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#