Replace the following function by its taylor polynomial of the given grade, and approximate the error in the given interval:
$f(x) = \sin(x) \textrm{ by } T_3(f,x,0) \textrm{ in } |x| \le\frac{1}{10}$
My solution and thoughts
We only need the first three derivatives:
$ f'(x) = \cos(x) \\ f''(x) = -\sin(x) \\ f'''(x) = -\cos(x) $
And by definition we know that
$ T_3(f,x,0) = \sum\limits_{k=0}^3 \frac{f^{(k)}(0)}{k!}x^k $
we get
$ = 0 + \frac{1}{1!}(x-0)^1 + 0 - \frac{1}{3!}(x-0)^3 $
Is this right so far? It looks suspiciously simple, which merely confuses me.
$ = x - \frac{x^3}{6} $
I am clueless when it comes to the error. In which points do I have to calculate $T_3(x)$ in order to get the error as
$ R_3(x) = sin(x) - T_3(x) $
in the given interval $[-\frac{1}{10};\frac{1}{10}]$?
Update
Oh, I'm reading in a book about the Lagrange representation of the error. Can I use it? $f(x)$ looks endlessly differentiable in $0$.