Re: Can it be done - easily?

*To*: mathgroup at smc.vnet.net*Subject*: [mg13298] Re: Can it be done - easily?*From*: Tobias Oed <tobias at physics.odu.edu>*Date*: Fri, 17 Jul 1998 03:18:49 -0400*Organization*: Old Dominion University*References*: <6od25q$hn9@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Barry Culhane wrote: > > Myself and two workmates are software developers. One guy wanted a > formula to calculate a result for the following equation... > Z = sum of X/Y where X is a fixed number, and Y ranges from A-B in > fixed steps... > i.e... X=10000 ; Y=100,200,300...1000 > i.e... Z= 10000/100 + 10000/200 + ... 10000/1000 = 292.896 > > He and I tried to figure out a simple formula to calculate it, but > couldn't. The third guy said it was *not* *possible* to derive a > formula - we think he's wrong, but can't prove it. MathCad can solve > it in the blink of an eye, even if the value of Y ranges from 1 to 1e6 > in steps of 1 !!! > > Can anyone come up with a simple formula to give a reasonably accurate > result? It is too slow to actually divide X by Y for each value of Y > as there may be 1000 or even 100,000 values of Y. > > Thanks in advance... > > Barry Culhane > > Schaffner Ltd, Limerick, IRELAND This is what mathematica says: ***** In[214]:= Sum[X/i,{i,A,B,C}] //InputForm Out[214]//InputForm= (X*(-PolyGamma[0, A/C] + PolyGamma[0, 1 + A/C + Floor[(-A + B)/C]]))/C I don't know what polygamma that's the online help: ***** In[215]:= ??PolyGamma PolyGamma[z] gives the digamma function psi(z). PolyGamma[n, z] gives the nth derivative of the digamma function. Attributes[PolyGamma] = {Listable, NHoldFirst, Protected} hope you can do something with this Tobias