Root Finding in MATLAB: Video Tutorial

This video tutorial shows how to solve equations numerically with the built-in functions of MATLAB. It covers polynomial equations, a single nonlinear equation, systems of nonlinear equations and systems of linear equations. It is part of the full Numerical Computations course, which you can find on this page.

The problem

A root of a function $f$ is a value of $x$ where the function becomes zero:

MATLAB has a different tool for each form of this problem. When $f$ is a polynomial, all of its roots can be found at once. When $f$ is a general nonlinear function, a solver starts from an initial guess and searches for one root. When there are several equations and several unknowns, $x$ and $f(x)$ become vectors. The last video treats the special case of a linear system, $Ax = b$, which strictly speaking is equation solving rather than root finding.

Watch online

The tutorial has four sections. They are embedded here as a playlist on YouTube.

Play playlistLoads from YouTube

What each video covers

Section 1: Roots of polynomials with roots

Finding roots of polynomials using roots (YouTube (opens in a new tab)). This first video also opens the whole course with a short overview of its topics.

MATLAB stores a polynomial as a vector of its coefficients, from the highest power down to the constant term. The polynomial

is the vector [an an-1 ... a1 a0]. The first example is the quadratic equation $x^2 - x - 1 = 0$, with the coefficient vector [1 -1 -1]. The video compares the output of roots with the quadratic formula, $x = (1 \pm \sqrt{5})/2$, and the two results agree (about $-0.618$ and $1.618$).

p = [1 -1 -1];
roots(p)

The second example is the cubic equation $x^3 - 3x^2 + 4x - 5 = 0$. Here roots returns one real root (about $2.2134$) and a pair of complex conjugate roots. The same call works for a polynomial of any degree.

Section 2: A nonlinear equation with fzero

Finding roots of nonlinear functions using fzero (YouTube (opens in a new tab)). The function fzero finds one root of a single equation in one unknown. It takes the function and an initial point $x_0$ where the search starts.

The example is the equation

The function is defined as an anonymous function with element-wise operators, so it also works with vectors. A plot made with ezplot shows a root close to 1, and fzero started from 0 returns about $0.9581$.

f = @(x) x.^2 - 5*exp(-x) + 1;
x = fzero(f, 0)

The video then shows why the starting point matters for an equation with many roots. With the cosine function, different starting points lead fzero to different roots, such as $-\pi/2$ or $\pi/2$. Starting from several points can reveal different roots, but it does not prove that every root has been found.

Section 3: Systems of nonlinear equations with fsolve

Solving systems of nonlinear equations using fsolve (YouTube (opens in a new tab)). The limit of fzero is one equation with one unknown. For a system $F(x) = 0$, where $x$ and $F$ are vectors, MATLAB offers fsolve.

The example system has two unknowns, $x$ and $y$:

First the unknowns are renamed $x_1$ and $x_2$, and every equation is written with zero on the right-hand side. The function then returns a column vector with one element per equation. Started from $(1, 1)$, fsolve reports that the equation is solved and returns $x_1 = 3$ and $x_2 = 5$. The second output gives the function value at the solution, which is almost zero.

F = @(x) [x(1)*x(2) - 5*x(2) + 10
          x(1)^3 - x(2)^2 - 2];
[x, fval] = fsolve(F, [1; 1])

Section 4: Systems of linear equations with linsolve

Solving systems of linear equations using linsolve (YouTube (opens in a new tab)). A linear system has the form $Ax = b$, where $A$ is an $n \times n$ matrix of coefficients, $b$ is the right-hand side and $x$ holds the unknowns. When $A$ is not singular, the solution is $x = A^{-1}b$.

The example system is:

The video solves it in three mathematically equivalent ways: with inv(A)*b, with the backslash operator A\b, and with the built-in function linsolve(A, b). All three give $x = 3$, $y = 2$ and $z = 1$. For computation, solve the system with A\b or linsolve rather than explicitly forming the inverse.

A = [1 -2 3; 1 4 -5; 5 -1 -1];
b = [2; 6; 12];
x = linsolve(A, b)

Which function to use

• roots: all roots of a polynomial, given its coefficients.
• fzero: one root of one nonlinear equation in one unknown, near a starting point.
• fsolve: a solution of a system of nonlinear equations, from a starting point. It is part of Optimization Toolbox, while the other three functions are in core MATLAB.
• linsolve or A\b: the solution of a system of linear equations.

For the classic iterative methods behind these tools, see Numerical Root Finding Methods in Python and MATLAB, which covers the bisection, secant and Newton-Raphson methods.

Check the result

Substitute each computed root into the original function. For the system examples, inspect fval or the residual A*x - b. A small residual shows how closely the equations are satisfied, while a poorly conditioned problem can still make the solution sensitive to small changes in the inputs. Treat each starting point as a separate experiment when a nonlinear system may have several solutions.