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Let $T_n(x)$ be the $n^{th}$ Maclaurin polynomial for $f(x) = e^x$. Use the error formula to determine a value of $n$ so that $\lvert T_n(2)−e^2\rvert < 10^{−4}.$

I haven't seen a problem like this before and I can't find any hints or examples in my textbook. Can anyone show me how to solve this? Please explain how to solve it, as I want to understand how to get the answer.

Side note, this was on a sheet of "fun extra problems" that my calculus teacher handed out. So it is not graded, but he said it might be helpful for the next exam.

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You will find an interesting presntation about the estimate of the remaminder in Taylor series here

Taylor expansion of $e^x$ is given by $$T_\infty=\sum_{n=0}^\infty \frac {x^n}{n!}$$ So, if you limit the expansion to $p$ terms, the error is given by $$\Delta_p=\sum_{n=p+1}^\infty \frac {x^n}{n!}<\sum_{n=p+1}^\infty \frac {x^n}{(p+1)!}=\frac{x^{p+1}}{(1-x) (p+1)!}$$ and you want $|\Delta_p|<\epsilon$.

In your case, $x=2$ so you search for $p$ such that $$|\Delta_p|=\frac{2^{p+1}}{ (p+1)!}<\epsilon$$ Let us try computing (this can be made easy if you notice that $\frac{|\Delta_{k+1}|}{|\Delta_{k}|}=\frac{2}{k+2}$). you will then find that the smallest value of $p$ which satisfies $|\Delta_{p}|<10^{-4}$ corresponds to $p=10$.

Let us check using $$T_p=\sum_{n=0}^p \frac {2^n}{n!}$$ The table below gives the value of $p$ and the corresponding value of $T_p$ as well as the error with the exact value of $(e^2-T_p)$.

$$\left( \begin{array}{ccc} p & T_p & (e^2-T_p) \\ 2 & 5.00000 & 2.38906 \\ 3 & 6.33333 & 1.05572 \\ 4 & 7.00000 & 0.389056 \\ 5 & 7.26667 & 0.122389 \\ 6 & 7.35556 & 0.0335005 \\ 7 & 7.38095 & 0.00810372 \\ 8 & 7.38730 & 0.00175451 \\ 9 & 7.38871 & 0.000343577 \\ 10 & 7.38899 & 0.0000613899 \\ 11 & 7.38905 & 0.0000100832 \end{array} \right)$$

You will probably soon learn about more sophisticated methods for the evaluation of the remainder but the basic idea is here.

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