Problem 2.6: Determination of orthogonal functions
Consider the family of functions
ℑ
=
{
1
,
cos
(
ω
x
)
,
cos
(
2
ω
x
)
,
sin
(
ω
x
)
,
sin
(
2
ω
x
)
}
{\displaystyle \Im =\left\{{1,\cos(\omega x),\cos(2\omega x),\sin(\omega x),\sin(2\omega x)}\right\}}
(4.1)
on the interval [0,T], where T=
2
Π
/
ω
{\displaystyle 2\Pi /\omega }
A) Construct
Γ
(
ℑ
)
{\displaystyle \Gamma (\Im )}
and observe its properties
B) Find
det
[
Γ
(
ℑ
)
]
{\displaystyle \det[\Gamma (\Im )]}
C) Is
ℑ
{\displaystyle \Im }
an orthogonal basis
Construct
Γ
(
ℑ
)
{\displaystyle \Gamma (\Im )}
:
Γ
(
ℑ
)
=
(
<
b
1
,
b
1
>
…
<
b
1
,
b
n
>
⋮
⋱
⋮
<
b
n
,
b
1
>
⋯
<
b
n
,
b
n
>
)
{\displaystyle \Gamma (\Im )=\left({\begin{array}{*{20}{c}}{<{b_{1}},{b_{1}}>}&\ldots &{<{b_{1}},{b_{n}}>}\\\vdots &\ddots &\vdots \\{<{b_{n}},{b_{1}}>}&\cdots &{<{b_{n}},{b_{n}}>}\end{array}}\right)}
(4.2)
where
b
1
=
1
b
2
=
cos
(
ω
x
)
b
3
=
cos
(
2
ω
x
)
b
4
=
sin
(
ω
x
)
b
5
=
sin
(
2
ω
x
)
{\displaystyle {\begin{array}{l}{b_{1}}=1\\{b_{2}}=\cos(\omega x)\\{b_{3}}=\cos(2\omega x)\\{b_{4}}=\sin(\omega x)\\{b_{5}}=\sin(2\omega x)\end{array}}}
In order to construct the matrix we must first define
<
b
i
,
b
j
>
{\displaystyle <{b_{i}},{b_{j}}>}
<
b
i
,
b
j
>=
∫
x
0
x
b
i
(
x
)
b
j
(
x
)
d
x
{\displaystyle <{b_{i}},{b_{j}}>=\int _{x_{0}}^{x}{{b_{i}}(x){b_{j}}(x)dx}}
(4.3)
Because multiplication of continuous functions is communicative it can be shown from equation 4.3 that
<
b
i
,
b
j
>=<
b
j
,
b
i
>
{\displaystyle <{b_{i}},{b_{j}}>=<{b_{j}},{b_{i}}>}
(4.4)
And therefore
Γ
(
ℑ
)
{\displaystyle \Gamma (\Im )}
is a symmetric matrix
We must now evaluate the terms of the matrix
<
b
1
,
b
1
>=<
b
1
,
b
1
>=
∫
0
T
1
⋅
1
d
x
=
x
|
0
T
=
T
<
b
1
,
b
2
>=<
b
2
,
b
1
>=
∫
0
T
1
⋅
cos
(
ω
x
)
d
x
=
1
ω
sin
(
ω
x
)
|
0
T
=
0
<
b
1
,
b
3
>=<
b
3
,
b
1
>=
∫
0
T
1
⋅
cos
(
2
ω
x
)
d
x
=
1
2
ω
sin
(
2
ω
x
)
|
0
T
=
0
<
b
1
,
b
4
>=<
b
4
,
b
1
>=
∫
0
T
1
⋅
sin
(
ω
x
)
d
x
=
−
1
ω
cos
(
ω
x
)
|
0
T
=
0
<
b
1
,
b
5
>=<
b
5
,
b
1
>=
∫
0
T
1
⋅
sin
(
2
ω
x
)
d
x
=
−
1
2
ω
cos
(
2
ω
x
)
|
0
T
=
0
<
b
2
,
b
2
>=<
b
2
,
b
2
>=
∫
0
T
cos
(
ω
x
)
⋅
cos
(
ω
x
)
d
x
=
(
sin
(
2
ω
x
)
4
ω
+
x
2
)
|
0
T
=
T
2
<
b
2
,
b
3
>=<
b
3
,
b
2
>=
∫
0
T
cos
(
ω
x
)
⋅
cos
(
2
ω
x
)
d
x
=
(
3
sin
(
ω
x
)
+
sin
(
3
ω
x
)
6
ω
)
|
0
T
=
0
<
b
2
,
b
4
>=<
b
4
,
b
2
>=
∫
0
T
cos
(
ω
x
)
⋅
sin
(
ω
x
)
d
x
=
−
cos
2
(
ω
x
)
2
ω
|
0
T
=
0
<
b
2
,
b
5
>=<
b
5
,
b
2
>=
∫
0
T
cos
(
ω
x
)
⋅
sin
(
2
ω
x
)
d
x
=
−
2
cos
3
(
ω
x
)
3
ω
|
0
T
=
0
<
b
3
,
b
3
>=<
b
3
,
b
3
>=
∫
0
T
cos
(
2
ω
x
)
⋅
cos
(
2
ω
x
)
d
x
=
(
sin
(
4
ω
x
)
8
ω
+
x
2
)
|
0
T
=
T
2
<
b
3
,
b
4
>=<
b
4
,
b
3
>=
∫
0
T
cos
(
2
ω
x
)
⋅
sin
(
ω
x
)
d
x
=
cos
(
3
ω
x
)
−
3
cos
(
ω
x
)
6
ω
|
0
T
=
0
<
b
3
,
b
5
>=<
b
5
,
b
3
>=
∫
0
T
cos
(
2
ω
x
)
⋅
sin
(
2
ω
x
)
d
x
=
−
cos
(
4
ω
x
)
8
ω
|
0
T
=
0
<
b
4
,
b
4
>=<
b
4
,
b
4
>=
∫
0
T
sin
(
ω
x
)
⋅
sin
(
ω
x
)
d
x
=
(
−
sin
(
2
ω
x
)
4
ω
+
x
2
)
|
0
T
=
T
2
<
b
4
,
b
5
>=<
b
5
,
b
4
>=
∫
0
T
sin
(
ω
x
)
⋅
sin
(
2
ω
x
)
d
x
=
(
2
sin
3
(
ω
x
)
3
ω
)
|
0
T
=
0
<
b
5
,
b
5
>=<
b
5
,
b
5
>=
∫
0
T
sin
(
2
ω
x
)
⋅
sin
(
2
ω
x
)
d
x
=
(
−
sin
(
4
ω
x
)
8
ω
+
x
2
)
|
0
T
=
T
2
{\displaystyle {\begin{array}{l}<{b_{1}},{b_{1}}>=<{b_{1}},{b_{1}}>=\int _{0}^{T}{1\cdot 1dx=x}|_{0}^{T}=T\\<{b_{1}},{b_{2}}>=<{b_{2}},{b_{1}}>=\int _{0}^{T}{1\cdot \cos(\omega x)dx={\frac {1}{\omega }}\sin(\omega x)|_{0}^{T}=0}\\<{b_{1}},{b_{3}}>=<{b_{3}},{b_{1}}>=\int _{0}^{T}{1\cdot \cos(2\omega x)dx={\frac {1}{2\omega }}\sin(2\omega x)|_{0}^{T}=0}\\<{b_{1}},{b_{4}}>=<{b_{4}},{b_{1}}>=\int _{0}^{T}{1\cdot \sin(\omega x)dx=-{\frac {1}{\omega }}\cos(\omega x)|_{0}^{T}=0}\\<{b_{1}},{b_{5}}>=<{b_{5}},{b_{1}}>=\int _{0}^{T}{1\cdot \sin(2\omega x)dx=-{\frac {1}{2\omega }}\cos(2\omega x)|_{0}^{T}=0}\\<{b_{2}},{b_{2}}>=<{b_{2}},{b_{2}}>=\int _{0}^{T}{\cos(\omega x)\cdot \cos(\omega x)dx=\left({{\frac {\sin(2\omega x)}{4\omega }}+{\frac {x}{2}}}\right)|_{0}^{T}={\frac {T}{2}}}\\<{b_{2}},{b_{3}}>=<{b_{3}},{b_{2}}>=\int _{0}^{T}{\cos(\omega x)\cdot \cos(2\omega x)dx=\left({\frac {3\sin(\omega x)+\sin(3\omega x)}{6\omega }}\right)|_{0}^{T}=0}\\<{b_{2}},{b_{4}}>=<{b_{4}},{b_{2}}>=\int _{0}^{T}{\cos(\omega x)\cdot \sin(\omega x)dx=-{\frac {{{\cos }^{2}}(\omega x)}{2\omega }}|_{0}^{T}=0}\\<{b_{2}},{b_{5}}>=<{b_{5}},{b_{2}}>=\int _{0}^{T}{\cos(\omega x)\cdot \sin(2\omega x)dx=-{\frac {2{{\cos }^{3}}(\omega x)}{3\omega }}|_{0}^{T}=0}\\<{b_{3}},{b_{3}}>=<{b_{3}},{b_{3}}>=\int _{0}^{T}{\cos(2\omega x)\cdot \cos(2\omega x)dx=\left({{\frac {\sin(4\omega x)}{8\omega }}+{\frac {x}{2}}}\right)|_{0}^{T}={\frac {T}{2}}}\\<{b_{3}},{b_{4}}>=<{b_{4}},{b_{3}}>=\int _{0}^{T}{\cos(2\omega x)\cdot \sin(\omega x)dx={\frac {\cos(3\omega x)-3\cos(\omega x)}{6\omega }}|_{0}^{T}=0}\\<{b_{3}},{b_{5}}>=<{b_{5}},{b_{3}}>=\int _{0}^{T}{\cos(2\omega x)\cdot \sin(2\omega x)dx=-{\frac {\cos(4\omega x)}{8\omega }}|_{0}^{T}=0}\\<{b_{4}},{b_{4}}>=<{b_{4}},{b_{4}}>=\int _{0}^{T}{\sin(\omega x)\cdot \sin(\omega x)dx=\left({-{\frac {\sin(2\omega x)}{4\omega }}+{\frac {x}{2}}}\right)|_{0}^{T}={\frac {T}{2}}}\\<{b_{4}},{b_{5}}>=<{b_{5}},{b_{4}}>=\int _{0}^{T}{\sin(\omega x)\cdot \sin(2\omega x)dx=\left({\frac {2{{\sin }^{3}}(\omega x)}{3\omega }}\right)|_{0}^{T}=0}\\<{b_{5}},{b_{5}}>=<{b_{5}},{b_{5}}>=\int _{0}^{T}{\sin(2\omega x)\cdot \sin(2\omega x)dx=\left({-{\frac {\sin(4\omega x)}{8\omega }}+{\frac {x}{2}}}\right)|_{0}^{T}={\frac {T}{2}}}\\\end{array}}}
All values were checked with Wolframalpha
The Gram matrix then becomes
Γ
(
ℑ
)
=
(
T
0
0
0
0
0
T
2
0
0
0
0
0
T
2
0
0
0
0
0
T
2
0
0
0
0
0
T
2
)
{\displaystyle \Gamma (\Im )=\left({\begin{array}{*{20}{c}}T&0&0&0&0\\0&{\frac {T}{2}}&0&0&0\\0&0&{\frac {T}{2}}&0&0\\0&0&0&{\frac {T}{2}}&0\\0&0&0&0&{\frac {T}{2}}\end{array}}\right)}
(4.5)
As we can see the Gram matrix based constructed from this set of functions is a diagonal matrix
Finding
det
[
Γ
(
ℑ
)
]
{\displaystyle \det[\Gamma (\Im )]}
The determinant of a diagonal matrix is
det
(
A
)
=
∏
i
=
1
n
a
i
i
{\displaystyle {\begin{array}{l}\det(A)=\prod \limits _{i=1}^{n}{a_{ii}}\\\end{array}}}
(4.6)
Where
A
n
x
n
=
(
a
11
0
⋱
0
a
n
n
)
{\displaystyle {A_{nxn}}=\left({\begin{array}{*{20}{c}}{a_{11}}&{}&0\\{}&\ddots &{}\\0&{}&{a_{nn}}\end{array}}\right)}
Based on equation 4.6
det
[
Γ
(
ℑ
)
]
=
T
⋅
T
2
⋅
T
2
⋅
T
2
⋅
T
2
=
T
5
16
{\displaystyle \det[\Gamma (\Im )]=T\cdot {\frac {T}{2}}\cdot {\frac {T}{2}}\cdot {\frac {T}{2}}\cdot {\frac {T}{2}}={\frac {T^{5}}{16}}}
(4.7)
For the set to be an orthogonal basis the Gram matrix must be a diagonal matrix with a non-zero determinant. As we can see from equations 4.5 and 4.7 both of these criteria are satisfied. Thus the set of functions is an orthogonal basis.