Question

In Exercises \(7-18\), find the Maclaurin polynomial of degree \(n\) for the function. $$ f(x)=\sin x, \quad n=5 $$

Short answer

The Maclaurin polynomial of degree 5 for the function \( \sin x \) is \( P_5(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}\)

Step-by-step solution

Determine the first 6 derivatives of the function at zero

Because we are finding the Maclaurin polynomial of degree 5, we will need the first six derivatives at zero. The derivatives of \(\sin x\) follow a cycle: \(\sin x\), \(\cos x\), \(-\sin x\), \(-\cos x\), and so on. When evaluated at zero, the \(\sin\) terms will vanish, while \(\cos x\) evaluated at zero will yield 1 or -1, depending on the order of the derivative. So, the results are: \(f(0)=0, f'(0)=1, f''(0)=0, f'''(0)=-1, f''''(0)=0, f''''(0)=1 \)

Apply the Maclaurin series formula

The Maclaurin series is given by the formula: \(f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \frac{f''''(0)x^4}{4!} + .... \). We substitute the values of the derivatives at zero obtained in the previous step up to the 5th term.

Simplify the polynomial

After substitution, the Maclaurin polynomial simplifies to: \( P_5(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}\)