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Past exam of the mathematics course of the University of Cambridge
/
2026
/
iii
/
Paper 358
/
2
/
c
/
Solution
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Past exam of the mathematics course of the University of Cambridge
2026
iii
Paper 358
2
c
Created
2026-09-24
Updated
2026-09-24
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Let
H
0
=
−
d
2
/
d
x
2
+
x
2
. The Hermite expansion gives
⟨
H
0
f
,
f
⟩
=
∑
m
=
n
+
1
∞
(
2
m
+
1
)
∣
c
m
∣
2
=
∥
f
′
∥
2
+
∥
x
f
∥
2
.
(1)
Since
T
=
−
d
2
/
d
x
2
+
i
x
2
,
⟨
T
f
,
f
⟩
=
∥
f
′
∥
2
+
i
∥
x
f
∥
2
,
(2)
and therefore
⟨
T
f
,
f
⟩
=
(
i
−
1
)
∥
x
f
∥
2
+
m
=
n
+
1
∑
∞
(
2
m
+
1
)
∣
c
m
∣
2
.
(3)
Write
X
=
∥
x
f
∥
2
and
S
=
∑
m
=
n
+
1
∞
(
2
m
+
1
)
∣
c
m
∣
2
. Then
Re
⟨
T
f
,
f
⟩
=
S
−
X
,
Im
⟨
T
f
,
f
⟩
=
X
, and
Re
⟨
T
f
,
f
⟩
+
Im
⟨
T
f
,
f
⟩
=
S
≥
2
n
+
3.
(4)
If
∣
z
∣
≤
n
, then
∣
Re
z
+
Im
z
∣
≤
2
n
<
2
n
+
3
, so
∣
z
∣
≤
n
⟹
z
∈
/
W
(
Q
n
T
Q
n
∗
)
.
(5)
Every fixed compact
set
is eventually excluded from the tail
numerical ranges
. Part (
b
) therefore implies
W
e
(
T
)
=
∅
.
(6)
Solved by
gpt-5
.
6
-sol high.
Ancestors
(11)
c
2
Paper 358
iii
2026
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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