Taylor series

  • #1
I can't work out how to calculate the Taylor series for

[tex] \frac{1}{|R-r|} [/tex]

when R>>r, but they are both vectors. We were told to expand in r/R but I did the step below and I'm not sure where to go from there

I got to

[tex] \frac{1}{R \sqrt{1 - (2R.r)/R^2 + (r^2)/(R^2)}} [/tex]

I also know the result in first order is

[tex] \frac{1}{R} + \frac{R.r}{R^3} [/tex]
 

Answers and Replies

  • #2
kuruman
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Simplify and rewrite your expression as
[tex]
\frac{1}{R \sqrt{1 - 2r/R + (r/R)^2}}
[/tex]

Now define a (small) quantity ε =r/R in terms of which you have

[tex]
\frac{1}{R \sqrt{1 - 2r/R + (r/R)^2}} =\frac{1}{R}f(\epsilon)
[/tex]

Where
[tex]f(\epsilon)=(1-2\epsilon+\epsilon^2)^{-1/2}[/tex]

Expand f(ε) in Taylor series the usual way, then replace ε with r/R.
 

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