## 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