Square root using Taylor series

Question

I would like to compute square root using Taylor series. I'm just learning about the series and I wrote a bit of code but I don't know why it doesn't work, maybe I shouldn't feed i to it? Please can anyone explain to me what I'm doing wrong?

I have the formula from http://en.wikipedia.org/wiki/Taylor_series#List_of_Maclaurin_series_of_some_common_functions

from math import sqrt

def factorial(n):
    result = 1
    for i in range(2, n+1):
        result *= i
    return result

def binomical(alpha, n):
    result = 1
    for i in range(0, n):
        result *= (alpha - i)
    return result / factorial(n)

for i in range(1, 10):
    x = sum(binomical(0.5, k) * i ** k for k in range(10))
    print x, sqrt(i)

About what point you want to take the Taylor's series? If you take it about `x=0` it is not defined. — John Alexiou, Jan 26 '13 at 14:48
What is surprising is that the error away from `x=1` gets larger the more terms are considered. — John Alexiou, Jan 26 '13 at 14:53
@ja72: about 1 to 10, that is I would like to calculate 10 square roots and compare them to standard function. — Tuom L., Jan 26 '13 at 14:54

score 4 · Accepted Answer · answered Jan 26 '13 at 14:48

There are two issues, one minor and one major. The minor is that the expansion is written in terms of (1+x)^alpha, not x^alpha, so your i**k should really be (i-1)**k. Doing this turns your output of

1.41920471191 1.0
5.234375 1.41421356237

where you can see how suspiciously close your answer for sqrt(1) is to sqrt(2) into

1.0 1.0
1.41920471191 1.41421356237

which is much better. Unfortunately the remaining terms still aren't very good:

5.234375 1.73205080757
155.677841187 2.0
2205.0 2.2360679775
17202.2201691 2.44948974278
91687.28125 2.64575131106
376029.066696 2.82842712475
1273853.0 3.0

and increasing the number of terms summed from 10 to 100 makes things even worse:

1.0 1.0
1.4143562059 1.41421356237
1.2085299569e+26 1.73205080757
3.68973817323e+43 2.0
9.21065601505e+55 2.2360679775
3.76991761647e+65 2.44948974278
2.67712017747e+73 2.64575131106
1.16004174256e+80 2.82842712475
6.49543428975e+85 3.0

But that's to be expected, because as the page you linked explains, this is only guaranteed to converge when the absolute value of x is less than 1. So we can do a good job of getting the roots of small numbers:

>>> i = 0.7
>>> sum(binomical(0.5, k) * (i-1) ** k for k in range(10))
0.8366601005565644
>>> i**0.5
0.8366600265340756

and we can try scaling things down to deal with other numbers:

>>> i0 = 123.0
>>> i = i0/(20**2)
>>> sum(binomical(0.5, k) * (i-1) ** k for k in range(50))
0.5545268253462641
>>> _*20
11.090536506925282
>>> i0**0.5
11.090536506409418

or take the Taylor series around a different point, etc.

The general take-away is that Taylor series have a radius of convergence -- possibly zero! -- within which they give the right results. The Wikipedia Taylor series page has a section on "Approximation and convergence" which covers this.

(P.S. No "c" in "binomial". :^)

Square root using Taylor series

1 Answers1