I'm learning SymPy now, and I wonder if there's a way to differentiate a function over one of its variables in the general form.
Consider this example:
There a vector, lets write its components as x_1, x_2, x_3. The length r of such vector will be r = sqrt(x_1^2 + x_2^2 + x_3^2).
I want to differentiate this vector over x_i, where I don't specify i, which should give something like x_i/r^3.
Is it possible to do this in SymPy?
Sorry for lack of equation rendering..
You can do this with IndexedBase. The result comes out in Kronecker delta functions which than simplify after substitution:
In [3]: x = IndexedBase('x')
In [4]: r = sqrt(x[1]**2 + x[2]**2 + x[3]**2)
In [5]: r
Out[5]:
_______________________
╱ 2 2 2
╲╱ x[1] + x[2] + x[3]
In [6]: i = Symbol('i')
In [7]: r.diff(x[i])
Out[7]:
δ ⋅x[1] + δ ⋅x[2] + δ ⋅x[3]
1,i 2,i 3,i
─────────────────────────────────
_______________________
╱ 2 2 2
╲╱ x[1] + x[2] + x[3]
In [8]: r.diff(x[i]).subs(i, 2)
Out[8]:
x[2]
──────────────────────────
_______________________
╱ 2 2 2
╲╱ x[1] + x[2] + x[3]
You can also do this for a vector of symbolic dimension:
In [9]: j = Symbol('j')
In [9]: N = Symbol('N')
In [10]: r = sqrt(Sum(x[i]**2, (i, 1, N)))
In [10]: r
Out[10]:
_____________
╱ N
╱ ___
╱ ╲
╱ ╲ 2
╱ ╱ x[i]
╱ ╱
╱ ‾‾‾
╲╱ i = 1
In [11]: r.diff(x[j])
Out[11]:
N
___
╲
╲ 2⋅δ ⋅x[i]
╱ i,j
╱
‾‾‾
i = 1
────────────────────────
_____________
╱ N
╱ ___
╱ ╲
╱ ╲ 2
2⋅ ╱ ╱ x[i]
╱ ╱
╱ ‾‾‾
╲╱ i = 1
In [12]: r.diff(x[j]).subs(N, 3).subs(j, 2)
Out[12]:
3
___
╲
╲ 2⋅δ ⋅x[i]
╱ 2,i
╱
‾‾‾
i = 1
────────────────────────
_____________
╱ 3
╱ ___
╱ ╲
╱ ╲ 2
2⋅ ╱ ╱ x[i]
╱ ╱
╱ ‾‾‾
╲╱ i = 1
In [13]: r.diff(x[j]).subs(N, 3).subs(j, 2).doit()
Out[13]:
x[2]
──────────────────────────
_______________________
╱ 2 2 2
╲╱ x[1] + x[2] + x[3]