# Taylor's Theorem in simple english

Hi guys, I'm having trouble conceptualizing Taylor's Theorem. I understand how to compute a taylor series, as well as figure out whether it converges or not. I also realize that since it is an approximation that there is a remainder that is the difference between it and the original function. I understand all that. My main issue of concern is with the algebraic notation of the remainder. In my book, which is written very poorly, it gives the remainder in two dfferent forms:
"If a function f is differentiable through order n+1 in an interval I containing c, then, for each x in I, there exists z between x and c such that
f(x) = f(c) + f'(c)(x-c) + ... + Rn(x)

where

Rn(x) = (fn+1z)/(n+1)! (x-c)n+1."
Forgive me for not knowing the coding and such. So, the equation(s) might look a bit sloppy.
The book goes on to say:
"One useful consequence of this theorem is that |Rnx| $$\leq$$ |x-c|n+1/(n+1)! max |fn+1z|"
This part does not make sense to me. Again, sorry for making the formulas appear sloppy. I do not quite understand what that last inequality means. And, in addition, I do not know what max |fn+1z| refers to. Thanks for any help, and for even reading this.

By max |fn+1z| I guess they mean the maximum value of |fn+1z| for z in the interval I. The usefulness of the inequality is that it gives a bound on the error in using a Taylor polynomial to approximate the function f.

Quick example: Take f(x) = exp(x), which has Taylor series 1 + x + x2/2 + ..., and let I be the interval [-1, 1]. The 2nd Taylor polynomial is T2(x) = 1 + x + x2/2, and the remainder is bounded by |R2(x)| ≤ (|x|3 / 3!) max |f3(z)|. Now f3(z) = exp(z), which has maximum value e on the interval I. Thus, |R2(x)| ≤ e/6 |x|3 for all x in I.

Since |x| ≤ 1 on I, you get a uniform bound |R2(x)| ≤ e/6 = 0.453.

When its an alternating taylor series, the maximum possible error is the next term from your approximation (the first term you dropped.).

When its nonnegative, you use the lagrange error bound. Where you need to find the max value and stuff.

HallsofIvy
$$f(x)= e^{-\frac{1}{x^2}}$$
if $x\ne 0$, 0 if x= 0. It can be shown that all derivatives exist and are equal to 0 at x= 0 so its Taylor series at x= 0 is just 0, but clearly the function is not 0 except at x= 0.