Message Boards Message Boards

0
|
7038 Views
|
2 Replies
|
0 Total Likes
View groups...
Share
Share this post:
GROUPS:

Typesetting Question

Posted 10 years ago
Hi All,
When copied from another notebook (not mine) the
expression calculates.  But when I type in a
version with the exact same appearance, it doesn't
work.
DSolve[{ff^(0,0,0,1)[r,\[THETA],\[PHI],t]==rhs,ff^(0,0,0,1)[0,0,0,0]==Sin[t]},
ff[r,\[THETA],\[PHI],t],{r,\[THETA],\[PHI],t}]
 The error msgs are:

Syntax::sntxf: "(" cannot be followed by "0,0,0,1)".
Syntax::tsntxi: "0,0,0,1" is incomplete; more input is needed.
Syntax::sntxi: Incomplete expression; more input is needed .

I've experimented with matching parentheses, but I haven't
been successful so far. What trick am I missing?

(M9, Linux 64)

thanks,
wb
Attachments:
2 Replies
Posted 10 years ago
Exactly, input the derivative as a partial, as below.
 In[1]:= D[ff[r, theta, phi, t], t]
 
 Out[1]=
 \!\(\*SuperscriptBox[\(ff\),
 TagBox[
 RowBox[{"(",
 RowBox[{"0", ",", "0", ",", "0", ",", "1"}], ")"}],
 Derivative],
 MultilineFunction->None]\)[r, theta, phi, t]
POSTED BY: David Keith
What appears as an exponent in the formated version of the code you posted (the image on teh bottom) is, in fact, not an exponent.  It is the formatted form for the partial derivative in question.  So, for example, if you execute this,
D[f[x, y], y]
 and then ask for it's InputForm, you will see the internal representation that you would need to use if you were to write it out in an InputForm:
InputForm[D[f[x, y], y]]

gives

Derivative[0, 1][f][x, y]
POSTED BY: David Reiss
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