How do you find the maclaurin series expansion of #sin^3 (x)#?
1 Answer
Feb 21, 2016
Explanation:
This is how you construct a Maclaurin Series.
#f(x) = f(0) + frac{f'(0)}{1!}x + frac{f''(0)}{2!}x^2 + frac{f^((3))(0)}{3!}x^3 + ...#
I will only do up to the first 3 non-zero terms.
#f(0) = sin^3 (0) =0#
#f'(x) = 3sin^2(x)cos(x)#
#f'(0) = 3sin^2(0)cos(0)#
#= 0#
#f''(0) = 0#
#f^((3))(0) = 6#
#f^((4))(0) = 0#
#f^((5))(0) = 60#
#f^((6))(0) = 0#
#f^((7))(0) = 546#
So, the series expansion is
#sin^3(x) = frac{f^((3))(0)}{3!}x^3 + frac{f^((5))(0)}{5!}x^5 + frac{f^((7))(0)}{7!}x^7 + O(x^8)#
#= frac{6}{6}x^3 + frac{60}{120}x^5 + frac{546}{5040}x^7 + O(x^8)#
#= x^3 + frac{1}{2}x^5 + frac{13}{120}x^7 + O(x^8)#
