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Today's Calculation Of Integral
2007 Today's Calculation Of Integral
212
Today's calculation of Integral 212
Today's calculation of Integral 212
Source: Kyoto University entrance exam 1978
June 16, 2007
calculus
integration
trigonometry
calculus computations
Problem Statement
For integers
k
(
0
≤
k
≤
5
)
k\ (0\leq k\leq 5)
k
(
0
≤
k
≤
5
)
, positive numbers
m
,
n
m,\ n
m
,
n
and real numbers
a
,
b
a,\ b
a
,
b
, let
f
(
k
)
=
∫
−
π
π
(
sin
k
x
−
a
sin
m
x
−
b
sin
n
x
)
2
d
x
f(k)=\int_{-\pi}^{\pi}(\sin kx-a\sin mx-b\sin nx)^{2}\ dx
f
(
k
)
=
∫
−
π
π
(
sin
k
x
−
a
sin
m
x
−
b
sin
n
x
)
2
d
x
,
p
(
k
)
=
5
!
k
!
(
5
−
k
)
!
(
1
2
)
5
,
E
=
∑
k
=
0
5
p
(
k
)
f
(
k
)
p(k)=\frac{5!}{k!(5-k)!}\left(\frac{1}{2}\right)^{5}, \ E=\sum_{k=0}^{5}p(k)f(k)
p
(
k
)
=
k
!
(
5
−
k
)!
5
!
(
2
1
)
5
,
E
=
∑
k
=
0
5
p
(
k
)
f
(
k
)
. Find the values of
m
,
n
,
a
,
b
m,\ n,\ a,\ b
m
,
n
,
a
,
b
for which
E
E
E
is minimized.
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