true for any integer n > 1 because
1Š D 1 < 2D 2Š 1ŠC 2Š D 3 < 6D 3Š 1ŠC 2Š C 3Š D 9 < 24D 4Š 1ŠC 2Š C 3Š C 4Š D 33 < 120 D 5Š 1ŠC 2Š C 3Š C 4Š C 5Š D 153 < 720 D 6Š (3.4) is a promising start.
Define S.n/ to be the statement
S.n/: n X jD1
3.2. Mathematical induction 97 To prove by induction, note that the first line of inequalities (3.4) shows that S.1/ is true. Now assume that k is an integer, k > 1, and that S.k/ is true, namely
IHYP: k X jD1
j Š < .kC 1/Š:
We must show thatPjkD1C1j Š < .kC 2/Š. The following calculations do the job: kC1 X jD1 j ŠD 0 @ k X jD1 j Š 1
AC .k C 1/Š peel off last term
< .kC 1/Š C .k C 1/Š by IHYP D 2.k C 1/Š
< .kC 2/.k C 1/Š 2 < kC 2 since k > 1 D .k C 2/Š :
Therefore S.kC 1/ is true, so S.n/ is true for all n > 1.
Example #3: solving a recurrence relation
An example of a recurrence relation is a0D 1
anD 2an 1C n 1 for n > 1.
(3.5)
It governs the iterative computation of the numbers a0; a1; a2; a3; : : :. The value a0D 1 is the initial condition, and then to get the successive values a1; a2; a3; : : :you just apply the rule anD 2an 1C n 1repeatedly: a0D 1 a1D 2a0C 1 1D 2.1/ C 0 D 2 a2D 2a1C 2 1D 2.2/ C 1 D 5 a3D 2a2C 3 1D 2.5/ C 2 D 12 :: : Question 102 What isa6?
To find a100and a1000and so forth, we would like a formula that allows us to jump right to a1000without computing the previous terms. Compute a few more terms and look for a pattern:
n 0 1 2 3 4 5 6 7
They seem to have something to do with powers of 2, specifically a0D 1 D 21 1 a1D 2 D 22 2 a2D 5 D 23 3 a3D 12 D 24 4 a4D 27 D 25 5 a5D 58 D 26 6 a6D 121 D 27 7 a7D 248 D 28 8: Our guess is that anD 2nC1 n 1holds for all n > 0.
To prove this by induction on n, first verify the formula when nD 0. The formula says a0D 20C1 0 1D 1. The recurrence defines a0D 1, so it is correct in this case.
Now assume k is an integer, k > 0, and that ak D 2kC1 k 1; this is IHYP. We must prove that akC1 D 2kC2 .kC 1/ 1, or equivalently that akC1D 2kC2 k 2. Here it is:
akC1 D 2akC k by the recurrence relation D 2.2kC1 k 1/C k by IHYP
D 2kC2 2k 2C k D 2kC2 k 2:
This proves that S.kC 1/ is true. Therefore anD 2nC1 n 1is true for all n > 0.
Example #4: solving a counting problem
Here’s a straightforward yet typical example of how you might use induction in combina- torics. You are trying to count the partitions of Œn into two blocks but don’t see how to jump directly to a formula. Instead, you define pnto be the number of partitions of Œn into two blocks, for n > 2. Using complete enumeration, you find p2 D 1, p3 D 3, p4 D 7, and p5 D 15. For example, the partitions of Œ3 into two blocks are
˚
f1g; f2; 3g ; ˚f2g; f1; 3g ; and ˚f3g; f1; 2g :
Then you discover a combinatorial proof of the identity pnD 2pn 1C 1, for n > 3. Question 103 Give the combinatorial proof.
Starting with p2D 1, you do some computation: p2D 1 p3D 2p2C 1 D 3 p4D 2p3C 1 D 7 p5D 2p4C 1 D 15 p6D 2p5C 1 D 31 p7D 2p6C 1 D 63:
3.2. Mathematical induction 99 The first four values agree with those you found by complete enumeration—good. Also, the pattern looks pretty clear:
pnD 2n 1 1 for n > 2.
In other words, you need to prove that the numbers defined by the recurrence relation p2D 1
pnD 2pn 1C 1 for n > 3 are really just the numbers pnD 2n 1 1, for n > 2.
Question 104 Give a proof by induction, like that of Example #3.
Strong mathematical induction
When using induction, sometimes the truth of S.k/ alone is not strong enough to imply the truth of S.kC 1/. In such cases we can try strong induction. In the induction hypothesis of strong induction, we assume the truth of S.j / for all j between the base value n0and the arbitrary integer k.
Theorem 3.2.3 (strong mathematical induction) Let n0 andn1 be integers,n0 6 n1, and suppose thatS.n/ is a statement involving the integer n. If the following two conditions are true, thenS.n/ is true for all n > n0:
Base case(s): S.n0/; : : : ; S.n1/ are true.
Inductive step: If k is an integer, k > n1, andS.j / is true for all j satisfying n0 6 j 6 k, then S.kC 1/ is true.
Notice that it may be necessary to verify that more than one statement is true in the base case. Since the assumptions are stronger than those of Theorem 3.2.1, essentially the same proof works. (The proof of Theorem 3.2.1 appears at the end of the section.)
Example #5: bounding terms of a recurrence relation
Consider another recurrence relation: L0 D 2 L1 D 1
LnD Ln 1C Ln 2 for n > 2.
(3.6)
This means L2D L1C L0D 3 and L3D L2C L1D 4 and so forth:
n 0 1 2 3 4 5 6 7
Ln 2 1 3 4 7 11 18 29
This is the well-known sequence of Lucas numbers which we will revisit a couple of times in the text. A formula for Lnis not obvious. Later in this chapter we develop a systematic technique that allows us to derive a formula.
But if at first you don’t succeed then lower your standards: sometimes just having an upper bound on the n-th term of a sequence is a useful thing. In this case, one easy upper bound is Ln< 2nwhich appears to hold for all n > 1. At least it is true for 1 6 n 6 7:
n 0 1 2 3 4 5 6 7
Ln 2 1 3 4 7 11 18 29
2n 1 2 4 8 16 32 64 128
(In fact the bound appears too generous. Exercise 9 asks you to prove a tighter bound.) To prove the upper bound, define the statement
S.n/: Ln< 2n:
Our proof is by strong induction. When n D 1, we have L1 D 1 by definition. Also, 21 D 2. It follows that L1 < 21and so S.1/ is true. When nD 2, we have L2 D 3 and 22 D 4. It follows that L
2< 22and so S.2/ is true. (In applying Theorem 3.2.3 we chose n0 D 1 and n1 D 2. The base case(s) portion requires us to show that S.1/ and S.2/ are true, which we just did.)
Now assume that k is an integer, k > 2, and that S.j / is true for all j satisfying 1 6 j 6 k, namely
IHYP: Lj < 2j for all j with 1 6 j 6 k. We must show that S.kC 1/ is true, namely LkC1< 2kC1. Now,
LkC1 D LkC Lk 1 by the recurrence relation < 2kC 2k 1 use IHYP
D 2k 1.2C 1/ D 2k 1 3 < 2k 1 22 D 2kC1:
Therefore LkC1 < 2kC1, and so S.k C 1/ is true. Therefore Ln < 2n for all integers n > 1.
It is very important to understand why we could apply the inductive hypothesis to both Lk and Lk 1in the second line of the calculation above. The reason is that because k is at least 2, then k 1is at least 1. Since the induction hypothesis assumes that S.j / is true for all j satisfying 1 6 j 6 k, we are safe in using both Lk < 2kand Lk 1< 2k 1. Question 105 What happens if you try to proveLn< 2nfor alln > 0 and do not verify the base cases?
Proof of the principle of mathematical induction
The proof of Theorem 3.2.1 uses an axiom called the well-ordering principle.
Axiom 3.2.4 (the well-ordering principle) A nonempty subset of integers that is bounded below contains a least element.
“Bounded below” means that there is some number L such that L 6 x for every integer x in the set. The well-ordering principle does not apply to, say, the set of even integers f0; ˙2; ˙4; : : :g because it is not bounded below.
3.2. Mathematical induction 101 Proof of Theorem 3.2.1:We prove by contradiction. Assume that the two conditions of the theorem are true and yet it is not the case that S.n/ is true for all n > n0. Consider the set of integers
fn0; n0C 1; n0C 2; : : :g:
There is at least one integer in this set for which the statement S is false. Collect all such integers into a set called F , and notice that each element of F is at least n0C 1. This is because S.n0/is true by the base case assumption.
This set F of integers is then nonempty and bounded below, and so the well-ordering principle tells us that it has a least element. Call it m, and notice that m > n0C 1. Then S.m/is certainly false but S.m 1/must be true. This is because m is the least integer in fn0; n0C 1; n0C 2; : : :g that makes the statement false, and also because m 1 > n0.
But the hypothesis of the theorem says then that S .m 1/C 1 D S.m/ is true, contradicting the fact that S.m/ is false! Therefore S.n/ is true for all n > n0.
Summary
Mathematical induction is a technique for proving a statement S.n/ that depends on an integer n. It requires two parts: verification of the base case and a proof of the inductive step. In the inductive step we prove that the truth of S.k/ implies the truth of S.kC 1/. Sometimes the truth of S.k/ alone does not imply the truth of S.kC1/, and so the principle of strong mathematical induction might work. In its inductive step, one assumes the truth of S.j / for all values of j satisfying j 6 k and then proves that S.kC 1/ is true.
Exercises
1. (a) Prove: for n > 0, 3n 1is divisible by 2. (b) Prove: for n > 0, 4n 1is divisible by 3. (c) Find a general theorem and prove it.
2. Let a and b be unequal integers. Prove: for n > 0, an bnis divisible by a b. 3. Prove: for n > 2, n Y jD2 1 1 j2 D nC 1
2n . The product notation means n Y jD2 1 1 j2 D 1 1 22 1 1 32 1 1 42 1 1 n2 :
4. Discover and prove formulas for each of the following products. (a) n Y jD1 1C 1 j (b) n Y jD2 1 1 j
5. Discover and prove a formula for the sum n X jD1
6. Conjecture and prove a formula for n X iD1 i X jD1
j. (You can call this the “Twelve Days of Christmas” formula because when n D 12 the sum equals the total number of gifts given in the song.)
7. Give a combinatorial proof: for n > 1,PjnD1j Š < .nC 1/Š.
Do so by asking a question and then performing a deliberate under- or over-count with one of the answers.
8. The work in (3.4), page 96, suggests thatPjnD1j Š 6 12.nC 1/Š might be true. (a) Prove this sharper inequality by induction.
(b) Give a combinatorial proof.
9. For the recurrence relation shown in (3.6), page 99, we proved Ln< 2nfor n > 1. (a) Prove the tighter inequality Ln6 1:7n. At what value of n should you start the
induction?
(b) What is so special about the number 1.7? Adjust your work in part (a) to create the tightest bound that you can.
10. Define a recurrence relation by a0 D a1 D a2D 1, and anD an 1C an 2C an 3 for n > 3. Prove: an61:9nfor all n > 0. Also, can you prove a tighter bound? 11. Define a0 D 1 and for n > 1, define an D nan 1C 1. Prove: For n > 0, an D
n X jD0
.n/j.
12. Prove: If n is an integer, n > 2, then either n is prime or else can be factored into a product of primes. (This is the fundamental theorem of arithmetic.)
13. Assume the truth of the following statement: if A and B are disjoint, finite sets, thenjA [ Bj D jAj C jBj. Prove the following by induction on n: for n > 2, if A1; A2; : : : ; Anare finite, pairwise disjoint sets, then
ˇ ˇ ˇ ˇ ˇ n [ iD1 Ai ˇ ˇ ˇ ˇ ˇD n X iD1 jAij:
14. Let n > 1. Prove that any 2n 2ncheckerboard with any one square removed can be completely tiled withL-shaped tiles. (The tiles take up three adjacent squares of the checkerboard in anL shape.)