Comparing Factors of Order Exp[I*x]

Question

I hope you are doing well.

If I have expressions stored in a list such as this $ \{ f_1'(y)(ie^{i x}) , f_2'(y)(2ie^{2 i x}) \}$, is there any way for Mathematica to compare the $e^{m x i}$ terms and determine which is the higher order in $x$, then return the higher ordered expression in the list? When determining which is the higher order, Mathematica should disregard the functions in front of each term.

I naively asked this question here: Limit at Infinity of Arbitrary Functions, forgetting I was working with complex exponentials. Finding the limit at infinity of the quotient of the expressions would have worked if they were normal exponentials and not complex exponentials.

Edit 1: I have an initial idea which piggy backs off my limits at infinity idea.

list = {(f_{1}'(y)*e^{i x}, f_{2}'(y)*e^{2 i x}}};
Factor1 = list[[1]];
Factor2 = list[[2]];

Factor1temp = Simplify[Factor1]./ {I*x->x}
Factor2temp = Simplify[Factor2]./ {I*x->x}

This returns

e^x f_{1}'(y)
e^{2 i x} f_{2}'(y)

The first expression the $i x$ got replaced with just $x$ but the second expression the $2 i x$ did not get changed to $2 x$. I can change the /. {I*x->x} to /. {2*I*x->x} but is there a way to tell Mathematica to ignore the integers out front and just replace the $ix$ to $x$.

Factor2temp = Simplify[Factor2]./ {2*I*x->2x}

returns

2 i e^{2 x} f_{2}'(y)

In any case now that they are no longer complex exponentials I can compare the terms by taking their limit at infinity.

Limit[Factor1temp/Factor2temp, x -> \[Infinity]]

This returns 0, therefore Factor2 is the higher ordered term. This one however,

Limit[Factor2temp/Factor1temp, x -> \[Infinity]]

returns

((i \[Infinty]) f_2 '(y))/f_1 '(y)

instead of $\infty$. Is there a way to make Mathematica output $\infty$ instead of what's given. The $f$ functions here are arbitrary but finite so the returned answer should give $\infty$.

Thank you for any help.

Answer 1

Not sure I understand correctly. Assuming that you want the largest prefactor of $x$ in the exponent,

L = {f1'[y][I E^(I x)], f2'[y][2 I E^(2 I x)]};
MaximalBy[L, # /. _[_. E^(b_. x)] -> b/I &]

(*    {Derivative[1][f2][y][2 I E^(2 I x)]}    *)

or if your definitions (which are a bit unclear) are rather

L = {f1'[y] (I E^(I x)), f2'[y] (2 I E^(2 I x))};
MaximalBy[L, # /. _. E^(b_. x) -> b/I &]

(*    {2 I E^(2 I x) Derivative[1][f2][y]}    *)

(using Default patterns)

Update

A more stable version that works even for $e^{0 i x}$ terms is to use the Exponent function:

L = Table[c[m] E^(I m x), {m, -3, 0}]
(*    {E^(-3 I x) c[-3], E^(-2 I x) c[-2], E^(-I x) c[-1], c[0]}    *)

MaximalBy[L, Exponent[#, E^(I x)] &]
(*    {c[0]}    *)