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Help with Confluent Hypergeometric Function

Posted 9 years ago

Can anyone tell me why this code wont evaluate the function? I'm using these functions to help me solve dirac particles in a potential well.

Hypergeometric1F1[ 1.99997 - (0.00729735 [Epsilon])/Sqrt[1 - [Epsilon]^2], 2.99995, 274.449 Sqrt[1 - [Epsilon]^2]]

If anyone would like, I can send the whole program. It isn't that long. I just need to set two equations together and solve numerically. However, I am getting funny answers that do not make sense and i think part of the problem are these functions.

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POSTED BY: Jacob Noon
5 Replies

Well, 1st in your expression there are square-brackets around Epsilon. That may give difficulties beacuse of the special use of square-brackets in Mathematica. If you delete them you get

In[2]:= Hypergeometric1F1[
 1.99997 - (0.00729735 Epsilon)/Sqrt[1 - Epsilon^2], 2.99995, 
 274.449 Sqrt[1 - Epsilon^2]]


Out[2]= Hypergeometric1F1[
 1.99997 - (0.00729735 Epsilon)/Sqrt[1 - Epsilon^2], 2.99995, 
 274.449 Sqrt[1 - Epsilon^2]]

So as remarked by Sean Epsilon is not defined. If you do that you get a (quite big) number

 In[4]:= Epsilon = .01; 
Hypergeometric1F1[  1.99997 - (0.00729735 Epsilon)/Sqrt[1 - Epsilon^2], 2.99995, 
  274.449 Sqrt[1 - Epsilon^2]]


 Out[4]= 1.11327*10^117

Regards, Hans

POSTED BY: Hans Dolhaine

The square brackets around Epsilon are actually just a common cut-and-copy problem.

They are supposed to be "[Epislon]" but the slash gets removed during cut-and-copy. The character "[Epislon]" is a valid character in Mathematica that looks like the actual greek character Epislon.

POSTED BY: Sean Clarke

Functions like Hypergeometric1F1 will evaluate with variables in their arguments if there's a simple transformation.

  1. Is that the case here? What are you expecting it to return? Are you sure there's a simple answer?

  2. Very often, these simplifications are symbolic and require that you not use floating point numbers. You have used floating point numbers. So it probably won't do any symbolic simplification. Symbolic transformations generally aren't safe to perform on finite precision numbers. You can convert from floating point to rational numbers (which are symbolic) using rationalize.

  3. You may need to use FunctionExpand or FullSimplify to search for your answer. FunctionExpand won't work in this case. FullSimplify typically requires that you list out assumptions with the Assumptions option to work well.

POSTED BY: Sean Clarke
Posted 9 years ago

Usually you can evaluate a Hypergeometric1F1 with variables in it and get an answer. When i hit shift enter, It just gives me back what i type in.

POSTED BY: Jacob Noon

I'm not sure I understand. Do you want to express the Hypergeometric1F1 as an infinite series?

Epsilon isn't defined anywhere. So it can't evaluate further.

POSTED BY: Sean Clarke
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