Hints and answers

12.10 HINTS AND ANSWERS
0
(a)
1
0
1
0
1
0
(c)
(b)
2
4
(d)
Figure 12.6 Continuations of exp(−x2 ) in 0 ≤ x ≤ 1 to give: (a) cosine terms
only; (b) sine terms only; (c) period 1; (d) period 2.
Sketch the graph of the function f(x), where
−x(π + x) for −π ≤ x < 0,
f(x) =
x(x − π)
for 0 ≤ x < π.
If f(x) is to be approximated by the first three terms of a Fourier sine series, what
values should the coefficients have so as to minimise E3 ? What is the resulting
value of E3 ?
12.10 Hints and answers
12.1
12.3
12.5
12.7
12.9
12.11
12.13
12.15
12.17
12.19
Note that the only integral of a sinusoid around a complete cycle of length L
that is not zero is the integral of cos(2πnx/L) when n = 0.
Only (c). In terms of the Dirichlet conditions (section 12.1), the others fail as
follows: (a)
(ii); (d) (ii); (e) (iii).
(i); (b)
n+1 −1
f(x) = 2 ∞
n sin nx; set x = π/2.
1 (−1)
(i) Series (a) from exercise 12.6 does not converge and cannot represent the
function y(x) = −1. Series (b) reproduces the square-wave function of equation
(12.8).
(ii) Series (a) gives the series for y(x) = −x − 12 x2 − 12 in the range −1 ≤ x ≤ 0
and for y(x) = x − 12 x2 − 12 in the range 0 ≤ x ≤ 1. Series (b) gives the series for
y(x) = x + 12 x2 + 12 in the range −1 ≤ x ≤ 0 and for y(x) = x − 12 x2 + 12 in the
range 0 ≤ x ≤ 1.
1
2
n
2 2 −1
f(x) = (sinh 1) 1 + 2 ∞
1 (−1) (1 + n π ) [cos(nπx) − nπ sin(nπx)] .
The series will converge to the same value as it does at x = 0, i.e. f(0) = 1.
See figure 12.6. (c) (i) (1 + e−1 )/2, (ii) (1 + e−1 )/2; (d) (i) (1 + e−4 )/2, (ii) e−1 .
(d) (i) The periods are both 2L; (ii) y0 /2.
So = π 2 /8. If Se = (2m)−2 then Se = 14 (Se + So ), yielding So − Se = π 2 /12 and
Se + So = π 2 /6.
(a) (π/4)(π/2−|θ|); (b) (πθ/4)(π/2−|θ|/2) from integrating (a). (c) Even function;
average value L2 /3; y(0) = 0; y(L) = L2 ; probably y(x) = x2 . Compare with the
worked example in section
12.5. n
2 2
cosh x = (sinh 1)[1 + 2 ∞
twice
n=1 (−1) (cos nπx)/(n
π +n1)] and after 2integrating
1
2
this form must be recovered. Use x = 3 +4 (−1) (cos nπx)/(n π 2 )] to eliminate
the quadratic term arising from the constants of integration; there is no linear
term.
C±(2m+1) = ∓2i/[(2m + 1)π];
|Cn |2 = (4/π 2 ) × 2 × (π 2 /8); the values n = ±1,
±3 contribute > 90% of the total.
431
FOURIER SERIES
12.21
12.23
12.25
cn = [(−1)n sinh π]/[π(1 + n2 )]. Having set x = 0, separate out the n = 0 term
and note that (−1)n = (−1)−n .
(π 2 − 8)/16.
(b) All an and αn are zero; bn = 2(−1)n+1 /(nπ) and βn = 4/(nπ). You will need
the result quoted in exercise 12.19.
432