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the following sum represent Madelung constant of ionic crystal is not converget but it is possible to sum by different method, the sum is symetric for interger positive and negative so then $$\sum _{m=1}^{\infty } \sum _{k=1}^{\infty } \sum _{j=1}^{\infty } \frac{(-1)^{j+k+m} \left(j^2+k^2+m^2\right)}{\left(j^2+k^2+m^2\right)^{3/2}}=\sum _{m=-\infty }^{-1} \sum _{k=-\infty }^{-1} \sum _{j=-\infty }^{-1} \frac{(-1)^{j+k+m} \left(j^2+k^2+m^2\right)}{\left(j^2+k^2+m^2\right)^{3/2}}$$ the result using Mathematica as follow

Timing[NSum[(-1)^(j + k + m)/(j^2 + k^2 + m^2)^2^(-1), {k, 1, Infinity}, {j, 1, Infinity}, {m, 1, Infinity}, WorkingPrecision -> 10]]
{145.703125`, -0.1324763008136454975`5.743297493360561}

Timing[NSum[(-1)^(j + k + m)/(j^2 + k^2 + m^2)^2^(-1), {k, -Infinity, -1}, {j, -Infinity, -1}, {m, -Infinity, -1}, WorkingPrecision -> 10]]
{216.53125`, -0.1324763008136454975`5.743297493360561}

$$M=\sum _{m=-\infty }^{\infty } \sum _{k=-\infty }^{\infty } \sum _{j=-\infty }^{\infty } \frac{(-1)^{j+k+m} \left(j^2+k^2+m^2\right)}{\left(j^2+k^2+m^2\right)^{3/2}}=-1.74756459463318219\text{..}$$ using NSum as usually gives $$\sum _{m=1}^{\infty } \sum _{k=1}^{\infty } \sum _{j=1}^{\infty } \frac{(-1)^{j+k+m} \left(j^2+k^2+m^2\right)}{\left(j^2+k^2+m^2\right)^{3/2}}=-0.13247\text{...}$$ and $$\sum _{m=-\infty }^{-1} \sum _{k=-\infty }^{-1} \sum _{j=-\infty }^{-1} \frac{(-1)^{j+k+m} \left(j^2+k^2+m^2\right)}{\left(j^2+k^2+m^2\right)^{3/2}}=-0.13247\text{...}$$ Could you help me understand this contradiction?

Sum[(-1)^(m + k + j)*((j^2 + k^2 + m^2)/(j^2 + k^2 + m^2)^(3/2)), {m, -Infinity, Infinity}, {k, -Infinity, Infinity}, {j, -Infinity, Infinity}]
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  • $\begingroup$ Please add the codes you have tried. $\endgroup$ May 13 at 11:33
  • $\begingroup$ You mention that you used NSum, however in the code you provide the command is Sum. Is the problem with NSum or Sum? Or both? $\endgroup$ May 13 at 12:26
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the triple sum M converges only conditionally and very slowly . All three indices run from -Infinity to Infinity . To convert it into a form where they run from 1 to Infinity, you have to divide the ranges of each index into three parts : from - Infinity to - 1, 0, and 1 to Infinity . The symmetry of the summand makes the sums running from -Infinity to -1 equal to the sum running from 1 to Infinity. Don't forget the summand with index=0! If you do this for all three indices you get for a finite stop index n eight triple sums, twelve double sums and six simple sums, all running from 1 to Infinity:

Sum[If[m == 0 && k == 0 && j == 0, 0, (-1.)^(m + k + j)/
Sqrt[j^2 + k^2 + m^2]], {m, -n, n}, {k, -n, n}, {j, -n, n}] = 
8*Sum[(-1.)^(m + k + j)/Sqrt[j^2 + k^2 + m^2], 
{m, 1, n}, {k, 1, n}, {j, 1, n}] + 
12*Sum[(-1.)^(k + j)/Sqrt[j^2 + k^2], {k, 1, n}, {j, 1, n}] + 
6*Sum[(-1.)^j/Sqrt[j^2], {j, 1, n}]

Both sides of the equation yield the same value . You didn't take into account the partial sums correctly.

Addendum:

From Borwein, Lattice Sums then and now, Eq. (2.9 .9) I find:

Sum[(-1)^(k + j)/Sqrt[j^2 + k^2], {k, 1, Infinity}, {j, 1,Infinity}] = 
Log[2] + ((Sqrt[2] - 1)/2)*Zeta[1/2]*(Zeta[1/2, 1/4] - Zeta[1/2, 3/4])
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First of all, when we ask you to provide the code you used, we mean the code which actually produced the claimed discrepancy of the two sums. This we ask in order to be able to test what you did and point out what went wrong.

Before I go on, and for other users not to be confused since you are providing no references, this is the Wiki related page with explanations and clarifications.

Anyway, let's continue with the main computational question. The above was just a brief comment.

There is NO difference in the two sums. This you can verify using the following:

I am using

$Version
"12.0.0 for Linux x86 (64-bit) (April 7, 2019)"

And then I run

In[10]:= NSum[(-1)^(m + k + 
      j)*((j^2 + k^2 + m^2)/(j^2 + k^2 + m^2)^(3/2)), {m, 1, 
   Infinity}, {k, 1, Infinity}, {j, 1, Infinity}] // Chop
NSum[(-1)^(m + k + 
      j)*((j^2 + k^2 + 
       m^2)/(j^2 + k^2 + m^2)^(3/
        2)), {m, -Infinity, -1}, {k, -Infinity, -1}, {j, -Infinity, \
-1}] // Chop

Out[10]= -0.132476

Out[11]= -0.132476

Both of the above yield $-0.132476$ as is obvious. Let me note at this point that Chop is taking care of a very unimportant factor which is equal to $2.76482 \times 10^{-13} i$.

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  • $\begingroup$ thanks @DiSp0sablE_H3r0 I get the same result but the total sum gives by Matematica it is 2 X -0.132476 and it is not fit with the M value could expain that ?? $\endgroup$
    – capea
    May 13 at 15:58
  • $\begingroup$ You can find some more background material for the total sum in the wiki page. I am not an expert and this is why I read briefly the wiki article. On that page, it is stated that this number, the - 1.7..., is wrong even though it is widely used. $\endgroup$ May 13 at 16:02

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