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scipy.optimize.

# approx_fprime

scipy.optimize.approx_fprime(xk, f, epsilon=np.float64(1.4901161193847656e-08), *args)

  Finite difference approximation of the derivatives of a scalar or vector-valued function.

  If a function maps from \(R^n\) to \(R^m\), its derivatives form an m-by-n matrix called the Jacobian, where an element \((i, j)\) is a partial derivative of f[i] with respect to `xk[j]`.

  Parameters:
    xkarray_like

      The coordinate vector at which to determine the gradient of f.

    fcallable

      Function of which to estimate the derivatives of. Has the signature `f(xk, *args)` where xk is the argument in the form of a 1-D array and args is a tuple of any additional fixed parameters needed to completely specify the function. The argument xk passed to this function is an ndarray of shape (n,) (never a scalar even if n=1). It must return a 1-D array_like of shape (m,) or a scalar.

      Suppose the callable has signature `f0(x, *my_args, **my_kwargs)`, where `my_args` and `my_kwargs` are required positional and keyword arguments. Rather than passing `f0` as the callable, wrap it to accept only `x`; e.g., pass `fun=lambda x: f0(x, *my_args, **my_kwargs)` as the callable, where `my_args` (tuple) and `my_kwargs` (dict) have been gathered before invoking this function.

      Changed in version 1.9.0: f is now able to return a 1-D array-like, with the \((m, n)\) Jacobian being estimated.

    epsilon{float, array_like}, optional

      Increment to xk to use for determining the function gradient. If a scalar, uses the same finite difference delta for all partial derivatives. If an array, should contain one value per element of xk. Defaults to `sqrt(np.finfo(float).eps)`, which is approximately 1.49e-08.

    *argsargs, optional

      Any other arguments that are to be passed to f.

  Returns:
    jacndarray

      The partial derivatives of f to xk.

  See also

  `check_grad`

    Check correctness of gradient function against approx_fprime.

  Notes

  The function gradient is determined by the forward finite difference formula:

               f(xk[i] + epsilon[i]) - f(xk[i])
      f'[i] = ---------------------------------
                          epsilon[i]

  Examples

      >>> import numpy as np
      >>> from scipy import optimize
      >>> def func(x, c0, c1):
      ...     "Coordinate vector `x` should be an array of size two."
      ...     return c0 * x[0]**2 + c1*x[1]**2

      >>> x = np.ones(2)
      >>> c0, c1 = (1, 200)
      >>> eps = np.sqrt(np.finfo(float).eps)
      >>> optimize.approx_fprime(x, func, [eps, np.sqrt(200) * eps], c0, c1)
      array([   2.        ,  400.00004208])