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Re: Series expansion of Lambert series

  • To: mathgroup at smc.vnet.net
  • Subject: [mg128013] Re: Series expansion of Lambert series
  • From: Bob Hanlon <hanlonr357 at gmail.com>
  • Date: Sat, 8 Sep 2012 03:12:08 -0400 (EDT)
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Your first example can be written in closed form.Once there, Series
works as usual.

f[x_] = Assuming[{0 <= x < 1},
  Limit[Simplify[Sum[x^n/(1 - x^n), {n, 1, m}]], m -> Infinity]]

(Log[1 - x] + QPolyGamma[0, 1, x])/
   Log[x]

g4[x_] = Series[f[x], {x, 0, 4}] // Normal

x + 2*x^2 + 2*x^3 + 3*x^4

g10[x_] = Series[f[x], {x, 0, 10}] // Normal

x + 2 x^2 + 2 x^3 + 3 x^4 + 2 x^5 + 4 x^6 + 2 x^7 + 4 x^8 + 3 x^9 +
 4 x^10

Plot[{
  Tooltip[f[x], "f[x]"],
  Tooltip[g4[x], "g4[x]"],
  Tooltip[g10[x], "g10[x]"]}, {x, 0, 0.85}]

In the second example, write the series as a finite series and Series
is not needed.

e1[m_][x_] = Sum[x^n/(1 + n!), {n, 0, m}];

e1[4][x]

1/2 + x/2 + x^2/3 + x^3/7 + x^4/25

Plot[{
  Tooltip[e1[4][x], "e1[4][x]"],
  Tooltip[e1[10][x], "e1[10][x]"]},
 {x, 0, 5}, PlotRange -> All]


Bob Hanlon


On Fri, Sep 7, 2012 at 4:54 AM, Dr. Wolfgang Hintze <weh at snafu.de> wrote:
> Define for 0<=x<1 the Lambert series
>
> f[x_] := Sum[ x^n/(1-x^n), {x,1,oo}]
>
> How do I get Mathematica to calculate the first few terms of a series
> expansion like
>
> Series[f[x],{x,0,4}] ?
>
> Also a series like
>
> e1[x_] := Sum[x^n/(1+n!), {n,0,oo}]
>
> does not seem to be expandable with Series.
>
> BTW same thing in 8.0 and 5.2
>
> Regards,
> Wolfgang
>



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