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:

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