dstefan
07-01-2008, 06:08 PM
Chapter 5, Problem 3
Find the Taylor series expansion of the functions \ln(1-x^2) and \frac{1}{1-x^2} around the point 0, using the Taylor series expansions of \ln(1-x) and \frac{1}{1-x}.
Solution:
Recall that
\ln{(1-x)} = - \sum_{k=1}^{\infty} \frac{x^k}{k} ~= - x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \dots~, ~\forall~ x \in [-1, 1)
\frac{1}{1-x} = \sum_{k=0}^{\infty} x^k = 1 + x + x^2 + x^3 + \dots, \forall x \in (-1, 1)
By substituting x^2 for x in the Taylor expansions above, where |x| < 1, we find that
\ln{(1-x^2)} = - \sum_{k=1}^{\infty} \frac{x^{2k}}{k} ~= - x^2 - \frac{x^4}{2} - \frac{x^6}{3} - \frac{x^8}{4} - \dots~, ~\forall~ x \in (-1, 1)
\frac{1}{1-x^2} = \sum_{k=0}^{\infty} ~x^{2k} = 1 + x^2 + x^4 + x^6 + \dots, \forall~ x \in (-1, 1)
Find the Taylor series expansion of the functions \ln(1-x^2) and \frac{1}{1-x^2} around the point 0, using the Taylor series expansions of \ln(1-x) and \frac{1}{1-x}.
Solution:
Recall that
\ln{(1-x)} = - \sum_{k=1}^{\infty} \frac{x^k}{k} ~= - x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \dots~, ~\forall~ x \in [-1, 1)
\frac{1}{1-x} = \sum_{k=0}^{\infty} x^k = 1 + x + x^2 + x^3 + \dots, \forall x \in (-1, 1)
By substituting x^2 for x in the Taylor expansions above, where |x| < 1, we find that
\ln{(1-x^2)} = - \sum_{k=1}^{\infty} \frac{x^{2k}}{k} ~= - x^2 - \frac{x^4}{2} - \frac{x^6}{3} - \frac{x^8}{4} - \dots~, ~\forall~ x \in (-1, 1)
\frac{1}{1-x^2} = \sum_{k=0}^{\infty} ~x^{2k} = 1 + x^2 + x^4 + x^6 + \dots, \forall~ x \in (-1, 1)