Taylor Approximation Proof for P(r) using Series Expansion

[essif]

[SOLVED] Taylor approximation

Homework Statement

I have an exact funktion given as:

$$P(r)=1-e^{\frac{-2r}{a}}(1+\frac{2r}{a}+\frac{2r^2}{a^2})$$

I need to prove, by making a tayler series expansion, that:
$$P(r)\approx \frac{3r^3}{4a^4}$$

When $$r \prec \prec a$$

The Attempt at a Solution

I am lost when it comes to these Taylor approximations. It should be a fairly easy problem, but don't know how to handle it.
Some help on how to do this would be appreciated.

 

[CompuChip]

In general, the Taylor series of a function f(r) around 0 is given by
$$f(r) = \sum_{n = 0}^\infty \frac{1}{n!} f^{(n)}(0) r^n = f(0) + f'(0) r + 1/2 f''(0) r^2 + 1/6 f'''(0) r^3 + 1/24 f''''(0) r^4 + \cdots$$
where $f^{(n)}$ denotes the nth derivative. Usually, one stops the expansion after a certain number of terms (e.g. $f(r) \approx f(0)$ is the "first order expansion", including the first derivative term gives the "second order expansion", etc.*)

So basically, all you need to do for this problem is construct the Taylor series (you'll need to calculate the derivatives) and terminate it at a proper point. Or alternatively, if you know the expansion of the exponential, you can plug it in, expand the brackets, and again terminate it somewhere (that is, neglect all powers $r^n, r^{n+1}, r^{n+2}$, etc. for an nth order expansion).

*) I also see people calling f(0) the zero'th order expansion, f'(0) r the first order, etc. Sometimes that is confusing but so be it[/size]

 

[essif]

Thanks for the quick reply

By expanding the exponential to fourth order I was able to find the desired result