Question

Use the definition to find the Taylor series (centered at \(c\) ) for the function. $$ f(x)=\sin 2 x, \quad c=0 $$

Short answer

The Taylor series for \(\sin(2x)\) centered at 0 is \[ \sum_{n=0}^{\infty} \frac{(-1)^n * (2x)^{2n+1}}{(2n+1)!} \]

Step-by-step solution

Identify the Function and Center

Identify the function as \(f(x) = \sin(2x)\) and the center of the Taylor series as \(c=0\).

Calculate Derivatives

Calculate the first few derivatives of the function evaluated at 0. \n The first few derivatives are: \n\n \(f'(x) = 2\cos(2x)\) \nEvaluation at \(x=0\) gives \(f'(0) = 2\),\n\n \(f''(x) = -4\sin(2x)\) \nEvaluation at \(x=0\) gives \(f''(0) = 0\),\n\n \(f'''(x) = -8\cos(2x)\) \nEvaluation at \(x=0\) gives \(f'''(0)= -8\), \n\n \(f''''(x) = 16\sin(2x)\) \nEvaluation at \(x=0\) gives \(f''''(0)= 0\), and so on.

Recognize Pattern

Recognize the pattern in the derivatives and their evaluations, the derivatives cycle every four steps. Rather than a random series of coefficients, they return to the original state over and over (2, 0, -8, 0, 32, 0, -128, 0, etc.).

Formulate the Taylor Series

Formulate the Taylor series using the pattern observed. The general expression for Taylor series is: \(f(x) = \sum_{n=0}^{\infty}\frac{f^n(c) * (x - c)^n}{n!}\) \nSubstituting the observed pattern, the Taylor series for \(\sin(2x)\) centered at 0 becomes: \[ f(x) = 2x - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} - \frac{(2x)^7}{7!} + ... = \sum_{n=0}^{\infty} \frac{(-1)^n *(2x)^{2n+1}}{(2n+1)!} \]