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 >
- References:
- Series expansion of Lambert series
- From: "Dr. Wolfgang Hintze" <weh@snafu.de>
- Series expansion of Lambert series