I need to define a function fun, and then re-define this function iteratively. The code is given at the end.

First, a function fun[x_, y_, d_] is defined, which is a polynomial in $x$ and $y$, and $d$ is the degree of this polynomial.

My goal is to modify fun according to some of its coefficients, see the definition fun2[x_, y_, d_] for example.

The problem is that, these coefficients are $0$ if $d$ is not substituted by a "real" number, like $d = 1$. See the code between the definitions of fun and fun2.

Just after the definition of fun2, I compute fun2[x, y, d] and it is the same as fun1[x, y, d]. The reason is that $d$ is not assigned a value. But fun2[x, y, 1] gives the desired answer, which is different from fun[x, y, 1].

The problem is that I want to repeat this process many times, say

fun3 = fun2 + Coefficient[fun2, x^2]
fun4 = fun3 + Coefficient[fun3, x^3]
fun"d" = fun2 + Coefficient[fun"d-1", x^(d-1)]

(This is just an example, the real process I need is far more complicated and can not be deined in one go. In this example, one can just use Sum[Coefficient...].)

Of course I don't want introduce so many functions. I want to define fun and modify it in a for loop. But the code below shows that Coefficient[fun[x, y, d], ...] is always $0$, since $d$ is not assigned a value.

I can't use a variable like temp in this for loop, because

For [i =...
  temp = fun[x, y, d]
  temp = temp + Coefficient[temp, x^i]

will just give fun[x, y, d], as Coefficient[temp, x^i] is always $0$ as mentioned above.

I need a suggestion on how to do such work more elegantly.

screen shot

  • 2
    $\begingroup$ Having code that can be pasted into a notebook makes it more likely that users will try to help. To convert the sigma notation to Sum, you can "Convert To > InputForm" from the "Cell" menus before copying, if that's a problem. $\endgroup$
    – Michael E2
    Aug 9, 2013 at 12:17

2 Answers 2


Try the following code:

fun[n_Integer?Positive][x_, y_, d_Integer] := 
  fun[n - 1][x, y, d] + Coefficient[fun[n - 1][x, y, d], x, n - 1 ];
fun[0][x_, y_, d_Integer] := 
  Sum[(i + j) x^i y^j Subscript[a, i, j], {i, d}, {j, d}];
Table[fun[n][x, y, 1], {n, 0, 2}] // TableForm
  • 2
    $\begingroup$ You might want some pattern restrictions, to prevent evaluation in cases where d is just the symbol d: E.g., fun[n_Integer?Positive][x_, y_, d_Integer] :=... $\endgroup$
    – Michael E2
    Aug 9, 2013 at 16:20
  • $\begingroup$ @MichaelE2 updated code to include your suggestion. $\endgroup$
    – Hector
    Aug 9, 2013 at 21:42
  • $\begingroup$ @Hecotr, thanks. It works. By the way, do you know if there is a more sophisticated way to define recursive functions, like a C programming. For example f[x_,n_]:= { exp1; exp2; exp3; return f[x,n-1]+exp3 }? $\endgroup$
    – user565739
    Aug 10, 2013 at 7:07
  • $\begingroup$ @user565739, "Sophisticated" is in the eye of the beholder … faster might be a better metric. In any case, the following code might satisfy you: f[x_, n_Integer?Positive] := g[f[x, n - 1]] $\endgroup$
    – Hector
    Aug 10, 2013 at 13:45

Another way (using same definition of f[x,y,d]:

  nestpol[n_Integer, d_Integer] := 
     NestList[{#[[1]] + 1, #[[2]] + Coefficient[#[[2]], x^(#[[1]])]} &, 
{1, f[x, y, d]}, n][[;; , 2]]

note (i) if just want the last value just change NestList to Nest (ii) in Hector's answer (based on no constant term in polynomial) f[0][x,y,d]=f[1][x,y.d], so nestpol[N,d] yields the same result as Table[fun[n][x, y, 1], {n, 1,N}]


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