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

with radius of convergence

Start from the geometric series:

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:

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To find the Maclaurin series expansion of ( f(x) = \ln|1+x^5| ), you can use the properties of logarithmic functions and the Maclaurin series expansion for ( \ln(1+x) ). First, we rewrite the function as:

[ f(x) = \ln|1+x^5| = \ln(1+x^5) ]

Then, we use the Maclaurin series expansion for ( \ln(1+x) ), which is:

[ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots ]

Now, substitute ( x^5 ) for ( x ) in the series expansion:

[ \ln(1+x^5) = x^5 - \frac{(x^5)^2}{2} + \frac{(x^5)^3}{3} - \frac{(x^5)^4}{4} + \cdots ]

[ = x^5 - \frac{x^{10}}{2} + \frac{x^{15}}{3} - \frac{x^{20}}{4} + \cdots ]

Therefore, the Maclaurin series expansion of ( f(x) = \ln|1+x^5| ) is:

[ f(x) = x^5 - \frac{x^{10}}{2} + \frac{x^{15}}{3} - \frac{x^{20}}{4} + \cdots ]

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