Author: Gaurav Sharma

Sicp Exercise 1.41

Exercise 1.41.  Define a procedure double that takes a procedure of one argument as argument and returns a procedure that applies the original procedure twice. For example, if inc is a procedure that adds 1 to its argument, then (double inc) should be a procedure that adds 2. What value is returned by

(((double (double double)) inc) 5)

 

 

Sicp Exercise 1.38

Exercise 1.38.  In 1737, the Swiss mathematician Leonhard Euler published a memoir De Fractionibus Continuis, which included a continued fraction expansion for e – 2, where e is the base of the natural logarithms. In this fraction, the Ni are all 1, and the Di are successively 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, .... Write a program that uses your cont-frac procedure from exercise 1.37 to approximate e, based on Euler’s expansion.

 

 

Continued Fraction Tail Recursive

An infinite continued fraction is an expression of the form

The infinite continued fraction expansion with the Ni and the Di all equal to 1 produces 1/, where is the golden ratio.
One way to approximate an infinite continued fraction is to truncate the expansion after a given number of terms. Such a truncation — a so-called k-term finite continued fraction — has the form

Suppose that n and d are procedures of one argument (the term index i) that return the Ni and Di of the terms of the continued fraction.

Define a procedure cont-frac such that evaluating (cont-frac n d k) computes the value of the k-term finite continued fraction. Check your procedure by approximating 1/ using

(cont-frac (lambda (i) 1.0)
(lambda (i) 1.0)
k)

for successive values of k.

 

Sicp Exercise 1.37

Exercise 1.37. a. An infinite continued fraction is an expression of the form

As an example, one can show that the infinite continued fraction expansion with the Ni and the Di all equal to 1 produces 1/, where is the golden ratio (described in section 1.2.2). One way to approximate an infinite continued fraction is to truncate the expansion after a given number of terms. Such a truncation — a so-called k-term finite continued fraction — has the form

Suppose that n and d are procedures of one argument (the term index i) that return the Ni and Di of the terms of the continued fraction. Define a procedure cont-frac such that evaluating (cont-frac n d k) computes the value of the k-term finite continued fraction. Check your procedure by approximating 1/ using

(cont-frac (lambda (i) 1.0)
(lambda (i) 1.0)
k)

for successive values of k. How large must you make k in order to get an approximation that is accurate to 4 decimal places?

b. If your cont-frac procedure generates a recursive process, write one that generates an iterative process. If it generates an iterative process, write one that generates a recursive process.

Sicp Exercise 1.36

Exercise 1.36. Modify fixed-point so that it prints the sequence of approximations it generates, using the newline and display primitives shown in exercise 1.22. Then find a solution to xx = 1000 by finding a fixed point of x log(1000)/log(x). (Use Scheme’s primitive log procedure, which computes natural logarithms.) Compare the number of steps this takes with and without average damping. (Note that you cannot start fixed-point with a guess of 1, as this would cause division by log(1) = 0.)