APM3711 Assignment 02 Solutions 2026
UNISA
Module: APM3711 – Numerical Methods for Differential
Equations
DUE DATE: 12 JUNE 2026
, Question 1
Given
𝑓(𝑥) = √1 + 𝑥
with power series
𝑥 𝑥2 𝑥3
𝑓(𝑥) = 1 + − + +⋯
2 8 16
The degree 2 truncated power series is
𝑥 𝑥2
𝑃2 (𝑥) = 1 + −
2 8
(a) Express 𝑷𝟐 (𝒙)in terms of Chebyshev polynomials 𝑻𝟎 , 𝑻𝟏 , 𝑻𝟐 .
The first three Chebyshev polynomials are
𝑇0 (𝑥) = 1
𝑇1 (𝑥) = 𝑥
𝑇2 (𝑥) = 2𝑥 2 − 1
From
𝑇2 (𝑥) = 2𝑥 2 − 1
we obtain
𝑇2 (𝑥) + 1
𝑥2 =
2
Substitute into 𝑃2 (𝑥):
1 1 𝑇2 (𝑥) + 1
𝑃2 (𝑥) = 1 + 𝑇1 (𝑥) − ( )
2 8 2
1 1 1
= 1 + 𝑇1 (𝑥) − 𝑇2 (𝑥) −
2 16 16
Combine constants:
UNISA
Module: APM3711 – Numerical Methods for Differential
Equations
DUE DATE: 12 JUNE 2026
, Question 1
Given
𝑓(𝑥) = √1 + 𝑥
with power series
𝑥 𝑥2 𝑥3
𝑓(𝑥) = 1 + − + +⋯
2 8 16
The degree 2 truncated power series is
𝑥 𝑥2
𝑃2 (𝑥) = 1 + −
2 8
(a) Express 𝑷𝟐 (𝒙)in terms of Chebyshev polynomials 𝑻𝟎 , 𝑻𝟏 , 𝑻𝟐 .
The first three Chebyshev polynomials are
𝑇0 (𝑥) = 1
𝑇1 (𝑥) = 𝑥
𝑇2 (𝑥) = 2𝑥 2 − 1
From
𝑇2 (𝑥) = 2𝑥 2 − 1
we obtain
𝑇2 (𝑥) + 1
𝑥2 =
2
Substitute into 𝑃2 (𝑥):
1 1 𝑇2 (𝑥) + 1
𝑃2 (𝑥) = 1 + 𝑇1 (𝑥) − ( )
2 8 2
1 1 1
= 1 + 𝑇1 (𝑥) − 𝑇2 (𝑥) −
2 16 16
Combine constants:
