MathGroup Archive 2005

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Problem with transformation rule of a function

  • To: mathgroup at
  • Subject: [mg53488] Re: [mg53467] Problem with transformation rule of a function
  • From: "Wolf, Hartmut" <Hartmut.Wolf at>
  • Date: Fri, 14 Jan 2005 08:54:35 -0500 (EST)
  • Sender: owner-wri-mathgroup at

>-----Original Message-----
>From: Alain Cochard [mailto:alain at] 
To: mathgroup at
>Sent: Thursday, January 13, 2005 9:12 AM
>To: mathgroup at
>Subject: [mg53488] [mg53467] Problem with transformation rule of a function
>I define an expression:
>     In[1]:= expr=M[1][t] + M[2][t] + Integrate[M[1][t],t] + 
>Integrate[M[2][t],t] + D[M[1][t],t] + D[M[2][t],t];
>     Out[1]= Integrate[M[1][t], t] + Integrate[M[2][t], t] + 
>M[1][t] + M[2][t] + 
>     >    (M[1])'[t] + (M[2])'[t]
>and then I try 2 transformation rules on this expression:
>     In[2]:= vers1=expr/.{M[1][t]->f[t], M[2][t]->0}
>     Out[2]= f[t] + Integrate[f[t], t] + (M[1])'[t] + (M[2])'[t]
>In this first one, I get the output I expect for the function 
>and integration terms, but not for the derivative ones.
>     In[3]:= vers2=expr/.{M[1]->f, M[2]->0}
>     Out[3]= 0[t] + f[t] + Integrate[0[t], t] + 
>Integrate[f[t], t] + f'[t]
>In this second version, I get these 0[t] terms for the 
>function and integration terms, with which I further have to 
>deal with to achieve what I want:
>     In[4]:= %/.{0[t]->0}
>     Out[4]= f[t] + Integrate[f[t], t] + f'[t]
>I would first like to understand why the derivation and 
>integration terms are not treated in an identical way, and 
>then I would like to know if there is a more elegant way to do 
>what I want in a single step.
>Thanks in advance,

To understand why vers1 one doesn't work we have to note

In[2]:= D[M[1][t], t] // FullForm
Out[2]//FullForm= Derivative[1][M[1]][t]

So vers2 seems to be more appropriate.  The drawback is that Matematica
doesn't know that 0[t] === 0 for all t.

Telling that with your second substitution is completely ok, so

In[15]:= expr /. {M[1] -> f, M[2] -> 0} /. 0[t] -> 0
Out[15]= f[t] + Integrate[f[t], t] + Derivative[1][f][t]

You may however make a definition in general for (e.g. integer) constant
functions (telling Mathematica what you mean with 0[t]):

In[28]:= Unprotect[Integer]

In[29]:= x_Integer[_] := x

In[30]:= Protect[Integer]

In[31]:= expr /. {M[1] -> f, M[2] -> 0}
Out[31]= f[t] + Integrate[f[t], t] + Derivative[1][f][t]

Hartmut Wolf

  • Prev by Date: Re: Re: Mathematica is not very clever
  • Next by Date: Re: Mathematica Graphics output in an ASP.NET WebApplication
  • Previous by thread: Re: Problem with transformation rule of a function
  • Next by thread: size of typeset boxes