Trying to obtain the derivative of →u=[−1,1] using the following code:
u = matrix(1,2,[-1, 1])
r = derivative(u,x); r
I get a scalar value 0.
Although according the following relation it should be a 2-dimensional zero vector.
d→udx=ddx[−1,1]=[ddx(−1),ddx(1)]=[0,0]
Why does it happen? In the case it's a bug where could I report it?
Thanks
Look at the help of derivative. It works on symbolic functions, polynomials, and symbolic expressions. Your variable u is not a function, so it is not really being considered a two dimensional symbolic expression or function. If you do
sage: u(x,y) = matrix(1,2,[-1,1])
sage: derivative(u, x)
(x, y) |--> 0then you can see that it is considering it as a two dimensional function. In the second case, you have a function of one variable in the variable x.
The alternative fix is to work in the symbolic ring:
sage: u = matrix(SR, [-1,1])
sage: derivative(u,x)
[0 0]Thanks, the fix with a symbolic ring is helpful. In my problem the →u really is a function, just happens to be a constant at this occasion.
Anyway, I reckon that the first example is probably a bug. The derivative of n-dimensional vector should have n-dimensions again no matter to which ring it belongs to. Would you know how to report that, please?
This is now ticket 15067
Asked: 11 years ago
Seen: 1,694 times
Last updated: Aug 19 '13
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Compared to: a(x) = function('a',x) b(x) = function('b',x) u = matrix(1,2,[a,b]) r = derivative(u,x);r Which gives a vector as expected: [x |--> D[0](a)(x) x |--> D[0](b)(x)]