numpy second derivative of a ndimensional array

Question

I have a set of simulation data where I would like to find the lowest slope in n dimensions. The spacing of the data is constant along each dimension, but not all the same (I co

Answer 1

You can see the Hessian Matrix as a gradient of gradient, where you apply gradient a second time for each component of the first gradient calculated here is a wikipedia link definig Hessian matrix and you can see clearly that is a gradient of gradient, here is a python implementation defining gradient then hessian :

import numpy as np
#Gradient Function
def gradient_f(x, f):
  assert (x.shape[0] >= x.shape[1]), "the vector should be a column vector"
  x = x.astype(float)
  N = x.shape[0]
  gradient = []
  for i in range(N):
    eps = abs(x[i]) *  np.finfo(np.float32).eps 
    xx0 = 1. * x[i]
    f0 = f(x)
    x[i] = x[i] + eps
    f1 = f(x)
    gradient.append(np.asscalar(np.array([f1 - f0]))/eps)
    x[i] = xx0
  return np.array(gradient).reshape(x.shape)

#Hessian Matrix
def hessian (x, the_func):
  N = x.shape[0]
  hessian = np.zeros((N,N)) 
  gd_0 = gradient_f( x, the_func)
  eps = np.linalg.norm(gd_0) * np.finfo(np.float32).eps 
  for i in range(N):
    xx0 = 1.*x[i]
    x[i] = xx0 + eps
    gd_1 =  gradient_f(x, the_func)
    hessian[:,i] = ((gd_1 - gd_0)/eps).reshape(x.shape[0])
    x[i] =xx0
  return hessian

As a test, the Hessian matrix of (x^2 + y^2) is 2 * I_2 where I_2 is the identity matrix of dimension 2