University of Florida/Egm4313/s12.team11.R7

Intermediate Engineering Analysis Section 7566 Team 11 Due date: April 25, 2012.

R7.1
Solved by Andrea Vargas

Problem Statement
Verify (4)-(5) pg 19-9: (4) $$\langle \phi_i, \phi_j \rangle = 0\!$$ for $$i \ne j\!$$ (5) $$\langle \phi_j, \phi_j \rangle =\frac{L}{2}\!$$ for $$i = j\!$$

Solution
From the lecture notes on p.19-9 we know that: (2) $$\phi_i=\sin(\omega_i x)\!$$ (3) $$\langle \phi_i, \phi_j \rangle =\int_0^L \phi_i(x) \phi_j(x) dx\!$$ Then, we have $$\langle \phi_i, \phi_j \rangle =\int_0^L \phi_i(x) \phi_j(x) dx=\int_0^L sin(\omega_i x) \sin(\omega_j x)dx\!$$

Since the period of $$\sin(x)\!$$ is $$ 2\pi\!$$ and we know that $$p=2L\!$$, we can assume that $$L=\pi\!$$ for simplicity.

We can compute the result of the previous integral using Wolfram ALpha (Mathematica software). The following is the input given to the software: int from 0 to pi sin(ax)sin(bx)

where a is $$\omega_i\!$$ and b is $$\omega_j\!$$. The software generates the following answer:

$$\int_0^L \sin(ax)sin(bx)= \frac{b \sin(\pi a)\cos(\pi b)-a \cos(\pi a)\sin(\pi b)}{a^2-b^2}\!$$ Substituting for $$ \omega_i \!$$ and $$ \omega_j\!$$: $$\int_0^L \sin(ax)sin(bx)= \frac{\omega_j \sin(\pi \omega_i)\cos(\pi \omega_j)-\omega_i \cos(\pi \omega_i)\sin(\pi \omega_j)}{\omega_i^2-\omega_j^2}\!$$ We also know that $$\sin(c\pi)=0\!$$ where c is any integer. We can cancel any terms with $$\sin(\pi \omega_i)\!$$ or $$\sin(\pi \omega_j)\!$$ as they are equal to zero. Then, we can verify: (4) $$\langle \phi_i, \phi_j \rangle = 0\!$$ for $$i \ne j\!$$

Similarly we can verify (5). We have $$\langle \phi_j, \phi_j \rangle =\int_0^L \phi_j(x) \phi_j(x) dx=\int_0^L sin(\omega_j x) \sin(\omega_j x)dx\!$$

Here, we will keep the integration boundaries as $$0\rightarrow L\!$$ to be consistent with the problem statement.

We can compute the result of the previous integral using Wolfram ALpha (Mathematica software). The following is the input given to the software: int 0 to L (sin(ax))^2

where a is $$\omega_j\!$$. The software generates the following answer:

$$\int_0^L \sin(ax)^2=\frac{L}{2} - \frac{\sin(2aL)}{4a}\!$$ Substituting for $$ \omega_j \!$$: $$\int_0^L \sin(ax)^2=\frac{L}{2} - \frac{\sin(2L\omega_j)}{4\omega_j}\!$$ We know from the previous explanation that $$L=\pi\!$$.So,we can apply the same assumption as before that $$\sin(c\pi)=0\!$$ where c is any integer. This allows us to cancel any terms with $$sin(2L\omega_j)\!$$ as they are equal to zero. Then, we can verify: (5) $$\langle \phi_j, \phi_j \rangle = \frac{L}{2}\!$$ for $$i = j\!$$

R7.2
Solved by Francisco Arrieta

Problem Statement
Plot the truncated series $$ u(x,t)=\sum_{j=1}^{n}a_j\cos C\omega_j t \sin \omega_j x \!$$ with $$ n=5 \!$$ and for:

$$ t=\alpha P_1=\alpha \frac{2\pi}{C\omega_1}=\alpha \frac{2L}{C} \!$$

$$ \alpha=0.5, 1, 1.5, 2 \!$$

Solution
Using:

$$ \left\{\begin{matrix} f(x)=x(x-2) \\ g(x)=0 \\ C=3 \\ L=4

\end{matrix}\right. \!$$

Then:

$$ a_j=\frac{2}{L}\int_{0}^{L}f(x)\sin \omega_j x dx \!$$


 * $$ =2\left [ \frac{(-1)^j-1}{\pi^3j^3} \right ] \!$$

$$ \therefore a_j=0 \!$$ for all even values of j

Plugging back to the truncated series:

$$ u(x,t)=\sum_{j=1}^{n}2\left [ \frac{(-1)^j-1}{\pi^3j^3} \right ]\cos [C \frac{j\pi}{L} \alpha \frac{2L}{C}] \sin[\frac{j\pi}{L}x] \!$$


 * $$ =\sum_{j=1}^{n}2\left [ \frac{(-1)^j-1}{\pi^3j^3} \right ]\cos (\alpha j2\pi) \sin(\frac{j\pi}{2}x) \!$$

For $$ n=5 \!$$ :

$$ u(x,t)=[\frac{-4}{\pi^3}\cos (2\pi \alpha) \sin(\frac{\pi x}{2})]+[\frac{-4}{27\pi^3}\cos (6\pi \alpha) \sin(\frac{3\pi x}{2})]+[\frac{-4}{125\pi^3}\cos (10\pi \alpha) \sin(\frac{5\pi x}{2})] \!$$

When $$ \alpha=0.5 \!$$ :



When $$ \alpha=1 \!$$ :



When $$ \alpha=1.5 \!$$ :



When $$ \alpha=2 \!$$ :



--Egm4313.s12.team11.arrieta (talk) 06:20, 22 April 2012 (UTC)

Problem Statement
Find (a) the scalar product, (b) the magnitude of $$f\!$$ and $$g\!$$ ,(c) the angle between $$f\!$$ and $$g\!$$ for:

1) $$f(x)=cos(x), \ g(x)=x \ for -2 \le x \le 10\!$$

2) $$f(x)=\frac{1}{2}(3x^2-1), \ g(x)=\frac{1}{2}(5x^3-3x) \ for -1 \le x \le 1\!$$

Part 1
solved by Kyle Gooding

Scalar Product
$$=\int_a^bf(x)g(x) \ dx\!$$

$$=\int_{-2}^{10}x\cos(x) \ dx\!$$

Using integration by parts;

$$=[x\sin(x)+\cos(x)]_{-2}^{10} $$

$$=-7.68\!$$

Magnitude
$$\| f \|=^{1/2}=\int_{a}^{b} f^2(x) \ dx\!$$

$$=\int_{-2}^{10} [\cos(x)]^2 \ dx\!$$

$$=[.5(x+\sin(x)\cos(x)|_{-2}^{10}]^{1/2}$$

$$\| f \| =2.457\!$$

$$\| g \|=\int_{a}^{b} g^2(x) \ dx\!$$

$$=\int_{-2}^{10}x^2 \ dx $$ $$=[x^{3}/3]_{-2}^{10}$$

$$\| g \| =\frac{1008}{3}\!$$

Angle Between Functions
$$cos(\theta)=\frac{}{\| f \| \| g \|}\!$$

$$cos(\theta)=\frac{-7.68}{\frac{1008}{3}(2.457)}$$

$$\theta=89.47$$ The two functions are nearly orthogonal.

Part 2
solved by Luca Imponenti

Scalar Product
$$=\int_a^bf(x)g(x) \ dx\!$$

$$=\int_{-1}^{1}[\frac{1}{2}(3x^2-1)][\frac{1}{2}(5x^3-3x)] \ dx\!$$
 * $$=\int_{-1}^{1}\frac{1}{4}(15x^5-4x^3+3x) \ dx\!$$
 * $$=\left. \frac{1}{4}(\frac{15}{6}x^6-x^4+\frac{3}{2}x^2)\right|_{-1}^{1}\!$$
 * $$=\frac{1}{4}[(\frac{15}{6}1^6-1^4+\frac{3}{2}1^2)-(\frac{15}{6}(-1)^6-(-1)^4+\frac{3}{2}(-1)^2)]\!$$

Since all exponents are even, everything in brackets cancels out

$$=0\!$$

Magnitude
$$\| f \|=^{1/2}=\int_{a}^{b} f^2(x) \ dx\!$$
 * $$=\int_{-1}^{1} [\frac{1}{2}(3x^2-1)]^2 \ dx\!$$
 * $$=\int_{-1}^{1} \frac{1}{4}(9x^4-6x^2+1) \ dx\!$$
 * $$=\left. \frac{1}{4}(\frac{9}{5}x^5-2x^3+x)\right|_{-1}^{1}\!$$
 * $$=\frac{1}{4}[(\frac{9}{5}1^5-2(1)^3+1)-(\frac{9}{5}(-1)^5-2(-1)^3+(-1))]\!$$
 * $$=\frac{1}{4}[\frac{4}{5}-(-\frac{4}{5})]\!$$

$$\| f \| =\frac{2}{5}\!$$

$$\| g \|=\int_{a}^{b} g^2(x) \ dx\!$$
 * $$=\int_{-1}^{1} [\frac{1}{2}(5x^3-3x)]^2 \ dx\!$$
 * $$=\int_{-1}^{1} \frac{1}{4}(25x^6-30x^4+9x^2) \ dx\!$$
 * $$=\left. \frac{1}{4}(\frac{25}{7}x^7-6x^5+3x^3)\right|_{-1}^{1}\!$$
 * $$=\frac{1}{4}[(\frac{25}{7}1^7-6(1)^5+3(1)^3)-(\frac{25}{7}(-1)^7-6(-1)^5+3(-1)^3)]\!$$
 * $$=\frac{1}{4}[\frac{4}{7}-(-\frac{4}{7})]\!$$

$$\| g \| =\frac{2}{7}\!$$

Angle Between Functions
$$cos(\theta)=\frac{}{\| f \| \| g \|}\!$$

Since $$=0\!$$ the two functions are orthogonal

$$\theta=90\!$$

R7.4
Solved by Gonzalo Perez

Problem Statement
Sketch or graph $$ f(x) \! $$ which for $$ -\pi < x < \pi \! $$ is given as follows:

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

Solution
The MATLAB code shown below was used to developed the graph of $$ f(x) = \left | x \right | \! $$:





Problem Statement
Sketch or graph $$ f(x) \! $$ which for $$ -\pi < x < \pi \! $$ is given as follows:

$$ f(x)=\left\{\begin{matrix} x, if: -\pi<x<0\\

\pi - x, if: 0<x<\pi \end{matrix}\right. \! $$

Solution
The MATLAB code shown below was used to developed the graph of the piecewise function $$ f(x)=\left\{\begin{matrix} x, if: -\pi<x<0\\

\pi - x, if: 0<x<\pi \end{matrix}\right. \! $$:





Solved by Jonathan Sheider

Problem Statement
Find the Fourier series of the given function which is assumed to have a period of $$ 2 \pi \!$$. Show the details of your work. Sketch or graph the partial sums up to that including $$ cos(5x) \!$$ and $$ sin(5x) \!$$

Given: $$f(x) = |x| \!$$

Solution
The Fourier series of a function with a period of $$ p = 2\pi \!$$ is defined: $$ f(x) = a_{0} + \sum_{n=1}^{\infty } (a_{n}cos(nx) + b_{n}sin(nx))\!$$ Where: $$ a_{0} = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) dx \!$$ $$ a_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) cos(nx) dx \!$$ $$ b_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) sin(nx) dx \!$$ This particular function given in the problem can be also defined in a piecewise manner over this interval, namely: $$ f(x) = \left\{\begin{matrix} -x \text{ if } -\pi \leq x \leq 0\\x \text{ if } 0 \leq x \leq \pi \end{matrix}\right. \!$$ Calculating the first term $$ a_{0} \!$$: $$ a_{0} = \frac{1}{2\pi} \left ( \int_{-\pi}^{0} -x dx + \int_{0}^{\pi} x dx \right) \!$$ $$ a_{0} = \frac{1}{2\pi} \left ( \left [\frac{-x^2}{2} \right ]_{-\pi}^{0} + \left [\frac{x^2}{2} \right ]_{0}^{\pi} \right)\!$$ $$ a_{0} = \frac{1}{2\pi} \left ( - \left(\frac{-\pi^2}{2}\right ) + \left (\frac{\pi^2}{2}\right)\right) \!$$ $$ a_{0} = \frac{1}{2\pi} (\pi^2) \!$$ $$ a_{0} = \frac{\pi}{2} \!$$ Calculating the coefficient $$ a_{n} \!$$: $$ a_{n} = \frac{1}{\pi} \left( \int_{-\pi}^{0} -x cos(nx) dx + \int_{0}^{\pi} x cos(nx) dx \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left( - \int_{-\pi}^{0} x cos(nx) dx + \int_{0}^{\pi} x cos(nx) dx \right ) \!$$ Using integration by parts with the following substitutions: $$ u = x \!$$ and therefore $$ du = dx \!$$ $$ dv = cos(nx)dx \!$$ and therefore $$ v = \frac{1}{n}sin(nx) \!$$ This yields for the integral: $$ a_{n} = \frac{1}{\pi} \left( -\left[\frac{1}{n} x sin(nx) - \int \frac{1}{n}sin(nx)dx \right]_{-\pi}^{0} + \left[\frac{1}{n} x sin(nx) - \int \frac{1}{n}sin(nx)dx \right]_{0}^{\pi} \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left( -\left[\frac{1}{n} x sin(nx) + \frac{1}{n^2}cos(nx) \right]_{-\pi}^{0} + \left[\frac{1}{n} x sin(nx) + \frac{1}{n^2}cos(nx) \right]_{0}^{\pi} \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left( -\left[\frac{1}{n^2} - \left (\frac{1}{n} (-\pi) sin(-n\pi) + \frac{1}{n^2}cos(-n \pi) \right) \right] + \left[\frac{1}{n} \pi sin(n\pi) + \frac{1}{n^2}cos(n\pi) - \frac{1}{n^2} \right] \right ) \!$$ Note that for all n = 1,2,3... : $$ sin(n\pi) = 0 \!$$ as $$ sin(\pi) = sin(2\pi) = sin(3\pi) ... = 0 \!$$ therefore these terms are evaluated as zero, which yields: $$ a_{n} = \frac{1}{\pi} \left( -\left[\frac{1}{n^2} - \left (0 + \frac{1}{n^2}cos(-n \pi) \right) \right] + \left[0 + \frac{1}{n^2}cos(n\pi) - \frac{1}{n^2} \right] \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left( -\frac{1}{n^2} + \frac{1}{n^2}cos(-n \pi) +  \frac{1}{n^2}cos(n\pi) - \frac{1}{n^2} \right ) \!$$ Note that $$ cos(-x) = cos(x) \!$$ therefore: $$ a_{n} = \frac{1}{\pi} \left( -\frac{1}{n^2} + \frac{1}{n^2}cos(n \pi) +  \frac{1}{n^2}cos(n\pi) - \frac{1}{n^2} \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left( -\frac{2}{n^2} + \frac{2}{n^2}cos(n \pi) \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left (\left ( -\frac{2}{n^2} \right )(1 - cos(n \pi) \right )\!$$ $$ a_{n} = -\frac{2}{n^2\pi} \left (1 - cos(n \pi) \right )\!$$ To evaluate the term  $$ (1-cos(n\pi)) \!$$: Note that $$ cos(n\pi) = -1 \!$$ for odd n values as $$ cos(\pi) = cos(3\pi) = cos(5\pi) ... = -1 \!$$ And that $$ cos(n\pi) = 1 \!$$ for even n values as $$ cos(2\pi) = cos(4\pi) = cos(6\pi) ... = 1 \!$$. Therefore, it can be concluded that for odd n values: $$ (1-cos(n\pi)) = 1-(-1) = 2 \!$$ And for even n values: $$ (1-cos(n\pi) = 1-(1) = 0 \!$$ Therefore, for the coefficient $$ a_{n} \!$$, all even terms will equal zero while all odd terms will have a multiplier of 2, this yields (for all odd n values): $$ a_{n} = -\frac{4}{n^2\pi} \text{ for  } n = 1,3,5... \!$$ Calculating the coefficient $$ b_{n} \!$$: $$ b_{n} = \frac{1}{\pi} \left( \int_{-\pi}^{0} -x sin(nx) dx + \int_{0}^{\pi} x sin(nx) dx \right ) \!$$ $$ b_{n} = \frac{1}{\pi} \left( - \int_{-\pi}^{0} x sin(nx) dx + \int_{0}^{\pi} x sin(nx) dx \right ) \!$$ Using integration by parts with the following substitutions: $$ u = x \!$$ and therefore $$ du = dx \!$$ $$ dv = sin(nx)dx \!$$ and therefore $$ v = \frac{-1}{n}cos(nx) \!$$ This yields for the integral: $$ b_{n} = \frac{1}{\pi} \left( -\left[\frac{-1}{n} x cos(nx) - \int \frac{-1}{n}cos(nx)dx \right]_{-\pi}^{0} + \left[\frac{-1}{n} x cos(nx) - \int \frac{-1}{n}cos(nx)dx \right]_{0}^{\pi} \right ) \!$$ $$ b_{n} = \frac{1}{\pi} \left( -\left[\frac{-1}{n} x cos(nx) + \frac{1}{n^2}sin(nx) \right]_{-\pi}^{0} + \left[\frac{-1}{n} x cos(nx) + \frac{1}{n^2}sin(nx) \right]_{0}^{\pi} \right ) \!$$ $$ b_{n} = \frac{1}{\pi} \left( -\left[0 - \left (\frac{-1}{n} (-\pi) cos(-n\pi) + \frac{1}{n^2}sin(-n \pi) \right) \right] + \left[\frac{-1}{n} \pi cos(n\pi) + \frac{1}{n^2}sin(n\pi) - 0 \right] \right ) \!$$ Note that for all n = 1,2,3... : $$ sin(n\pi) = 0 \!$$ as $$ sin(\pi) = sin(2\pi) = sin(3\pi) ... = 0 \!$$ therefore these terms are evaluated as zero, which yields: $$ b_{n} = \frac{1}{\pi} \left( -\left[0 - \left (\frac{-1}{n} (-\pi) cos(-n\pi) + 0 \right) \right] + \left[\frac{-1}{n} \pi cos(n\pi) + 0 \right] \right ) \!$$ $$ b_{n} = \frac{1}{\pi} \left( \left[\frac{1}{n} (\pi) cos(-n\pi) \right] + \left[\frac{-1}{n} \pi cos(n\pi) \right] \right ) \!$$ Note that $$ cos(-x) = cos(x) \!$$ therefore: $$ b_{n} = \frac{1}{\pi} \left( \frac{1}{n} \pi cos(n\pi) + \frac{-1}{n} \pi cos(n\pi)  \right ) \!$$ $$ b_{n} = \frac{1}{\pi} \left( 0 \right ) \!$$ $$ b_{n} = 0 \!$$ Therefore there will be no $$ sin(nx) \!$$ terms in the Fourier representation. In conclusion, the Fourier series representation for the given function is as follows: $$ f(x) = \frac{\pi}{2} - \frac{4}{\pi} cos(x) - \frac{4}{3^2 \pi} cos(3x) - \frac{4}{5^2 \pi} cos(5x) + ... \!$$                                     $$ f(x) = \frac{\pi}{2} - \frac{4}{\pi}\left (cos(x) + \frac{1}{9} cos(3x) + \frac{1}{25} cos(5x) + ... \right ) \!$$ A graph of the function, and the Fourier series for $$ n = 1,3,5 \!$$ is shown below: --Egm4313.s12.team11.sheider (talk) 06:00, 22 April 2012 (UTC)

Problem Statement
Find the Fourier series of the given function which is assumed to have a period of $$ 2 \pi \!$$. Show the details of your work. Sketch or graph the partial sums up to that including $$ cos(5x) \!$$ and $$ sin(5x) \!$$

Given: $$ f(x) = \left\{\begin{matrix} -x \text{ if } -\pi < x < 0\\\pi - x \text{ if } 0 < x < \pi \end{matrix}\right. \!$$

Solution
The Fourier series of a function with a period of $$ p = 2\pi \!$$ is defined: $$ f(x) = a_{0} + \sum_{n=1}^{\infty } (a_{n}cos(nx) + b_{n}sin(nx))\!$$ Where: $$ a_{0} = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) dx \!$$ $$ a_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) cos(nx) dx \!$$ $$ b_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) sin(nx) dx \!$$ Calculating the first term $$ a_{0} \!$$: $$ a_{0} = \frac{1}{2\pi} \left ( \int_{-\pi}^{0} x dx + \int_{0}^{\pi} (\pi - x) dx \right) \!$$ $$ a_{0} = \frac{1}{2\pi} \left ( \left [\frac{x^2}{2} \right ]_{-\pi}^{0} + \left [\pi x - \frac{x^2}{2} \right ]_{0}^{\pi} \right)\!$$ $$ a_{0} = \frac{1}{2\pi} \left ( - \left(\frac{\pi^2}{2}\right ) + \left (\pi^2 - \frac{\pi^2}{2}\right)\right) \!$$ $$ a_{0} = \frac{1}{2\pi} \left ( -\frac{\pi^2}{2} + \pi^2 - \frac{\pi^2}{2}\right) \!$$ $$ a_{0} = \frac{1}{2\pi} \left (\pi^2 - \pi^2 \right) \!$$ $$ a_{0} = 0 \!$$ Calculating the coefficient $$ a_{n} \!$$: $$ a_{n} = \frac{1}{\pi} \left( \int_{-\pi}^{0} x cos(nx) dx + \int_{0}^{\pi} (\pi - x) cos(nx) dx \right ) \!$$ Using integration by parts with the following substitutions for the integral $$ \int_{-\pi}^{0} x cos(nx) dx \!$$: $$ u = x \!$$ and therefore $$ du = dx \!$$ $$ dv = cos(nx)dx \!$$ and therefore $$ v = \frac{1}{n}sin(nx) \!$$ Using integration by parts with the following substitutions for the integral $$ \int_{0}^{\pi} (\pi - x) cos(nx) dx \!$$: $$ u = \pi - x \!$$ and therefore $$ du = -dx \!$$ $$ dv = cos(nx)dx \!$$ and therefore $$ v = \frac{1}{n}sin(nx) \!$$ This yields for the overall expression: $$ a_{n} = \frac{1}{\pi} \left( \left[\frac{1}{n} x sin(nx) - \int \frac{1}{n}sin(nx)dx \right]_{-\pi}^{0} + \left[\frac{1}{n} (\pi - x) sin(nx) - \int -\frac{1}{n}sin(nx)dx \right]_{0}^{\pi} \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left( \left[\frac{1}{n} x sin(nx) + \frac{1}{n^2}cos(nx) \right]_{-\pi}^{0} + \left[\frac{1}{n} (\pi - x) sin(nx) - \frac{1}{n^2}cos(nx) \right]_{0}^{\pi} \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left( \left[\frac{1}{n^2} - \left (\frac{1}{n} (-\pi) sin(-n\pi) + \frac{1}{n^2}cos(-n \pi) \right) \right] + \left[-\frac{1}{n^2}cos(n\pi) + \frac{1}{n^2} \right] \right ) \!$$ Note that for all n = 1,2,3... : $$ sin(n\pi) = 0 \!$$ as $$ sin(\pi) = sin(2\pi) = sin(3\pi) ... = 0 \!$$ therefore these terms are evaluated as zero, which yields: $$ a_{n} = \frac{1}{\pi} \left( \left[\frac{1}{n^2} - 0 - \frac{1}{n^2}cos(-n \pi) \right] + \left[-\frac{1}{n^2}cos(n\pi) + \frac{1}{n^2} \right] \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left(\frac{1}{n^2} - \frac{1}{n^2}cos(-n \pi) - \frac{1}{n^2}cos(n\pi) + \frac{1}{n^2} \right ) \!$$ Note that $$ cos(-x) = cos(x) \!$$ therefore: $$ a_{n} = \frac{1}{\pi} \left(\frac{1}{n^2} - \frac{1}{n^2}cos(n \pi) - \frac{1}{n^2}cos(n\pi) + \frac{1}{n^2} \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left(\frac{2}{n^2} - \frac{2}{n^2}cos(n \pi) \right ) \!$$ $$ a_{n} = \frac{1}{\pi} \left( \left (\frac{2}{n^2} \right)(1 - cos(n \pi)) \right ) \!$$ $$ a_{n} = \frac{2}{n^2\pi} \left (1 - cos(n \pi) \right )\!$$ To evaluate the term $$ (1-cos(n\pi)) \!$$: Note that $$ cos(n\pi) = -1 \!$$ for odd n values as $$ cos(\pi) = cos(3\pi) = cos(5\pi) ... = -1 \!$$ And that $$ cos(n\pi) = 1 \!$$ for even n values as $$ cos(2\pi) = cos(4\pi) = cos(6\pi) ... = 1 \!$$. Therefore, it can be concluded that for odd n values: $$ (1-cos(n\pi)) = 1-(-1) = 2 \!$$ And for even n values: $$ (1-cos(n\pi)) = 1-(1) = 0 \!$$ Therefore, for the coefficient $$ a_{n} \!$$, all even terms will equal zero while all odd terms will have a multiplier of 2, this yields (for all odd n values): $$ a_{n} = \frac{4}{n^2\pi} \text{ for  } n = 1,3,5... \!$$ Calculating the coefficient $$ b_{n} \!$$: $$ b_{n} = \frac{1}{\pi} \left( \int_{-\pi}^{0} x sin(nx) dx + \int_{0}^{\pi} (\pi - x) sin(nx) dx \right ) \!$$ Using integration by parts with the following substitutions for the integral $$ \int_{-\pi}^{0} x sin(nx) dx \!$$: $$ u = x \!$$ and therefore $$ du = dx \!$$ $$ dv = sin(nx)dx \!$$ and therefore $$ v = \frac{-1}{n}cos(nx) \!$$ Using integration by parts with the following substitutions for the integral $$ \int_{0}^{\pi} (\pi - x) sin(nx) dx \!$$: $$ u = \pi - x \!$$ and therefore $$ du = -dx \!$$ $$ dv = sin(nx)dx \!$$ and therefore $$ v = \frac{-1}{n}cos(nx) \!$$ This yields for the overall expression: $$ b_{n} = \frac{1}{\pi} \left( \left[\frac{-1}{n} x cos(nx) - \int \frac{-1}{n}cos(nx)dx \right]_{-\pi}^{0} + \left[\frac{-1}{n}(\pi - x) cos(nx) - \int -\frac{-1}{n}cos(nx)dx \right]_{0}^{\pi} \right ) \!$$ $$ b_{n} = \frac{1}{\pi} \left( \left[\frac{-1}{n} x cos(nx) + \frac{1}{n^2}sin(nx) \right]_{-\pi}^{0} + \left[\frac{-1}{n}(\pi- x) cos(nx) - \frac{1}{n^2}sin(nx) \right]_{0}^{\pi} \right ) \!$$ $$ b_{n} = \frac{1}{\pi} \left( \left[0 - \left (\frac{-1}{n} (-\pi) cos(-n\pi) + \frac{1}{n^2}sin(-n \pi) \right) \right] + \left[- \frac{1}{n^2}sin(n\pi) - \left( \frac{-1}{n}\pi \right) \right] \right ) \!$$ $$ b_{n} = \frac{1}{\pi} \left( - \frac{1}{n} \pi cos(-n\pi) - \frac{1}{n^2}sin(-n \pi) - \frac{1}{n^2}sin(n\pi) + \frac{1}{n}\pi   \right ) \!$$ Note that for all n = 1,2,3... : $$ sin(n\pi) = 0 \!$$ as $$ sin(\pi) = sin(2\pi) = sin(3\pi) ... = 0 \!$$ therefore these terms are evaluated as zero, which yields: $$ b_{n} = \frac{1}{\pi} \left( - \frac{1}{n} \pi cos(-n\pi) -0 - 0 + \frac{1}{n}\pi   \right ) \!$$ $$ b_{n} = \frac{1}{\pi} \left(\frac{1}{n}\pi - \frac{1}{n} \pi cos(-n\pi) \right ) \!$$ Note that $$ cos(-x) = cos(x) \!$$ therefore: $$ b_{n} = \frac{1}{\pi} \left(\frac{1}{n}\pi - \frac{1}{n} \pi cos(n\pi) \right ) \!$$ $$ b_{n} = \frac{1}{\pi} \left( \left(\frac{\pi}{n} \right)(1-cos(n\pi)) \right ) \!$$ $$ b_{n} = \frac{1}{n} \left(1-cos(n\pi) \right ) \!$$ To evaluate the term $$ (1-cos(n\pi)) \!$$: Note that $$ cos(n\pi) = -1 \!$$ for odd n values as $$ cos(\pi) = cos(3\pi) = cos(5\pi) ... = -1 \!$$ And that $$ cos(n\pi) = 1 \!$$ for even n values as $$ cos(2\pi) = cos(4\pi) = cos(6\pi) ... = 1 \!$$. Therefore, it can be concluded that for odd n values: $$ (1-cos(n\pi)) = 1-(-1) = 2 \!$$ And for even n values: $$ (1-cos(n\pi)) = 1-(1) = 0 \!$$ Therefore, for the coefficient $$ b_{n} \!$$, all even terms will equal zero while all odd terms will have a multiplier of 2, this yields (for all odd n values): $$ b_{n} = \frac{2}{n} \text{ for  } n = 1,3,5... \!$$ In conclusion, the Fourier series representation for the given function is as follows: $$ f(x) = 0 + \left (\frac{4}{\pi} cos(x) + \frac{4}{3^2 \pi} cos(3x) + \frac{4}{5^2 \pi} cos(5x) + ... \right ) + \left (2sin(x) + \frac{2}{3} sin(3x) + \frac{2}{5} sin(5x) + ... \right ) \!$$

$$ f(x) = \frac{4}{\pi} \left (cos(x) + \frac{1}{9} cos(3x) + \frac{1}{25} cos(5x) + ... \right ) + 2 \left (sin(x) + \frac{1}{3} sin(3x) + \frac{1}{5} sin(5x) + ... \right ) \!$$ A graph of the function, and the Fourier series for $$ n = 1,3,5 \!$$ is shown below: --Egm4313.s12.team11.sheider (talk) 06:00, 22 April 2012 (UTC)

R7.5
Solved by Daniel Suh

Problem Statement
Consider the following, $$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\int_{0}^{p}\phi _{2j-1}(x) \cdot \phi _{2k-1}(x)dx\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\int_{0}^{p}\sin {j\omega x} \cdot \sin {k\omega x} \; dx\!$$

with $$j\neq k\; and\; j,k = 1,2,...\!$$, and $$p=2\pi, j=2, k=3\!$$

1. Find the integration with the given data.

2. Confirm the results with Matlab's trapz command for the trapezoidal rule.

Trigonometric Identities
Angle Sum and Difference Identities

$$(1) \cos{(a+b)} = \cos{a}\cos {b} - \sin {a}\sin{b}\!$$ $$(2) \cos{(a-b)} = \cos{a}\cos {b} + \sin {a}\sin{b}\!$$

Rearrange

$$(1) \sin {a}\sin{b} = \cos{a}\cos {b}-\cos{(a+b)}\!$$ $$(2) \cos{a}\cos {b} = \cos{(a-b)} - \sin{a}\sin {b}\!$$

Substitute and Combine

$$\sin {a}\sin{b} = \cos{(a-b)} - \sin{a}\sin {b} - \cos{(a+b)}\!$$ $$2\sin {a}\sin{b} = \cos{(a-b)} - \cos{(a+b)}\!$$ $$\sin {a}\sin{b} = \frac{1}{2}\cos{(a-b)} - \frac{1}{2}\cos{(a+b)}\!$$

Utilize Trig Identities
$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\int_{0}^{p}\sin {jwx} \cdot \sin {kwx} \; dx\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\int_{0}^{2\pi}\sin {2\omega x} \cdot \sin {3\omega x} \; dx\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\int_{0}^{2\pi}\frac{1}{2}\cos {(2\omega x - 3\omega x)} - \frac{1}{2}\cos{(2\omega x + 3\omega x)} \; dx\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\frac{1}{2}\int_{0}^{2\pi}\cos {(-\omega x)} \; dx - \frac{1}{2}\int_{0}^{2\pi}\cos{(5\omega x)} \; dx\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle=\frac{1}{2}\sin {(-\omega x)}|_{0}^{2\pi} - \frac{1}{2}\sin{(5\omega x)}|_{0}^{2\pi}\!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle= (0-0) - (0-0) \!$$

$$\left \langle \phi_{2j-1}, \phi_{2k-1} \right \rangle= 0 \!$$

Part 2
>> X = 0:2*pi/100:2*pi;

>> Y = sin(2*X).*sin(3*X);

>> Z = trapz(X,Y)

Z =

2.9490e-017