1

Is it possible that meijerG function contain a negative value (i.e. is {-1,0,0})? I tried both Mathematica and Matlab to compute this meijerG function but they generate an error that this meijerG is not defined for the given parameters`.

Here is my code:

D = (0.6);
lg1 = lg2 = 1;
G = evalin(symengine, sprintf('meijerG([[0], []], [[-1,0,0], []],%f)',(D/(lg1*lg2))));
CD = -((2*D)/(lg1*lg2*(log(4))))*G;

Here I have also attached the image of the function from the text.

function

einpoklum
  • 118,144
  • 57
  • 340
  • 684
Baidal Kocham
  • 21
  • 3

2 Answers2

0

From the documentation of meijerG:

No pair of parameters ai - bj, i = 1, …, n. j = 1, …, m, should differ by a positive integer [...] . Otherwise, meijerG returns an error.

Complex numbers are valid for any coefficent; however in you case you have a0-b0 = 1 which is forbidden.

Bentoy13
  • 4,886
  • 1
  • 20
  • 33
  • Thank you, Bentoy, for your explanation. It cleared a lot of confusions. So from your comment, I can presume that the original function is incorrect? above I have also attached the image of the function from the text. I would really appreciate if you can comment on that and suggest some thing? error? – Baidal Kocham Mar 17 '17 at 11:55
  • I'm sorry, I don't know anything about this special function. I cannot help you more. Maybe try to look for the relationship between meijer G and Bessel K_0 functions, you may find the correct parameters... – Bentoy13 Mar 20 '17 at 17:23
0

I quickly looked at that. If one expand log2(1+x) into Taylor series, substitute \gamma->x^2, then integral would be

S K0(x) x^m dx = 2^(m-1) G((m+1)/2)^2

see here for details. G is gamma-function and for argument like (k+1/2) it is expressed via binomial coeff times sqrt(\pi), see here for details.

After all that you have infinite sum of terms with polynomials over lambdas and b and some binomial coefs and \pi etc. Whether it could be summed or not - I don't know...

Severin Pappadeux
  • 18,636
  • 3
  • 38
  • 64