What is the power series of $f(x)= ln(5-x)^2$ ? What is its radius of convergence?

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Answer 1 · 2018-03-03T20:26:17

$ln(5-x)^2 = 2ln5 -2/5sum_(n=0)^oo x^(n+1)/(5^n(n+1))$

with radius of convergence $R=5$ .

Start from the sum of a geometric series:

$sum_(n=0)^oo x^n = 1/(1-x)$

with radius of convergence $R=1$ .

Integrating term by term, we obtain a series with the same radius of convergence:

$sum_(n=0)^oo int_0^x t^n = int_0^x dt/(1-t)$

$sum_(n=0)^oo x^(n+1)/(n+1) = -ln(1-x)$

Now using the properties of logarithms we have:

$ln(5-x)^2 = 2ln(5(1-x/5)) =2ln5 +2ln(1-x/5)$

and substituting $x/5$ to $x$ in the series above we get:

$ln(5-x)^2 = 2ln5 -2/5sum_(n=0)^oo x^(n+1)/(5^n(n+1))$

converging for $abs (x/5) < 1$ , that is with radius of convergence $R=5$

What is the power series of f(x)= ln(5-x)^2f(x)= ln(5-x)^2? What is its radius of convergence?