From: jos.bergervoet@nxp.com   
      
   On 19/05/25 4:06 PM, MM wrote:   
   > On Sat, 25 May 2019 10:14:13 +0100, mfreeman1943 wrote:   
   >   
   >> On Friday, May 24, 2019 at 3:29:19 PM UTC-7, MM wrote:   
   >>> Hello all.   
   >>> Straight to the point, stemming from a problem concerning a phi^3   
   >>> potential.   
   >>>   
   >>> How to find out the series expansion for s = s(t) given by the implicit   
   >>> equation:   
   >>>   
   >>> 18 t + s(1+s) (1+2s) =0   
   >>>   
   >>> There actually exists a closed expression for series s(t) = sum_k s_k   
   >>> t^k   
   >>   
   >> Since t is a cubic in s, you could use the solution of the cubic to get   
   >> s(t) - no series necessary.   
   >   
   > I need the series expansion, that's precisely the point. t> 0.   
      
   You could start in the usual way with Mathematica:   
      
   <0. Plot   
    them to be sure: *)   
      
   Plot[{s2}, {t,-2,2}] (* and the same for the others *)   
      
   (* Now let's expand in a point a>0, up to 7th power: *)   
      
   ser = Series[s2, {t,a, 7}]   
      
   (* or did you want the expansion to be around t=0 ? *)   
      
   ser = Series[s2, {t,0, 7}]   
      
   (* Anyhow, you can now ask for the series coefficients: *)   
      
   SeriesCoefficient[ser, 0]   
   SeriesCoefficient[ser, 1]   
    (* ... *)   
   SeriesCoefficient[ser, 6]   
   SeriesCoefficient[ser, 7]   
      
      
   >   
   > I don't see any obvious way to find out the generic s_k .   
      
   If you want a closed form expression for any k then the above   
   approach does not immediately help..   
      
   And if you want at least to have reasonably short expressions,   
   you will have to stick to the a=0 case. For general a, the 7th   
   coefficient, for (t-a)^7, spans more than 1000 lines of text,   
   whereas the 6th already needs a few pages:   
      
    2/3   
   Coeff[6]= (6428310336 3 (1 + I Sqrt[3])   
      
    3   
    (-245268 Sqrt[3] a + 69536102400 Sqrt[3] a +   
      
    5 7   
    5996269108297728 Sqrt[3] a - 900884751948397412352 Sqrt[3] a +   
      
    9   
    14010559662301476556898304 Sqrt[3] a +   
      
    11   
    301696925355848164725529509888 Sqrt[3] a +   
      
    2 2 2   
    143 Sqrt[-1 + 34992 a ] - 338162688 a Sqrt[-1 + 34992 a ] +   
      
    4 2   
    3565569466368 a Sqrt[-1 + 34992 a ] +   
      
    6 2   
    5632312076917014528 a Sqrt[-1 + 34992 a ] -   
      
    8 2   
    169643528674590528110592 a Sqrt[-1 + 34992 a ] -   
      
    10 2   
    2793490049591186710421569536 a Sqrt[-1 + 34992 a ])) /   
      
    2 11/2 2 19/3   
    ((-1 + 34992 a ) (-324 a + Sqrt[3] Sqrt[-1 + 34992 a ]) ) -   
      
    1/3   
    (6428310336 3 (1 - I Sqrt[3])   
      
    3   
    (-59940 Sqrt[3] a + 91409741568 Sqrt[3] a -   
      
    5 7   
    4491912250146816 Sqrt[3] a - 351824814370380054528 Sqrt[3] a +   
      
    9   
    13222983823296363969380352 Sqrt[3] a +   
      
    11   
    63820503440660188691938934784 Sqrt[3] a -   
      
    2 2 2   
    143 Sqrt[-1 + 34992 a ] - 255161664 a Sqrt[-1 + 34992 a ] +   
      
    4 2   
    46332812021760 a Sqrt[-1 + 34992 a ] +   
      
    6 2   
    1327185513742860288 a Sqrt[-1 + 34992 a ] -   
      
    8 2   
    130878830945781380284416 a Sqrt[-1 + 34992 a ] -   
      
    10 2   
    590930587413520265666101248 a Sqrt[-1 + 34992 a ])) /   
      
    2 11/2 2 17/3   
    ((-1 + 34992 a ) (-324 a + Sqrt[3] Sqrt[-1 + 34992 a ]) )   
      
      
      
      
   --   
   Jos   
      
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