Exercise 7.1

Rewrite the function print_n from Section 5.8 using iteration instead of recursion.

Exercise 7.2

Encapsulate this loop in a function called square_root that takes a as a parameter, chooses a reasonable value of x, and returns an estimate of the square root of a.

Exercise 7.3

To test the square root algorithm in this chapter, you could compare it with math.sqrt. Write a function named test_square_root that prints a table like this:

1.0 1.0           1.0           0.0
2.0 1.41421356237 1.41421356237 2.22044604925e-16
3.0 1.73205080757 1.73205080757 0.0
4.0 2.0           2.0           0.0
5.0 2.2360679775  2.2360679775  0.0
6.0 2.44948974278 2.44948974278 0.0
7.0 2.64575131106 2.64575131106 0.0
8.0 2.82842712475 2.82842712475 4.4408920985e-16
9.0 3.0           3.0           0.0

The first column is a number, a; the second column is the square root of a computed with the function from Section 7.5; the third column is the square root computed by math.sqrt; the fourth column is the absolute value of the difference between the two estimates.

Exercise 7.4

The built-in function eval takes a string and evaluates it using the Python interpreter. For example:

>>> eval('1 + 2 * 3')
7
>>> import math
>>> eval('math.sqrt(5)')
2.2360679774997898
>>> eval('type(math.pi)')
<type 'float'>

Write a function called eval_loop that iteratively prompts the user, takes the resulting input and evaluates it using eval, and prints the result. It should continue until the user enters 'done', and then return the value of the last expression it evaluated.

Exercise 7.5

The mathematician Srinivasa Ramanujan found an infinite series that can be used to generate a numerical approximation of $\pi$ Write a function called estimate_pi that uses this formula to compute and return an estimate of $\pi$. It should use a while loop to compute terms of the summation until the last term is smaller than 1e-15 (which is Python notation for $10^{−15}$). You can check the result by comparing it to math.pi.

$$ \frac{1}{\pi} = \frac{2\sqrt{2}}{9801}\sum_{k=0}^{\infty} \frac{(4k)!(1103 + 26390k)}{(k!)^4 396^{4k}} $$