APM3711 Assignment 2 Solutions Due 12 June 2026 |Numerical Methods II|

UNIVERSITY OF SOUTH AFRICA
Department of Mathematical Sciences


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APM3711: Numerical Meth-
ods for Differential Equations
Assignment 02 — Semester 1, 2026

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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



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,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




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,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].




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, 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




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