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I am performing a complicated summation in which fa is a Sum of a[j] over j, and terms in fa end up containing a BesselJ, a Hypergeometric0F1, or neither.

t[j_] := ((-1)^j)*(j - 1)!/(2*v^j);
DD[i_, j_] := t[i]*t[j]*F + (i + j)*G/v;
h[j_] := t[j]*(H + j*K/v);
m2[j_] = 4*W*(Z^(j - 1))*(v^(3*j - 1))/((j - 1)!^3);
a = Expand[h[i]*m2[i]*DD[i, j]*m2[j]*h[j] /. {i -> i + 1, j -> j + 1}];
fa = Sum[a, {j, 0, Infinity}];

I would like to create three separate functions a1, a2, and a3 respectively contain the terms from a that sum to BesselJ, Hypergeometric0F1, or neither. In the end, a1+a2+a3 should equal a. How can I do this?

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  • $\begingroup$ Unless you know a priori the coefficients of BesselJ and Hypergeometric0F1, it seems unlikely that you can find a unique decomposition. $\endgroup$ – bbgodfrey Feb 27 '15 at 0:47
  • $\begingroup$ @bbgodfrey It should certainly be possible to take one term at a time, Sum it, see what the result was, and add the term to the correct working function accordingly. $\endgroup$ – Jerry Guern Feb 27 '15 at 0:57
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fa = Sum[#, {j, 0, Infinity}] & /@ (List @@ a);
{bssTerms, hyperTerms} = (Position[fa, #] & /@ {_BesselJ, _Hypergeometric0F1})[[All, All, 1]]
indepTerms = Complement[Range@Length@fa, Join @@ {bssTerms, hyperTerms}]

For example:

Sum[a[[#]], {j, 0, Infinity}] & /@ hyperTerms

Mathematica graphics

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  • $\begingroup$ Perfect! And I learned some new syntax too. Thanks! $\endgroup$ – Jerry Guern Feb 27 '15 at 20:47

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