Let f(x) =e^(x/2) . If the second degree Taylor polynomial for f about x = 0 is used to approximate f on the interval [0, 2], what is the Lagrange error bound for the maximum error on the interval [0, 2]?

Question

Let f(x) =e^(x/2)  . If the second degree Taylor polynomial for f about x = 0 is used to approximate f on the interval [0, 2], what is the Lagrange error bound for the maximum error on the interval [0, 2]?

anonymous · Answer

$$e ^{x/2}$$

anonymous · Answer

By Definition the error is
$$\Large 
\frac{x^3
   f^{(3)}(c)}{3!}=\frac{1}{48
   } e^{c/2} x^3
$$
What is its maximum on the interval [0,2]

anonymous · Answer

The maximum occurs  when c=x=2$$
\Large 
  \frac{1}{48
   } e^{2/2} 2^3\approx 0.453047


$$

anonymous · Answer

I'm confused, 1/48?

anonymous · Answer

$$
\Large
f^{(3)}(x)=\frac{e^{x/2}}{8}\
8 (3!)=8(6)=48

$$

anonymous · Answer

oh, I see. Thank you very much :]

anonymous · Answer

YW