A Taylor approximation

Suppose we write the estimate

Use Taylor’s remainder theorem to provide an upper bound for the error in that estimate.

Recall that the remainder from the Taylor Theorem is:
\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}.

n=5 and x=3 and x_0=0

and f(x)=e^x so f^6(x)=e^x

So \frac{e^{(\xi)}}{(6)!}(3)^{6}. Since the largest possible value for \xi is 3, the upper bound for the error is:
\frac{e^{(3)}}{(6)!}(3)^{6}

Taylor’s remainder theorem states that |T_n(x)-f(x)|=|R_n(x)| \leq \frac{|x-x_0|^{n+1}}{(n+1)!} max (f^{n+1}(\xi)), for some \xi \in [x_0,x]=[0,3]. Since we are expanding about the origin x_0=0 also f^{n+1}(x)=e^x for all n. This simplifies our expression R_n(x)\leq \frac{|x|^{n+1}}{(n+1!)}e^{\xi}, the value of \xi which maximizes this function is \xi = 3. So R_5(3) \leq \frac{3^6}{6!}e^3 \approx e^3. If this is correct than you would need many more terms to be sure that the error was negligible the actual error is 2.68

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