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I am trying to create an expansion of the form

f[x,y]->Sum[Cn[x] y^n,{n,1,order}]

To replace the function f by the coefficients cn. Since the function f appears in a differential equation, it is useful to get this to work as a set of replacement rules for a pure function, such that the function f and all its derivatives are replaced. Something like this works:

f->((c0[#1]+c1[#1] #2+c2[#1]#2^2)&)

Curiously, this expression doesn't

f->(( Sum[c[[n]][#1] #2^n,{n,1,order}] )&)

(where c is a pre-defined vector of coefficients). The index n is not recognized in the context of defining the pure function (though the same expression defining f[x,y] would work).

Similarly, if I have several functions and I am trying to define a collection of these replacement rules, this also does not work:

Table[ f[[i]]->f->((c0[[i]][#1]+c1[[i]][#1] #2+c2[[i]][#1]#2^2)&), {i,1,whatever}]

Somehow defining and manipulating a collection of pure functions with an index needs a different syntax. I am curious as for the logic of why this does not work, maybe then I'll understand how to fix it.

Edit: Apologies for the vague question. I think the logic that I was missing is related to attributes of pure functions I was trying to use in replacement rules, as pointed out by acl below. This is related to this question: Function in Table

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  • $\begingroup$ One problem is that you are doing the sum with iterators {n,0,order} which goes from n=0 up to n=order. But with n=0, C[[n]] is not defined. So maybe C[[n+1]] would fix it. $\endgroup$ – bill s Mar 23 '13 at 10:08
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There's a number of reasons your code doesn't work (all related to attributes; basically, Function has attribute HoldAll). I think this does what you want:

MapThread[
 Rule,
 Transpose@Table[
   {f[n], With[{n = n}, c[[n]][#1] #2^n &]},
   {n, 1, order}
   ]
 ]

Mathematica graphics

How does this work? With[{n=n},body] is what fixes the problem, as it literally inserts n into the body of the Function, so the HoldAll attribute never gets to see the n; it just sees the explicit value.

To illustrate what With does here:

Block[{y = 2}, y*x &]
Block[{y = 2}, With[{y = y}, y*x &]]

Mathematica graphics

So, here, in the first case the y in the function doesn't evaluate to 2, then the block exits and we're done. In the second case, y gets literally replaced by 2. That is, you may think of With[{y=2},body] as a search-and-replace of y by 2 in the body before anything else happens (before, i.a., evalution of body).

Hope this clarifies things a bit.

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  • $\begingroup$ Thanks, the attributes of the function is the point I was missing. Thanks for the link, I will take a look. $\endgroup$ – Moshe Mar 23 '13 at 20:06
  • $\begingroup$ note that I used f[i] instead of f[[i]], because if f is not defined, trying to take a part of it gives an error. replace [] with [[]] if this is what you need $\endgroup$ – acl Mar 23 '13 at 20:06
  • $\begingroup$ That was a point I caught (fixed by predefining the coefficient vector), but the attributes and the use of with is likely the point I missed. My question was vague because I didn't want to simply fix the code, I wanted to understand what I am missing since the syntax seemed to me like it ought to work. Thanks for pointing me the right direction. $\endgroup$ – Moshe Mar 23 '13 at 20:10
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    $\begingroup$ @Moshe I tried to explain it a bit more here, hope it's clearer. $\endgroup$ – acl Mar 23 '13 at 20:19
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I'm not really sure about what are you trying to do with those rules, but the following works:

c = Table[#/n! &, {n, 0, 10}];
f[order_] := ((Sum[c[[n + 1]][#1] #2^n, {n, 0, order}]) &)

Plot3D[{f[5][x, y], f[10][x, y]}, {x, -10, 10}, {y, -10, 10}, 
       PlotStyle -> {{Opacity[0.6], Red}, {Opacity[0.6], Blue}}]

Please remember to start your symbols with lowercase chars, Cis a system's defined name.

Mathematica graphics

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  • $\begingroup$ Thanks for your answer. I don't think I have been very clear, I tried to modify my question to emphasize the point I am not getting, how to define and use a collection of pure functions. $\endgroup$ – Moshe Mar 23 '13 at 19:03
  • $\begingroup$ @Moshe I don't understand your question because you seem to be defining RULES (->) and not pure functions $\endgroup$ – Dr. belisarius Mar 23 '13 at 19:16

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