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

Question

Isabella Ackerman · Accepted Answer

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

with radius of convergence #R=1#.

Start from the geometric series: #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) #

Aurora Alvarado · Answer

undefined 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:

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