How do you find the maclaurin series expansion of #f(x)= x / (1-x^4)#?

Question

Aurora Alvarado · Accepted Answer

Use the Maclaurin series for #1/(1-t)# and substitution to find:

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

The Maclaurin series for #1/(1-t)# is #sum_(n=0)^oo t^n# since #(1-t) sum_(n=0)^oo t^n = sum_(n=0)^oo t^n - t sum_(n=0)^oo t^n = sum_(n=0)^oo t^n - sum_(n=1)^oo t^n = t^0 = 1# Substitute #t = x^4# to find: #1/(1-x^4) = sum_(n=0)^oo x^(4n)# Multiply by #x# to get: #x/(1-x^4) = sum_(n=0)^oo x^(4n+1)# This is a geometric series with common ratio #x^4#, hence the series converges when #abs(x) < 1# and the radius of convergence is #1#.

Isabella Ackerman · Answer

undefined To find the Maclaurin series expansion of $ f(x) = \frac{x}{1-x^4} $, you can express the function as a geometric series and then expand it term by term.

The geometric series expansion for $ \frac{1}{1 - x} $ is $ 1 + x + x^2 + x^3 + \ldots $ with a common ratio of $ x $, valid for $ |x| < 1 $.

To adapt this for $ \frac{x}{1 - x^4} $, you need to rewrite $ x^4 $ as $ (x^2)^2 $, and then replace $ x $ with $ x^2 $ in the geometric series. This yields:

$$ \frac{x}{1 - x^4} = x \cdot \frac{1}{1 - x^4} $$

$$ = x \cdot \frac{1}{1 - (x^2)^2} $$

Now, replace $ x^2 $ with $ -x^2 $ in the geometric series:

$$ = x \cdot \left(1 + (-x^2) + (-x^2)^2 + (-x^2)^3 + \ldots\right) $$

$$ = x - x^3 + x^6 - x^9 + \ldots $$

This is the Maclaurin series expansion of $ f(x) = \frac{x}{1 - x^4} $.