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Appendix: All Solutions to the Problems
(1 − x)
2
= 1 − 2x + x
2
(1 − x)
4
= (1 − 2x + x
2
)
2
= 1 + 4x
2
+ x
4
− 4x − 4x
3
+ 2x
2
= 1 − 4x + 6x
2
− 4x
3
+ x
4
there is
y 1 = a 0 (1 − 2x + x
2
) + a 1 x(1 − 2x + x
2
)
The square is
y
2
1 = [a 0 (1 − 2x + x
2
) + a 1 x(1 − 2x + x
2
)]
2
= a
2
0 (1 − 4x + 6x
2
− 4x
3
+ x
4
)
+ a
2
1 x
2
(1 − 4x + 6x
2
− 4x
3
+ x
4
) + 2a 0 a 1 x(1 − 4x + 6x
2
− 4x
3
+ x
4
)
The second derivative of y 1 with respect to x is
d
2 y 1
dx 2 = y 1
= 2a 0 − 4a 1 + 6a 1 x = 2(a 0 − 2a 1 + 3a 1 x)
The square is
y
2
1 = 4(a
2
0 + 4a
2
1 + 9a
2
1 x
2
− 4a 0 a 1 + 6a 0 a 1 x − 12a
2
1 x)
+ 4[a
2
0 + a
2
1 (4 − 12x + 9x
2
) − a 0 a 1 (4 − 6x)]
Substituting the above expression into the original functional, we get
J [a 0 , a 1 ] =
1
0
{4x
3
[a
2
0 + (4 − 12x + 9x
2
)a
2
1 − (4 − 6x)a 0 a 1 ]
+ 100x(1 − 4x + 6x
2
− 4x
3
+ x
4
)a
2
0
+ 100x
3
(1 − 4x + 6x
2
− 4x
3
+ x
4
)a
2
1
+ 200x
2
(1 − 4x + 6x
2
− 4x
3
+ x
4
)a 0 a 1
− 20x(1 − 2x + x
2
)a 0 − 20x
2
(1 − 2x + x
2
)a 1 }dx
The integral result is
J [a 0 , a 1 ] =
13
3
a
2
0 +
53
70
a
2
1 +
284
105
a 0 a 1 −
5
3
a 0 −
2
3
a 1
Deriving a 0 and a 1 respectively in the above expression, we get
∂ J
∂a 0
=
26
3
a 0 +
284
105
a 1 −
5
3
= 0
∂ J
∂a 1
=
284
105
a 0 +
53
35
a 1 −
2
3
= 0