Keep recieving errors as I attempt to use a function to replace the power function, namely when raising to 1/2

Question

For an assignment, I must create a program with a function that replaces the "power function" assuming that it is broken for whatever reason. The current model I have works pretty decently, but I'm struggling to get it to work when raising to a 1/2, 1/4, etc. (essentially performing a square root function). I know that the issue lies within floats not cooperating with range, but I personally don't know how to avoid this

def power (value, exp):
    num = 1.0 
    if exp>0:
        for function in range(exp):
            num = num * value 
    elif exp<0:
        for function in range(-exp):
            num = num / value
    else:
        num = 1
    return num

number = float(raw_input("Please enter a the base number you wish to raise to a power (no fractions, decimal allowed): "))
exponent = float(raw_input("Please enter the power you wish to raise the base number to (no fractions, decimal allowed): "))

print "Your base number was:", number
print "You rasied this number to the power of:", exponent
print "The ending product of this operation is:", power(number, exponent)

I get that you're trying to replace `pow()`, but is the `**` operator really off limits? — Matt Hall, Feb 09 '18 at 12:43
@tobias_k , if the exp was 1/2 it would fall under the first if statement, as this is a positive value. — Christian, Feb 09 '18 at 12:46
@kwinkunks Sadly yes, the point of it is to test our logical reasoning, not our use of python functions — Christian, Feb 09 '18 at 12:46
@Christian your `elif` could simply be `num = 1/power(value, -exp)` if you are assuming integer exponents up to that point. — Ma0, Feb 09 '18 at 12:47
Is [using logarithms](https://en.wikipedia.org/wiki/Exponentiation#Powers_via_logarithms) okay? — tobias_k, Feb 09 '18 at 12:48
Unless the exponent is an irrational number, it's a fraction, e.g. two integers.Knowing that you can avoid floats in `range()`. — Psytho, Feb 09 '18 at 13:01

tobias_k · Answer 1 · 2018-02-10T10:09:42.897

Fractional powers are a bit tricky, but you can use the fact that b^n/d = (bⁿ)^1/d = ^d√ bⁿ. See here for details. So you could write a recursive algorithm that

reduces the exponent until it is a float between 0 and 1
approximates a fraction n/d representing that float, e.g. using fractions or this
find the dth root of the n power, e.g. using simple binary search

Code:

def power(b, e):
    # the simple cases
    if e == 0: return 1
    if b in (0, 1): return b
    if e < 0: return 1 / power(b, -e)
    if e >= 1: return b * power(b, e - 1)
    # fractions: b^(n/d) == d-/ b^n
    n, d = float_to_fraction(e)  # see linked answer
    return root(power(b, n), d)

def root(x, r):
    if x < 0: raise ValueError("No roots of negative numbers!")
    low, high = 0, x
    cur = (low + high) / 2
    while high - low > 0.001:
        if power(cur, r) < x:
            low = cur
        else:
            high = cur
        cur = (low + high) / 2
    return cur

Example:

>>> b, e = 1.23, -4.56
>>> b**e
0.38907443175580797
>>> power(b, e))
0.38902984218983117

Or you could just use logarithms, although that might be considered cheating:

>>> math.exp(e * math.log(b))
0.38907443175580797

Addendum: Of course, the same could be implemented iteratively instead of using recursion and would probably be faster this way, but that should not really be a concern here and I like the elegance of the recursive definition. Also, there are certainly better ways to compute the nth root, but again, that's not really the point of the question. — tobias_k, Feb 10 '18 at 12:48

Keep recieving errors as I attempt to use a function to replace the power function, namely when raising to 1/2

1 Answers1