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Then we can obtain $\frac{\partial }{\partial y}\left(J(y)z\right)$ by deriving derivatives with respect to the right hand side.
With the new jac block class added in #76, the jac-vector product shall be divided into three categories:
if jac block is a matrix, then we have $J_{ij}$@ $z_i$;
if jac block is a diagonal matrix, then we have $\mathrm{diag}(J_{ij})$* $z_j$, where $\mathrm{diag}(J_{ij})$ is the vector of the diagonal entries of $J_{ij}$.
if jac block is a column vector, the we have $J_{ij}$* $z_j$.
The text was updated successfully, but these errors were encountered:
In some solvers, we need the derivative
where the Hessian tensor
We can avoid calculating the Hessian tensor by first multiplying out the Jac blocks so that we have
Then we can obtain$\frac{\partial }{\partial y}\left(J(y)z\right)$ by deriving derivatives with respect to the right hand side.
With the new jac block class added in #76, the jac-vector product shall be divided into three categories:
The text was updated successfully, but these errors were encountered: