Message Boards Message Boards

0
|
3195 Views
|
2 Replies
|
2 Total Likes
View groups...
Share
Share this post:

[?] Difference between DawsonF[5] and DawsonF[5.0] integration?

Posted 7 years ago

What's the difference between DawsonF[5] and DawsonF[5.0]? The former doestn't calculate. Thanks.

In[815]:= DawsonF[5]

Out[815]= DawsonF[5]

In[816]:= DawsonF[5.0]

Out[816]= 0.102134

In[820]:= DawsonF[1 + I]

Out[820]= DawsonF[1 + I]

In[819]:= DawsonF[1.0 + I]

Out[819]= 0.990373 - 0.638873 I

enter image description here

POSTED BY: Jacques Ou
2 Replies
Posted 7 years ago

I get it. Thanks a lot.

POSTED BY: Jacques Ou

Also E^2 does not calculate, just as DawsonF[5] and Dawson[1+I]. They are exact symbolic numbers for which no automatic calculations are done, if you evaluate them in isolation. However, you may try to find different symbolic expressions, for example with

DawsonF[5] // FunctionExpand
DawsonF[1 + I] // FunctionExpand

The floating point in Dawson[5.0] triggers numeric algorithms. You can do it also with N[DawsonF[5]].

POSTED BY: Gianluca Gorni
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