# Multiplying expressions within a list of pure functions

Let's say I have a list of pure functions that are nice algebraic expressions: say

l = {(#1 - 1)&, (#1^2 + #1)&, (#1^3 - 1)&}


What's an easy way to get a pure function that will give the product of these expressions? For example, in the above I'd like to get a function f with

f = (#1 - 1)(#1^2 + #1)(#1^3 - 1)&


I've tried (Times @@ Identity @@@ l)&, but this just gives (#1 - 1)(#1^2 + #1)(#1^3 - 1). Essentially, it seems my difficulty is converting a List of Function expressions to a Function[Times[...]] expression, and I can't see how to "strip off" the functions without messing up the referencing of the Slot expressions inside.

Thanks for any help.

• @PinguinDirk, I was presuming that the OP wanted to keep it as a pure function. Jul 4 '13 at 8:28
• I misread anyway, actually I am not too sure what the OP actually wants... maybe Function[x, Times @@ Through[l[x]]]? But thanks for pointing that out, @Jonathan Shock Jul 4 '13 at 8:36
• @PinguinDirk Your suggestion gives what I want as well; the use of Through is clever. Thanks for your help! Jul 4 '13 at 8:49

Mapping First over your list will strip off the Function head. Then multiply them together with Apply[Times... and finally Apply[Function... makes the result a pure function. Use the final Apply so the argument to Function is evaluated first as Function has the attribute HoldAll.

Apply[Function,{Apply[Times, Map[First,l]]}]


Gives...

(-1+#1) (#1+#1^2) (-1+#1^3)&

• Thanks so much! This works perfectly! Jul 4 '13 at 8:45
• This can also be written: Function @@ {Times @@ First /@ l} Jul 4 '13 at 9:47

Here is another option:

l = {(#1 - 1) &, (#1^2 + #1) &, (#1^3 - 1) &};


(#1 - 1) (#1^3 - 1) (#1^2 + #1) &


This has the benefit of not evaluating the body of the functions. For example:

l = {(#1 - 1) &, (#1^2 + #1) &, (Print["!"]; #1^3 - 1) &};


(#1 - 1) (Print["!"]; #1^3 - 1) (#1^2 + #1) &

Note that the Print statement remains. Compare with:
Function @@ {Times @@ First /@ l}

(-1 + #1) (#1 + #1^2) (-1 + #1^3) &