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Generation QR Code ISO/IEC18004 in Software Basic Concepts

EXAMPLE 68
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f[x_]=
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g[x_]= x + 2 x + 3; h1[x_]= f[x]+ g[x] 3+ x + 2 x + x2 h2[x_]= f[x] g[x] -3 + x - 2 x - x2 h3[x_]= f[x]g[x] x (3 + 2 x + x2)
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Basic Concepts
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h4[x_]= f[x]/g[x] x 3 + 2 x + x2 h5[x_]= f[g[x]] 3 + 2 x + x2 h6[x_]= g[f[x]] 3+ 2 x + x
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The composition of two or more functions can be accomplished with the Composition command. Note that Composition is a functional operation and as such, its arguments are functions, f, not f[x].
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Composition[f1, f2, f3, ...] constructs the composition f1 o f2 o f3...
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EXAMPLE 69
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f[x_] =
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g[x_] = x2 + 2 x + 3; h1 = Composition[f, g]; h1[x] 3 + 2 x + x2 h2 = Composition[g, f]; h2[x] 3+ 2 x + x
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If we wish to compute the composition of a function with itself we could, of course, use Composition[f, f], Composition[f, f, f], and so forth. A more convenient tool is Nest or NestList.
Nest[f, expression, n] applies f to expression successively n times. NestList[f, expression, n] applies f to expression successively n times and returns a list of all the intermediate calculations from 0 to n. (Lists are discussed in detail in 3.)
EXAMPLE 70
f[x_]= x2; Nest[f, x, 5] x32 NestList[f, x, 5] {x, x2, x4, x8, x16, x32} Nest[f, 2 x + 3, 5] (3 + 2 x)32 NestList[f, 2 x + 3, 5] {3 + 2 x, (3 + 2 x)2, (3 + 2 x)4, (3 + 2 x)8, (3 + 2 x)16, (3 + 2 x)32}
EXAMPLE 71 The function Framed[symbol] draws a frame around symbol. We can use Nest and NestList
to show the effect of repetitive framing.
Basic Concepts
Nest[Framed,x,10]
NestList[Framed,x,10]
SOLVED PROBLEMS
2.72 If f(x) = sin x + 2 cos x and g(x) = 2 sin x 3 cos x, construct ( f + g) (x), ( f g) (x), ( f g) (x), and ( f /g) (x) and evaluate them at /2.
SOLUTION
f[x_]= Sin[x]+ 2 Cos[x]; g[x_]= 2 Sin[x] 3 Cos[x]; h1[x_]= f[x]+ g[x] Cos[x]+ 3 Sin[x] h2[x_]= f[x] g[x] 5 Cos[x] Sin[x] h3[x_]= f[x] g[x] (2 Cos[x]+ Sin[x])( 3 Cos[x]+ 2 Sin[x]) h4[x_]= f[x]/g[x] 2 Cos[x] + Sin[x] 3 Cos[x] + 2 Sin[x] h1[ /2] 3 h2[ /2] 1 h3[ /2] 2 h4[ /2] 1 2
2.73 Let f ( x ) = 1 + x . Compute ( f o f o f o f o f )(x).
SOLUTION
f[x_]= 1 + x ; Nest[f, x, 5] 1+ 1+ 1+ 1+ 1+ x
Basic Concepts
2.74 Let f ( x ) = 1 . Let f n ( x ) = (14 o ...o f )( x ) . Evaluate f ( x ), f 2 ( x ), f 3 ( x ), f 4 ( x ), and f 5 ( x ) . Then evaluate fof 24 3 1+ x
f (1), f 2 (1), f 3 (1), . . . , f 20 (1). What do you observe Convert to a decimal form and approximate lim f n (1) .
SOLUTION
f[x_]=
1 1+ x
NestList[f, x, 5] 1 1 1 1 1 , , , x,1 + x , 1 1 1 1 1+ 1+ 1+ 1+ 1 1 1 1+ x 1+ 1+ 1+ 1 1+ x 1+ 1+ 1 1+ x 1+ 1 1+ x
NestList[f, 1, 20] 10 946 5 1,1,2,3,5, 8 ,13,21,55, 89 ,144,233,377,610, 987 ,1597,2584,4181, 6765 , 3 2 3 5 8 13 21 34 89 144 233 377 610 987 1597 2584 4181 6765 10 946 17 711
The numerators (and denominators) appear to be terms of the Fibonacci sequence. NestList[f, 1, 20] //N
{1.,0.5,0.666667,0.6,0.625,0.615385,0.619048,0.617647,0.618182,0.617978, 0.618056,0.618026,0.618037,0.618033,0.618034,0.618034,0.618034, 0.618034,0.618034,0.618034,0.618034}
The numbers appear to be approaching a limit of approximately 0.618034.
2.75 If x is an approximation to a , it can be shown that 1 x + a is a better approximation. (This is a
special case of Newton s method.) Use NestList to observe the first 10 approximations obtained in computing 3 , starting with x = 100.
SOLUTION
a = 3; f[x_] = 1 x + a ; 2 x NestList[f, 100, 10] // N {100., 50.015, 25.0375, 12.5787, 6.40858, 3.43835, 2.15543, 1.77363, 1.73254, 1.73205, 1.73205}
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