In Calculus 2 you used Taylor's inequality to bound errors. Taylor's theorem says exactly what the error is, up to one unknown point, and it is the mean value theorem taken to higher order. (The case $n = 0$ is the MVT itself: $f(x) = f(a) + f'(c)(x - a)$.)
Key idea
Taylor's theorem. If $f$ has $n + 1$ derivatives on an interval containing $a$ and $x$, then $$f(x) = P_n(x) + \frac{f^{(n+1)}(c)}{(n + 1)!}(x - a)^{n+1}$$ for some $c$ between $a$ and $x$, where $P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x - a)^{k}$. If $|f^{(n+1)}| \le M$ between $a$ and $x$, then $|f(x) - P_n(x)| \le \frac{M|x - a|^{n+1}}{(n + 1)!}$.
Why it works
Fix $x \ne a$ and choose the number $K$ so that $g(t) = f(t) - P_n(t) - K(t - a)^{n+1}$ has $g(x) = 0$. By construction $g(a) = g'(a) = \cdots = g^{(n)}(a) = 0$. Rolle's theorem on $[a, x]$ gives $c_1$ with $g'(c_1) = 0$; Rolle again on $[a, c_1]$ (since $g'(a) = 0$) gives $c_2$ with $g''(c_2) = 0$; after $n + 1$ steps, $g^{(n+1)}(c) = 0$ for some $c$ between $a$ and $x$. But $g^{(n+1)}(t) = f^{(n+1)}(t) - (n + 1)!\,K$, because $P_n$ has degree $n$. So $K = \frac{f^{(n+1)}(c)}{(n + 1)!}$, which is the formula. $\blacksquare$
Worked examples
Worked example: How many terms for $e$?
Approximating $e = e^{1}$ by $P_n(1) = \sum_{k=0}^{n} \frac{1}{k!}$, using $e^{c} < 3$ for $0 \le c \le 1$, find the smallest $n$ with error bound below $0.001$.
- The bound is $\frac{3}{(n + 1)!}$.
- $n = 5$: $\frac{3}{720} \approx 0.0042$, too big. $n = 6$: $\frac{3}{5040} \approx 0.0006$.
Answer: $n = 6$ (or more).
Worked example: A sine error
Bound the error of $\sin x \approx x - \frac{x^{3}}{6}$ at $x = \frac{1}{2}$.
- This is $P_3$ (the $x^{2}$ and $x^{4}$ terms are $0$, so it is also $P_4$). Using $P_4$: $|R_4| \le \frac{|x|^{5}}{5!}$ since every derivative of $\sin$ is at most $1$ in size.
- $\frac{(1/2)^{5}}{120} = \frac{1}{3840}$.
Answer: Error at most $\frac{1}{3840} \approx 0.00026$.
Worked example: Finding the point $c$
For $f(x) = x^{4}$, $a = 0$ and $n = 2$, find $c$ with $f(x) = P_2(x) + \frac{f'''(c)}{3!}x^{3}$ at $x = 6$.
- $P_2 = 0$ (all derivatives up to order $2$ vanish at $0$), and $f'''(t) = 24t$.
- So $6^{4} = \frac{24c}{6}\cdot 6^{3}$, that is, $1296 = 864c$.
Answer: $c = \frac{3}{2} = \frac{6}{4}$, which is between $0$ and $6$.
Common mistake
The wrong derivative, or a guessed $c$. The remainder after $P_n$ uses the $(n + 1)$st derivative. And since $c$ is unknown, you cannot plug in $c = a$ or $c = x$; bound $|f^{(n+1)}|$ over the whole interval between $a$ and $x$ instead. For $e^{x}$ on $[0, 1]$ that bound is $e < 3$, not $e^{0} = 1$.