# Taylor's Theorem in simple english

1. May 5, 2009

### InterNational

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.

2. May 6, 2009

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.

3. May 10, 2009

### physicsnoob93

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.

4. May 11, 2009

### HallsofIvy

Staff Emeritus
Be careful about one thing: if a function has derivatives of all orders, we can, of course, form its Taylor series. but even if that series converges for all x, it does not necessarily converge to that function!

A function is said to be (real) analytic at a point if it is differentiable of all orders at that point (so its Taylor series exists at that point) and there exists some neighborhood of the point in which that series converges to the function.

An example of a function that has derivatives of all orders but is NOT analytic at a point is
$$f(x)= e^{-\frac{1}{x^2}}$$
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.

Last edited: May 11, 2009