## Maclaurin formula: finding a delta for a given error

Approximate the function $\displaystyle f(x)=\sin(x)$ using the corresponding Maclaurin polynomial: $\displaystyle P_5(x)$, in a bound $\displaystyle \epsilon(0,\delta)$. Determine a value of $\displaystyle \delta>0$, so that the rest $\displaystyle R_5(x)$ verifies $\displaystyle |R_5(x)|<0.0005$ for all $\displaystyle x\in{\epsilon(0,\delta)}$

Well, the first thing that puzzles me a bit is that the order for $\displaystyle P_5$ and $\displaystyle R_5$ is the same. I assume that it is a mistake, and that the polynomial must be $\displaystyle P_4$, or the rest $\displaystyle R_6$. I have chosen to leave the degree of the polynomial as it was and turn up one grade to the rest.

So I have to find a delta for which any value within the environment $\displaystyle (0,\delta)$ is less than the error 0.0005

So I did:

$\displaystyle R_6(x)=|\displaystyle\frac{\sin(\alpha)x^6}{6!}|<0 .0005$ $\displaystyle 0<\alpha<x$

So I must find x:

$\displaystyle R_6(x)=|\sin(\alpha)x^6|<0.36$ $\displaystyle 0<\alpha<x$

I must find x that satisfies the equation: $\displaystyle \sin(x)x^6<0.36$

I don't know how to go ahead.

Bye there.