Jacobian of scalar matrix

조회 수: 1 (최근 30일)
Allison . 2014년 3월 5일
편집: Oliver Woodford . 2014년 3월 5일
I have a function of two variables, V(x1, x2) that's something like:
syms x1 x2
V = 0.5*(x1^2 + x2^2)
The partial derivatives V_x1 and V_x2 (with respect to x1 and x2) should be
V_x1 = x1
V_x2 = x2
This can be verified by something like jacobian(V, [x1 x2]).
I'm in the situation where I only know V numerically as a matrix of scalars, for example, on a grid:
[x1,x2] = meshgrid(-1:.1:1, -1:.1:1)
V = 0.5*(x1.^2+x2.^2) % used just for generating the scalar version of V
I'm interested in obtaining V_x1 and V_x2 as above. Obviously, I can no longer use the jacobian command because V is no longer an analytic function. I'm currently doing something like:
Vx1 = gradient(V, 1)
Vx2 = gradient(V, 2)
However, I get different numerical values than if I do the Jacobian by hand and evaluate it on the mesh grid. If I calculate the difference between Vx1 and the exact solution (which is x1),
norm(Vx1 - x1)
I'm expecting zero but it is not zero. If I plot Vx1, I see that it visually looks very similar to the exact solution.
Am I using the wrong function with gradient()? How do I explain the discrepancy between its result and the analytical result?

답변 (2개)

Roger Stafford
Roger Stafford 2014년 3월 5일
편집: Roger Stafford 님. 2014년 3월 5일
If you use discrete values for the variables x1 and x2, only an approximation to the partial derivatives can be achieved. That is true whether you use 'gradient' or some other higher order method. You can only get formulas for the exact derivatives by using symbolic methods, which is to say, by using analytical methods.
Also I believe you called 'gradient' improperly. It should have been:
[V_x1,V_x2] = gradient(V,dx1,dx2);
where dx1 and dx2 are the respective interval lengths (both .1 in your example.)

Oliver Woodford
Oliver Woodford 2014년 3월 5일
편집: Oliver Woodford 님. 2014년 3월 5일
gradient() computes partial derivatives approximately, using finite differences. It can't compute the exact partial derivatives because it has no knowledge of the equation used to compute the values it's given. This explains the discrepancy between what it outputs and the analytic solution.
If you want a function to compute the analytic partial derivatives, you either need to code it yourself, or use the Symbolic Toolbox functions to automatically create a function which does it, given the original cost function. Either way, the function will need to be given the values x1 and x2, rather than V.


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