Find the Maclaurin series for f(x) = ln((1 + x)/(1- x)) and determine its radius

Question

Find the Maclaurin series for f(x) = ln((1 + x)/(1- x)) and determine its radius of convergence. Use the first 4 terms of the series to approximate ln 3.

Explanation / Answer

(a) Note that, using the laws of logarithms, you can write: f(x) = ln[(1 + x)/(1 - x)] = ln(1 + x) - ln(1 - x). By the infinite geometric series: 1/(1 - x) = sum(n=0 to infinity) x^n. Replacing x with -x yields: 1/[1 - (-x)] = sum(n=0 to infinity) (-x)^n ==> 1/(1 + x) = sum(n=0 to infinity) [(-1)^n * x^n]. Then, integrating both sides with respect to x yields: ln(1 + x) = sum(n=0 to infinity) [(-1)^n * x^(n + 1)]/(n + 1) = x - x^2/2 + x^3/3 - x^4/4 + x^5/5 + ... . Then, replacing x with -x here gives: ln[1 + (-x)] = sum(n=0 to infinity) [(-1)^n * (-x)^(n + 1)]/(n + 1) ==> ln(1 - x) = -x * sum(n=0 to infinity) x^n/(n + 1). = -x - x^2/2 - x^3/3 - x^4/4 - x^5/5 + ... . Subtracting the two two series gives: ln[(1 + x)/(1 - x)] = ln(1 + x) - ln(1 - x) = (x - x^2/2 + x^3/3 - x^4 + x^5/5 + ...) - (-x - x^2/2 - x^3/3 - x^4/4 - x^5/5 + ...) = 2x + 2x^3/3 + 2x^5/5 + ... . = 2*sum(n=0 to infinity) x^(n + 1)/(2n + 3).

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