Taylor approximation

  • Thread starter essif
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  • #1
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[SOLVED] Taylor approximation

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]


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

  • #2
CompuChip
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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
 
  • #3
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Thanks for the quick reply

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

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