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You may have noticed a long pause in the calculation of this number. To see this more precisely, we will time the operation. (Your times may be slightly different, depending upon your computer.)
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f[35]//Timing
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Intermediate calculations have not been stored. Each computation of f[n] necessitates the computation of f[n 2] and f[n 1], each of which causes all values of f down to f[3] to be computed. Since each intermediate value of f is computed recursively based upon the values of f[1] and f[2], the result is that it takes an extremely large number of iterations to compute f[35]. To eliminate this problem, we can store each value of f in memory as it is computed. The values can then be recalled almost instantaneously.
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f[1] = 1; f[2] = 1; f[n_] f[n] = f[n 2] + f[n 1] f[35]//Timing
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This causes Mathematica to store each f[n] value. Type f after computing f[35] to confirm this.
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Functions of two or more variables can be defined in an analogous manner. The syntax is self-explanatory.
EXAMPLE 66 EXAMPLE 67
f[x_, y_] = x2 + y3; f[2, 3] 31 f[3, 2] 17
g[x_, y_, z_] = x + y * z; g[2, 3, 4] 14
SOLVED PROBLEMS
2.65 Define f(x) to be the polynomial x5 + 3x4 7x2 + 2 and compute f(2).
SOLUTION
f[x_] = x5 + 3x4 7x2 + 2 2 7x2 + 3x4 + x5 f[2] 54 x 2.66 Let f ( x ) = x 2 18 3 x if x 0 if 0 < x 3 if x > 3
Sketch the graph of f(x) for 6 x 6.
SOLUTION
f[x_] x /; x 0 f[x_] x2 /; 0 < x 3 f[x_] 18 3x /; x > 3 Plot[f[x], {x, 6, 6}]
Basic Concepts
2.67 If f(x) is defined on an interval [a, b], the periodic extension of f with period T = b a is the function F such that
if a x b f (x) F (x) = f ( x T ) otherwise
Let f(x) = x2 if 1 x 1. Plot the periodic extension of f with period 2 from x = 0 to x = 10.
SOLUTION
f[x_]= x2; F[x_] f[x]/; 1 x 1 F[x_] F[x 2]/; x > 1 Plot[F[x],{x, 0, 10}]
1.0 0.8 0.6 0.4 0.2
f (1) = 1 f (2) = 2 2.68 Define the function f(n): f (3) = 3 f (n) = f (n 3) + f (n 2) + f (n 3) if n 4
Compute f(20).
SOLUTION
Clear[f] f[1]= 1; f[2]= 2; f[3]= 3; f[n_] f[n]= f[n 3]+ f[n 2]+ f[n 1]; f[20] 101 902
2.69 Define a function that represents the distance from the point (x, y) to (3, 4) and compute the value of the function at the point ( 5, 3).
SOLUTION
f[x_, y_]= (x 3)2 + (y 4)2 ; f[5, 2] 2 10
Basic Concepts
2.70 Define a function that represents the distance between the points (x1, y1) and (x2, y2) and use it to compute the distance from (2, 3) to (8, 11).
SOLUTION
d[x1_, y1_, x2_, y2_]= (x2 x1)2 + (y2 y1)2 ; d[2, 3, 8, 11] 10
2.71 The area enclosed by a triangle whose sides have length a, b, and c is given by Heron s formula:
K = s(s a)(s b)(s c) b where s = a + 2 + c . Express the area of a triangle as a function of a, b, and c and compute the area of the triangle whose sides are (a) 3, 4, 5 and (b) 5, 9, 12
SOLUTION
s = a + b + c; 2 k[a_, b_, c_]= s (s a)(s b)(s c) ; k[3, 4, 5] 6 k[5, 9, 12] 4 26
2.11 Operations on Functions
If f and g are two functions with the same domain, D, we define their sum, difference, product, and quotient pointwise, that is, (f + g )(x) = f(x) + g(x) (f g )(x) = f(x) g(x) (fg)(x) = f(x) g(x) (f/g)(x) = f(x)/g(x) for all x in D for all x in D for all x in D for all x in D for which g(x) 0
If x is a number in the domain of g such that g(x) is in the domain of f, we define the composite function fog: (f o g)(x) = f(g(x) ) The function g o f can be defined in a similar manner. The following example illustrates how to construct these functions.
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