How do you find the maclaurin series expansion of #f(x) = ln abs(1+x^5)#?

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Answer 1 · 2018-07-31T07:50:49

# ln abs(1+x^5) = sum_(n=0)^oo (-1)^nx^(5n+5)/(n+1) #

with radius of convergence #R=1#.

#1/(1-q) = sum_(n=0)^oo q^n #

having radius on convergence #R=1#.
Let now #q=-x^5#:

#1/(1+x^5) = sum_(n=0)^oo (-x^5)^n= sum_(n=0)^oo (-1)^nx^(5n) #

Still with #R=1# because #abs(-x^5) <= 1 => abs x <=1#
Multiply by #x^4#:

#x^4/(1+x^5) = sum_(n=0)^oo (-1)^nx^(5n+4) #

In the interior of the interval of convergence we can integrate term by term obtaining a series with at least the same radius of convergence:

#int_0^x t^4/(1+t^5)dx = sum_(n=0)^oo (-1)^n int_0^x t^(5n+4)dt #

#1/5 int_0^x (d(1+t^5))/(1+t^5)dx = sum_(n=0)^oo (-1)^n[ t^(5n+5)/(5n+5)]_0^x #

#1/5 ln abs(1+x^5) = sum_(n=0)^oo (-1)^nx^(5n+5)/(5n+5) #

# ln abs(1+x^5) = sum_(n=0)^oo (-1)^nx^(5n+5)/(n+1) #