Concept preview

What you will be able to do

1. 01State the Lagrange error bound $|R_n(x)|\le\dfrac{M}{(n+1)!}\,|x-a|^{n+1}$, where $M$ is any bound on $|f^{(n+1)}|$ over the interval between the centre $a$ and the input $x$.
2. 02Apply the bound to guarantee a maximum error for a given Taylor approximation, choosing a valid $M$ for the $(n+1)$-th derivative and computing the ceiling numerically.
3. 03Determine the smallest degree $n$ needed so that a Taylor approximation is guaranteed accurate to a stated tolerance, by solving the inequality $\dfrac{M}{(n+1)!}|x-a|^{n+1}\le \text{tolerance}$.

Define the remainder as the exact gap between the function and its Taylor polynomial: $R_n(x)=f(x)-T_n(x)$. This is the true error you commit by using $T_n$ in place of $f$. The Lagrange form of the remainder states that there exists some point $c$ between $a$ and $x$ for which $R_n(x)=\dfrac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$. You cannot generally find that $c$, but you do not need to. If you know that $|f^{(n+1)}|$ never exceeds some number $M$ on the whole interval between $a$ and $x$, then replacing $f^{(n+1)}(c)$ by its worst case gives the Lagrange error bound: $$|R_n(x)|\le\dfrac{M}{(n+1)!}\,|x-a|^{n+1}.$$