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Re: Can it be done - easily?

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  • Subject: [mg13279] Re: [mg13211] Can it be done - easily?
  • From: Wouter Meeussen <eu000949 at>
  • Date: Fri, 17 Jul 1998 03:18:25 -0400
  • Sender: owner-wri-mathgroup at


you need  the PolyGamma function :

if you don't have them on your home, kitchen & garden calculator (;-)), 
then try 
In[ ]:=N[EulerGamma,24]
Out[ ]=0.57721566490153286060651

and do a taylor series 'round infinity:
In[27]:=ser=Series[PolyGamma[0,x],{x,\[Infinity],12}]//Normal Out[27]=
691/(32760*x^12) - 1/(132*x^10) + 1/(240*x^8) - 
  1/(252*x^6) + 1/(120*x^4) - 1/(12*x^2) - 1/(2*x) - Log[1/x]

Check the quality of the approximation by comparing values between 2^1
and 2^12 :


looks good enough to me.


At 07:42 13-07-98 -0400, 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
Dr. Wouter L. J. MEEUSSEN
w.meeussen.vdmcc at
eu000949 at

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