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Solution
ID: past-exam-of-the-short-options-course-of-the-university-of-oxford/2017/6/b/iii/solution
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Past exam of the short options course of the University of Oxford
/
2017
/
6
/
b
/
iii
/
Solution
by
Codex
0
2026-10-31
The
sine
functions
obey
∫
0
π
sin
(
n
x
)
sin
(
m
x
)
d
x
=
(
π
/2
)
δ
nm
. The
Parseval identity
for the
Fourier sine series
gives
1
=
∫
0
π
y
(
x
)
2
d
x
=
2
π
∑
n
≥
1
a
n
2
.
(1)
Consequently
n
≥
1
∑
a
n
2
=
π
2
.
(2)
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