To be completely honest I am a litte lost.

Let us say a variable x is defined as y = 1/x. I have one function

f[x_]:= x^2

which I would like to multiply into another function

g[y_]:= Exp[y] + y^2 + 3 y 

Without having to change all y into x. The way I'd do it is

G[y_]:= f[1/y][Exp[y] + y^2 + 3 y]

Is this a correct approach? Are there smarter ways to do it?

  • 2
    $\begingroup$ Use parentheses ((...)) to group expressions, not square brackets ([...]) - those are only for function arguments in Mathematica. Also, why not use f[1/y]*g[y]? $\endgroup$
    – Lukas Lang
    Jan 9, 2019 at 17:22
  • $\begingroup$ Yes I can do that. I just wanted to make sure that just putting f[1/y] inside there wouldn't cause any issues. Thank you ! $\endgroup$ Jan 9, 2019 at 20:03

1 Answer 1


There are quite a few ways to accomplish what you want to do. Here are some.

f[x_] := x^2
g[y_] := Exp[y] + y^2 + 3 y

The simplest is probably is what Lukes suggested.

fg1[y_] := f[1/y] g[y]

or as pure function

fg2 = f[1/#] g[#] &;

However, in this case, since it is easy to define x as the inverse function of y,

x[y_] := 1/y

Composition ( @* ) can also be used.

fg3[y_] := (f@*x)[y] g[y]


fg4 = f@*x@# g@#&;

The last definition is a good example of the Wolfram Language's terse functional style. Some consider it elegant; others consider it obscure.

You can be satisfy yourself that all the definitions represent the same function by evaluating

    Plot[#1[t], {t, 0, 10}, PlotStyle -> Directive[AbsoluteThickness[#2], #3]] &, 
    {{fg1, fg2, fg3, fg4}, {17, 11, 7, 2}, {Red, Green, Orange, Black}}],
  ImageSize -> Large]


  • 1
    $\begingroup$ You seem to have forgotten a space in before g in the definition of fg4 $\endgroup$
    – Lukas Lang
    Jan 9, 2019 at 20:18
  • $\begingroup$ @LukasLang. Good catch. I have made the required edit. $\endgroup$
    – m_goldberg
    Jan 9, 2019 at 21:43

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