Self composition of a function with symbolic variable

Processing math: 100%

sberner
11 ●3

slelievre
17809 ●22 ●163 ●352 http://carva.org/samue...

I have a function in one variable with a placeholder variable, and would like to find the nth iterate. For example, using:

$f(z) = z^3 + a z^2$ , a function in variable $z$ with placeholder coefficient $a$ .

then I would like to see

$f(f(z)) = (z^3 + az^2)^3 + a(z^3 + az^2)^2$

However I am struggling to perform this in Sage.

f(f) returns (a*z^2 + z^3)*z^2 + z^3, so it substitutes for $a$ instead of $z$ , and declaring f(f(z)) returns the function only in terms of $z$ .

How can we self compose functions with a placeholder variable for a coefficient?

add a comment

1 Answer

Sort by » oldest newest most voted

answered 4 years ago

tmonteil
27483 ●32 ●205 ●519 http://wiki.sagemath.o...

What is wrong here :

sage: var('a')
a
sage: f(z)=z^3+a*z^2
sage: f(f(z))
(a*z^2 + z^3)^3 + (a*z^2 + z^3)^2*a

link

Comments

Ahh, my mistake was that I was also declaring z as a variable. Why don't I need to do this? Does sage "know" that this is over the complex field?

sberner ( 4 years ago )

No, the construction f(z)= is not Pythonic, so there is some Sage preparsing that does the job of defining the symbol z :

sage: preparse('f(z)=z^3+a*z^2')
'__tmp__=var("z"); f = symbolic_expression(z**Integer(3)+a*z**Integer(2)).function(z)'

tmonteil ( 4 years ago )

add a comment

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

Stats

Asked: 4 years ago

Seen: 584 times

Last updated: Oct 24 '21

Self composition of a function with symbolic variable savecancel

1 Answer

Comments