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public class TestMutualRecursion { public static void main(String[] args) { String fmt1 = "%18s%18s%18s%n"; String fmt2 = "%18.13f%18.13f%18.13f%n"; System.out.printf(fmt1, "s(x) ", "Math.sin(x) ", "error "); for (double x = 0.0; x < 1.0; x += 0.1) { System.out.printf(fmt2, s(x), Math.sin(x), Math.sin(x) - s(x)); } System.out.printf(fmt1, "c(x) ", "Math.cos(x) ", "error "); for (double x = 0.0; x < 1.0; x += 0.1) { System.out.printf(fmt2, c(x), Math.cos(x), c(x) - Math.cos(x)); } } public static double s(double x) { if (-0.005 < x && x < 0.005) { return x - x*x*x/6; // basis } return 2*s(x/2)*c(x/2); // recursion } public static double c(double x) { if (-0.005 < x && x < 0.005) { return 1.0 - x*x/2; // basis } return 1 - 2*s(x/2)*s(x/2); // recursion } }

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and G ( z ) =

MUTUAL RECURSION When a function calls itself, it is called direct recursion. Another form of recursion is when a function calls other functions that call other functions that eventually call the original function. This is called indirect recursion. Its most common form is when two functions call each other. This is called mutual recursion. (See Figure 9.6.)

H ( z ) = 0.4 -t0 . 8 ~ -- 0.5z-' '

The output is:

b(k) = a ( p --k ) fork = 0 , 1 ,

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The sine and cosine functions from trigonometry can be defined in several different ways, and there are several different algorithms for computing their values. The simplest (although not the most efficient) is via mutual recursion. It is based upon the identities: sin2 = 2sin cos cos2 = 1 2(sin )2 and the two Taylor polynomials: sin x x x 3/6

.. . , p - 1 , and b ( p ) = 1 .

Math.sin(x) error 0.0000000000000 0.0000000000000 0.0998334166468 0.0000000000005 0.1986693307951 0.0000000000009 0.2955202066613 0.0000000000069 0.3894183423087 0.0000000000018 0.4794255386042 0.0000000000051 0.5646424733950 0.0000000000120 0.6442176872377 0.0000000000015 0.7173560908995 0.0000000000027 0.7833269096275 0.0000000000043 0.8414709848079 0.0000000000063 Math.cos(x) error 1.0000000000000 0.0000000000000 0.9950041652780 0.0000000000000 0.9800665778412 0.0000000000002 0.9553364891256 0.0000000000021 0.9210609940029 0.0000000000007 0.8775825618904 0.0000000000028 0.8253356149097 0.0000000000082 0.7648421872845 0.0000000000013 0.6967067093472 0.0000000000027 0.6216099682707 0.0000000000054 0.5403023058681 0.0000000000098 This works because on each recursive call x is divided by 2, and eventually it reaches the basis criterion (-0.005 < x && x < 0.005), which stops the recursion. s(x) 0.0000000000000 0.0998334166464 0.1986693307941 0.2955202066544 0.3894183423069 0.4794255385991 0.5646424733831 0.6442176872362 0.7173560908969 0.7833269096232 0.8414709848016 c(x) 1.0000000000000 0.9950041652781 0.9800665778414 0.9553364891277 0.9210609940036 0.8775825618932 0.8253356149179 0.7648421872857 0.6967067093499 0.6216099682760 0.5403023058779

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Hmin(z) FlR with each zero having even order (i.e., Hmin(z) G 2 ( z ) is = where G ( z )is a minimum phase system).

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

2 H ( z )= 1 - az-I'

CHAP. 9]

2 ) and x2(n) = aS(n - I )

g ( l ) = - 1.2 and g(2) = -2.

9.1 9.2 9.3 9.4 9.5 A recursive function must have two parts: its basis and its recursive part. Explain what each of these is and why it is essential to recursion. How many recursive calls will the call f(10) to the recursive factorial function (Example 9.2 on page 165) generate How many recursive calls will the call fib(6) to the recursive Fibonacci function (Example 9.6 on page 167) generate What are the advantages and disadvantages of implementing a recursive solution instead of an iterative solution What is the difference between direct recursion and indirect recursion

2 -2

The output is:

(h) Q ( z ) = '

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