I can't seem to find a math object to allow custom root of a number; I've found sqrt but what if I'm interested in taking the cube(3) or dodeca(12) root?

Is there an object for this?

If it does exist I'd imagine it would follow a similar syntax of 'pow'.

[expr pow($f1, 1/$f2)]
where $f1 is the inlet, and $f2 is the power the inlet is a result of.
sorry, some things are hard to describe for me in english language....

edit: an example: [expr pow($f1, 1/3)] would compute the 3rd root of the inlet.

That was it, thank you. I was not using proper syntax.

Toxonic,

for me the syntax doesn't compute... I don't get an error reported, but the result always shows 0. Now considering that I'm a math disaster I guess the mistake is with me. From this set

16
|
expr pow($f1, 1/2)
|
0

I shouldn't get '0' as a result but 4 - no?

This is actually a weird syntax issue with [expr]. In most objects in Pd, every number is treated as a float. However, [expr] is based off of C expressions. In C, when an expression is compiled, if it looks like an integer it's treated like an integer. 1 and 2 are integers, and therefore the expression 1/2 will be compiled as an integer i.e., 1/2 = .5, int(.5) = 0. So your expression becomes pow($f1, 0). (By the way, this always gives me 1, as it should, not 0).

To fix this, just add a decimal point to one of those numbers. For example:

[expr pow($f1, 1/2.)]

Seeing the decimal point after the 2 tells [expr] that it should be treated as a float. This will give you the right answer. As a side note, this is also how Max does it with all objects using numbers, which makes it even weirder to see in Pd.

yeah, indeed. i must concede, i didn't try it the way of my second example, but with a second inlet for expr like in my first example [expr pow($f1, 1/$f2)] it works fine.
so, like maelstorm said, it works fine, or this should work too:
[expr pow($f1, float(1)/2)]