# assign function to an existing function

Hi,

Consider the following code:

`sage.var('R')`

`rho = sage.function('rho')(R)`

`r = rho(R=R).function(R)`

Is there a way to assign some function to `rho`

such that `r`

is modified? For instance, if I do:

`y = sage.function('y')(R)`

`rho = sage.desolve(sage.diff(y,R) - y(R), dvar=y, ivar=R, ics=[0,1]).function(R)`

->`R |--> e^R`

Then I get:

`r(0)`

->`rho(0)`

`rho(0)`

->`1`

So `rho`

is of course modified, but not `r`

.

Thanks!

Martin

Sure, in Python (and hence Sage) you can assign anything to

`r`

. But we would need to see a more specific example. Also, be aware that some commands have a side effect, butreturn`None`

, so you have to be careful about that - for instance, the bare command`show(plot())`

might give an empty plot to show, but the actual Python value returned is`None`

, not what you might be looking for.Thanks! Just updated my question. Any idea?

Ah, this is more of a Python question, which I'll let others answer. By the way, you don't need the

`sage`

bits unless you are importing Sage explicitly in Python. Maybe you are.Thanks! Yes, I prefer python standalone (makes it easier to use other python libraries) with

`import sage.all as sage`

(makes it where each function comes from).