1
$\begingroup$

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,

  • 0
    @MarcF:See my comment $u$nder my answer.2012-07-31

1 Answers 1

1

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.

  • 0
    @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.2012-07-31