MANE 3351
Lecture 12
Classroom Management
Agenda
- Test 1 not graded; some people still need to take
- Homework 3 Assignment
- Lab 6 Assignme
- Secant Method; last root finding method
Calendar
| Date | Lecture | Lab |
|---|---|---|
| 10/7 | Secant Method, Homework 3 | Lab 6 |
| 10/9 | Trapezoid Rule | Lab 6, continued |
| 10/14 | Simpson's Rule, Homework 4 | Lab 7 |
| 10/16 | Romberg Integration | Lab 7, continued |
| 10/21 | No Class, Dr. Timmer on ABET Visit | No Class, Dr. Timmer on ABET Visit |
| 10/23 | Gaussian Quadrature, Homework 5 | Lab 8 |
| 10/28 | Numerical Differentiation (not on Test 2) | Lab 8, continued |
| 11/4 | Numerical Integration (not on Test 2) | Lab 9 |
| 11/6 | Linear Algebra, part 1 (not on Test 2) | Lab 9, continued |
| 11/11 | Test 2 (Root Finding and Numerical Integration) | No Lab |
Resources
Handouts
Lecture 12 Content
- Today's topic is the secant method.
- The secant method does not utilize derivative information as Newton's method does.
- The secant method is also similar to the false position method.
- The secant method is the last root finding method covered
Limit Definition of the Derivative
Recall the limit definition of the derivative1

Geometric Inspiration
Cheney and Kincaid (2004)3 demonstrate the geometric inspiration for secant method

Secant Method
- The formula is simply
\[
x_{n+1}=x_n - g(x_n)\frac{x_n-x_{n-1}}{g(x_n)-g(x_{n-1})}
\]
Example Problem Used for Newton's Method
- Consider a new function, \(f(x)=e^x+2^{-x}+2\cos(x)-6=0\)

Pseudo-code
Brin (2020)2provides the following pseudo-code

Convergence
- Brin (2020)2 study the performance of Secant Method
- The secant method converges with order \(\frac{1+\sqrt{5}}{2}=1.62\)
- Not quite as fast as Newton's Method (quadratic)
Similarity to False Position
- Secant method: \(x_{n+1}=x_n - g(x_n)\frac{x_n-x_{n-1}}{g(x_n)-g(x_{n-1})}\)
- False Position method: \(x_r=x_u-\frac{f(x_u)(x_l-x_u)}{f(x_l)-f(x_u)}\)
- Secant method is not guaranteed to bracket result when starting
Speed of Convergence
Chapra and Canale (2015)4 compared the performance of various root finding methods

Secant Code
import math
def f(x):
return (math.exp(x)+2**(-x)+2*math.cos(x)-6)
N=100
tol=0.0005
x0=6.0
x1=5.0
# step 1
y0=f(x0)
y1=f(x1)
# counter is additional
counter=0
for j in range(N+1):
counter=counter+1
# step 3
x=x1-y1*((x1-x0)/(y1-y0))
if math.fabs(x-x1)<tol:
print("the root is {} with value {}, required {} steps".format(x,f(x),counter))
break
# step 5
x0=x1
y0=y1
x1=x
y1=f(x1)
print("completed")
-
https://math.hmc.edu/calculus/hmc-mathematics-calculus-online-tutorials/single-variable-calculus/limit-definition-of-the-derivative/ ↩
-
Brin, L, (2020), Tea Time Numerical Analysis: Experiences in Mathematics, 3rd edition ↩↩
-
Cheny, W., and Kincaid, D., (2004), Numerical Mathematics and Computer, 5th edition ↩
-
Chapra, S., and Canale, R., (2015), Numerical Methods for Engineers, 7th edition ↩