Re: Can it be done - easily?

*To*: mathgroup at smc.vnet.net*Subject*: [mg13258] Re: Can it be done - easily?*From*: Paul Abbott <paul at physics.uwa.edu.au>*Date*: Fri, 17 Jul 1998 03:17:51 -0400*Organization*: University of Western Australia*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. He's wrong -- sort of. There is a formula, but it is not that simple: In[1]:= f[x_,a_,b_,d_] = x Sum[1/(a + i*d), {i, 0, (b-a)/d}] Out[1]= b a x (PolyGamma[0, 1 + -] - PolyGamma[0, -]) d d ----------------------------------------- d For your parameter values, x=10000, a=100, b=1000, d=100, the exact sum is In[2]:= f[10000,100,1000,100] Out[2]= 36905 ----- 126 with numerical value, In[3]:= N[%] Out[3]= 292.897 > 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 !!! But Mathematica can give you the exact formula for any X and Y! Also, using NSum (numerical summation) instead of Sum, we find In[4]:= Nf[x_,a_,b_,d_] := x NSum[1/(a + i*d), {i, 0, (b-a)/d}] In[5]:= Nf[x,1,10^6,1]//Timing Out[5]= {0.14 Second,14.3927 x} which is the correct result for any value of x. > 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. Obviously, X is a constant that can should be factored out of the computation (as is done above). For appropriate parameter ranges, NSum uses the Euler-Maclaurin formula which greatly speeds up the computation. There is a simple asymptotic formula which you can use: For large n, the harmonic series Sum[1/i, {i, 1, n}] is well approximated by 1 1 1 1 1 5 (EulerGamma - Log[-]) + --- - ----- + ------ + O[-] n 2 n 2 4 n 12 n 120 n where EulerGamma is 0.577215664901532860606512090... Inserting n = 10^6 into this truncated asymptotic series, one gets 14.3927 which agrees with the above computation using NSum. Cheers, Paul ____________________________________________________________________ Paul Abbott Phone: +61-8-9380-2734 Department of Physics Fax: +61-8-9380-1014 The University of Western Australia Nedlands WA 6907 mailto:paul at physics.uwa.edu.au AUSTRALIA http://www.pd.uwa.edu.au/~paul God IS a weakly left-handed dice player ____________________________________________________________________