# Bounding approximation error for Taylor polynomial

I've got this problem:

Let $f(x) = e^x$. If we aproximate $f(x)$ by $P_4(x)$ in $x_0 = 0$ at $(-r, r)$, find $r \gt 0$ so that the error in the approximation is $\lt 10^{-5}$

What I did is:

1) $R_4(x) = |\displaystyle\frac{f^{(5)}_{(x)}}{5!} (x - 0)^5| = \frac{|f^{(5)}_{(x)}|}{120} |x|^5$, where $R_4(x)$ is the error in the approximation.

2) Now, because $f^{(n)}_{(0)} = 1 \forall n \in \mathbb{N}$ ($f^{(n)}$ is the nth-derivative of $f$), I can write: $R_4(x) = \displaystyle\frac{1}{120} |x|^5 \lt 10^{-5} = \frac{1}{10^5}$

3) Then I solved the inequality: $R_ 4(x) = |x|^5 < \frac{120}{10^5} \to |x| < \frac{\sqrt[5]{120}}{10}$, which leds me to $r = \frac{\sqrt[5]{120}}{10}$

I'm not pretty sure if I'm doing things right here, specially in step 2. Is the error well bounded? If not, any hint about how to solve this kind of problems will be very appreciated. Thanks a lot for your time!

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I think the $R_4(x) = \frac{|f'(\xi)|}{120}x^5$ , where $\xi\in(-r,r)$, thus the bound shall be $e^{r}$, not 1. –  Yimin Feb 15 '13 at 17:27
Usually to show the fifth derivative of $f(x)$ you would write it $f^{(5)}(x)$ without making the $(x)$ a subscript. –  Ross Millikan Feb 15 '13 at 17:41
@Yimin: you are correct and that is a good answer. Please post it that way so we can upvote it. –  Ross Millikan Feb 15 '13 at 17:42

$R_4(x) = \frac{f'(\xi)}{120}x^5$ ,where $\xi\in(-r,r)$ and $R_4(x)$ is the error.

Since $f'(\xi) = e^{\xi}$ ,thus it is bounded by $e^{r}$ on $(-r,r)$, thus

$R_4(x)\le \frac{e^r}{120}x^5$, as required, $R_4(x)\le 10^{-5}$, i.e

$\frac{e^r}{120}r^5\le 10^{-5}$, and solve this you can get the $r$.

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Now I see what was wrong.. thank you @Yimin for your answer! –  Lucas Feb 16 '13 at 14:13