Question

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

Short answer

The Maclaurin polynomial of degree 5 for the function \(f(x)=e^{-x}\) is \(P_5(x) = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \frac{x^5}{5!}\).

Step-by-step solution

Compute the derivatives of the function

Since the Maclaurin series requires us to compute the nth derivative of the function, start by computing the first derivative of \(f(x) = e^{-x}\), which equals \(-e^{-x}\). Then, proceed to compute the second, third, fourth, and fifth derivatives, following the same pattern. The second derivative equals \(e^{-x}\), third derivative equals \(-e^{-x}\), fourth derivative equals \(e^{-x}\), and the fifth derivative equals \(-e^{-x}\).

Compute the coefficients of the Taylor series

The coefficients of the Taylor series are determined by the nth derivative at zero divided by \(n!\). Hence, the coefficients become \(f(0)/1!, f''(0)/2!, f'''(0)/3!, f''''(0)/4!, f'''''(0)/5!\). After substituting the necessary values, the coefficients respectively equal 1, -1, 1, -1, 1, and -1.

Write out the Maclaurin polynomial

After taking each coefficient, place it in front of the corresponding degree of x in the polynomial. Hence, the Maclaurin polynomial of degree 5 for the function \(f(x)=e^{-x}\) will be \(P_5(x) = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \frac{x^5}{5!}\).