Created on 2024-06-20Asked by Amelia Torres (Solvelet student)
Find the Maclaurin series for the function f(x)=sin(x).
Solution
To find the Maclaurin series for f(x)=sin(x), we use the definition of the Maclaurin series: f(x)=n=0∑∞n!f(n)(0)xn 1. Compute the derivatives of f(x)=sin(x) at x=0: f(x)=sin(x)⟹f(0)=0f′(x)=cos(x)⟹f′(0)=1f′′(x)=−sin(x)⟹f′′(0)=0f′′′(x)=−cos(x)⟹f′′′(0)=−1f(4)(x)=sin(x)⟹f(4)(0)=0 2. Identify the pattern of the non-zero derivatives: f(n)(0)={0(−1)(n−1)/2if n is evenif n is odd 3. Write the series expansion: f(x)=n=0∑∞n!(−1)(n−1)/2xnfor odd n 4. Simplify the sum for the odd terms: f(x)=n=0∑∞(2n+1)!(−1)nx2n+1 Therefore, the Maclaurin series for f(x)=sin(x) is: sin(x)=n=0∑∞(2n+1)!(−1)nx2n+1Solved on Solvelet with Basic AI Model
Some of the related questions asked by Samuel Harris on Solvelet
DefinitionSo a Maclaurin series is a Taylor series that has a function f(x) and is expanded at x=0. It is represented by f(x) = f(0) + f'(0)x + 2! f′′(0)x2+3! f′′′(0)x3+⋯. The Maclaurin series is an infinite polynomial that approximates the function around 0. The Maclaurin series for ex, for example, is 1+x+2! x2+3! x3+⋯.
