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The information I'm given is that we are approximating $f(8)$ with a second degree Taylor polynomial for f centered around $a = 10$. Assume that the $abs(f'''(x)) \lt 3$ for $x$ on the interval $[6, 11]$.

This is all the information I have. The only way I know how to approximate the error of a Taylor polynomial is using the formula $f(x) - pn(x) =( \frac{(f^{(n+1)})(c)}{(n+1)!})(x - a)^{(n+1)}$ where $c$ is on the interval $[a, x]$, and $pn(x)$ is the n-th degree Taylor polynomial approximation of $f$ and the weird looking $f^{(n+1)}(c)$ is suppose to be the n+1th derivative of$f$ of c.

So I tried filling in the formula like this : $f(8) - p2(8)$ = (blank / 3!)(8-10)^3 and I'm trying to figure out what blank is suppose to be. So the interval should be $[10, 8]$ which is empty and besides that I have no idea of what to put where blank is.

Richard J
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jtht
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1 Answers1

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You can use the bound on the magnitude of the 3rd derivative, $3$, for the worst case for the bound. See this wikipedia article for details.

If $a$ is the center of the taylor series, then on the interval $(a-r,a+r)$, you can bound the remainder $R_k(x)$ for the $k$-th order polynomial as $M \frac{|x-a|^{k+1}}{(k+1)!} \leq M \frac{r^{k+1}}{(k+1)!}$ where $|f^{(k+1)}(x)| \leq M$ on the interval.

Batman
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  • How do I pick r and do you know why I get the interval [10, 8]? – jtht Feb 14 '14 at 07:01
  • You only need the third derivative between $a$ and $x$, since the point $c$ of the remainter formula will be inside this segement. That is why consideration of $[8,10]$ is sufficient, and the bound for $[6,11]$ is of course also valid on the smaller interval. – Lutz Lehmann Feb 14 '14 at 11:55