I'm trying to solve coupled equations by using iteration. There are two functions $G$ and $H$ that I need to find. In the code, G and H are actually lists of numbers. I start by using a simpler solution for G and H and then iterate to get the actual solutions. I have two functions newG[G,H] and newH[G,H] that get the previous $G_{i-1}$ and $H_{i-1}$ as input and give the new functions $G_i$ and $H_i$. When my problem had only one function the easiest way to iterate was to use Nest.


where NewG is a function that takes the previous G and gives the new function back and steps is just the number of steps in the iteration. Now that I'm working with two variables I have two functions NewG and NewH, but I don't really know how to modify the Nest command to make this work. Any comments are welcome! and please ask if you need more information from me.


1 Answer 1


You can use Through wrapped in Apply to generalize your Nest command to multiple functions:

(* dummy implementations of newG/newH *)
newG[G[i_], H[i_]] := G[i + 1]
newH[G[i_], H[i_]] := H[i + 1]

(* the actual iteration *)
Nest[Apply[Through[{newG, newH}@##] &], {G[0], H[0]}, 5]
(* {G[5], H[5]} *)

Here, the Apply is used to pass multiple arguments to the functions, and the Through is used to apply multiple functions to the arguments. The chain of evaluations is essentially:

Apply[Through[{newG, newH}@##] &][{G[0], H[0]}]
Through[{newG, newH}@##] &[G[0], H[0]]
Through[{newG, newH}[G[0], H[0]]]
{newG[G[0], H[0]], newH[G[0], H[0]]}
{G[1], H[1]}
  • $\begingroup$ Hello, Thanks for your answer! I edited my question to add an important piece of information I had left behind. G and H are lists of numbers. Do you think what you wrote would still work if G and H are lists and not individual numbers? Thanks again! $\endgroup$ Sep 28, 2019 at 23:05
  • $\begingroup$ @P.C.Spaniel the code in the answer should work for any type of arguments - I've just used G[i]/H[i] to demonstrate how it's working $\endgroup$
    – Lukas Lang
    Sep 28, 2019 at 23:07

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