mission21-template.rkt

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0 0 17 3 37 #"; CS1101S --- Programming Methodology"
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0 0 17 3 1 #";"
0 0 4 29 1 #"\n"
0 0 17 3 12 #"; Mission 21"
0 0 4 29 1 #"\n"
0 0 17 3 1 #";"
0 0 4 29 1 #"\n"
0 0 17 3 69
(
 #"; Note that written answers are commented out to allow us to run you"
 #"r"
) 0 0 4 29 1 #"\n"
0 0 17 3 45 #"; code easily while grading your problem set."
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0 0 4 29 1 #"\n"
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0 0 4 29 1 #"\n"
0 0 17 3 10 #";;;;;;;;;;"
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0 0 17 3 10 #"; Task 1 ;"
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0 0 17 3 10 #";;;;;;;;;;"
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0 0 15 3 6 #"define"
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0 0 15 3 6 #"define"
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0 0 19 3 3 #"   "
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0 0 19 3 1 #" "
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0 0 20 3 2 #"10"
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0 0 19 29 1 #"\n"
0 0 17 3 10 #";;;;;;;;;;"
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0 0 17 3 10 #"; Task 2 ;"
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0 0 17 3 10 #";;;;;;;;;;"
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0 0 19 29 1 #"\n"
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0 0 15 3 6 #"define"
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(
 #"; Explain how this definition works and how the coefficients in the "
 #"series are generated"
) 0 0 19 29 1 #"\n"
0 0 17 3 101
(
 #"; By doing this we are defining y = e^x as the solution to the integ"
 #"ral equation y = 1 + integrate(y)"
) 0 0 19 29 1 #"\n"
0 0 17 3 103
(
 #"; where integrate(y) is the integral of y with respect to x from 0 t"
 #"o x. The definition expands as such"
) 0 0 19 29 1 #"\n"
0 0 17 3 65
#"; where int-n is the integrate-n function and : means stream-cons"
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0 0 19 29 1 #"\n"
0 0 17 3 37 #"; exp-series = 1 : (int-1 exp-series)"
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0 0 17 3 49 #";            = 1 : (int-1 (1 : int-1 exp-series))"
0 0 19 29 1 #"\n"
0 0 17 3 49 #";            = 1 : 1/1 : int-2 (int-1 exp-series)"
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0 0 17 3 61
#";            = 1 : 1/1 : int-2 (int-1 (1 : int-1 exp-series))"
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0 0 17 3 65
#";            = 1 : 1/1 : int-2 (1/1 : (int-2 (int-1 exp-series)))"
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#";            = 1 : 1/1 : 1/2 : (int-3 (int-2 (int-1 exp-series)))"
0 0 19 29 1 #"\n"
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(
 #";            = 1 : 1/1 : 1/2 : 1/6 : (int-4 (int-3 (int-2 (int-1 exp"
 #"-series))))"
) 0 0 19 29 1 #"\n"
0 0 17 3 47 #";            = 1 : 1/1 : 1/2 : 1/6 : 1/24 : ..."
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0 0 19 29 1 #"\n"
0 0 17 3 10 #";;;;;;;;;;"
0 0 19 29 1 #"\n"
0 0 17 3 10 #"; Task 3 ;"
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0 0 17 3 10 #";;;;;;;;;;"
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0 0 17 3 22 #";(define jw-exp-series"
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0 0 17 3 38 #";  (integrate-series jw-exp-series 1))"
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0 0 19 29 1 #"\n"
0 0 17 3 31 #";(eval-stream jw-exp-series 10)"
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0 0 19 29 1 #"\n"
0 0 17 3 72
(
 #"; Explain why this definition does not work while the previous one w"
 #"orks"
) 0 0 19 29 1 #"\n"
0 0 17 3 109
(
 #"; The integrate-series function is not a special form and hence it i"
 #"s strict in all its arguments. Hence when"
) 0 0 19 29 1 #"\n"
0 0 17 3 111
(
 #"; jw-exp-series is evaluated jw-exp-series (second argument of integ"
 #"rate-series) must be evaluated, causing an "
) 0 0 19 29 1 #"\n"
0 0 17 3 108
(
 #"; infinite loop. In the first definition exp-series is defined using"
 #" cons-stream which is lazy in its second"
) 0 0 19 29 1 #"\n"
0 0 17 3 82
(
 #"; argument, hence when evaluating exp-series, evaluation of exp-seri"
 #"es is delayed."
) 0 0 19 29 1 #"\n"
0 0 19 29 1 #"\n"
0 0 17 3 10 #";;;;;;;;;;"
0 0 19 29 1 #"\n"
0 0 17 3 10 #"; Task 4 ;"
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0 0 17 3 10 #";;;;;;;;;;"
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0 0 15 3 6 #"define"
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