Egm4313.s12.team11.R7

Report 7

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)

R7.3

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

$<f,g>=\int_a^bf(x)g(x) \ dx\!$

$<f,g>=\int_{-2}^{10}x\cos(x) \ dx\!$

Using integration by parts;

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

                      $<f,g>=-7.68\!$

Magnitude

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

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

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

                       $\| f \| =2.457\!$

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

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

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

Angle Between Functions

$cos(\theta)=\frac{<f,g>}{\| f \| \| g \|}\!$

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

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

Part 2

solved by Luca Imponenti

Scalar Product

$<f,g>=\int_a^bf(x)g(x) \ dx\!$

$<f,g>=\int_{-1}^{1}[\frac{1}{2}(3x^2-1)][\frac{1}{2}(5x^3-3x)] \ dx\!$

$=\int_{-1}^{1}\frac{1}{4}(15x^5-4x^3+3x) \ dx\!$

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

$=\frac{1}{4}[(\frac{15}{6}1^6-1^4+\frac{3}{2}1^2)-(\frac{15}{6}(-1)^6-(-1)^4+\frac{3}{2}(-1)^2)]\!$

Since all exponents are even, everything in brackets cancels out

                      $<f,g>=0\!$

Magnitude

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

$=\int_{-1}^{1} [\frac{1}{2}(3x^2-1)]^2 \ dx\!$

$=\int_{-1}^{1} \frac{1}{4}(9x^4-6x^2+1) \ dx\!$

$=\left. \frac{1}{4}(\frac{9}{5}x^5-2x^3+x)\right|_{-1}^{1}\!$

$=\frac{1}{4}[(\frac{9}{5}1^5-2(1)^3+1)-(\frac{9}{5}(-1)^5-2(-1)^3+(-1))]\!$

$=\frac{1}{4}[\frac{4}{5}-(-\frac{4}{5})]\!$

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

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

$=\int_{-1}^{1} [\frac{1}{2}(5x^3-3x)]^2 \ dx\!$

$=\int_{-1}^{1} \frac{1}{4}(25x^6-30x^4+9x^2) \ dx\!$

$=\left. \frac{1}{4}(\frac{25}{7}x^7-6x^5+3x^3)\right|_{-1}^{1}\!$

$=\frac{1}{4}[(\frac{25}{7}1^7-6(1)^5+3(1)^3)-(\frac{25}{7}(-1)^7-6(-1)^5+3(-1)^3)]\!$

$=\frac{1}{4}[\frac{4}{7}-(-\frac{4}{7})]\!$

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

Angle Between Functions

$cos(\theta)=\frac{<f,g>}{\| f \| \| g \|}\!$

Since $<f,g>=0\!$ the two functions are orthogonal

                         $\theta=90\!$

R7.4

Solved by Gonzalo Perez

K 2011 pg.482 pb. 6

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

K 2011 pg.482 pb. 9

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

K 2011 p.482 pb. 12

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)

K 2011 p.482 pb. 13

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.

Solution

Part 1

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