Given a function f(y), where f, y or both can be vectors, calculate the derivative (gradient, Jacobian) df/dy.
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Given a function f(y), where f, y or both can be vectors, calculate the derivative (gradient, Jacobian) df/dy.
Calculation is done using numerical differencing, which should be considered a last resort for cases in which the analytic derivative is unavailable. (Note that you can obtain an analytic gradient automatically from the source code for f using automatic differentiation methods like complex step derivatives, ADIFOR, etc.).
- Theory and Implementation
The SimTK::Differentiator class uses methods adapted from the book Practical Optimization by Gill, Murray, and Wright (1981), section 8.6 (339ff) and Numerical Recipies in C++ 2nd ed. (2002) section 5.7 (192ff). Here is a summary:
- We want to differentiate a function f(y) whose estimated relative accuracy eps is known (e.g. eps=1e-6). (We'll treat y as a scalar here but for vector y this is done for one element yi at a time.)
- We need to know what perturbation h to use for calculating an estimate of df/dy that optimally balances roundoff error (h too small) with truncation error (h too big).
- First guess at h depends on the order of the numerical method: either forward difference (1st order) or central difference (2nd order). For 1st order, h0=eps^(1/2); for 2nd order h0=eps^(1/3).
- Now we have to make sure that we can compute y+h reliably. If y is very large, we can not allow h to be too small. We calculate a scaled perturbation h1=h0*max(y, 0.1). The 0.1 allows a small y to pull down the step size a little; but it is dangerous to go much lower because a very small y might just be zero plus noise.
- Finally, the step size should be exactly representable as a power of 2. Conceptually, this is just h=(y+h1)-y although one must be careful to stop the compiler from cleverly "simplifying" this expression. Differentiator uses a C++ volatile variable for that purpose.
Then the derivative, gradient element, or Jacobian column is computed as df/dy=[f(x+h)-f(x)]/h (1st order) or df/dy=[f(x+h)-f(x-h)]/(2h) (2nd order).
1.8.8 
