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Groups > comp.soft-sys.math.mathematica > #2879
| From | "Nasser M. Abbasi" <nma@12000.org> |
|---|---|
| Newsgroups | comp.soft-sys.math.mathematica |
| Subject | Re: Dt@x@1 |
| Date | 2011-06-01 10:57 +0000 |
| Organization | Steven M. Christensen and Associates, Inc and MathTensor, Inc. |
| Message-ID | <is55ug$d2e$1@smc.vnet.net> (permalink) |
| References | <is4tgc$b3t$1@smc.vnet.net> |
On 6/1/2011 1:33 AM, Chris Chiasson wrote: > Why does Dt@x@1 return zero? Dt[ x[1] ] The last evaluation of x[1] in the call gave 1, and this is what is left and is passed to Dt and then Dt[1] returns zero. You can see that by ----------------------- In[65]:= Remove["Global`*"] TracePrint@Dt [ x[1] ] During evaluation of In[65]:= Dt[x[1]] During evaluation of In[65]:= Dt During evaluation of In[65]:= x[1] During evaluation of In[65]:= x During evaluation of In[65]:= 1 During evaluation of In[65]:= 0 Out[66]= 0 ------------------------------ But what I do not know, is why when typing In[84]:= x[1] one gets back Out[84]= x[1] and so, now it did not 'evaluate' to 1 like it seems to have done inside Dt call above. I can see that x[1] should return x[1], but I am not sure why x[1] ended 1 inside the Dt call as shown by the trace above and not when typing it on the top level. This is for the experts to explain, something to do with how different evaluation rules works in different contexts I suppose. --Nasser
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Dt@x@1 Chris Chiasson <chris.chiasson@gmail.com> - 2011-06-01 08:33 +0000
Re: Dt@x@1 "Nasser M. Abbasi" <nma@12000.org> - 2011-06-01 10:57 +0000
Re: Dt@x@1 Chris Chiasson <chris.chiasson@gmail.com> - 2011-06-02 11:15 +0000
Re: Dt@x@1 magma <maderri2@gmail.com> - 2011-06-02 11:14 +0000
Re: Dt@x@1 Chris Chiasson <chris.chiasson@gmail.com> - 2011-06-02 11:14 +0000
Re: Dt@x@1 Roland Franzius <roland.franzius@uos.de> - 2011-06-02 23:12 +0000
Re: Dt@x@1 Chris Chiasson <chris.chiasson@gmail.com> - 2011-06-04 10:20 +0000
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