Status: Accepted
Owner: asmeurer
CC: mattpap
Labels: Type-Defect Priority-Medium Integration Polynomial
New issue 2132 by asmeurer: Derivative of RootSum
http://code.google.com/p/sympy/issues/detail?id=2132
diff does not work at all with RootSum. This makes it impossible to
reverse integrate() whenever it returns that. For example:
In [16]: a = RootSum(z**5 - z**2 + 1, Lambda(z, z*log(x - z)))
In [17]: a
Out[17]:
⎛ 5 2 ⎞
RootSum⎝z - z + 1, Λ(z, z⋅log(x - z))⎠
In [18]: a.diff(x)
Out[18]:
d ⎛ ⎛ 5 2 ⎞⎞
──⎝RootSum⎝z - z + 1, Λ(z, z⋅log(x - z))⎠⎠
dx
In [19]: a.diff(z)
Out[19]:
d ⎛ ⎛ 5 2 ⎞⎞
──⎝RootSum⎝z - z + 1, Λ(z, z⋅log(x - z))⎠⎠
dz
But the derivative of x is just a rational function:
In [27]: b = (-5+4*x)/(1-x+x**5)
In [28]: b
Out[28]:
-(5 - 4⋅x)
──────────
5
1 - x + x
In [29]: integrate(b, x)
Out[29]:
⎛ 5 ⎞
RootSum⎝t - t + 1, Λ(t, t⋅log(x - t))⎠
The derivative with respect to z should be even easier: it's just 0, since
z is a dummy variable.
Now, fixing diff(z) should be trivial, but I don't know exactly how to
compute the derivative with respect to x. This requires a bit more
algebraic machinery than I currently know of. Mateusz, do you know how
this might be done? I know that it can be done, because Maple can do it:
a:=sum(r*log(x - r), r=RootOf(_Z**5 - _Z + 1));
-----
\
)
a := / r ln(x - r)
-----
/ 5 \
r = RootOf\_Z - _Z + 1/
diff(a, x);
-5 + 4 x
----------
5
1 - x + x
At any rate, it should at least return RootOf(poly, Lambda_expr.diff(x)).
--
You received this message because you are subscribed to the Google Groups
"sympy-issues" group.
To post to this group, send email to [email protected].
To unsubscribe from this group, send email to
[email protected].
For more options, visit this group at
http://groups.google.com/group/sympy-issues?hl=en.
