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and the solution is given by
˛
0:024;
ˇ
1:196;
0:338;
which gives the quadratic approximation
p.t/
D
0:024
C
1:196 t
0:338 t
2
:
(5.131)
Example 5.8.
Let
y.t/
D
e
t
e
t
(5.132)
be defined on 0
t
1.Then(
5.128
) takes the form
0
1
0
1
0
1
e
C
e
1
2
2e
1
5e
1
C
e
4
1 1=2 1=3
1=2 1=3 1=4
1=3 1=4 1=5
˛
ˇ
@
A
@
A
D
@
A
(5.133)
and the coefficients are given by
˛
0:019;
ˇ
1:782;
0:531:
The quadratic least squares approximation of (
5.132
) is therefore given by
p.t/
D
0:019
C
1:782 t
C
0:531 t
2
:
(5.134)
Example 5.9.
Let us consider
1
10
cos.t /
y.t/
D
t
2
C
(5.135)
for 0
t
1. The linear system (
5.128
) takes the form
0
1
0
1
0
1
1
3
1
10
1 1=2 1=3
1=2 1=3 1=4
1=3 1=4 1=5
˛
ˇ
C
sin.1/
@
A
@
A
D
@
A
3
20
1
10
1
10
C
cos.1/
C
sin.1/
1
5
1
5
1
10
C
cos.1/
sin.1/
and the solution is given by
˛
0:100;
ˇ
0:004;
0:957:
