Problem 2.6: Determination of orthogonal functions
Consider the family of functions
(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 )} :
(4.2)
where
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.
![{\displaystyle \det[\Gamma (\Im )]}](../../6f7ee128f1304dd1bf90947ff713555d36f501ae.svg)
![{\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}}}](../../7bc5a12b055635bdcf0679413f124e55a5af25ad.svg)