f(x) = ((e^x) - 1)/ x if x =/= 0
f(x) = 1 if x = 0
Find the Taylor Polynomial of degree n for f at 0 and an estimate for the remainder term.
Detailed answer would be greatly appreciated but all help is much appreciated.
That's not true at all =/ the derivatives of that function are NOT 0 at x=0 - just because you obtained the piece of information that the function is defined there, that doesn't mean the limit corresponding to the dervative is 0 there. The definition of derivatives deals with how the function changes as x changes - it doesn't care how the function was defined.
Edit - I suppose you could keep differentiating and apply L'Hopital's to find the derivatives. But they still won't be zero.
And the question isn't badly worded, either. They meant they want the series centered at x=0, so the form will be $\displaystyle a_n x^n$, instead of $\displaystyle a_n (x-c)^n$
@OP: In this specific case, you start with the expansion for e^x:
$\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{n!}$
Subtract 1:
$\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n!}$ (notice how i simply adjusted the index to ommit the constant term)
then divide by x:
$\displaystyle \sum_{n=1}^{\infty} \frac{x^{n-1}}{n!}$
or equivalently
$\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{(n+1)!}$
Can you work from there?
For the remainder, unfortunately I haven't worked too much with that so I can't confidently provide an answer, but this might help:
Taylor's theorem - Wikipedia, the free encyclopedia