# Tag Info

## Hot answers tagged series-expansion

4

The bug is fixed at least since v12.2: Series[Pochhammer[1 + n, n], {n, Infinity, 1}] (* E^SeriesData[n, Infinity, {-1 + 2*Log + Log[n]}, -1, 3, 1]* SeriesData[n, Infinity, {Sqrt, 0, -1/12*1/Sqrt}, 0, 4, 2] *) Plot[{(2^(2 n) Gamma[1/2 + n])/Sqrt[π], % // Normal} // Evaluate, {n, 0, 2}, PlotStyle -> {Automatic, Dashed}]

3

\$Version (* "12.3.0 for Mac OS X x86 (64-bit) (May 10, 2021)" *) Clear["Global`*"] x << 1 and x >> 1 represent Asymptotic (v12.1 or later) behavior. For x << 1 Asymptotic[f[x], {x, 0, 4}] (* 1 - x - x^3 *) In earlier versions Series[f[x], {x, 0, 4}] // Normal (* 1 - x - x^3 *) For x >> 1 Asymptotic[f[x], {x, ...

3

Maybe this post answers your question: https://math.stackexchange.com/questions/51770/taylor-expansion-at-infinity Expansion at large x is equivalent to expansion at x-->1/x for small x. f[x_] := ((1 + x (1 - Sqrt[1 + x^2]))^2 - x + x^3 (1 - Sqrt[1 + x^2])^2)/(1 + x^2 (1 - Sqrt[1 + x^2])^2) Normal[Series[f[x], {x, Infinity, 4}]] Normal[Series[f[1/x]...

Only top voted, non community-wiki answers of a minimum length are eligible