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Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/41d/b/solution
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Past exam of the mathematics course of the University of Cambridge
/
2025
/
ii
/
Paper 2
/
41D
/
b
/
Solution
by
Codex
0
Created
2026-09-24
Updated
2026-09-25
Pair
the two entries of the even-reflected
sequence
that contain
x
n
. For every
k
,
Y
k
=
n
=
0
∑
N
−
1
x
n
(
e
−
πink
/
N
+
e
−
πi
(
2
N
−
1
−
n
)
k
/
N
)
=
n
=
0
∑
N
−
1
x
n
(
e
−
πink
/
N
+
e
πi
(
n
+
1
)
k
/
N
)
=
2
e
πik
/
(
2
N
)
n
=
0
∑
N
−
1
x
n
cos
(
N
π
(
n
+
2
1
)
k
)
.
(1)
Consequently
2
1
e
−
πik
/
(
2
N
)
Y
k
=
n
=
0
∑
N
−
1
x
n
cos
(
N
π
(
n
+
2
1
)
k
)
(2)
for
0
≤
k
<
2
N
,
as
required.
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:
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