MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS

Transcription

1 1 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 1 INSTITIÚID TEICNEOLAÍOCHTA CHEATHARLACH INSTITUTE OF TECHNOLOGY CARLOW MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 1 Introduction Recurrence relations, recursive algorithms, and mathematical induction are closely related. In all three, prior instances of the current case are assumed known. A recurrence relation uses prior values in a sequence to complete the current value. A recursive algorithm uses smaller instances of the current input to process the current input. The inductive step in a proof by mathematical induction assumes the truth of prior instances of the statement to prove the truth of the current statement. The Principle of Mathematical Induction In order to prove that certain results hold for all positive integers, the Principle of Mathematical Induction is often employed. These results may, for example, be summation formulae or results from number theory relating to divisibility. We will also illustrate how the principle is used to establish the truth of inequalities. Firstly, some preliminary information..1 Integers and Natural Numbers An integer is a whole number. We denote the the set (or collection) of integers by Z. (The word for number in German is Zahl.) Z = {... 3,, 1, 0, 1,, 3,...}

2 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS We specify important subsets of this set. The non-negative integers are Z + = {0, 1,, 3,...}. The non-positive integers are Z = {0, 1,, 3,...}. The positive integers are Z + 0 = {1,, 3, 4,...} The negative integers are Z 0 = { 1,, 3, 4,...} The natural numbers are the positive integers 1,,3,4,... N = {1,, 3, 4,...} In some books 0 appears in this set. Note that the terms natural number and positive integer refer to the same objects.. The Well-Ordered Principle Every nonempty set of positive integers contains a smallest member. So, clearly, the set N is well ordered as are the sets Z + and Z + 0. However, the sets Q+ 0 or R+ 0 do not have a smallest element and hence are not well ordered. The principle of mathematical induction is based on this property of subsets..3 Introduction to the Principle of Mathematical Induction For each n N, let S n, denote the sum of the first n (positive) odd numbers. Calculating S 1, S, S 3, S 4, S 5 we find S 1 = 1 = 1 S = = 4 S 3 = = 9 S 4 = = 16 S 5 = = 5 You may notice a pattern beginning to emerge. Does this pattern continue? Suppose that we see whether or not the pattern continues to S 6. Adding up, we find S 6 = = 36

3 3 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 3 We are thus led to the conjecture that S n = n for all natural numbers n? Can we prove this? If so, how? Merely testing the proposition for a few values of n, no matter how many, cannot in itself suffice to prove that the proposition holds for all natural numbers n. Some propositions may turn out to be true in a very large number of cases, and yet fails for others. Such a proposition is the following: n < 1, 000, 000, 000 This proposition holds for a very large number of natural numbers n, (for 999,999,999 of them to be precise), yet it obviously fails to hold for all natural numbers n. A strategy for proving that some conjecture holds for all natural numbers n is the Principle of Mathematical Induction. Theorem 1 Let P (n) denote some proposition. i. Verify statement P (n) is true for the smallest value of n (usually 1). ii. Assume the statement P (n) is true for some value k < n and prove it also true for k + 1. iii. If P (1) holds, then by point ii, P () holds and so on up to n. To understand the justification for the Principle of Mathematical Induction, consider the following. For each natural number n, let P (n) denote a proposition (that is either true or false). We suppose that we have proved that P (1) is true, and that if P (k) is true then P (k + 1) is true. Now P (1) is true. If P (1) is true then P () is true. Moreover P (1) is true. Therefore P () is true. If P () is true then P (3) is true. Moreover P () is true. Therefore P (3) is true. If P (3) is true then P (4) is true. Moreover P (3) is true. Therefore P (4) is true. : : : : : : If P (n ) is true then P (n 1) is true. Moreover P (n ) is true.

4 4 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 4 Therefore P (n 1) is true. If P (n 1) is true then P (n) is true. Moreover P (n 1) is true. Therefore P (n) is true. The pattern exhibited in these statements should convince you that P (n) is true for any natural number n, no matter how large..4 Examples Summation Formulae Example We claim that n i = n(n + 1) for all n N, where n i = n We prove this result using the Principle of Mathematical Induction. Let P (n) = n(n + 1) We can verify statement P (n) is true for the smallest value of n i.e., n = 1. So P (1) = 1(1 + 1) = 1 Therefore P (1) is true. Now assume the statement P (n) is true for some value k < n and prove it also true for k + 1. So we assume the following statement to be true P (k) = k i = k(k + 1) We must now prove that the following statement is true k+1 P (k + 1) = i = (k + 1)(k + )

5 5 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 5 We can do so as follows k+1 P (k + 1) = i = = = = = = [ ] k + (k + 1) k i + (k + 1) k(k + 1) + (k + 1) k(k + 1) + (k + 1) (k + 1)(k + ) k+1 i We can now conclude from the Principle of Mathematical Induction that P (n) is true for all natural numbers, which is the result we set out to prove. Example We claim that for all n N, where n i = n(n + 1)(n + 1) 6 n i = n We prove this result using the Principle of Mathematical Induction. Let P (n) = n(n + 1)(n + 1) 6 We can verify statement P (n) is true for the smallest value of n i.e., n = 1. So P (1) = 1(1 + 1)( + 1) 6 = 1 Therefore P (1) is true. Now assume the statement P (n) is true for some value k < n and prove it also true for k + 1. So we assume the following statement to be true P (k) = k i = k(k + 1)(k + 1) 6

6 6 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 6 We must now prove that the following statement is true We can do so as follows k+1 P (k + 1) = i = k+1 P (k + 1) = i = = (k + 1)(k + )(k + 3) 6 [ k ] + (k + 1) k i + (k + 1) k(k + 1)(k + 1) = + (k + 1) 6 k(k + 1)(k + 1) + 6(k + 1) = [ 6 ] (k + 1) k(k + 1) + 6(k + 1) = 6 = (k + 1)(k + 7k + 6) 6 (k + 1)(k + )(k + 3) = 6 k+1 = i We can now conclude from the Principle of Mathematical Induction that P (n) is true for all natural numbers, which is the result we set out to prove. Example We claim that for all n N, where n i(i + 3) = n(n + 1)(n + 5) 3 n i(i + 3) = n(n + 3) We prove this result using the Principle of Mathematical Induction. Let P (n) = n(n + 1)(n + 5) 3 We can verify statement P (n) is true for the smallest value of n i.e., n = 1. So

7 7 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 7 P (1) = 1(1 + 1)(1 + 5) 3 = 4 Therefore P (1) is true. Now assume the statement P (n) is true for some value k < n and prove it also true for k + 1. So we assume the following statement to be true P (k) = k i(i + 3) = k(k + 1)(k + 5) 3 We must now prove that the following statement is true k+1 P (k + 1) = i(i + 3) = (k + 1)(k + )(k + 6) 3 We can do so as follows k+1 P (k + 1) = i(i + 3) = = [ ] k(k + 3) + (k + 1)(k + 4) k i(i + 3) + (k + 1)(k + 4) k(k + 1)(k + 5) = + (k + 1)(k + 4) 3 k(k + 1)(k + 5) + 3(k + 1)(k + 4) = [ 3 ] (k + 1) k(k + 5) + 3(k + 4) = 3 = (k + 1)(k + 8k + 1) 3 (k + 1)(k + )(k + 6) = 3 k+1 = i(i + 3) We can now conclude from the Principle of Mathematical Induction that P (n) is true for all natural numbers, which is the result we set out to prove.

8 8 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 8 Example We claim that for all n N, where n 5 i i = 5 ( ) (4n 1)5 n n 5 i i = n n We prove this result using the Principle of Mathematical Induction. Let P (n) = 5 n n. We can verify statement P (n) is true for the smallest value of n i.e., n = 1. So P (1) = = 5. Therefore P (1) is true. Now assume the statement P (n) is true for some value k < n and prove it also true for k + 1. So we assume the following statement to be true P (k) = k 5 i i = 5 ( ) (4k 1)5 k We must now prove that the following statement is true k+1 P (k + 1) = 5 i i = 5 ( ) (4(k + 1) 1)5 k We can do so as follows k+1 P (k + 1) = 5 i i = k 5 i i + 5 k+1 (k + 1) = 5 ( ) (4k 1)5 k k+1 (k + 1) 16 = 5 ) ((4k 1)5 k (k + 1)5 k 16 = 5 ( ) (0k + 15)5 k = 5 ( ) (4k + 3)5 k = 5 ( ) (4(k + 1) 1)5 k = k+1 5 i i We can now conclude from the Principle of Mathematical Induction that P (n) is true for all natural numbers, which is the result we set out to prove.

9 9 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 9 Exercise i Prove using the Principle of Mathematical Induction n i 3 = n (n + 1) 4 ii Prove using the Principle of Mathematical Induction n (i 1) = n(n 1)(n + 1) 3 iii Prove using the Principle of Mathematical Induction n i(i + ) = n(n + 1)(n + 7) 6

10 10 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 10 3 Difference Equations 3.1 Introduction Definition A sequence is a function which maps the natural numbers to the real numbers. f : N R Let a particular sequence be defined by the following general term u n = 5. n 1 To generate the first 4 terms of this sequence we have u 1 = = 5. 0 = 5 u = 5. 1 = 5. 1 = 10 u 3 = = 5. = 0 u 4 = = 5. 3 = 40 Hence the sequence is 5, 10, 0, 40,... Exercise For the sequence defined by the following general term u n = n n + 1 find the terms u 1, u, u 3, u 4, u 5. Many sequences have had explicit formulas for their terms. That is, sequences for which we could write the n th or general term as u n = f(n) for some known function f. For example u n = n (n 1) then it is an easy matter to compute explicitly, say, u 4 = 7 or u 5 = 9. In such cases we are able to compute any given term in the sequence without reference to any other terms in the sequence. However, it is often the case in applications that we do not begin with an explicit formula for the terms of a sequence; rather, we may know only some relationship between the various terms.

11 11 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS Recurrence Relations Definition A recurrence relation is a mathematical relationship expressing f n as some combination of f i with i < n. Example The most famous example of a recurrence relation is the one defining the Fibonacci numbers, f n = f n 1 + f n for n 3 and with f 1 = f = 1. The Fibonacci numbers are computed as follows: f 3 = f + f 1 = = f 4 = f 3 + f = + 1 = 3 f 5 = f 4 + f 3 = 3 + = 5 f 6 = f 5 + f 4 = = 8 f 7 = f 6 + f 5 = = 13 f 8 = f 7 + f 6 = = 1 f 9 = f 8 + f 7 = = 34 So we get the following sequence 1, 1,, 3, 5, 8, 13, 1, 34, 55,... Remark In his book, written in 10, the mathematician Fibonacci ( ) posed a problem concerning the growth of the number of rabbits in a certain area. The problem can be phased as follows: A young pair of rabbits, one of each sex, is placed on an island. Assuming that rabbits do not breed until they are two months old and after they are two months old, each pair of rabbits produces another pair each month, how many pairs are there after n months. Exercise Compute the first five terms of the following recurrence relations: i u n = u n for n 1 with u 1 = 1. ii u n = 3u n 1 + u n for n 1 with u 1 = 1 and u = 1. iii u n = u n 1 + u n for n with u 1 = u = 1.

12 1 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 1 Definition Consider the sequence u 1, u, u 3,... A function f : Z + Z + is recursively defined if u k = f(u k 1, u k,..., u, u 1 ) A difference equation may be defined as a rule which expresses each member of the sequence, from some point on, in terms of the previous member of the sequence. u k u k 1 = f(u k,..., u, u 1 ) Remark For the sequence u 1, u, u 3, u 4... we have a first order recurrence relation u n = f(u n 1 ) and first order difference equation u n u n 1 = f(u n 1 ) Example For the first order recurrence relation u n = n(u n 1 ) (n =, 3, 4,...) and the initial condition u 1 = 1 determine the sequence as far as u 5. Solution: In the recurrence relation we allow n =, 3, 4, 5 to get u = (u 1 ) = 1 = u 3 = 3(u ) = 3 = 1 u 4 = 4(u 3 ) = 4 1 = 576 u 5 = 5(u 4 ) = =

13 13 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 13 Note A recurrence relation may be converted to a difference equation by subtracting u n 1 from both sides. Correspondently a difference equation may be converted to a recurrence relation by moving the u n 1 to the other side of the equation. For the recurrence relation u n = n(u n 1 ) we have the corresponding difference equation u n u n 1 = n(u n 1 ) u n 1 Remembering that the L.H.S. is now the difference between a member of the sequence and the previous member of the sequence with u 1 = 1 we have u u 1 = (u 1 ) u 1 = 1 1 = 1 u 3 u = 3(u ) u = 3 = 10 u 4 u 3 = 4(u 3 ) u 3 = = 564 u 5 u 4 = 5(u 4 ) u 4 = = which again yields the sequence 1,, 1, 576, Recurrance Relation Dif f erence Equation u n = f(u n 1 ) u n u n 1 = f(u n 1 )

14 14 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 14 Example For the second order recurrence relation u n = u n 1 + (u n ) 3 (n 3) with initial conditions u 1 = 1 and u = determine the solution as far as u 5. Solution: In the recurrence relation we allow n = 3, 4, 5 to get u 3 = u + (u 1 ) 3 = = 5 u 4 = u 3 + (u ) 3 = = 18 u 5 = u 4 + (u 3 ) 3 = = 161 Notice that as the order increases the number of required initial conditions increase correspondently. Also we will only consider first order recurrence relations and corresponding first order difference equations. Exercise For the first order recurrence relation u n = u n 1 + (u n 1 ) 3 and the initial condition u 1 = 1 determine the sequence as far as u 5. Write down the corresponding difference equation and again determine the solution as far as u Solving Recurrence Relations To solve a recurrence relation involving the sequence u 1, u, u 3,... is to find an explicit formula for the general term u n. The process of repeatedly substituting old values back into the recurrence relation to produce new ones is called iteration. It is clear that this process will eventually produce u n for any value of u. While iteration has the advantage of being repetitive, and therefore easy to apply, it has drawbacks. For example, if you wished to calculate u 100, say, by iteration then you would also have to write down all the preceding members of the sequence u 1, u, u 3,..., u 98, u 99 regardless if you wanted them or not. For some recurrence relations it is possible to find a simple formula giving the solution u n as a function of u. Such a formula is said to provide a closed-form solution of the recurrence relation and

15 15 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 15 enables u 100, say, to be calculated directly, without any need to calculate all the preceding members of the sequence. The following examples illustrate how a closed form solution may be guessed and then checked. Example Guess a formula for the solution of the first order recurrence relation u n = 1 + u n u n 1 (n ) with initial conditions u 1 = 0. Solution: First calculate u 1, u, u 3, u 4. From the recurrence relation (or corresponding difference equation) and the initial condition it follows by iteration that these four numbers are respectively 0, 3, 8, 15. These numbers look very close to the perfect squares 1, 4, 9, 16. More precisely u 1 = 1 1 u = 1 u 3 = 3 1 u 4 = 4 1 This leads to the guess u n = n 1 for all n 1. Remark We can show that the formula u n = n 1 satisfies both the initial condition and the difference equation as follows: At n = 1 we get the initial condition u 1 = 0. At n = n 1 we get u n 1 = (n 1) 1. Substituting into the difference equation we get u n = 1 + u n u n 1 = (n 1) + (n 1) as n 1 = (n 1) + (n 1) = n 1 = u n Thus the difference equation is satisfied.

16 16 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS Linear Difference Equations with Constant Coefficients Now that we have defined a recurrence relation (and corresponding difference equation) and sought to find a solution of such equations we should proceed by classifying different types of difference equations and consider how to determine their solution in a more structured way. In ordinary algebra, we call y = mx + c a linear equation. Certain difference equations look a little like this, namely those of the form u n = αu n 1 + β and are examples of first-order linear difference equations. Here, α and β are constants. They may, in more advanced cases, take the form of functions. So this is a first-order linear difference equation with constant coefficients. Definition If the terms of the difference equation that involve sequence variables (i.e., which contain u s n ) involve no product of sequence variables, nor functions of sequence variables, then we call the difference equation linear. Otherwise the difference equation is termed non-linear. Note that the restriction apply only to the terms involving the sequence variable and not to any added terms which don t involve the sequence variables. To formalize the definition of the order of a difference equation we have the following definition: Definition If from the highest subscript of the difference equation you subtract the lowest subscript the resulting number is called the order of the difference equation. The following are first-order linear difference equations with constant coefficients u n = u n 1 u n = 3u n The following are second-order linear difference equations with constant coefficients u n = 3u n + n u n = 5u n + u n 1 u n = 3u n 4u n 1 6 The following are non-linear difference equations with constant coefficients u n = (u n 1 ) u n = (u n )(u n 1 )

17 17 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS First-Order Linear Difference Equation u n = f(n) u n = f(u n 1 ) u n u n 1 = f(u n 1 ) Theorem For a first-order linear difference equation of the form: where u 1 = C a solution is of the form u n = αu n 1 u n = Cα n 1 Example For the first-order linear difference equation u n 3u n 1 = 0 with initial condition u 1 = 5 notice that it is equivalent to the difference equation u n = 3u n 1 and has a solution is given as u n = 5.(3) n

18 18 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS Second-Order Linear Difference Equation u n = f(n) u n = f(u n, u n 1 ) u n u n 1 = f(u n, u n 1 ) Theorem 3 For a second-order linear difference equation of the form: u n+ αu n+1 + βu n = 0 where α and β are constants, a solution is of the form u n = l.a n + m.b n where a and b are the solutions of the quadratic equation (characteristic equation) x αx + β = 0. Example To solve the second-order difference equation u n+ 8u n u n = 0 given that u 1 = 1 and u = 7 it has characteristic equation x 8x + 15 = 0. This equation has solutions x 1 = 5 and x = 3. Hence the solution of this second-order difference equation is of the form u n = l.5 n + m.3 n Now evaluating this expressions at each of the initial conditions we get

19 19 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 19 u 1 = 1 l m.3 1 = 1 5l + 3m = 1 u = 7 l.5 + m.3 = 7 5l + 9m = 7 Now solving these two simultaneous equations * for l and m yieldsl = 1 and m =. So, the solution of the difference equation is u n =.3 n 5 n As an exercise check that this is the correct solution. Exercise Show that u n = 4.3 n + n is a solution of the difference equation u n+ 5u n+1 + 6u n = 0 Exercise Solve the difference equation where u 1 = 9 and u = 41. 3u n+ 11u n+1 + 6u n = 0 Exercise The solution of the difference equation u n+ 9u n u n = 0 is given as u n = a.7 n n. Find the value of a if u = 41.

20 0 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS The Fibonacci Sequence Return again to the most famous example of a recurrence relation that defines the Fibonacci numbers, for n 1 and with f 1 = f = 1. f n+ = f n+1 + f n The Fibonacci numbers are computed as follows: f 3 = f + f 1 = = f 4 = f 3 + f = + 1 = 3 f 5 = f 4 + f 3 = 3 + = 5 f 6 = f 5 + f 4 = = 8 f 7 = f 6 + f 5 = = 13 f 8 = f 7 + f 6 = = 1 f 9 = f 8 + f 7 = = 34 So we get the following sequence 1, 1,, 3, 5, 8, 13, 1, 34, 55,... Each term in the sequence (after the second) is the sum of the two that immediately proceed it. Now to find an explicit formulae for the Fibonacci sequence notice that the sequence is linear second-order difference equation f n+ f n+1 f n = 0 with f 1 = 1 and f = 1 it has characteristic equation x x 1 = 0. This equation has solutions x 1 = 1 + 5, x = 1 5 Hence the solution of this second-order difference equation is of the form ( ) n ( f n = l. + m. ) n Now evaluating this expressions at each of the initial conditions we get

21 1 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 1 ( ) ( ) f 1 = 1 l. + m. = 1 ( ) ( ) f = 1 l. + m. = 1 Solving these two equations for l and m yields l = 1 5, m = 1 5 Therefore an explicit formula for the Fibonacci sequence is f n = 1 5. ( ) n 1 ( ) n Exercise The most famous example of a recursive defined sequence is the one defining the Fibonacci sequence, f n = f n 1 + f n for n 3 and with f 1 = f = 1. i Prove, using the Principle of Mathematical Induction, that n fi = f n f n+1 for all n N. ii Prove, using the Principle of Mathematical Induction, that for all n N. n f i = f n+ 1

22 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 3.6 Recursive Algorithms A recursive function is a function that invokes itself. A recursive algorithm is an algorithm that contains a recursive function. Recursion is a powerful, elegant, and a natural way to solve a large class of problems. A problem in this class can be solved using a divide-and-conquer technique in which the problem is decomposed into problems of the same type as the original problem. Each subproblem, in turn, is decomposed further until the process yields subproblems that can be solved in a straightforward way. Finally, solutions to the subproblems are combined to obtain a solution to the original problem. An example of a recursive function is n factorial, i.e., f(n) = n! = n(n 1)(n ) = n(n 1)! This representation shows how to decompose the original problem (to compute n!) into increasingly simpler subproblems of computing (n 1)!, and computing (n )! etc., until the process reaches the straightforward problem of computing 0!. The solutions to these subproblems can then be combined, by multiplying, to solve the original problem. So, for example, the problem of computing 5! is reduced to computing 4!; the problem of computing 4! is reduced to computing 3!; and so on until the computing of 0!. P roblem Solution 0! 1 1! 1.0!!.1! 3! 3.! 4! 4.3! 5! 5.4! A recursive algorithm that computes factorials is as follows: Algorithm: Factorial n Input: n, an integer greater than or equal to zero. Output: n! 1. factorial(n) {. if (n==0) 3. return 1 4. return n*factorial(n-1) 5. }

23 3 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 3 We can use the Principle of Mathematical Induction to prove that the algorithm correctly returns the correct value of n!. At n = 0, the algorithm returns the correct value for 0! = 1. Assume algorithm correctly returns the value of (n 1)!, n N. Now we must prove that algorithm correctly returns the value of n!. If the function correctly returns the value of (n 1)!, then line 4., the function correctly computes the value n.(n 1)! = n!. Therefore, by the Principle of Mathematical Induction, the algorithm correctly returns the value of n!, n N. Remark There is an important link between mathematical induction and recursive algorithms. The basic step of a proof by mathematical induction corresponds to the basic case(s) of a recursive function, and the inductive step of a proof corresponds to the part of a recursive function where the function calls itself. Example A robot can take steps of 1 meter or meters. Write a recursive algorithm to calculate the number of ways the robot can walk n meters. Firstly, some analysis Distance Sequence of Steps N umber of ways to walk , 1 or 3 1, 1, 1 or 1, or, , 1, 1, 1 or 1, 1, or 1,, 1 or, 1, 1, or, 5 Let walk(n) denote the number of ways the robot can walk n meters. We have observed that walk(1) = 1, walk() =

24 4 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 4 Now suppose that n >. The robot can begin by taking a step of 1 meter or a step of meters. If the robot begins by taking a 1-meter step, a distance of n 1 meters remains. By definition the remainder of the walk can be completed in walk(n 1) ways. Similarly, if the robot begins by taking a -meter step, a distance of n meters remains. By definition the remainder of the walk can be completed in walk(n ) ways. Since the walk must begin with a 1-meter step or a -meter step, all of the ways to walk n meters are accounted for. We obtain the formula So, for example, walk(n) = walk(n 1) + walk(n ) walk(4) = walk(3) + walk() = 3 + = 5 We can write a recursive algorithm to compute walk(n) by translating this equation directly into an algorithm. This algorithm computes the function defined by 1, n = 1 walk(n) =, n = walk(n 1) + walk(n ) n >. Algorithm: Robot Walking Input: n, an integer greater than or equal to zero. Output: walk(n) 1. walk(n) {. if (n==1 n==) 3. return n 4. return walk(n-1)+walk(n-) 5. } Example i Use the formulas s 1 =, s n = s n 1 + n, n, to write a recursive algorithm that computes s n = n

25 5 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 5 ii Give a proof, using mathematical induction, that your algorithm from part i is correct. i Input: Natural Number n. Output: n 1. sum(s,n) {. if (n==1) 3. return 4. return sum(n-1)+n 5. } ii Using the Principle of Mathematical Induction we can prove that the algorithm is correct. Let sum(n) = n We can verify statement sum(n) is true for the smallest value of n i.e., n = 1. So sum(1) = Therefore, if n = 1, we correctly return. Now assume the statement sum(k 1) is true for some value k < n and prove it also true for k. So we assume the following statement to be true sum(k 1) = (k 1) We must now prove that the following statement is true sum(k) = k We can do so as follows sum(k) = (k 1) + k = sum(k 1) + k which is the correct value. Hence sum(k 1) sum(k) We can now conclude from the Principle of Mathematical Induction that P (n) is true for all natural numbers, which is the result we set out to prove.

26 6 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 6 Exercise i Write a recursive algorithm to find the maximum of a finite sequence of numbers. ii Give a proof, using mathematical induction, that your algorithm from part i is correct. Input: The sequence s 1, s,...s n and the length n of the sequence. Output: The maximum value of the sequence. 1. findmax(s, n) {. if (n==1) 3. return s 1 4. x = findmax(s, n 1) 5. if(x > s n ) 6. return x 7. else 8. return s n 9. } Example Consider the following recursive algorithm Input: Natural number n Output: Sequence 1. if n=1 then. u else 4. u n 1 + u n u n 1 i Trace the algorithm as far as u 5. ii By inspection, conjecture a non-recursive formula for the n th term of the sequence. iii Use the Principle of Mathematical Induction to prove the formula from part ii is correct. iv Using the non-recursive formula from part ii, write an iterative algorithm to output the first n terms of the sequence.

27 7 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 7 Firstly, u 1 = 0, hence u = 1 + u u 1 = = 3 u 3 = 1 + u u = = 8 u 4 = 1 + u u 3 = = 15 u 5 = 1 + u u 4 = = 4... So, our sequence is, 0, 3, 8, 15, 4,... By inspection, we conjecture the formula u n = f(n) = n 1 for all n 1. Using the Principle of Mathematical Induction we can prove that this formula is correct. Let P (n) = n 1. We can verify statement P (n) is true for the smallest value of n i.e., n = 1. So P (1) = 1 1 = 0. Therefore P (1) is true. Now assume the statement P (n) is true for some value k < n and prove it also true for k + 1. So we assume the following statement to be true P (k) = k 1 We must now prove that the following statement is true We can do so as follows P (k + 1) = (k + 1) 1 P (k + 1) = 1 + u k u k = 1 + k k 1 = 1 + k 1 + k = 1 + k 1 + k = k + k = (k + 1) 1 = P (k + 1) We can now conclude from the Principle of Mathematical Induction that P (n) is true for all natural numbers, which is the result we set out to prove. Finally, using this non-recursive formula from part ii, we can write an iterative algorithm to output the first n terms of the sequence.

28 8 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 8 Input: Natural number n Output: Sequence 1. for n=1 to n. u n n 1 Example Consider the following recursive algorithm Input: Natural number n Output: Sequence 1. if n=1 then. u if n= then 4. u 7 5. else 6. u n 8u n 1 15u n i Trace the algorithm as far as u 5. ii Determine a non-recursive formula for the n th term of the sequence. iii Using the non-recursive formula from part ii, write an iterative algorithm to output the first n terms of the sequence. Firstly, u 1 = 1 and u = 7, hence u 3 = 8u 15u 1 = 8.( 7) 15.(1) = 71 u 4 = 8u 3 15u = 8.( 71) 15.( 7) = 463 u 5 = 8u 4 15u 3 = 8.( 463) 15.( 71) = So, our sequence is, 1, 7, 71, 463, 639,... Now u n = 8u n 1 15u n u n 8u n u n = 0

29 9 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 9 This is equivalent to the following second order linear difference equation u n+ 8u n u n = 0 Using Theorem 3. this equation may be solved to yield the non-recursive formula f(n). Firstly, its has characteristic equation x 8x + 15 = 0 This equation has solutions x 1 = 5 and x = 3 Hence the solution of this second-order difference equation is of the form u n = l.5 n + m.3 n Now evaluating this expressions at each of the initial conditions we get u 1 = 1 l m.3 1 = 1 5l + 3m = 1 u = 7 l.5 + m.3 = 7 5l + 9m = 7 Now solving these two simultaneous equations * for l and m yields l = 1, m = So, the solution of the difference equation is u n = f(n) =.3 n 5 n Finally, using this non-recursive formula from part ii, we can write an iterative algorithm to output the first n terms of the sequence. Input: Natural number n Output: Sequence 1. for n=1 to n. u n.3 n 5 n

30 30 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 30 Exercise Consider the following recursive algorithm Input: Natural number n Output: Sequence 1. if n=1 then. u if n= then 4. u else 6. u n 10u n 1 16u n i Trace the algorithm as far as u 4. ii Determine a non-recursive formula for the n th term of the sequence. iii Using the non-recursive formula from part ii, write an iterative algorithm to output the first n terms of the sequence.

31 31 CHAPTER 6. MATHEMATICAL INDUCTION AND DIFFERENCE EQUATIONS 31 Contents 1 Introduction 1 The Principle of Mathematical Induction 1.1 Integers and Natural Numbers The Well-Ordered Principle Introduction to the Principle of Mathematical Induction Examples Summation Formulae Difference Equations Introduction Recurrence Relations Solving Recurrence Relations Linear Difference Equations with Constant Coefficients First-Order Linear Difference Equation Second-Order Linear Difference Equation The Fibonacci Sequence Recursive Algorithms