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How would you write the general solution, I'm assuming something like a sum, of:

$$ \frac{d^j}{d\sigma^j}\exp[0.5(\alpha-\sigma)^2] $$

Regards,

J. M. ain't a mathematician
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MarcF
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1 Answers1

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You can have your solution in terms of The MeijerG function. It is a very general special function

$$ \left( -1 \right) ^{\frac{3}{2}\,j}{2}^{\frac{1}{2}\,j} G^{1, 2}_{2, 3}\left(-\frac{\left( \alpha-\sigma \right) ^{2}}{2}\, \Big\vert\,^{-\frac{1}{2}\,j, -\frac{1}{2}\,j+\frac{1}{2}}_{-\frac{1}{2}\,j, \frac{1}{2}, 0}\right) $$

Note that, the above formula is a unified formula. It gives derivatives of real order (including the integer order), if $j>0$, anti-derivatives of real order (including the integer order) if $j<0$, and the original function if $j=0$ of the function $ \mathrm{e}^{\frac{1}{2}(\alpha-\sigma)^2}\,.$

If you simplify the above Meijer G-function, you can get the answer in terms of the hypergeometric function

$$ {\frac { \left( \alpha-\sigma \right) ^{-j} \left( -1 \right) ^{j} {\mathrm{F} (1/2,1;\,-1/2\,j+1,-1/2\,j+1/2;\,1/2\, \left( \alpha-\sigma \right) ^{2})} }{\Gamma \left( -j+1 \right) }} \,.$$

But the above formula has a deficiency, since you need to deal with the poles of the gamma function in the denominator. This is due to the limitations of the hypergeometric function.

Mhenni Benghorbal
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  • @J.M:You can get it in terms of the hypergeometric function, but the problem is you need to deal with the poles of the gamma function as you see. This is one of the deficiencies of the hypergeometric function. It is one of the main reasons has made me to deal with the MeijerG function. – Mhenni Benghorbal Jul 31 '12 at 12:35
  • I'd never come across the MeijerG before, it's pretty cool. I apologise for not being more specific, I should have said that 'j' is Real and Integer, which probably allows for something less 'general'. My end goal is a numerical implementation as part of a larger physical model. – MarcF Jul 31 '12 at 13:22
  • @MarcF:The formula I gave in the answer gives you what you want. If you put j a positive integer you get derivatives of integer order and if you put j any positive real number, you get real order derivatives. Try to use mathematics software like Maple or Mathematica, both have the MeijerG function implemented. I have already explained in my answer why we can not simplify the function in terms of the hypergeometric function which is a les general function. – Mhenni Benghorbal Jul 31 '12 at 14:01
  • I've played around with your solution in Mathematica and although of the correct result form, for odd j it produces a complex result?? – MarcF Jul 31 '12 at 15:06
  • @MarcF:I used Maple to get this formula. There is a difference between maple and mathematica in the way they implemented the MeijerG function. But always, we can convert one to the other. So, try to use maple. – Mhenni Benghorbal Jul 31 '12 at 15:37