Is the reason why we are multiplying the Hessian in the quadratic term by (x - x_0) is so that when we perform the dot product, we get out the "quadratic" that we are looking for?
kinematics
Would this typically work as expected if our basis isn't e.g. orthonormal?
birb
Are there \nabla^3, \nabla^4, etc. that are used in this sort of multivariable Taylor Series approximation as well?
derk
In approximating a general f(u), is it typically sufficient to approximate up to the "quadratic" hessian component of the Taylor Series? When, why, and how might we incorporate cubic, quartic, etc. pieces to the approximation.
Concurrensee
if we include \nabla^3, \nabla^4, would it be helpful to have a more accurate result?
Is the reason why we are multiplying the Hessian in the quadratic term by (x - x_0) is so that when we perform the dot product, we get out the "quadratic" that we are looking for?
Would this typically work as expected if our basis isn't e.g. orthonormal?
Are there \nabla^3, \nabla^4, etc. that are used in this sort of multivariable Taylor Series approximation as well?
In approximating a general f(u), is it typically sufficient to approximate up to the "quadratic" hessian component of the Taylor Series? When, why, and how might we incorporate cubic, quartic, etc. pieces to the approximation.
if we include \nabla^3, \nabla^4, would it be helpful to have a more accurate result?