I am trying to evaluate the gradient of an expression with respect to a vector, whose elements are expressions, not variables. Let’s say the expression looks something like this :
I want to take the Gradient of H as follows:
As you can see, Mathematica doesn’t evaluate the Gradient. Any suggestions? Thanks in advance!
You can replace either by inspection or solving for the new variables:
h = (t1 - t2)^2 + (t2 - t3)^2 + (t1 - t3)^2 + (m1*(ob1 - on))^2/
m1 + (m2*(ob2 - on))^2/m2 + (m3*(ob3 - on))^2/m3;
Simplify[h /.
Solve[{e1 == t1 - t2, e2 == t2 - t3,
p1 == m1*(ob1 - on), p2 == m2*(ob2 - on), p3 == m3*(ob3 - on)},
{t2, t3, ob1, ob2, ob3}][[1]]]
Grad[%, {e1, e2, p1, p2, p3}]
Your three components of eta are not independent.
Thanks a lot!
But the thing is, I would like to have in the output of the gradient in terms of {ob1,ob2, ob3}
For example, Grad[h,{p1,p2,p3}] should result in {2*(ob1-on),2*(ob2-on),2*(ob3-on)} and not in {2*p1/m1,2*p2/m2,2*p3/m3}
Any ideas?
Thanks again!
The clean way is to start with H in terms of eta and p from the beginning. However, you can replace back:
h = (t1 - t2)^2 + (t2 - t3)^2 + (t1 - t3)^2 + (m1*(ob1 - on))^2/
m1 + (m2*(ob2 - on))^2/m2 + (m3*(ob3 - on))^2/m3;
Simplify[h /.
Solve[{e1 == t1 - t2, e2 == t2 - t3, p1 == m1*(ob1 - on),
p2 == m2*(ob2 - on), p3 == m3*(ob3 - on)}, {t2, t3, ob1, ob2,
ob3}][[1]]]
Grad[%, {e1, e2, p1, p2, p3}] /. {e1 -> t1 - t2, e2 -> t2 - t3,
p1 -> m1*(ob1 - on), p2 -> m2*(ob2 - on), p3 -> m3*(ob3 - on)}
You have to make up your mind on the three components of eta.
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