Lecture 14

E21 Computer Engineering Fundamentals

Emad Masroor

October 22, 2026

Bisection Method

Convergence rates

At each subsequent bisection step, the bisection window is halved.

General approach to root-finding

In all root-finding approaches, we:

• start with a guess of where the root might be,
• work iteratively, repeatedly applying the same formula many times
  • the number of iterations it takes to arrive at a solution is an important measure of how good an approach is.
  • in practice, this is either a while or for loop.
• hopefully converge to one root.

The Secant Method

Secant Method — Zoomed in

Secant Method — Zoomed in again

Summary of Secant Method

Start with two estimates of the root.

Evaluate the function at both \(x_1\) and \(x_2\)

Draw a straight line between the two points. This is an estimate of the function’s behavior.

Locate the point \((x_3,0)\) where this line crosses the x-axis. Call this \(x_3\).

Let \((x_3,f_3)\) be a new point. Forget about \((x_1,f_1)\).

Repeat the process above, using the points \((x_2,f_2)\) and \((x_3,f_3)\)

Writing a program for the Secant method

from numpy import linspace
def f(x):
    return -(x+1.5)*(x+2.5)

x1 = -2
x2 = -1

fx1 = f(x1)
fx2 = f(x2)

# Find equation of straight line
# thru (x1,fx1) and (x2,fx2)

m = (fx2-fx1)/(x2-x1)

print(f"Slope of the line btwn 2 points is {m}")

# To find y-intercept, use one of the points
# y = mx + c
# fx1 = m*x1 + c
# c = fx1 - m*x1
c = fx1 - m*x1

# equation of straight line is y = mx + c

# Now find the point where this line intscts
# the x-axis

# y = mx + c
# 0 = m*x_intercept + c
# x_intercept = - c/m
x_int = -c/m

print(f"The line intersects x-axis at {x_int}")

Newton’s Method

Summary of Newton’s Method

Evaluate the function at your first guess.

Make a straight line tangent to \((x_1,f_1)\)

Find location where this line intersects x-axis. This is the 2nd guess, call it \(x_2\).

Repeat.

Comparing Newton’s method with Secant method

• Newton’s method for finding roots of \(f(x)\) requires that you know \(f'(x)\) already.
• Secant method requires starting with two points instead of one.
• If two points very close to each other are chosen for the Secant method, the two methods are nearly identical

Equations for Newton’s method

Derive an equation for \(x_{n+1}\) given \(x_n\)

Equation of the tangent line is \(y = mx+c\),

where \(m = f'(x_n)\). So we get \[y = f'(x_n) x + c\]

and \(c\) can be found by substituting a known point into the equation:

\(\underbrace{f(x_n)}_{\text{y coordinate}} = f'(x_n) \underbrace{x_n}_{\text{x coordinate}} + c \implies c = f(x_n) - f'(x_n) x_n\).

Thus, the equation of the straight line is \(\displaystyle{\color{red}{y}} = \overbrace{f'(x_n)}^{m} {\color{red}{x}} + \overbrace{f(x_n) - f'(x_n) x_n}^{c}\)

The intersection of this line with the \(x\) axis occurs when \(y=0\), i.e.:

\(0 =f'(x_n) {\color{red}{x}} + f(x_n) - f'(x_n) x_n\)

Recognizing that the unknown \({\color{red}{x}}\) is \(x_{n+1}\), we get \[f'(x_n) x_n - f(x_n)= f'(x_n) {\color{red}{x}} \implies \frac{f'(x_n) x_n - f(x_n)}{f'(x_n)} = {\color{red}{x}}\]

\[\implies \boxed{x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}}\]

Writing a program for Newton’s method

Pitfalls of Root Finding

If a root-finding program does not find a solution, it could be because:

1. the equation \(f(x)=0\) has no solution
2. a solution exists but the program wasn’t able to find it
3. a solution is found but it’s not the desired solution
4. too many iterations are needed to arrive at the solution

A well-executed numerical method can solve problems 2-4 but cannot solve problem 1.

Example — Finding the wrong root

Suppose an equation is known to have multiple solutions, but only one of them is physically meaningful. A root-finding program will not necessarily find the solution we care about.

Example — Stuck in a flat place

In the neighborhood of zero-slope points, these methods run into problems.