Are vector functions supported in sage?
as in:
f1(r,phi,theta) = 1/r
f2(r,phi,theta) = 1
f3(r,phi,theta) = 1
F = vector([f1, f2, f3])
G=rot(F) # vector function
An ancient post seems to suggest that rot has to be defined with a positional python-argument like rot(*F).
Is there a way for "rot" to take the vector symbolically as an argument?
I don't understand the question. Are you asking if it's possible to use the special notation like f(x) = x^2 for the creation of vector valued functions? If so, the answer is currently no, but it could probably be added to the Sage preparser without much problem. Of course you can always define the function rot the ordinary way to take a vector and return a vector:
sage: rot = lambda v: vector([1/v[0],1,1])
sage: rot(vector([2,1,1]))
(1/2, 1, 1)
I usually think of a gradient as being a property of a single-variable function... but anyway, is this what you want?
sage: F.diff()
[(r, phi, theta) |--> -1/r^2 (r, phi, theta) |--> 0 (r, phi, theta) |--> 0]
[ (r, phi, theta) |--> 0 (r, phi, theta) |--> 0 (r, phi, theta) |--> 0]
[ (r, phi, theta) |--> 0 (r, phi, theta) |--> 0 (r, phi, theta) |--> 0]
sage: diff(F)
[(r, phi, theta) |--> -1/r^2 (r, phi, theta) |--> 0 (r, phi, theta) |--> 0]
[ (r, phi, theta) |--> 0 (r, phi, theta) |--> 0 (r, phi, theta) |--> 0]
[ (r, phi, theta) |--> 0 (r, phi, theta) |--> 0 (r, phi, theta) |--> 0]
Asked: 2012-10-01 15:48:52 +0200
Seen: 1,679 times
Last updated: Oct 02 '12
Copyright Sage, 2010. Some rights reserved under creative commons license. Content on this site is licensed under a Creative Commons Attribution Share Alike 3.0 license.
I think I know what you mean now. `sage: rot(r,phi,theta) = F` could in theory make a vector function, but it doesn't currently know what to do.