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Re: HypergeometricPFQ Question


$Version

"4.0 for Power Macintosh (April 20, 1999)"

Series[ HypergeometricPFQ[{1/2}, {3/2, 2}, -t^2] , {t, 0, 2} ]

\!\(\*
  InterpretationBox[
    RowBox[{"1", "-", \(t\^2\/6\), "+", 
      InterpretationBox[\(O[t]\^3\),
        SeriesData[ t, 0, {}, 0, 3, 1]]}],
    SeriesData[ t, 0, {1, 0, 
      Rational[ -1, 6]}, 0, 3, 1]]\)

However, back to the question of (0,0,1)

This is the partial derivative with respect to the third variable. 
The position of the "1" tells you which for which variable the partial 
derivative is taken. For example, look at

D[f[x, y, z], #] & /@ {x, y, z}

\!\(\*
  RowBox[{"{", 
    RowBox[{
      RowBox[{
        SuperscriptBox["f", 
          TagBox[\((1, 0, 0)\),
            Derivative],
          MultilineFunction->None], "[", \(x, y, z\), "]"}], ",", 
      RowBox[{
        SuperscriptBox["f", 
          TagBox[\((0, 1, 0)\),
            Derivative],
          MultilineFunction->None], "[", \(x, y, z\), "]"}], ",", 
      RowBox[{
        SuperscriptBox["f", 
          TagBox[\((0, 0, 1)\),
            Derivative],
          MultilineFunction->None], "[", \(x, y, z\), "]"}]}], "}"}]\)

InputForm[%]

{Derivative[1, 0, 0][f][x, y, z], Derivative[0, 1, 0][f][x, 
  y, z], Derivative[0, 0, 1][f][x, y, z]}

In general, if you see a notation that is unfamiliar, try looking at the 
InputForm. It is often clearer as to what function Mathematica is 
using and how to get Help.

Bob Hanlon

In a message dated 7/30/99 7:50:02 AM, mjlee at postech.ac.kr writes:

>    ------------------------------------------------------------------
>     In[3]:= Series[ HypergeometricPFQ[{1/2},{3/2,2},-t^2] , {t,0,2} ]
>
>     gives
>                                  (0,0,1)  1    3          2       3
>     Out[4]= 1 - HypergeometricPFQ       [{-}, {-, 2}, 0] t  + O[t]
>                                           2    2
>     ------------------------------------------------------------------
>
>     What does (0,0,1) mean?
>     The Mathematica Book does not have any explanation.
>


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