Using the $5$-th order Maclaurin polynomial of $f(x) = e^x$ to approximate $f(-1)$

Question

Find the 5th-order Maclaurin polynomial $P_5(x)$ for $f(x) = e^x$.

I got $$P_5(x) = 1 + x +\frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + O(x^6) $$

From this answer, I'm supposed to approximate $f(-1)$, correct to the fifth decimal place. Is it right to just put in $x= -1$ like below?

$$\begin{align} & 1 + (-1) + \frac{(-1)^2}{2} + \frac{(-1)^3}{6} + \frac{(-1)^4}{24} + \frac{(-1)^5}{120} + O((-1)^6) \\ =\; & 1 -1 + \frac12 - \frac16 + \frac1{24} - \frac1{120} + O((-1)^6) \\ =\; & 0.5 - 0.166666 + 0.041666 - 0.008333 + O((-1)^6) \\ =\; & 0.366667 \\ =\; & 0.36667 + O((-1)^6) \end{align}$$

Is this done right?

Answer 1

$O(x^6)$ is defined in terms of asymptotics as $x\to \infty$.

When you replace $x$ by $-1$ it loses all meaning.

What you need here really is an upper bound on the error, to make sure that your answer has the expected precision. Therefore the $O(x^6)$ formula is helpless.

Instead you can use the Remainder estimation theorem:

$|f(x)-P_n(x)|\leq \left|\frac{||f^{(n+1}||_\infty}{(n+1)!}x^{n+1}\right|$

and see whether the right-hand side is small enough.

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Edit: note that by computing a number of terms you may observe that at some point the first 5 decimals do not change any more, but the above estimation theorem gives a formal proof that it won't change.

Answer 2

Yes, it looks like your method for evaluating $f(-1)$ there is correct. I'd avoid the notation $O((-1)^6)$, and instead just write $$f(-1)\approx1 + (-1) + {(-1)^2\over2} + {(-1)^3\over6} + {(-1)^4\over24} + {(-1)^5\over120}$$ and so on. However it's worth pointing out that this isn't an approximation within 5 decimals, and you ought to use more terms in the approximating polynomial to get close enough.