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Taylor approximation

  • Thread starter essif
  • Start date
4
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[SOLVED] Taylor approximation

1. Homework Statement
I have an exact funktion given as:

[tex]P(r)=1-e^{\frac{-2r}{a}}(1+\frac{2r}{a}+\frac{2r^2}{a^2})[/tex]

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

When [tex]r \prec \prec a[/tex]


3. The Attempt at a Solution
I am lost when it comes to these Taylor approximations. It should be a fairly easy problem, but dont know how to handle it.
Some help on how to do this would be appreciated.
 

Answers and Replies

CompuChip
Science Advisor
Homework Helper
4,284
47
In general, the Taylor series of a function f(r) around 0 is given by
[tex]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[/tex]
where [itex]f^{(n)}[/itex] denotes the nth derivative. Usually, one stops the expansion after a certain number of terms (e.g. [itex]f(r) \approx f(0)[/itex] 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 [itex]r^n, r^{n+1}, r^{n+2}[/itex], 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
 
4
0
Thanks for the quick reply

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

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