APM3715 Assignment 01 Solutions 2026

UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology



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APM3715 ASSIGNMENT 01
Semester 1 Assignment 01 — 2026


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Module Code: APM3715

Module Name: Applied Mathematics

Assignment No.: 01

Due Date: 2026

Semester: Semester 1, 2026




Submitted in partial fulfilment of the requirements for APM3715
at the University of South Africa.

, UNISA | APM3715 Assignment 01 – 2026



Question 1: Secant Method — Solving x = cos x

Question: Solve the following expression using the Secant method:


x = cos x


Comment on your solution, approximations and round off errors.


1.1 Problem Setup


Rewriting the equation in standard root-finding form:


f (x) = cos x − x = 0



The Secant method uses two initial approximations and applies the recurrence:

xn − xn−1
xn+1 = xn − f (xn )
f (xn ) − f (xn−1 )


Initial guesses chosen: x0 = 0 and x1 = 1.


1.2 Iteration Table


Table 1: Secant Method Iterations for f (x) = cos x − x
Iteration xn f (xn ) |xn − xn−1 |
0 0.0000 1.0000 —
1 1.0000 −0.4597 1.0000
2 0.6847 0.0893 0.3153
3 0.7360 0.0051 0.0513
4 0.7391 ≈ 0.0000 0.0031



1.3 Step-by-Step Calculations


Step 1: Evaluate at the two initial points.



f (x0 ) = f (0) = cos(0) − 0 = 1 − 0 = 1

f (x1 ) = f (1) = cos(1) − 1 ≈ 0.5403 − 1 = −0.4597

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