Series expansion of a function up to linear terms

Question

I have the following:

\[CapitalSigma] = r^2 + a^2 Cos[\[Theta]]^2;
\[CapitalDelta] = r^2 - 2 M r + a^2 - k/3 r^2 (r^2 + a^2);
grr = \[CapitalSigma]/\[CapitalDelta];

and I want to obtain the expansion of grr up to linear terms in $M/r$ and $k r^2$. What I have done is to employ Series but I could not get the right answer which is $$1+\frac{2M}{r}-\frac{kr^2}{3}$$ Does anyone have an idea as to what I am missing here? Thanks

Does this answer your question? Multivariable Taylor expansion does not work as expected — Carl Woll, Aug 16 '21 at 16:30
That also bothers me @CarlWoll, although the instruction is to expand up to linear terms in $M/r$ and $kr^2$ only — user583893, Aug 16 '21 at 16:41
@CarlWoll, I am only expanding with respect to $r$ for large $r$. — user583893, Aug 16 '21 at 16:48
Is the required true? Limit[grr, r -> Infinity] produces 0. — user64494, Aug 16 '21 at 16:50

Ulrich Neumann · Answer 1 · 2021-08-16T18:56:10.017

0

Assuming k r^2, M/r, a/r to be small of same order, expansion gives nearly the result you're expecting:

grr /. {M -> r eps M/r , k -> 1/r^2 eps k r^2,a -> r  eps a /r } // Simplify  
Normal[Series[%, {eps, 0, 1}]] /. eps -> 1
(*1 + (2 M)/r + (k r^2)/3*)

edited Aug 16 '21 at 18:56

answered Aug 16 '21 at 18:43

Ulrich Neumann

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Pay your attention to the minus sign - k r^2/3 in the required expression. – user64494 Aug 16 '21 at 18:50
What a pity, thanks. – Ulrich Neumann Aug 16 '21 at 19:01
Your solution helps @UlrichNeumann, although the sign of the third term "should" be negative. – user583893 Aug 17 '21 at 01:40
Ok , I think you need to define the order of parameter a too. Otherwice it isn't possible to do the series expansion. – Ulrich Neumann Aug 17 '21 at 06:26

Series expansion of a function up to linear terms

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