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The leading-order power in your series should be the limit as $x \to 0$ of $x f'(x)/f(x)$. So you could do something like leadcoeff = Limit[x func'[x]/func[x], x->0] Series[func[x], {x, 0, leadcoeff+2}] I'm not sure if the extra step of taking the limit kills off any efficiencies you might gain from only calculating the extra terms, though. Examples: ...


4

There is another approach that works to all orders—and I expect would be useful for other expressions arising in 1-loop integrals in QFT. Since Sum[Pochhammer[1 - eps, n]/Pochhammer[3 - eps, n]/2^n, {n, 0, Infinity}] is the sum you are examining, i.e., Hypergeometric2F1[1, 1 - eps, 3 - eps, 1/2] you can perform the series expansion on the summand to ...


7

Here a one-liner twitterable solution: Expand@Normal@Series[Hypergeometric2F1[1, 1 - eps/2, 3 - eps, x], {eps,0,0}]/.x->(1/2) giving 4 - 4 Log[2]


5

Your reply to Jens's answer suggests that you do not want to manually replace variables. This function just automates the method in Jens answer... cme[hgm_, symb_: c] := Module[ { places, symbs }, places = Position[hgm, eps]; symbs = Array[symb, Length[places]]; FullSimplify[Normal[Series[ ReplacePart[hgm, (Rule @@ # &) /@ Transpose[{places, ...


7

This does indeed work flawlessly with Mathematica version 8. But in version 10, I had to resort to the following workaround: AbsoluteTiming[ Series[Normal[ Series[Hypergeometric2F1[1, 1 - eps1, 3 - eps, 1/2], {eps1, 0, 1}, {eps, 0, 1}]] /. eps1 -> (eps/2), {eps, 0, 0}]] (* ==> {0.016225, SeriesData[eps, 0, {2 (EulerGamma + PolyGamma[0, ...



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