User:Eml5526.s11.team6.tupsakhare/hwk2

=Problem 2.5=

Given
We have Eq. (2) from page 7-2 of mtg 7 ,

$$ \underline{b} _{i}  .\underline{P}(\underline{v})=0 $$

where  i =  1, 2 ,..., n

Find
Show that $$ \underline{w} .\underline{P}(\underline{v})=0 $$ ,

where $$  \underline{w}= \sum_{i=1}^{n}\alpha_{i} \underline{b} _{i}$$ ,

and $$ \forall (\alpha _{1},\alpha _{2},...,\alpha _{n})\in R^{n} $$

Solution
We have,

$$ \underline{b} _{i}  .\underline{P}(\underline{v})=0 $$

With i = 1 to n, we can then write -

$$ \underline{b}_{1}.\underline{P}(\underline{v})=0 $$

$$ \underline{b}_{2}.\underline{P}(\underline{v})=0 $$

$$ \underline{b}_{3}.\underline{P}(\underline{v})=0 $$

. . . . ..

. . . . ..

. . . . ..

. . . . ..

. . . . ..

$$ \underline{b} _{n}.\underline{P}(\underline{v})=0 $$

Now consider a collection of arbitrary numbers $$  (  \alpha _{1}, \alpha _{2} ,..., \alpha _{n}  ) $$ i.e.   \alpha _{i}      with i = 1 to n

Now successively multiplying above n equations, on both sides by corresponding  \alpha _{i}, we get the following new set of n equations -

With i = 1 to n, we can then write -

$$ \alpha _{1}. \underline{b}_{1} .\underline{P}(\underline{v})=0 $$

$$ \alpha _{2}. \underline{b}_{2}.\underline{P}(\underline{v})=0 $$

$$ \alpha _{3}. \underline{b}_{3}.\underline{P}(\underline{v})=0 $$

. . . . ..

. . . . ..

. . . . ..

. . . . ..

. . . . ..

$$ \alpha _{n}. \underline{b}_{n}.\underline{P}(\underline{v})=0 $$

Adding LHS and RHS respectively for above n equations, we get the following -

$$ \alpha _{1} \underline{b}_{1}.\underline{P}(\underline{v})+ $$ $$ \alpha _{2} \underline{b}_{2}.\underline{P}(\underline{v})+...+ $$ $$ \alpha _{n} \underline{b}_{n}.\underline{P}(\underline{v})=0 $$

which in index notation can be expressed as the following:

$$ \sum_{i=1}^{n}\alpha_{i} \underline{b}_{i} .\underline{P}(\underline{v})=0 $$

But as per previous declaration in 'Find' tab, we have stated -

$$ \underline{w} = \sum_{i=1}^{n}\alpha_{i} \underline{b}_{i}$$ ,

Hence we can now say that,

$$ \underline{w}.\underline{P}(\underline{v})=0 $$

This shows that required equation (2) of page 7- 2 mtg 7 can be used to get equation (2) of page 8-2 mtg 7

= Problem 2.6 Linearly Independent Basis Function.=

Find
1) Construct Γ5x5(F) and observe the proportionality of Γ.

2) Find det Γ(F)

3) Conclude whether F is orthogonal basis i.e. $$\Gamma _{ij}=\left \langle b_{i},b_{j} \right \rangle=\delta _{ij}$$

Given
$$F=\left \{ 1,cos(\omega x),cos(2\omega x),sin(\omega x),sin(2\omega x) \right \} , \Omega =[0,T], i=1,2$$

Solution
We have from reference of Mtg 10_Page 2

1) $$ \mathbf{\Gamma} _{5x5}=\int_{\Omega }{}b_{i}\cdot b_{j}\;dx=\int_{0}^{T}b_{i}\cdot b_{j}\;dx $$

Constructing a matrix of bi ˑ bj yields:

$$ \Rightarrow \begin{bmatrix} 1\cdot 1 & cos(\omega x)\cdot 1 & cos(2\omega x)\cdot 1 & sin(\omega x)\cdot 1 & sin(2\omega x)\cdot 1\\ 1\cdot cos(\omega x) & cos(\omega x)\cdot cos(\omega x) & cos(2\omega x)\cdot cos(\omega x) & sin(\omega x)\cdot cos(\omega x) & sin(2\omega x)\cdot cos(\omega x)\\ 1\cdot cos(2\omega x) & cos(\omega x)\cdot cos(2\omega x) & cos(2\omega x)\cdot cos(2\omega x) & sin(\omega x)\cdot cos(2\omega x) & sin(2\omega x)\cdot cos(2\omega x)\\ 1\cdot sin(\omega x) & cos(\omega x)\cdot sin(\omega x) & cos(2\omega x)\cdot sin(\omega x) & sin(\omega x)\cdot sin(\omega x) & sin(2\omega x)\cdot sin(\omega x)\\ 1\cdot sin(2\omega x) & cos(\omega x)\cdot sin(2\omega x) & cos(2\omega x)\cdot sin(2\omega x) & sin(\omega x)\cdot sin(2\omega x) & sin(2\omega x)\cdot sin(2\omega x) \end{bmatrix} $$

Multiplying out the above matrix we integrate the resulting terms within limits from 0 to T to get the $$ \mathbf{\Gamma} _{5x5} $$ :

$$\Rightarrow \begin{bmatrix} x & sin(\omega x) & sin(\omega x)cos(\omega x) & -cos(\omega x) & -\frac{1}{2}cos(2\omega x)\\ sin(\omega x) & \frac{1}{2}(x+sin(\omega x)cos(\omega x)) & \frac{1}{6}(3sin(\omega x)+sin(3\omega x)) & -\frac{1}{2}cos^2(\omega x) & -\frac{2}{3}cos^3(\omega x)\\ sin(\omega x)cos(\omega x) & \frac{1}{6}(3sin(\omega x)+sin(3\omega x)) & \frac{1}{8}(4x+sin(4\omega x)) & \frac{1}{6}(3cos(\omega x)-cos(3\omega x)) & -\frac{1}{8}cos(4\omega x))\\ -cos(\omega x)) & -\frac{1}{2}cos^2(\omega x) & \frac{1}{6}(3cos(\omega x)-cos(3\omega x)) & \frac{1}{2}(x-sin(\omega x)cos(\omega x)) & \frac{2}{3}sin^3(\omega x)\\ -\frac{1}{2}cos(2\omega x) & -\frac{2}{3}cos^3(\omega x) & -\frac{1}{8}cos(4\omega x)) & \frac{2}{3}sin^3(\omega x) & \frac{1}{8}(4x-sin(4\omega x)) \end{bmatrix} $$

By definition of Angular Frequency

$$2\pi =\omega T\rightarrow \omega=\frac{2\pi}{T}$$

Above is the resulting matrix to be integrated within limits of 0 to T.

And now integrating the results using a table of integrals yielded the following terms for the final integrated matrix.

$$\begin{array}{l} \int_0^T {1 \cdot 1dx = x} |_0^T = T \\ \int_0^T {1 \cdot \cos (\omega x)dx = \frac{1}{\omega }\sin (\omega x)|_0^T = 0} \\ \int_0^T {1 \cdot \cos (2\omega x)dx = \frac{1}\sin (2\omega x)|_0^T = 0} \\ \int_0^T {1 \cdot \sin (\omega x)dx = - \frac{1}{\omega }\cos (\omega x)|_0^T = 0} \\ \int_0^T {1 \cdot \sin (2\omega x)dx = - \frac{1}\cos (2\omega x)|_0^T = 0} \\ \int_0^T {\cos (\omega x) \cdot \cos (\omega x)dx = \left( {\frac + \frac{x}{2}} \right)|_0^T = \frac{T}{2}} \int_0^T {\cos (\omega x) \cdot \cos (2\omega x)dx = \left( {\frac} \right)|_0^T = 0} \\ \int_0^T {\cos (\omega x) \cdot \sin (\omega x)dx = - \frac|_0^T = 0} \\ \int_0^T {\cos (\omega x) \cdot \sin (2\omega x)dx = - \frac|_0^T = 0} \\ \int_0^T {\cos (2\omega x) \cdot \cos (2\omega x)dx = \left( {\frac + \frac{x}{2}} \right)|_0^T = \frac{T}{2}} \\ \int_0^T {\cos (2\omega x) \cdot \sin (\omega x)dx = \frac|_0^T = 0} \\ \int_0^T {\cos (2\omega x) \cdot \sin (2\omega x)dx = - \frac|_0^T = 0} \\ \int_0^T {\sin (\omega x) \cdot \sin (\omega x)dx = \left( { - \frac + \frac{x}{2}} \right)|_0^T = \frac{T}{2}} \\ \int_0^T {\sin (\omega x) \cdot \sin (2\omega x)dx = \left( {\frac} \right)|_0^T = 0} \\ \int_0^T {\sin (2\omega x) \cdot \sin (2\omega x)dx = \left( { - \frac + \frac{x}{2}} \right)|_0^T = \frac{T}{2}} \\ \end{array}$$
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Above integrals results were verified with Wolframalpha.

The $$ \mathbf{\Gamma} _{5x5} $$ matrix is thus :


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We note that this Gram matrix is a diagonal matrix.

2 ) Determinant of the above gram matrix $$ \mathbf{\Gamma} _{5x5} $$   :

Det $$ \mathbf{\Gamma} _{5x5} $$ = $$ \frac{T^5}{16} $$ < Shawn, please put the matlab figure here for calculation of determinant >

We note that the determinant is non- zero and hence we can say that the given functions form a family of Linearly Independent Basis Functions

3) We see that the $$ \mathbf{\Gamma} _{5x5} $$ derived is not equivalent to The Kronecker Delta.

Hence, we can conclude that $$ \mathbf{\Gamma} _{5x5} $$ is  NOT an ORTHOGONAL Basis.