The function $f$ has derivatives of all orders for all real numbers, and $f^{(4)}(x)=e^{\sin{x}}$. If the third-degree Taylor polynomial for $f$ about $x=0$ is used to approximate $f$ on the interval $[0,1]$, what is the Lagrange error bound for the maximum error on the interval $[0,1]$ ?

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The error is given by:

$$R_n(x)= \frac{f^{n+1}(z)}{(n+1)!}(x-c)^{n+1}$$

The value of $z$ that will maximize the error on the interval $[0,1]$ is $z=1$ since sine increases until $\pi/2$.

Therefore, the maximum error when $n=3$ (3rd degree polynomial), $c=0$ (centered around $x=0$) and $z=1$ is:

$$R_3(x)= \frac{f^{3+1}(1)}{(3+1)!}(1-0)^{3+1}$$ $$= \frac{f^4(1)}{4!}$$ $$= \frac{e^{\sin1}}{4!}$$ $$\approx \boxed{0.097}$$