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MathGroup Archive 2008

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pattern matching against the Dt function?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg85238] pattern matching against the Dt function?
  • From: "Adam M." <adamm at san.rr.com>
  • Date: Mon, 4 Feb 2008 03:06:40 -0500 (EST)

Hello,

I'm having trouble matching the Dt function with a pattern, even though 
it works for all other functions I've tried.

(*It doesn't match Dt[b] here.*)
In[101]:= {f[a], Dt[b]} /. Dt[n_]->n
Out[101]= {f[a], Dt[b]}

(*But it matches f[a] with no problem.*)
In[100]:= {f[a], Dt[b]} /. f[n_]->n
Out[100]= {a, Dt[b]}

(*In a process of elimination, I tried another built-in function, D, and 
it worked fine.*)
In[99]:= {f[a], D[b]} /. D[n_]->n
Out[99]= {f[a], b}

(*I tried another function more than one character long, Sin, and that 
works.*)
In[102]:= {f[a], Sin[b]} /. Sin[n_]->n
Out[102]= {f[a], b}

(*The full forms all seem to follow the same pattern.*)
In[103]:= Sin[b] // FullForm
Out[103]//FullForm= Sin[b]

In[104]:= Dt[b] // FullForm
Out[104]//FullForm= Dt[b]

In[105]:= f[b] // FullForm
Out[105]//FullForm= f[b]

(*I thought it might be related to the evaluation of the Dt function, so 
I tried Holding it. No luck.*)
In[111]:= {f[a], Hold[Dt[b]]} /. Dt[n_]->n
Out[111]= {f[a], Hold[Dt[b]]}

(*However, it has no problem matching f[a] in a Hold.*)
In[112]:= {Hold[f[a]], Dt[b]} /. f[n_]->n
Out[112]= {Hold[a], Dt[b]}

(*I tried looking at the attributes to find out if there was something 
special about the Dt function, but it doesn't seem like it...*)
In[116]:= Attributes[Sin]
Out[116]= {Listable, NumericFunction, Protected}

In[117]:= Attributes[D]
Out[117]= {Protected, ReadProtected}

In[118]:= Attributes[Dt]
Out[118]= {Protected}

I've read every section on pattern matching in the documentation center, 
and I'm at a complete loss to explain why I can't seem to match the Dt 
function with the Dt[n_] pattern when I can match these other functions. 
I know I can use the _Dt pattern to match it, but then I don't get 
control over matching the arguments...

Thank you.

Very curious,
-- Adam M.


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