I'm trying some transformations and they are getting complicated. It's common in physics to replace functions by symbols so you term get shorter.

Lets say I have transformation

xr[A_,B_] := K A + L B + KK A^2 + LL B^2,

where A = A[x_] and B = B[x_]

I want to input some random values of x. But for the code to be readable I want to substitute the quadratic part as

H =  KK A^2 + LL B^2

How do I go about this? I don't want to define H as a function, since that would make me have to write

xr[A,B] = xr[A,B,H] = xr[A,B,H[A,B]] = xr[A[x],B[x],H[A[x],B[x]]]

which just gets unwieldy.

Essentially I just want to be able to write H such that mathematica would understand that when I define

xr[A,B] := K A + L B + H

Output of sending a x value to it would be

xr[A[x],B[x]] = K A[x] + L B[x] + KK A[x]^2 + LL B[x]^2.

Hopefully you can understand what I'm trying to do. Thank you for your help.

Edit: This is just an example of function. Of course It wouldn't be useful in this case and I also know there are reasons why this specific type substitution wouldn't be a healthy practice but what are the alternatives.

  • 1
    $\begingroup$ Why can't you just define: H[A_, B_] = KK A^2 + LL B^2; xr[A_, B_] := K A + L B + H[A, B] ? $\endgroup$ Dec 11, 2021 at 9:56
  • 1
    $\begingroup$ Try this: xr[A_, B_] := k A + l*B + kk A^2 + ll B^2; xr[A, B] /. kk A^2 + ll B^2 -> H which returns H + A k + B l . $\endgroup$ Dec 11, 2021 at 12:45
  • $\begingroup$ Thank you very much $\endgroup$
    – Nitaa a
    Dec 12, 2021 at 13:07

1 Answer 1


You can display the unevaluated functions in any desired format.

Format[A[x_]] := A;
Format[B[x_]] := B;

xr[a_, b_] := K a + L b + KK a^2 + LL b^2


xr[A[x], B[x]]

(* K A + KK A^2 + L B + LL B^2 *)

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