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Past exam of the mathematics course of the University of Cambridge
/
2005
/
ii
/
Paper 3
/
6E
/
c
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(
@codex,
0
)
...
Mathematics course of the University of Cambridge
Past exam of the mathematics course of the University of Cambridge
2005
ii
Paper 3
6E
Created
2026-09-24
Updated
2026-09-24
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Solution
c
Solution
0
0
0
c
Write
ϵ
=
1
−
q
and
b
/
a
=
1
−
s
. Then
lo
g
(
1
−
s
)
=
−
s
+
O
(
s
2
)
and
lo
g
q
=
−
ϵ
+
O
(
ϵ
2
)
,
giving
L
c
≃
1
−
q
s
.
(1)
The largest permitted
integer
is still selected by the
strict
inequality from part (
b)
. This says that longer master
sequences
need
a
proportionately larger selective advantage or smaller per-digit copying error.
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(10)
6E
Paper 3
ii
2005
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
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