Message Boards Message Boards

0
|
5077 Views
|
2 Replies
|
3 Total Likes
View groups...
Share
Share this post:

Obtain an Nth derivative with HypergeometricU?

Hello community, I am trying to create a code for an Nth derivative that has a confluent hypergeometric U function, but I must be doing something wrong... see my attempt:

This is the equation I'm trying to reproduce:

im0

Simplify[D[
   Exp[-x^2*Erf[(Sqrt[2*Pi])*x]*Erfi[(Sqrt[2*Pi])*x]], {x, 
    n}]] // TraditionalForm
Simplify[2^n*
   Exp[-x^2*Erf[Sqrt[2*Pi]*x]*
     Erfi[Sqrt[2*Pi]*x]]*(-x*Erf[Sqrt[2*Pi]*x]*Erfi[Sqrt[2*Pi]*x])^
    n*(x^2*Erf[Sqrt[2*Pi]*x]*Erfi[Sqrt[2*Pi]*x])^(-n/2)*
   HypergeometricU[-n/2, 1/2, 
    x^2*Erf[Sqrt[2*Pi]*x]*Erfi[Sqrt[2*Pi]*x]]] // TraditionalForm

im1

For example if I use:

n = 2

Using the "n" defined in the equations:

a = Simplify[
  D[Exp[-x^2*Erf[(Sqrt[2*Pi])*x]*Erfi[(Sqrt[2*Pi])*x]], {x, n}]]

im2

b = Simplify[
  2^n*Exp[-x^2*Erf[Sqrt[2*Pi]*x]*
     Erfi[Sqrt[2*Pi]*x]]*(-x*Erf[Sqrt[2*Pi]*x]*Erfi[Sqrt[2*Pi]*x])^
    n*(x^2*Erf[Sqrt[2*Pi]*x]*Erfi[Sqrt[2*Pi]*x])^(-n/2)*
   HypergeometricU[-n/2, 1/2, 
    x^2*Erf[Sqrt[2*Pi]*x]*Erfi[Sqrt[2*Pi]*x]]]

im3

Table[N@Limit[b/a, x -> i], {i, 0, 4}]

im4

Both sides of the equation should be the same (equal), but they are not, as we can see by making the relation “b/a” and giving values for "x" in the table above. If they were equal the table should be {1,1,1,1,1}. Is it right or not?

As I am not a specialist in this type of derivative operation yet, I decided to ask for help in the community.

What am I doing wrong? Or am I missing some concept?

Thank you very much.

POSTED BY: Claudio Chaib
2 Replies

It's looks like a bug.

Mathematica 12.0 return input, so he doesn't know the answer.

Regards M.I.

POSTED BY: Mariusz Iwaniuk

Mariusz, thanks for your time answering me.

Yes, I agree with you. Mathematica 12.0 does not give a generalized answer to the Nth derivative of this equation. The original equation I was trying to reproduce was initially evaluated by Wolfram/Alpha, so I suppose the right side of the equality (the response generated with N cases) it's not correct. I think only the left side evaluated by defining a specific "n" is more likely to be correct using successive Mathematica derivatives. Therefore, I believe the error was initially generated by the Wolfram/Alpha response.

POSTED BY: Claudio Chaib
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract