# Maclaurin expansion and error

1. Dec 15, 2014

### kidsasd987

my professor told me any n times differentiable function can be approximated by macularine/taylor expansion.

is that true? As far as I know, if the function is approximated at point a, the approximation is valid if we pick a point near a.

however, if we assume that we picked a point of a+infinite=b, how do we know the approximation can we conclude the approximation is valid or not?

For example, if we approximate a function with macularine series, and picked the point b=infinite for x, shouldn't we check the ratio test first?

ex.lima,n→∞(n+1)(a)*(x-a)(n+1)/(n+1)!]/[ƒ(n)(a)*(x-a)(n)/(n)!]

simplyfying this equation gives

lima,n→∞n+1(a)*(xn+1+ .... +an+1)]/[ƒn(a)*(n+1)*(xn....+an]

if it converges for every continuous smooth n time differetiable function, can anyone provie me proof?[/SUB]

Last edited: Dec 15, 2014
2. Dec 15, 2014

### Staff: Mentor

That's Maclaurin...
A Maclaurin series or a Taylor series has an interval of convergence, centered at a. Some series converge only at a, and others converge on an interval (a - r, a + r). Still others converge over the entire real number line.
If we're evaluating the series at a point within its interval of convergence.

3. Dec 15, 2014

### kidsasd987

yeah.. that was weird cause I found macularine converges but taylor somewhat behaves differently as x,a,n three variables reach to infinity

4. Dec 15, 2014

### Staff: Mentor

A Maclaurin series is a Taylor series. A Taylor series is an infinite polynomial in powers of x - a. A Maclaurin series is an infinite polynomial in powers of x - 0. There is no such thing as a "macularine" series.
I have no idea what you're trying to say here. In a Taylor series, a is fixed. If x is outside the interval of convergence then of course the series will fail to converge. The same is true for a Maclaurin series.

5. Dec 17, 2014

### FactChecker

You may have to be patient. You will only be able to completely understand convergence of Taylor series when you start into complex analysis. That is because the radius of convergence of a Taylor series is only as far as the nearest singularity in the complex plane.