Civil Engineering 2 Mathematics Autumn 2011
Solutions 3
P∞
= x, odd about x = 0, therefore an = 0R ⇒ Ff = n=1 bn sin nx.
1.(i) f (x)
R
π
π
π
1
bn = π1 −π x sin nx dx = πn
{[−x cos nx]−π + −π cos nx dx} = − 2 cosn nπ =
2(−1)n+1
.
n
f (x) is continuous for −π < x < π so
Ff (x) = x =
∞
X
2(−1)n+1
sin nx, |x| < π; .
n
n=1
Though from the expression of Ff we have Ff (π) = Ff (−π) = 0 so Ff = f only
for |x| < π.
P∞
(ii) f (x) = x2 , even about x = 0 ⇒ bn = 0 ⇒ Ff = 12 a0 + n=1 an cos nx.
R
R
n
2
π
π
a0 = π1 −π x2 dx = 2π3 , an = π1 −π x2 cos nx dx = (by parts twice) = 4(−1)
n2 .
f (x) is continuous for −π < x < π and Ff (π) = Ff (−π) = π 2 , so series is valid
for |x| ≤ π
Ff (x) = x2 =
∞
X
π2
(−1)n
cos nx
+4
3
n2
n=1
for |x| ≤ π.
P∞
(iii) fR(x) = sinh x, odd about x = 0 ⇒ an = 0 ⇒ FRf (x) = n=1 bn sin nx.
π
π
π
bn = π1 −π sinh x sin nx dx = π1 {[cosh x sin nx]−π − n −π cosh x cos nxdx} =
R
2
π
π
n
− nπ [sinh x cos nx]−π − nπ −π sinh x sin nxdx ⇒ (1+n2 )bn = − 2n
π (−1) sinh π ⇒
2n
n+1
bn = (−1)
π(1+n2 ) sinh π. f (x) is continuous for −π < x < π and Ff (π) =
Ff (−π) = 0, so series is valid for |x| < π :
Ff (x) = sinh x =
∞
2 sinh π X
n
(−1)n+1
sin nx,
π
1 + n2
n=1
1
for |x| < π.
cos αx is even about x = 0 so bn = 0 and f (x) = Ff (x) = 12 a0 +
n=1 an cos nx. The expansion is valid for | x |≤ π by continuity and by
looking at Rthe periodic extension of cos(αx).
π
απ
a0 = π1 −π cos αx dx = 2 sin
απ .
R
Rπ
1 π
1
cos(n + α)x + cos(n − α)x dx
an = π −π cos αx cos nx dx = 2π
−π
h
iπ
sin(n+α)x
sin(n−α)x
sin(απ+nπ)
1
απ
= cosπnπ sin
= 2π
+ n−α
= π1
− sin(απ−nπ)
n+α
n+α
n−α
n+α −
−π
P
∞
2α sin απ
sin απ
sin απ
= (−1)n 2α π(α
+ n=1 (−1)n π(α
2 −n2 ) . So cos αx =
2 −n2 ) cos nx,
π
for |x| ≤ π. As α → m, all terms of the series → 0 except for am :
m
2α sin απ
am = limα→m (−1)
To evaluate limit above you can use L’Hopital’s
π(α2 −m2 )
P∞2.
rule, so limα→m
(−1)m 2α sin απ
π(α2 −m2 )
=
(−1)m 2
π
limα→m
sin απ+απ cos απ
2α
=
(−1)m 2 mπ(−1)m
π
2m
1. Thus, am = 1. This means that the Fourier Series converges to cos mx as
α → m.
3. The cosine series converges to f (x) for 0 ≤ x ≤ 2 (because (x − 1)2 is
continuous and also P
the periodic extension of its even extension is continuous),
∞
so (x − 1)2 = 12 a0 + n=1 an cos nπx , 0 ≤ x ≤ 2 .
R2
a0 = 0 (x − 1)2 dx = 2/3,
R2
16
an = 0 (x − 1)2 cos nπx
2 dx = (by parts twice) = n2 π 2 when n is even and 0
when n is odd.
P∞
⇒ (x − 1)2 = 13 + π42 n=1 cos(nπx) n12 , 0 ≤ x ≤ 2 where we used the fact
that even numbers
2n.
R 2 can be written as P
P∞
∞
Parseval ⇒ 0 [f (x)]2 dx = 12 a20 + n=1 a2n ⇒ 2/5 = 2/9 + n=1 16/(n4 π 4 )
⇒
∞
X
π4 2 2
1
π4
=
.
( − )=
4
16 5 9
90
n
n=1
4. f (x) even about x = 0Pand continuous. Also its periodic extension is
∞
continuous. ⇒ Ff (x) = 12 a0 n+ n=1 an cos nx.
o
Rπ
R0
Rπ
a0 = π1 −π f (x) dx = π1 −π (1 + πx ) dx + 0 (1 − πx ) dx = 1,
nR
o
Rπ
Rπ
0
an = π1 −π (1 + πx ) cos nx dx + 0 (1 − πx ) cos nx dx = π2 0 (1 − πx ) cos nx
= (by parts) = − n22π2 ((−1)n − 1),
so an = 0 (n even), 4/(nπ)2 (n odd).
P∞
1
4
1
Therefore Ff = f (x) = 2 + π2 k=0 (2k+1)
2 cos(2k + 1)x for x ∈ [−π, π].
P
∞
1
4
1
Putting x = 0 : f (0) = 1 = 2 + π2 k=0 (2k+1)2
⇒
∞
X
k=0
π2
1
=
.
2
8
(2k + 1)
2
=
sin απ
n−α
P∞
5. Sine series: Ff = n=1 bn sin nx.
Rπ
bn = π2 0 x(π−x) sin nx dx = by parts =
4
n
πn3 (1−(−1) )
=
Ff (π) = Ff (0) = 0 so
⇒ x(π − x) =
0
8
πn3
n even
n odd
∞
8 X sin (2k − 1)x
, 0 ≤ x ≤ π.
π
(2k − 1)3
k=1
The expansion is valid on [0, π] because the odd and periodic (of period 2π)
extension of f is everywhere continuous.
Rπ
2
Cosine series: a0 = π2 0 x(π − x) dx = π3 ,
R
−4/n2 n even
2 π
2
n
an = π 0 x(π−x) cos nx dx = (by parts) = − n2 (1+(−1) ) =
0
n odd
P∞
π2
4
Ff (x) = 6 − k=1 (2k)2 cos 2kx
⇒ x(π − x) =
∞
π 2 X cos 2kx
, 0 ≤ x ≤ π.
−
k2
6
k=1
The expansion is valid on [0, π] because the even and periodic (of period 2π)
extension of f is everywhere continuous.
P∞ 1
π2
(a) put x = 0 in FCS to get
k=1 k2 = 6 .
(b) put x = π/2 in FCS to get
(c) put x = π/2 in FSS to get
π2
4
π2
4
=
=
π2
6 −
8
π
P∞
P∞
k=1
(−1)k
k2
⇒
π2
12
sin(2k−1)π/2
(2k−1)3
⇒
k=1
=
P∞
k=1
P∞
(−1)k+1
.
k2
(−1)k+1
k=1 (2k−1)3
Applying Parseval’s formula to the FSS we have:
P∞
R
P∞
P∞
2 π 2
8 2
1
1
2
2
n=1 Bn = π
k=1 (2k−1)6 ⇒
k=1 (2k−1)6 =
π 0 x (π − x)R dx =
π
because π 5 /30 = 0 x2 (π − x)2 dx.
P∞ 1
P∞
P∞
P∞ 1 π 6
1
1
1
∴ n=1 n6 ≡ n=1 (2n)
6 +
n=1 (2n−1)6 = 26
n=1 n6 + 960 .
P∞ 1
6
1
π
6
⇒
n=1 n6 = (π /960)/(1 − 26 ) = 945 .
=
π3
32
π6
960
RL
P∞
x
6. Sine series is n=1 bn sin(nπx/L) where bn = L2 0 (1 + L
) sin nπx
dx.
L
2
(1 − 2(−1)n ). The function is continIntegrating by parts we find: bn = nπ
uous and Ff (0) = Ff (L) = 0, so
1+
∞
nπx X
x
2
=
(1 − 2(−1)n ) sin
L n=1 nπ
L
only for x ∈ (0, L).
3
.