Suppose f(0) = 1, f'(0) = 1, and f''(0) = -1. Use a Taylor polynomial of degree two to approximate f $\left( \frac { 1 } { 2 } \right)$ . Is $f\left( \frac { 1 } { 2 } \right) \approx \frac { 11 } { 8 }$ the solution?

Question

Suppose f(0) = 1, f'(0) = 1, and f''(0) = -1. Use a Taylor polynomial of degree two to approximate f

\left( \frac { 1 } { 2 } \right)

. Is

f\left( \frac { 1 } { 2 } \right) \approx \frac { 11 } { 8 }

the solution?

Quizplus · Accepted Answer

The Taylor polynomial of degree two for f(x) is $f(x) \approx 1 + x - \frac { x^2 } { 2 }$. So, $f\left( \frac { 1 } { 2 } \right) \approx 1 + \frac { 1 } { 2 } - \frac { 1 } { 8 } = \frac { 11 } { 8 }$.

Suppose F(0) = 1, F'(0) = 1, and F''(0) = (12)\left( \frac { 1 } { 2 } \right)(21​)

Suppose F(0) = 1, F'(0) = 1, and F''(0) = $\left( \frac { 1 } { 2 } \right)$