One way of looking at the principle of mathematical induction is that it tells us that if we know the “first” case of a theorem and we can derive each other case of the theorem from a smaller case, then the theorem is true in all cases. However, the particular way in which we stated the theorem is rather restrictive in that it requires us to derive each case from the immediately preceding case. This restriction is not necessary, and removing it leads us to a more general statement of the principle of mathematical induction which people often call the strong principle of mathematical induction. It states:
In order to prove a statement about an integer nif we can 1. prove our statement whenn=b and
2. prove that the statements we get with n = b, n = b+ 1, . . .n=k−1 imply the statement withn=k,
then our statement is true for all integers n≥b.
367. What postage do you think we can make with five and six cent stamps? Do you think that there is a number N such that if n≥N, then we can make ncents worth of postage?
You probably see that we can make n cents worth of postage as long as n is at least 20. However, you didn’t try to make 26 cents in postage by working with 25 cents; rather you saw that you could get 20 cents and then add six cents to that to get 26 cents. Thus if we want to prove by induction that we are right that ifn≥20, then we can make ncents worth of postage, we are going to have to use the strong version of the principle of mathematical induction.
We know that we can make 20 cents with four five-cent stamps. Now we let k be a number greater than 20, and assume that it is possible to make any amount between 20 and k−1 cents in postage with five and six cent stamps. Now if k is less than 25, it is 21, 22, 23, or 24. We can make 21 with three fives and one six. We can make 22 with two fives and two sixes, 23 with one five and three sixes, and 24 with four sixes. Otherwise k−5 is between 20 andk−1 (inclusive) and so by our inductive hypothesis, we know that k−5 cents can be made with five and six cent stamps, so with one more five cent stamp, so cank cents. Thus by the (strong) principle of mathematical induction, we can make n cents in stamps with five and six cent stamps for eachn≥20.
B.2. STRONG INDUCTION 165 Some people might say that we really had five base cases,n= 20, 21, 22, 23, and 24, in the proof above and once we had proved those five consecutive base cases, then we could reduce any other case to one of these base cases by successively subtracting 5. That is an appropriate way to look at the proof. In response, a logician might say that it is also the case that, for example, by proving we could make 22 cents, we also proved that if we can make 20 cents and 21 cents in stamps, then we could also make 22 cents. We just didn’t bother to use the assumption that we could make 20 cents and 21 cents! On the other hand a computer scientist might say that if we want to write a program that figures out how to maken cents in postage, we use one method for the cases n= 20 to n= 24, and then a general method for all the other cases. So to write a program it is important for us to think in terms of having multiple base cases. How do you know what your base cases are? They are the ones that you solve without using the inductive hypothesis. So long as one point of view or the other satisfies you, you are ready to use this kind of argument in proofs.
368. A number greater than one is called prime if it has no factors other than itself and one. Show that each positive number is either a power of a prime (remember what p0 and p1 are) or a product of powers of prime numbers.
369. Show that the number of prime factors of a positive number n≥2 is less than or equal to log2n. (If a prime occurs to the kth power in a factorization of n, you can consider that power ask prime factors.) (There is a way to do this by induction and a way to do it without induction. It would be ideal to find both ways.)
370. One of the most powerful statements in elementary number theory is Euclid’s Division Theorem2. This states that if m and n are positive integers, then there are unique nonnegative intergers q and r with 0≤r < n, such thatm=nq+r. The numberq is called the quotient and the number r is called the remainder. In computer science it is
2
In a curious twist of language, mathematicians have long called The Division Algo- rithm or Euclid’s Division Algorithm. However as computer science has grown in impor- tance, the word algorithm has gotten a more precise definition: an algorithm is now a method to do something. There is a method (in fact there are more than one) to get the
qandrthat Euclid’s Division Theorem gives us, and computer scientists would call these methods algorithms. Your author has chosen to break with mathematical tradition and restrict his use of the word algorithm to the more precise interpretation as a computer scientist probably would. We aren’t giving a method here, so this is why the name used here is “Euclid’s Division Theorem.”
common to denote r by mmodn. In elementary school you learned how to use long division to find q and r. However, it is unlikely that anyone ever proved for you that for any pair of positive intgers,mand n, there is such a pair of nonnegative numbers q andr. You now have the tools needed to prove this. Do so. Online hint.
Appendix C
Exponential Generating
Functions
C.1
Indicator Functions
When we introduced the idea of a generating function, we said that the
formal sum ∞
X
i=0 aixi
may be thought of as a convenient way to keep track of the sequenceai. We
then did quite a few examples that showed how combinatorial properties of arrangements counted by the coefficients in a generating function could be mirrored by algebraic properties of the generating functions themselves. The monomials xi are called indicator polynomials. (They indicate the position of the coefficient ai.) One example of a generating function is given by
(1 +x)n= ∞ X i=0 n i ! xi.
Thus we say that (1 +x)n is the generating function for the binomial coef- ficients ni
. The notation tells us that we are assuming that only i varies in the sum on the right, but that the equation holds for each fixed integer n. This is implicit when we say that (1 +x)n is the generating function for
n i
, because we haven’t writtenianywhere in (1 +x)n, so it is free to vary. Another example of a generating function is given by
xn= ∞ X i=0 s(n, i)xi. 167
Thus we say that xn is the generating function for the Stirling numbers of the first kind,s(n, i). There is a similar equation for Stirling numbers of the second kind, namely
xn=
∞
X
i=0
S(n, i)xi.
However, with our previous definition of generating functions, this equation would not give a generating function for the Stirling numbers of the second kind, becauseS(n, i) is not the coefficient ofxi. If we were willing to consider the falling factorial powers xi as indicator polynomials, then we could say thatxn is the generating function for the numbers S(n, i) relative to these
indicator polynomials. This suggests that perhaps different sorts of indicator polynomials go naturally with different sequences of numbers.
The binomial theorem gives us yet another example. ◦371. Write (1 +x)n as a sum of multiples of xi
i! rather than as a sum of multiples of xi.
This example suggests that we could say that (1 +x)n is the generating function for the falling factorial powersni relative to the indicator polynomi- als xi!i. In general, a sequence of polynomials is called a family ofindicator polynomialsif there is one polynomial of each nonnegative integer degree in the sequence. Those familiar with linear algebra will recognize that this says that a family of indicator polynomials forms a basis for the vector space of polynomials. This means that each polynomial can be expressed as a sum of numerical multiples of indicator polynomials in one and only one way. One could use the language of linear algebra to define indicator polynomials in an even more general way, but a definition in such generality would not be useful to us at this point.