Question

In Taylor and fourier series, ,point out the range where the expansion provides a “good” approximation and please explain why.

Answer 1

f(x) =
∞ X n=0
an (∆x)n
= a0 + a1∆x + a2 (∆x)2 + a3 (∆x)3 + ... (2)

an =
1/ n!*dnf/ dxn
x=x0
. (5)

In order to nd the Taylor series expansion we need only to take derivatives of f(x) evaluated only at the point of reference, x = x0. Eqs. (2) and (5) dene the Taylor series expansion of a functions of a single variable. Functions which can be represented by a Taylor series are known as analytic functions. Notice from eq. (2) that as x → x0 and ∆x →0 the higher order (large n) terms in the power series expansion go to zero very quickly. Hence, if one is interested in f(x) suciently close to x0, a Taylor series expansion truncated to include only a few leading terms may often be sucient to approximate the function. Geometrically, we can think of this in terms of a “local” description of a function near x = x0.
f(x) = f(x0) |{z} constant
+∆xf0(x0) | {z } linear
+
(∆x)2 2! f00(x0) | {z } parabolic
+
(∆x)3 3! f000(x0) | {z } cubic
+... (6)
2
This shows that suciently close a point of interest that analytic functions are well approximated by constant plus a sloped, linear correction plus a parabolic correction plus .... Further away from a given reference point at x = x0, the less and less a function looks like a straight line. In order to get a better a approximation, you need functions with more wiggles (e.g. higher order polynomials).