UNIVERSITY OF SOUTH AFRICA
Department of Mathematical Sciences
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
APM3711: Numerical Meth-
ods for Differential Equations
Assignment 02 — Semester 1, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
APM3711
Module Code:
Numerical Methods for Differential Equa-
Module Name:
tions
02
Assignment Number:
12 June 2026
Due Date:
50
Total Marks:
Submitted in partial fulfilment of the requirements for APM3711 — UNISA 2026
,UNISA | APM3711 Assignment 02
Question 1: Chebyshev Polynomial Approximation [10 marks]
1(a) Express the degree-2 truncated power series in terms of Chebyshev polynomi-
als T0 (x), T1 (x), T2 (x) [6
marks]
Question: 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 truncated power series up to degree 2 is:
x x2
P (x) = 1 + −
2 8
The standard Chebyshev polynomials are:
T0 (x) = 1, T1 (x) = x, T2 (x) = 2x2 − 1
Step 1: Express x2 in terms of T2 (x).
Starting from T2 (x) = 2x2 − 1, make x2 the subject:
2x2 = T2 (x) + 1
T2 (x) + 1
x2 =
2
Step 2: Substitute into P (x).
Note that x = T1 (x). Substituting:
1 1 T2 (x) + 1
P (x) = 1 + T1 (x) −
2 8 2
Page 2 of 17
,UNISA | APM3711 Assignment 02
Step 3: Expand the bracket.
1 T2 (x) 1
P (x) = 1 + T1 (x) − −
2 16 16
Step 4: Collect the constant terms.
1 1 1
P (x) = 1− + T1 (x) − T2 (x)
16 2 16
15 1 1
P (x) = + T1 (x) − T2 (x)
16 2 16
Step 5: Replace the constant using T0 (x) = 1.
15 1 1
P (x) = T0 (x) + T1 (x) − T2 (x)
16 2 16
1(b) Compare the maximum errors of the two approximations on [−1, 1] [4 marks]
Question: 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 P2 (x)
The truncated power series retains terms up to x2 :
x x2
P2 (x) = 1 + −
2 8
x3
The first neglected term is . On the interval [−1, 1], the maximum of |x3 | is 1, attained at
16
x = ±1. Therefore:
x3 1
EP = max = = 0.0625
x∈[−1,1] 16 16
Page 3 of 17
,UNISA | APM3711 Assignment 02
Error of the Chebyshev approximation
From part (a), the Chebyshev expansion is:
15 1 1
P (x) = T0 (x) + T1 (x) − T2 (x)
16 2 16
The Chebyshev economization truncates by discarding the highest-degree Chebyshev term.
1
The coefficient of T2 (x) is − .
16
Since |Tn (x)| ≤ 1 for all x ∈ [−1, 1], the maximum error introduced by discarding T2 (x) is
bounded by the magnitude of its coefficient:
1 1
EC = − = = 0.0625
16 16
Key Distinction
Comparison: Both approximations yield the same maximum error of 0.0625 on
[−1, 1]. This occurs because for a degree-2 truncation, the Chebyshev economization
produces the same bound as the direct power series truncation. The advantage of
Chebyshev economization becomes more pronounced when reducing from higher-degree
polynomials, where the minimax property of Chebyshev polynomials gives a uniformly
smaller error than naive truncation.
EP = 0.0625 EC = 0.0625
Both approximations have the same maximum error on [−1, 1].
Page 4 of 17
, UNISA | APM3711 Assignment 02
Question 2: Eigenvalues, Eigenvectors, and Diagonalizability [12 marks]
Consider the matrices:
2 1 3 0
A1 = , A2 =
1 2 0 1
2(a) Determine the eigenvalues by solving the characteristic equation [4 marks]
Question: Determine the eigenvalues of each matrix by solving the characteristic equation.
Solution:
Eigenvalues of A1
Form the characteristic matrix A1 − λI:
2−λ 1
A1 − λI =
1 2−λ
Set the determinant equal to zero:
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
λ1 = 3, λ2 = 1
Page 5 of 17
Department of Mathematical Sciences
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
APM3711: Numerical Meth-
ods for Differential Equations
Assignment 02 — Semester 1, 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
APM3711
Module Code:
Numerical Methods for Differential Equa-
Module Name:
tions
02
Assignment Number:
12 June 2026
Due Date:
50
Total Marks:
Submitted in partial fulfilment of the requirements for APM3711 — UNISA 2026
,UNISA | APM3711 Assignment 02
Question 1: Chebyshev Polynomial Approximation [10 marks]
1(a) Express the degree-2 truncated power series in terms of Chebyshev polynomi-
als T0 (x), T1 (x), T2 (x) [6
marks]
Question: 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 truncated power series up to degree 2 is:
x x2
P (x) = 1 + −
2 8
The standard Chebyshev polynomials are:
T0 (x) = 1, T1 (x) = x, T2 (x) = 2x2 − 1
Step 1: Express x2 in terms of T2 (x).
Starting from T2 (x) = 2x2 − 1, make x2 the subject:
2x2 = T2 (x) + 1
T2 (x) + 1
x2 =
2
Step 2: Substitute into P (x).
Note that x = T1 (x). Substituting:
1 1 T2 (x) + 1
P (x) = 1 + T1 (x) −
2 8 2
Page 2 of 17
,UNISA | APM3711 Assignment 02
Step 3: Expand the bracket.
1 T2 (x) 1
P (x) = 1 + T1 (x) − −
2 16 16
Step 4: Collect the constant terms.
1 1 1
P (x) = 1− + T1 (x) − T2 (x)
16 2 16
15 1 1
P (x) = + T1 (x) − T2 (x)
16 2 16
Step 5: Replace the constant using T0 (x) = 1.
15 1 1
P (x) = T0 (x) + T1 (x) − T2 (x)
16 2 16
1(b) Compare the maximum errors of the two approximations on [−1, 1] [4 marks]
Question: 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 P2 (x)
The truncated power series retains terms up to x2 :
x x2
P2 (x) = 1 + −
2 8
x3
The first neglected term is . On the interval [−1, 1], the maximum of |x3 | is 1, attained at
16
x = ±1. Therefore:
x3 1
EP = max = = 0.0625
x∈[−1,1] 16 16
Page 3 of 17
,UNISA | APM3711 Assignment 02
Error of the Chebyshev approximation
From part (a), the Chebyshev expansion is:
15 1 1
P (x) = T0 (x) + T1 (x) − T2 (x)
16 2 16
The Chebyshev economization truncates by discarding the highest-degree Chebyshev term.
1
The coefficient of T2 (x) is − .
16
Since |Tn (x)| ≤ 1 for all x ∈ [−1, 1], the maximum error introduced by discarding T2 (x) is
bounded by the magnitude of its coefficient:
1 1
EC = − = = 0.0625
16 16
Key Distinction
Comparison: Both approximations yield the same maximum error of 0.0625 on
[−1, 1]. This occurs because for a degree-2 truncation, the Chebyshev economization
produces the same bound as the direct power series truncation. The advantage of
Chebyshev economization becomes more pronounced when reducing from higher-degree
polynomials, where the minimax property of Chebyshev polynomials gives a uniformly
smaller error than naive truncation.
EP = 0.0625 EC = 0.0625
Both approximations have the same maximum error on [−1, 1].
Page 4 of 17
, UNISA | APM3711 Assignment 02
Question 2: Eigenvalues, Eigenvectors, and Diagonalizability [12 marks]
Consider the matrices:
2 1 3 0
A1 = , A2 =
1 2 0 1
2(a) Determine the eigenvalues by solving the characteristic equation [4 marks]
Question: Determine the eigenvalues of each matrix by solving the characteristic equation.
Solution:
Eigenvalues of A1
Form the characteristic matrix A1 − λI:
2−λ 1
A1 − λI =
1 2−λ
Set the determinant equal to zero:
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
λ1 = 3, λ2 = 1
Page 5 of 17
