## Contents

- Gradient of the Rosenbrock function at [1,1], the global minimizer
- The Hessian matrix at the minimizer should be positive definite
- Gradient estimation using gradest - a function of 5 variables
- Simple Hessian matrix of a problem with 3 independent variables
- A semi-definite Hessian matrix
- Directional derivative of the Rosenbrock function at the solution
- Directional derivative at other locations
- Jacobian matrix of a scalar function is just the gradient
- Jacobian matrix of a linear system will reduce to the design matrix
- The jacobian matrix of a nonlinear transformation of variables

    % Multivariate calculus demo script

    % This script file is designed to be used in cell mode
    % from the matlab editor, or best of all, use the publish
    % to HTML feature from the matlab editor. Older versions
    % of matlab can copy and paste entire blocks of code into
    % the Matlab command window.

    % Typical usage of the gradient and Hessian might be in
    % optimization problems, where one might compare an analytically
    % derived gradient for correctness, or use the Hessian matrix
    % to compute confidence interval estimates on parameters in a
    % maximum likelihood estimation.

## Gradient of the Rosenbrock function at [1,1], the global minimizer

    rosen = @(x) (1-x(1)).^2 + 105*(x(2)-x(1).^2).^2;
    % The gradient should be zero (within floating point noise)
    [grad,err] = gradest(rosen,[1 1])

    grad =
       3.6989e-20            0
    err =
       1.4545e-18            0

## The Hessian matrix at the minimizer should be positive definite

    H = hessian(rosen,[1 1])
    % The eigenvalues of h should be positive
    eig(H)

    H =
              842         -420
             -420          210
    ans =
          0.39939
           1051.6

## Gradient estimation using gradest - a function of 5 variables

    [grad,err] = gradest(@(x) sum(x.^2),[1 2 3 4 5])

    grad =
                2            4            6            8           10
    err =
       1.2533e-14   2.5067e-14   3.4734e-14   5.0134e-14    2.836e-14

## Simple Hessian matrix of a problem with 3 independent variables

    [H,err] = hessian(@(x) x(1) + x(2)^2 + x(3)^3,[1 2 3])

    H =
         0     0     0
         0     2     0
         0     0    18
    err =
                0            0            0
                0   4.6563e-15            0
                0            0   3.3318e-14

## A semi-definite Hessian matrix

    H = hessian(@(xy) cos(xy(1) - xy(2)),[0 0])
    % one of these eigenvalues will be zero (approximately)
    eig(H)

    H =
               -1            1
                1           -1
    ans =
               -2
      -3.8795e-07

## Directional derivative of the Rosenbrock function at the solution

This should be zero. Ok, its a trivial test case.

    [dd,err] = directionaldiff(rosen,[1 1],[1 2])

    dd =
         0
    err =
         0

## Directional derivative at other locations

    [dd,err] = directionaldiff(rosen,[2 3],[1 -1])

    % We can test this example
    v = [1 -1];
    v = v/norm(v);
    g = gradest(rosen,[2 3]);

    % The directional derivative will be the dot product of the gradient with
    % the (unit normalized) vector. So this difference will be (approx) zero.
    dot(g,v) - dd

    dd =
           743.88
    err =
       1.5078e-12
    ans =
       1.5916e-12

## Jacobian matrix of a scalar function is just the gradient

    [jac,err] = jacobianest(rosen,[2 3])

    grad = gradest(rosen,[2 3])

    jac =
              842         -210
    err =
       2.8698e-12   1.0642e-12
    grad =
              842         -210

## Jacobian matrix of a linear system will reduce to the design matrix

    A = rand(5,3);
    b = rand(5,1);
    fun = @(x) (A*x-b);

    x = rand(3,1);
    [jac,err] = jacobianest(fun,x)

    disp 'This should be essentially zero at any location x'
    jac - A

    jac =
          0.81472      0.09754      0.15761
          0.90579       0.2785      0.97059
          0.12699      0.54688      0.95717
          0.91338      0.95751      0.48538
          0.63236      0.96489      0.80028
    err =
       3.9634e-15   3.8376e-16   5.4271e-16
       3.9634e-15   8.8625e-16   5.0134e-15
       7.0064e-16   3.0701e-15   5.3175e-15
         3.76e-15   4.8542e-15   2.4271e-15
       3.0701e-15   4.3417e-15   3.9634e-15
    This should be essentially zero at any location x
    ans =
                0  -3.1919e-16            0
       1.1102e-16  -1.6653e-16            0
       2.7756e-17            0  -1.1102e-16
      -1.1102e-16  -3.3307e-16   1.6653e-16
                0   2.2204e-16   1.1102e-16

## The jacobian matrix of a nonlinear transformation of variables

evaluated at some arbitrary location [-2, -3]

    fun = @(xy) [xy(1).^2, cos(xy(1) - xy(2))];
    [jac,err] = jacobianest(fun,[-2 -3])

    jac =
               -4            0
         -0.84147      0.84147
    err =
       2.5067e-14            0
       6.1478e-14   1.9658e-14