Taylor's Theorem For Error Approximation

Question

I'm trying to evaluate a function $f(t)$ with a given $t$ value to within 10$^{-5}$.

So, if I use Taylor's Theorem :

$f(a)+f′(a)1!(x−a)+f′′(a)2!(x−a)2+⋯+f(n)(a)n!(x−a)n.$

Would my $t$ value = $a, f(t) = f(a),$ and $ n= 5$? Meaning that I would have to evaluate the 5th derivative of my $f(x)$?

Answer 1

Use this:

Lagrange's Form of the Remainder: If $f^{(N+1)}$ is continuous on an open interval $I$ that contains $a$, and $x$ is in $I$, then there exists a number $c$ between $a$ and $x$ such that $$R_N(x)={f^{(N+1)}(c)\over (N+1)!}(x-a)^{N+1}.$$

You didn't tell us what your $f(x)$ was or the center of the series, $a$, or the $x$ value at which you want the approximation smaller than $10^{-5}$. But given those ingredients, you find the smallest positive integer $N$ which solves the inequality $$ |R_N(x)|<10^{-5}. $$

In practice, you can replace $|f^{(N+1)}(c)|$ with $\max_{x\in I}|f^{(N+1)}(x)|$ since the latter is an upper bound on the former. This will provide an upper bound on $|R_N(x)|$.