Egm4313.s12.teamboss/R7

Report 7

Problem R7.1: Determining Orthogonality

Statement

Determine orthogonality of:

$\langle \phi_i , \phi_j \rangle = 0 \ \text{ for } \ i \ne j$ $\langle \phi_j , \phi_j \rangle = \frac{L}{2} \ \text{ for } \ i=j$

(1.0)

Where: $\displaystyle \phi_i (x) = \sin(\omega_i x)$, $\displaystyle \phi_j (x) = \sin(\omega_j x)$, $\displaystyle \omega_i = \frac{i \pi}{L}$, and $\displaystyle \omega_j = \frac{j \pi}{L}$.

Solution

The definition of a scalar product is:

$\langle \bar f , \bar g \rangle := \int_0^L \bar f(x) \ \bar g(x) \, dx$

(1.1)

So, substituting the problem statement values into Eq. (1.0):

$\langle \phi_i , \phi_j \rangle := \int_0^L \sin(\omega_i x) \ \sin(\omega_j x) \, dx$

(1.2)

and:

$\langle \phi_j , \phi_j \rangle := \int_0^L \sin(\omega_j x) \ \sin(\omega_j x) \, dx$

(1.3)

Solving Eq. 1.2 first:

$\langle \phi_i , \phi_j \rangle = \int_0^L \sin(\omega_i x) \ \sin(\omega_j x) \, dx$ $= \frac {1}{2} \int_0^L \cos((\omega_i - \omega_j)x)-\cos((\omega_i+\omega_j)x) \, dx$ $= \frac {1}{2(\omega_i - \omega_j)}[\sin((\omega_i - \omega_j)L)-\sin(0)]-\frac {1}{2(\omega_i + \omega_j)}[\sin((\omega_i + \omega_j)L)-\sin(0)]$ $= \frac {1}{2L(i \pi - j \pi)}\sin(i \pi - j \pi)-\frac {1}{2L(i \pi + j \pi)}\sin(i \pi + j \pi)$

Because $\displaystyle i$ and $\displaystyle j$ are both integers, the sin function will always be a periodic integer varying by $\displaystyle \pi$, meaning that $\sin(i \pi - j \pi)=0$ and $\sin(i \pi + j \pi)=0$. Thus proving:

$\langle \phi_i , \phi_j \rangle = 0 \ \text{ for } \ i \ne j$

(1.0.a)

Now solving for Eq. 1.3:

$\langle \phi_j , \phi_j \rangle = \int_0^L \sin(\omega_j x) \ \sin(\omega_j x) \, dx$ $= \frac {1}{2} \int_0^L \cos((\omega_j - \omega_j)x)-\cos((\omega_j+\omega_j)x) \, dx$ $= \frac {1}{2} \int_0^L \cos((j \pi - j \pi)\frac{x}{L})-\cos((j \pi+j \pi)\frac {x}{L}) \, dx$ $= \frac {1}{2} \int_0^L \cos(0)-\cos(\frac {2jx \pi}{L}) \, dx$ $= \frac {1}{2} \int_0^L 1-\cos(\frac {2jx \pi}{L}) \, dx$ $= \frac {L}{2}-\frac {L}{2j \pi}\sin(2j \pi)$

Again, because $\displaystyle j$ is an integer, the sin function will always be a periodic integer varying by $\displaystyle 2j \pi$, meaning that $\sin(2j \pi)=0$. Thus proving:

$\langle \phi_j , \phi_j \rangle = \frac{L}{2} \ \text{ for } \ i=j$

(1.0.b)

Author

Solved and Typed By ---Egm4313.s12.team1.durrance (talk) 16:49, 24 April 2012 (UTC)
Reviewed By - Egm4313.s12.team1.armanious (talk) 23:48, 24 April 2012 (UTC)

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Problem R7.2: Truncated Series

Statement

Plot the truncated-series (3) p19-12 with n=5, and for:

$\displaystyle t=\alpha p_1 = \alpha \frac{2\pi}{c \omega_1} = \alpha \frac {2L}{c}$

(2.0)

$\displaystyle \alpha = .5, 1, 1.5, 2$

(2.1)

Solution

Let it be stated first that (3) p.19-12 is equal to:

$\displaystyle u(x,t) = \sum{j=1}{n} a_j cos(c \omega _j t) sin(\omega _j x)$

(2.2)

And in this series representation, c=3, L=2, and $a_j$. is equaled to this:

$\displaystyle a_j=\left\{\begin{matrix} 0 for j=2m (even)\\ -4/( \pi^3j^3)for j=2m+1 (odd) \end{matrix}\right.$

(2.3)

With this in mind, we are prepared to create these truncated-series.
For $\alpha =.5$

$\displaystyle u(x,t) = [a_1 cos(c \omega _1 \alpha \frac{2L}{c})sin(\omega _1 x)]$

(2.4)

$\displaystyle +[a_2 cos(c \omega _2 \alpha \frac{2L}{c})sin(\omega _2 x)]$

$\displaystyle +[a_3 cos(c \omega _3 \alpha \frac{2L}{c})sin(\omega _3 x)]$

$\displaystyle + [a_4 cos(c \omega _4 \alpha \frac{2L}{c})sin(\omega _4 x)]$

$\displaystyle +[a_5 cos(c \omega _5 \alpha \frac{2L}{c})sin(\omega _5 x)]$

These simplify below as shown, after plugging in 3 for c, 10, for L, .5 for alpha, and the value of $a_j$ as determined from (2.2).

$\displaystyle u(x,t) = [\frac{-4}{\pi ^3} cos(\omega _1 (.5) (2*10))sin(\omega _1 x)]$

(2.5)

$\displaystyle +[\frac{-4}{\pi ^3 * 3^3}cos(\omega _3 (.5)(2*10))sin(\omega _3 x)]$

$\displaystyle +[\frac{-4}{\pi ^3 * 5^3}cos(\omega _5 (.5)(2*10))sin(\omega _5 x)]$

Making $\omega$ equal to $2 \pi$, the graph appears as shown below.

Alpha = .5

500px

For $\alpha =1$

These simplify below as shown, after plugging in 3 for c, 10, for L, 1 for alpha, and the value of $a_j$ as determined from (2.2).

$\displaystyle u(x,t) = [\frac{-4}{\pi ^3} cos(\omega _1 (1) (2*10))sin(\omega _1 x)]$

(2.6)

$\displaystyle +[\frac{-4}{\pi ^3 * 3^3}cos(\omega _3 (1)(2*10))sin(\omega _3 x)]$

$\displaystyle +[\frac{-4}{\pi ^3 * 5^3}cos(\omega _5 (1)(2*10))sin(\omega _5 x)]$

Making $\omega$ equal to $2 \pi$, the graph appears as shown below.

Alpha = 1

500px

For $\alpha =1.5$

These simplify below as shown, after plugging in 3 for c, 10, for L, 1.5 for alpha, and the value of $a_j$ as determined from (2.2).

$\displaystyle u(x,t) = [\frac{-4}{\pi ^3} cos(\omega _1 (1.5) (2*10))sin(\omega _1 x)]$

(2.7)

$\displaystyle +[\frac{-4}{\pi ^3 * 3^3}cos(\omega _3 (1.5)(2*10))sin(\omega _3 x)]$

$\displaystyle +[\frac{-4}{\pi ^3 * 5^3}cos(\omega _5 (1.5)(2*10))sin(\omega _5 x)]$

Making $\omega$ equal to $2 \pi$, the graph appears as shown below.

Alpha = 1.5

500px

For $\alpha =2$

These simplify below as shown, after plugging in 3 for c, 10, for L, 2 for alpha, and the value of $a_j$ as determined from (2.2).

$\displaystyle u(x,t) = [\frac{-4}{\pi ^3} cos(\omega _1 (2) (2*10))sin(\omega _1 x)]$

(2.8)

$\displaystyle +[\frac{-4}{\pi ^3 * 3^3}cos(\omega _3 (2)(2*10))sin(\omega _3 x)]$

$\displaystyle +[\frac{-4}{\pi ^3 * 5^3}cos(\omega _5 (2)(2*10))sin(\omega _5 x)]$

Making $\omega$ equal to $2 \pi$, the graph appears as shown below.

Alpha = 2

500px

Author

Solved and Typed By - Egm4313.s12.team1.silvestri (talk) 19:01, 23 April 2012 (UTC)
Reviewed By - Egm4313.s12.team1.armanious (talk) 19:40, 25 April 2012 (UTC)

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Problem R7.3 Scalar Product, Magnitude, and Angles of Functions

Statement

Find the scalar product <f,g>, the magnitude of f and g, and the angle between f and g for (see R7.3 Lect. 11-1 pg. 8):

1) $f(x)=\cos x, g(x)=x, \text{for} -2\leq x \leq 10\!$
2) $f(x)=\frac12(3x^2-1), g(x)=\frac12(5x^3-3x), \text{for} -1\leq x \leq +1\!$

Solution

The scalar product, the function equivalent of the vector dot product, of two functions can be found in the following way:

$\displaystyle <f,g>:=\int^b_af(x)g(x)dx$

(3.0)

The magnitude of a function is defined as the square root of the scalar product between the function and itself:

$\displaystyle ||f||:=<f,f>^{1/2}=\left [\int^b_af^2(x)dx \right ]^{1/2}$

(3.1)

The cosine of the angle between two functions can be found in the following way:

$\displaystyle \cos \theta = \frac{<f,g>}{||f||||g||}$

(3.2)

Note that these equations are very similar to their vector counterparts.
Part 1

$\displaystyle f(x)=\cos x, g(x)=x, \text{for} -2\leq x \leq 10$

(3.3)

The scalar product of f and g can be found using (3.0):

$\displaystyle <f,g>=\int_{-2}^{10}x\cos (x) dx=\left [ x\sin(x)+cos(x) \right ]_{-2}^{10}$

(3.4)

Solving this yields the scalar product:

$\displaystyle (10\sin10+\cos10)-(-2\sin(-2)+\cos(-2))=-7.68$

(3.5)

To find the magnitude of f, the scalar product of f with itself must first be found:

$\displaystyle <f,f>=\int_{-2}^{10}\cos^{2}(x)dx=\int_{-2}^{10}\left (\frac12\cos(2x)+\frac12 \right )dx=\left [ \frac14\sin(2x)+\frac12x \right ]_{-2}^{10}$

(3.6)

Solving this yields:

$\displaystyle \left [ \frac14\sin(2x)+\frac12x \right ]_{-2}^{10}=\left ( \frac14\sin(20)+5 \right )-\left ( \frac14\sin(-4)-1 \right )=6.04$

(3.7)

Therefore the magnitude of f is found to be:

$\displaystyle ||f||=<f,f>^{1/2}=\sqrt{6.04}=2.46$

(3.8)

A similar approach can be taken for the magnitude of g:

$\displaystyle <g,g>=\int_{-2}^{10}x^2dx=\left [ \frac{x^3}{3} \right ]_{-2}^{10}$

(3.9)

$\displaystyle \left [ \frac{x^3}{3} \right ]_{-2}^{10}=\frac{1000}3-\frac{-8}{3}=\frac{1008}{3}=336$

(3.10)

$\displaystyle ||g||=<g,g>^{1/2}=\sqrt{336}=18.33$

(3.11)

Using the above information with (3.2), the cosine of the angle between f and g is found to be

$\displaystyle \cos \theta = \frac{<f,g>}{||f||||g||}=\frac{-7.68}{(2.46)(18.33)}=-0.1705$

(3.12)

Therefore, the angle between the functions f and g is

$\displaystyle \theta =\cos^{-1}(-0.1705)=1.74rad=99.8^{\circ}$

(3.13)

Part 2

$\displaystyle f(x)=\frac12(3x^2-1), g(x)=\frac12(5x^3-3x), \text{for} -1\leq x \leq +1$

(3.14)

The scalar product of f and g can be found using (3.0):

$\displaystyle <f,g>=\int_{-1}^{1}\frac14\left ( 3x^2-1 \right )\left ( 5x^3-3x \right ) dx=\frac14\int_{-1}^{1}\left ( 15x^5-14x^3+3x \right ) dx=\frac14\left [ \frac{15}{6}x^6-\frac{14}{4}x^4+\frac32x^2 \right ]_{-1}^{1}$

(3.15)

Solving this yields the scalar product:

$\displaystyle \frac14\left [ \frac52x^6-\frac72x^4+\frac32x^2 \right ]_{-1}^{1}=\frac14\left ( \frac12-\frac12 \right )=0$

(3.16)

To find the magnitude of f, the scalar product of f with itself must first be found:

$\displaystyle <f,f>=\int_{-1}^{1}\left (\frac12\left (3x^2-1 \right )\right)^2 dx=\frac14\int_{-1}^{1}\left (9x^4-6x^2+1 \right )dx=\frac14\left [ \frac95x^5-2x^3+x \right ]_{-1}^{1}$

(3.17)

Solving this yields:

$\displaystyle \frac14\left [ \left( \frac95-2+1\right )-\left (-\frac95+2-1 \right ) \right ] =\frac14\left ( \frac45+\frac45 \right )=\frac25$

(3.18)

Therefore the magnitude of f is found to be:

$\displaystyle ||f||=<f,f>^{1/2}=\sqrt{0.4}=0.632$

(3.19)

A similar approach can be taken for the magnitude of g:

$\displaystyle <g,g>=\int_{-1}^{1}\left (\frac12\left (5x^3-3x \right )\right)^2 dx=\frac14\int_{-1}^{1}\left (25x^6-15x^4+9x^2 \right )dx=\frac14\left [ \frac{25}{7}x^7-3x^5+3x^3 \right ]_{-1}^{1}$

(3.20)

$\displaystyle \frac12 \left (\frac{25}{7}-3+3 \right ) =\frac{25}{14}$

(3.21)

$\displaystyle ||g||=<g,g>^{1/2}=\sqrt{\frac{25}{14}}=1.336$

(3.22)

Using the above information with (3.2), the cosine of the angle between f and g is found to be

$\displaystyle \cos \theta = \frac{<f,g>}{||f||||g||}=\frac{0}{(0.632)(1.336)}=0$

(3.23)

Because the scalar product of the two functions is zero, the two functions are orthogonal. That is:

$\displaystyle \theta =\frac{\pi}{2}rad=90^{\circ}$

(3.24)

Author

Solved and Typed By - Egm4313.s12.team1.armanious (talk) 23:48, 21 April 2012 (UTC)
Reviewed By - --Egm4313.s12.team1.stewart (talk) 02:22, 25 April 2012 (UTC)

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Problem R7.4 The Fourier Series and Plotted Functions

Statement

From Lecture 11 Pg. 8 Do K 2011 p.482 pb.6,9,12,13. Problems 6 and 12 ask to graph the function and the Fourier series of the function below (4.1).

$\displaystyle f(x)=\left | x \right |$

(4.1)

Problems 9 and 13 ask to graph the function and Fourier series of the function below (4.2).

$\displaystyle f(x)=\begin{Bmatrix} x\;\; if \; -\pi <x<0\\ \pi -x\;\; if \; 0<x<\pi \end{Bmatrix}$

(4.2)

Solution

For Question number 6, plotted using Wolfram Alpha.

The Function 4.1

500px

For Question number 9, plotted using Matlab.

The Function 4.2

500px

Question 12 asks for the Fourier series of equation 4.1, $f(x)=\left | x \right |\!$ which is an even function because $f(-x)=+f(x)\!$, and because it is even $b_{n}=0\!$. So the Fourier looks like this:

$\displaystyle f(x)=a_0+\sum_{n=1}^{\infty}a_ncos\frac{n\pi x}{L}$

(4.3)

In which $L= \pi \!$ and:

$\displaystyle a_0=\frac{1}{2\pi}\int_{-\pi}^{0}-xdx+\frac{1}{2\pi}\int_{0}^{\pi}xdx$

(4.4)

$\displaystyle a_{n}=\frac{1}{\pi }\int_{-\pi}^{\pi}f(x)cos(nx)dx$

(4.5)

$\displaystyle b_{n}=\frac{1}{\pi }\int_{-\pi}^{\pi}f(x)sin(nx)dx=0$

(4.6)

These simplify to:

$\displaystyle a_0=\frac{1}{2\pi}\frac{\pi^{2}}{2}+\frac{1}{2\pi}\frac{\pi^{2}}{2}=\frac{\pi}{2}$

(4.7)

$\displaystyle a_n=\frac{1}{\pi}\int_{-\pi}^{0}-xcos(nx)dx+\frac{1}{\pi}\int_{-\pi}^{0}xcos(nx)dx$

(4.8)

$\displaystyle a_n=\frac{-4}{n^{2}\pi}$

(4.9)

$\displaystyle f(x)=\frac{\pi}{2}+\sum_{n-1}^{\infty }\frac{4}{n^{2}\pi}cos\frac{n\pi x}{L}$

(4.10)

The final answer:

$\displaystyle f(x)=\frac{\pi}{2}-\frac{4}{\pi}(cosx+\frac{cos3x}{9}+\frac{cos5x}{25}+...)$

(4.11)

Plotted with Wolfram Aplpha

Fourier Series of Function 4.1

500px

Now the Fourier series of equation 4.2, we need to divide the function into two parts. First let's evaluate $f(x)=x\; if\; - \pi <x<0 \!$, which is an odd function because $f(-x)=-f(x)\!$. For an odd function $a_{0}=a_{n}=0\!$ so:

$\displaystyle f(x)=\sum\limits_{n=1}^{\infty }{\left[ {{b}_{n}}\sin (\frac{n \pi x}{L}) \right]}=\sum\limits_{n=1}^{\infty }{\left[ {{b}_{n}}\sin (nx) \right]}$

(4.12)

$\displaystyle b_{n}=\frac{1}{\pi}\int_{-\pi}^{0}xsin(nx)=\frac{1}{\pi}[\frac{sin(nx)-nxcos(nx)}{n^{2}}]=\frac{-cos(-\pi n)}{n}$

(4.13)

Which simplifies to:

$\displaystyle b_{n}=[1,\frac{-1}{2},\frac{1}{3},\frac{-1}{4},\frac{1}{5},\frac{-1}{6}...]$

(4.14)

The Fourier Series for just the top part of equation 4.2

$\displaystyle f(x)=sinx-\frac{sin2x}{2}+\frac{sin3x}{3}-\frac{sin4x}{4}+\frac{sin5x}{5}...$

(4.15)

Now the bottom part of equation 4.2 is neither even nor odd so we must add all the components and at the end combine the solution with the Fourier series for the top part of equation 4.2

$\displaystyle f(x)=a_{0}+\sum_{n=1}^{\infty }a_ncos(\frac{n \pi x}{L})+\sum_{n=1}^{\infty }b_nsin(\frac{n \pi x}{L})=a_{0}+\sum_{n=1}^{\infty }a_{n}cos(nx)+\sum_{n=1}^{\infty }b_{n}sin(nx)$

(4.16)

$\displaystyle a_{0}=\frac{1}{2 \pi}\int_{0}^{\pi}(\pi-x)dx=\frac{1}{2 \pi }[\pi x-\frac{x^2}{2}]=\frac{\pi}{4}$

(4.17)

$\displaystyle a_{n}=\frac{1}{ \pi}\int_{0}^{\pi}(\pi-x)cos(nx)dx=\frac{1}{ \pi }[\frac{n(\pi-x)sin(nx)-cos(nx)}{n^2}]=\frac{1-cos(\pi n)}{\pi n^2}$

(4.18)

$\displaystyle a_{n}=[\frac{2}{\pi},0,\frac{2}{6\pi},0,\frac{2}{25\pi},0,\frac{2}{49 \pi}...]$

(4.19)

$\displaystyle b_{n}=\frac{1}{\pi}\int_{0}^{\pi}(\pi-x)sin(nx)dx=\frac{1}{\pi}[\frac{sin(nx)+n(\pi-x)cos(nx)}{n^2}]=\frac{1}{n}$

(4.20)

$\displaystyle b_{n}=[1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6}...]$

(4.21)

The Fourier Series of $f(x)= \pi -x\; if\; 0<x< \pi \!$.

$\displaystyle f(x)=a_{0}+\sum_{n=1}^{\infty }a_{n}cos(nx)+\sum_{n=1}^{\infty }b_{n}sin(nx)=\frac{\pi}{4}+[\frac{2cosx}{\pi}+\frac{2cos3x}{6\pi}+\frac{2cos5x}{25\pi}...]+[1+\frac{sinx}{2}+\frac{sin2x}{3}+\frac{sin3x}{4}+\frac{sin4x}{5}+\frac{sin5x}{6}...]$

(4.22)

Combining with the Fourier Series of $f(x)= x\; if\; -\pi <x<0 \!$:

$\displaystyle f(x)=\frac{\pi}{4}+[\frac{2cosx}{\pi}+\frac{2cos3x}{6\pi}+\frac{2cos5x}{25\pi}...]+[sinx+\frac{sin2x}{2}+\frac{sin3x}{3}+\frac{sin4x}{4}+\frac{sin5x}{5}+...]+[sinx-\frac{sin2x}{2}+\frac{sin3x}{3}-\frac{sin4x}{4}+\frac{sin5x}{5}...]$

(4.23)

The total series and final answer:

$\displaystyle f(x)=\frac{\pi}{4}+[\frac{2cosx}{\pi}+\frac{2cos3x}{6\pi}+\frac{2cos5x}{25\pi}...]+[2sinx+\frac{2sin3x}{3}+\frac{2sin5x}{5}...]$

(4.24)

Plotted with Wolfram Alpha

Fourier Series of Function 4.2

500px

Author

Solved and Typed By - --Egm4313.s12.team1.stewart (talk) 02:22, 25 April 2012 (UTC)
Reviewed By - Egm4313.s12.team1.essenwein (talk) 16:50, 25 April 2012 (UTC)

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Problem R7.5 Approximation by Trigonometric Series

Statement

Consider equation (5.0) below:

$\displaystyle <\phi_{2j-1},\phi_{2k-1}> = \int_{0}^{P}\sin{j\omega x}\cdot \sin{k\omega x}dx = 0$

(5.0)

1) Find the exact integration of equation (5.0) with $P = 2\pi\!$, $j = 2\!$, $k = 3\!$.
2) Confirm the result with Matlab's trapz command for the trapezoidal rule.

Solution

1) Substituting the given values, we get equation (5.1) below:

$\displaystyle <\phi_{3},\phi_{5}> = \int_{0}^{2\pi}\sin{2\omega x}\cdot \sin{3\omega x}dx = 0$

(5.1)

Using the product and sum trigonometry identities, we can evaluate the definite integral:

$\displaystyle \int_{0}^{2\pi}\sin{2\omega x}\cdot \sin{3\omega x}dx = \frac{1}{2}\int_{0}^{2\pi}(\cos{\omega x}-\cos{5\omega x})dx = 0$

(5.2)

$\displaystyle \rightarrow [\frac{1}{2\omega}{\sin{\omega x} - \frac{1}{5}\sin{5\omega x}}]_0^{2\pi} = 0$

(5.3)

Also, $\omega = \frac{2\pi}{P} = 1\!$, and substituting this into equation (5.3) allows the definite integral to be solved:

$\displaystyle [\frac{1}{2}{\sin{x} - \frac{1}{5}\sin{5x}}]_0^{2\pi} = 0$

(5.4)

2) The following is the code for evaluating the definite integral in equation (5.1) in Matlab:

EDU>> X = 0:pi/100:2*pi;
EDU>> Y = sin(2*X).*sin(3*X);
EDU>> Z = trapz(X,Y)

Z =

 -2.2633e-17

EDU>> plot(X,Y)

The returned value of the evaluated integral in Matlab is essentially equal to zero, and it can be seen in the graph that the areas under the curve cancel each other out.

Author

Solved and Typed By - Egm4313.s12.team1.wyattling
Reviewed By - Egm4313.s12.team1.rosenberg

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Contributing Members

Team Contribution Table
Problem Number	Lecture	Assigned To	Solved By	Typed By	Proofread By
7.1	R7.1 Lect. 19c-1 pg. 10	Jesse Durrance	Jesse Durrance	Jesse Durrance	George Armanious
7.2	R7.2 Lect. 19c-1 pg. 13	Emotion Silvestri	Emotion Silvestri	Emotion Silvestri	George Armanious
7.3	R7.3 Lect. 11-1 pg. 8	George Armanious	George Armanious	George Armanious	Chris Stewart
7.4	R7.4 Lect. 11-1 pg.8	Chris Stewart	Chris Stewart	Chris Stewart	Eric Essenwein
7.5	R7.5 Lect. 12-1 pg. 8	Wyatt Ling	Wyatt Ling	Wyatt Ling	Steven Rosenberg