Mfrhtyj

Given a function $f(x)$, how do I calculate it's value with given error range?

For example, how do I calculate $g(-0.1)$ with an error $< 0.001$ when:

$$g(x)=sin(x)+cos(x)$$

With
$$
f(x)=f(0)+frac{f^{(1)}(0)}{1!}x+cdots+frac{f^{(n)}(0)}{n!}x^n+R_n(x)
$$

Use the Lagrange form for the remainder
$$
R_n(x)=frac{f^{(n+1)}(c)}{(n+1)!}x^{(n+1)}
$$
where $c$ is between $x$ and $0$.

The derivatives of $g(x)$ are always $pmsin(x)pm cos(x)$ and so
$$
| f^{(n+1)}(c) | le 2
$$
for any $c$.

Then considering the magnitude of the error, find n such that
$$
2 frac{(0.1)^{(n+1)}}{(n+1)!} lt 0.001
$$

The Taylor series of $g(x) = sin(x) + cos(x)$ is
$$
1+x-frac{x^2}{2!}-frac{x^3}{3!}+frac{x^4}{4!}+frac{x^5}{5!}- ...
$$
You want to evaluate the function $g(-0.1)$ such that it's error is smaller than $0.001$.
From the Taylor series you can see that the third term $frac{x^3}{3!}$ will be of that order, because $frac{-0.1^3}{6} sim 0.001$. The next term in the series will always be an order smaller so the error will be of the order of $0.0001$ or $mathcal{O}(x^4).$

What you are left with is
$$
g(x) approx 1+x-frac{x^2}{2!}-frac{x^3}{3!}.
$$
Now you just fill in $-0.1$ to retrieve the value of the function in that point.