Answer to Question #100464 in Python for oladepo

Question

Write a function named test_sqrt that prints a table like the following using a while loop, where

"dià" is the absolute value of the diàerence between my_sqrt(a) and math.sqrt(a).

a = 1 | my_sqrt(a) = 1 | math.sqrt(a) = 1.0 | diff = 0.0

a = 2 | my_sqrt(a) = 1.41421356237 | math.sqrt(a) = 1.41421356237 | diff =

2.22044604925e-16

a = 3 | my_sqrt(a) = 1.73205080757 | math.sqrt(a) = 1.73205080757 | diff =

0.0

a = 4 | my_sqrt(a) = 2.0 | math.sqrt(a) = 2.0 | diff = 0.0

a = 5 | my_sqrt(a) = 2.2360679775 | math.sqrt(a) = 2.2360679775 | diff = 0.0

a = 6 | my_sqrt(a) = 2.44948974278 | math.sqrt(a) = 2.44948974278 | diff =

0.0

a = 7 | my_sqrt(a) = 2.64575131106 | math.sqrt(a) = 2.64575131106 | diff =

0.0

a = 8 | my_sqrt(a) = 2.82842712475 | math.sqrt(a) = 2.82842712475 | diff =

4.4408920985e-16

a = 9 | my_sqrt(a) = 3.0 | math.sqrt(a) = 3.0 | diff = 0.0

Modify your program so that it outputs lines for a values from 1 to 25 instead of just 1 to 9

Answer 1

def test_sqrt():
    a = 1
    while a<26:
        print('a =', a,'| my_sqrt(a) =',my_sqrt(a),'| math.sqrt(a) =', math.sqrt(a),'| diff =', abs(math.sqrt(a)-my_sqrt(a)))
        a = a + 1
    return 0