UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology
⋄
ASSIGNMENT 02
Semester 1, 2026
⋄
Module Code: APM3711
Module Name: Numerical Methods for Differential Equa-
tions
Assignment No.: 02
Due Date: 12 June 2026
Semester: Semester 1, 2026
Submitted in partial fulfilment of the requirements for APM3711
at the University of South Africa.
,UNISA | APM3711 Assignment 02
Question 1: Chebyshev Polynomial Approximation
Question 1(a): Express the truncated power series in terms of Chebyshev polynomi-
als
The function
f (x) = (1 + x)1/2
has the power series expansion
x x2 x3
1+ − + + ···
2 8 16
Express this polynomial up to x2 in terms of Chebyshev polynomials T0 (x), T1 (x), T2 (x).
Solution:
The polynomial truncated at degree 2 is:
x x2
P (x) = 1 + −
2 8
The first three Chebyshev polynomials are:
T0 (x) = 1, T1 (x) = x, T2 (x) = 2x2 − 1
Making x2 the subject from T2 (x):
T2 (x) + 1
T2 (x) = 2x2 − 1 =⇒ 2x2 = T2 (x) + 1 =⇒ x2 =
2
Also noting that x = T1 (x) and 1 = T0 (x). Substituting into P (x):
1 1 T2 (x) + 1
P (x) = 1 + T1 (x) −
2 8 2
Expanding:
1 T2 (x) 1
P (x) = 1 + T1 (x) − −
2 16 16
Combining constant terms:
1 1 1 15 1 1
P (x) = 1 − + T1 (x) − T2 (x) = + T1 (x) − T2 (x)
16 2 16 16 2 16
Page 1 of 14
,UNISA | APM3711 Assignment 02
Since T0 (x) = 1, the final Chebyshev representation is:
15 1 1
P (x) = T0 (x) + T1 (x) − T2 (x)
16 2 16
Question 1(b): Compare the maximum errors on [−1, 1]
Compare the maximum error on the interval [−1, 1] of the truncated power series (degree 2)
and the corresponding Chebyshev approximation.
Solution:
Error of the truncated power series
x x2 x3
The truncated series P2 (x) = 1 + − neglects all terms from onwards. The leading
2 8 16
neglected term is:
x3
16
On [−1, 1] we have |x3 | ≤ 1, so the maximum error bound for the truncated series is:
x3 1
EP = max = = 0.0625
x∈[−1,1] 16 16
Error of the Chebyshev approximation
From part (a), the Chebyshev series is:
15 1 1
P (x) = T0 (x) + T1 (x) − T2 (x)
16 2 16
The Chebyshev approximation is obtained by discarding the highest-degree Chebyshev term.
1
Here the last retained term involves T2 (x) with coefficient − 16 . Since |Tn (x)| ≤ 1 for all
x ∈ [−1, 1], the maximum error of the Chebyshev approximation is bounded by the absolute
value of the coefficient of the highest discarded term. Comparing at the same degree 2, the
approximation error is governed by the T2 coefficient:
1 1
EC = − · max |T2 (x)| = × 1 = 0.0625
16 x∈[−1,1] 16
Page 2 of 14
,UNISA | APM3711 Assignment 02
Table 1: Maximum Error Comparison on [−1, 1]
Approximation Maximum Error
1
Truncated Power Series EP = 16 = 0.0625
1
Chebyshev Approximation EC = 16 = 0.0625
Implementation Insight
Both approximations yield the same maximum error bound of 0.0625 on [−1, 1]. How-
ever, the Chebyshev expansion distributes the error more evenly across the interval
(equioscillation property), making it preferable in practice for a given degree, even when
the bound appears identical.
EP = 0.0625 EC = 0.0625
Both approximations have the same maximum error bound.
Page 3 of 14
, UNISA | APM3711 Assignment 02
Question 2: Eigenvalues, Eigenvectors and Diagonalization
Consider the matrices:
2 1 3 0
A1 = , A2 =
1 2 0 1
Question 2(a): Eigenvalues via the characteristic equation
Determine the eigenvalues of each matrix by solving the characteristic equation.
Solution:
Matrix A1
The characteristic equation is det(A1 − λI) = 0:
2−λ 1
A1 − λI =
1 2−λ
det(A1 − λI) = (2 − λ)(2 − λ) − (1)(1) = 0
(2 − λ)2 − 1 = 0
4 − 4λ + λ2 − 1 = 0
λ2 − 4λ + 3 = 0
Factorising:
(λ − 3)(λ − 1) = 0
Eigenvalues of A1 :
λ1 = 3, λ2 = 1
Page 4 of 14
College of Science, Engineering and Technology
⋄
ASSIGNMENT 02
Semester 1, 2026
⋄
Module Code: APM3711
Module Name: Numerical Methods for Differential Equa-
tions
Assignment No.: 02
Due Date: 12 June 2026
Semester: Semester 1, 2026
Submitted in partial fulfilment of the requirements for APM3711
at the University of South Africa.
,UNISA | APM3711 Assignment 02
Question 1: Chebyshev Polynomial Approximation
Question 1(a): Express the truncated power series in terms of Chebyshev polynomi-
als
The function
f (x) = (1 + x)1/2
has the power series expansion
x x2 x3
1+ − + + ···
2 8 16
Express this polynomial up to x2 in terms of Chebyshev polynomials T0 (x), T1 (x), T2 (x).
Solution:
The polynomial truncated at degree 2 is:
x x2
P (x) = 1 + −
2 8
The first three Chebyshev polynomials are:
T0 (x) = 1, T1 (x) = x, T2 (x) = 2x2 − 1
Making x2 the subject from T2 (x):
T2 (x) + 1
T2 (x) = 2x2 − 1 =⇒ 2x2 = T2 (x) + 1 =⇒ x2 =
2
Also noting that x = T1 (x) and 1 = T0 (x). Substituting into P (x):
1 1 T2 (x) + 1
P (x) = 1 + T1 (x) −
2 8 2
Expanding:
1 T2 (x) 1
P (x) = 1 + T1 (x) − −
2 16 16
Combining constant terms:
1 1 1 15 1 1
P (x) = 1 − + T1 (x) − T2 (x) = + T1 (x) − T2 (x)
16 2 16 16 2 16
Page 1 of 14
,UNISA | APM3711 Assignment 02
Since T0 (x) = 1, the final Chebyshev representation is:
15 1 1
P (x) = T0 (x) + T1 (x) − T2 (x)
16 2 16
Question 1(b): Compare the maximum errors on [−1, 1]
Compare the maximum error on the interval [−1, 1] of the truncated power series (degree 2)
and the corresponding Chebyshev approximation.
Solution:
Error of the truncated power series
x x2 x3
The truncated series P2 (x) = 1 + − neglects all terms from onwards. The leading
2 8 16
neglected term is:
x3
16
On [−1, 1] we have |x3 | ≤ 1, so the maximum error bound for the truncated series is:
x3 1
EP = max = = 0.0625
x∈[−1,1] 16 16
Error of the Chebyshev approximation
From part (a), the Chebyshev series is:
15 1 1
P (x) = T0 (x) + T1 (x) − T2 (x)
16 2 16
The Chebyshev approximation is obtained by discarding the highest-degree Chebyshev term.
1
Here the last retained term involves T2 (x) with coefficient − 16 . Since |Tn (x)| ≤ 1 for all
x ∈ [−1, 1], the maximum error of the Chebyshev approximation is bounded by the absolute
value of the coefficient of the highest discarded term. Comparing at the same degree 2, the
approximation error is governed by the T2 coefficient:
1 1
EC = − · max |T2 (x)| = × 1 = 0.0625
16 x∈[−1,1] 16
Page 2 of 14
,UNISA | APM3711 Assignment 02
Table 1: Maximum Error Comparison on [−1, 1]
Approximation Maximum Error
1
Truncated Power Series EP = 16 = 0.0625
1
Chebyshev Approximation EC = 16 = 0.0625
Implementation Insight
Both approximations yield the same maximum error bound of 0.0625 on [−1, 1]. How-
ever, the Chebyshev expansion distributes the error more evenly across the interval
(equioscillation property), making it preferable in practice for a given degree, even when
the bound appears identical.
EP = 0.0625 EC = 0.0625
Both approximations have the same maximum error bound.
Page 3 of 14
, UNISA | APM3711 Assignment 02
Question 2: Eigenvalues, Eigenvectors and Diagonalization
Consider the matrices:
2 1 3 0
A1 = , A2 =
1 2 0 1
Question 2(a): Eigenvalues via the characteristic equation
Determine the eigenvalues of each matrix by solving the characteristic equation.
Solution:
Matrix A1
The characteristic equation is det(A1 − λI) = 0:
2−λ 1
A1 − λI =
1 2−λ
det(A1 − λI) = (2 − λ)(2 − λ) − (1)(1) = 0
(2 − λ)2 − 1 = 0
4 − 4λ + λ2 − 1 = 0
λ2 − 4λ + 3 = 0
Factorising:
(λ − 3)(λ − 1) = 0
Eigenvalues of A1 :
λ1 = 3, λ2 = 1
Page 4 of 14
