How do you find the power series for f(x)=ln(1-3x^2) and determine its radius of convergence?

1 Answer
Andrea S.
Mar 14, 2018

ln(1-3x^2) = -sum_(k=0)^oo 3^(k+1)x^(2k+2)/(k+1)

Explanation:

Start from the sum of the geometric series:

sum_(k=0)^oo q^k = 1/(1-q)

For abs(q) <1

Let q=3x^2 so that we have:

sum_(k=0)^oo 3^kx^(2k) = 1/(1-3x^2)

and multiplying by 6x:

6sum_(k=0)^oo 3^kx^(2k+1) = (6x)/(1-3x^2)

The series is convergent for 3x^2 < 1, that is for x in (-1/sqrt3,1/sqrt3). Inside this interval we can integrate the series term by term and obtain a series with the same radius of convergence.

6sum_(k=0)^oo 3^kint_0^x t^(2k+1)dt = int_0^x (6tdt)/(1-3t^2)

6sum_(k=0)^oo 3^kx^(2k+2)/(2k+2) = int_0^x -(d(1-3t^2))/(1-3t^2)

6sum_(k=0)^oo 3^kx^(2k+2)/(2k+2) = -lnabs(1-3x^2)

and finally:

ln(1-3x^2) = -sum_(k=0)^oo 3^(k+1)x^(2k+2)/(k+1)

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