Speeding up single-variable series expansion

Question

I am facing the following problem: I want to expand the expression

exp=Tan[Sum[Subscript[a, k] x^((2 k + 1)/2), {k, 0, n}]];

in a series around $x=0$. Using

Series[exp /. n -> 2, {x, 0, 2}]

works as intended and gives a result within a moment.

The problem is expanding up to higher orders in $n$. Run times explode for $n>10$, with a wall time for $n=10$ of $88\, \mathrm{s}$ and for $n=11$ I was not able to get a result at all, having it run for more then 15 minutes, on an [email protected] machine. Of course computing higher derivatives of this expression gets harder with every order but given the rather simplistic form of the argument of $\tan$ I would expect it to work faster given the fact that there is nothing to fancy involved in terms of algebra.

Are there any tricks to speed up this expansion? I tried using assumptions like

$Assumptions = (And @@ (Element[Subscript[a, #], Reals] &) /@ 
Range[1, 15])

but without success.

In my despair I turned to my second computer algebra program - Maple - and it can expand this expression with its series() method much much faster. $n=20$ takes only around $.5\,\mathrm{s}$.

Which leave me with to possible conclusions: Either I am not aware of steps to take to enable Mathematica to use its Series[] method much faster or Mathematica's Series[] method is not very good.

I would really like to keep all computations in Mathematica since I need the expanded result for further processing.

Answer 1

By default, Series doesn't expand the coefficients of the series, and so the leaf count of the series can grow very quickly. For example:

LeafCount /@ Series[Tan[Sum[Subscript[a,k] x^((2 k+1)/2),{k,0,17}]], {x,0,17}][[3]]

{3,1,13,1,68,1,176,1,412,1,928,1,1990,1,4143,1,12045,1,31616,1,74275,1,163011,1,343993,1,709771,1,1445670,1,2922569,1,6810467}

Notice how the leaf count more than doubles for each nonzero term of the series. On the other hand, after simplification:

LeafCount /@ Simplify /@ Series[Tan[Sum[Subscript[a,k] x^((2 k+1)/2), {k,0,17}]],{x,0,17}][[3]]

{3,1,13,1,22,1,43,1,80,1,124,1,190,1,287,1,409,1,580,1,809,1,1105,1,1505,1,2033,1,2692,1, 3573}

The leaf count grows much more slowly. So, one idea is to compose the series by hand. For example:

tanSeries[n_] := Normal @ Series[Tan[x], {x, 0, n}]
argSeries[p_, n_] := Series[Sum[Subscript[a, k] x^(k+1/2), {k, 0, n}]^p, {x, 0, n}]

Here, argSeries is the series of $\arg ^p$ where arg is the argument of Tan, and tanSeries is the normal form of the series approximation of Tan. Since the leading term of argSeries[p, n] is $x^{p/2}$, to compute the desired series to order $n$ we need to have to keep twice as many terms of the Tan series. So:

r10 = tanSeries[20] /. x^p_. :> argSeries[p, 10]; //Timing
r20 = tanSeries[40] /. x^p_. :> argSeries[p, 20]; //Timing

{0.057843, Null}

{4.73507, Null}

The leaf count of this approach is still high:

LeafCount /@ r20[[3]]

{3, 1, 13, 1, 22, 1, 57, 1, 121, 1, 262, 1, 542, 1, 1049, 1, 1921, 1, 3363, 1, 5679, 1, 9322, 1, 14869, 1, 23094, 1, 35046, 1, 52162, 1, 76285, 1, 109893, 1, 155998, 1, 218370}

but not nearly as high as the naive application of Series in M11.1.1. I expect that this approach will scale better as higher order terms are needed.

Answer 2

When I try executing the following, I get answers rapidly

exp = Tan[Sum[Subscript[a, k] x^((2 k + 1)/2), {k, 0, n}]];
Table[Timing[Series[exp, {x, 0, n}];], {n, 2, 20}]

{{0., Null}, {0., Null}, {0., Null}, {0., Null}, {0., Null}, {0., 
  Null}, {0., Null}, {0., Null}, {0.028, Null}, {0.036, Null}, {0.04, 
  Null}, {0.048, Null}, {0.064, Null}, {0.092, Null}, {0.128, 
  Null}, {0.216, Null}, {0.4, Null}, {0.764, Null}, {1.444, Null}}

I am using

$Version
(* "11.1.1 for Linux x86 (64-bit) (April 18, 2017)" *)

Perhaps I've misunderstood your question, or you are using an older version?