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) $⟨{\varphi }_i,{\varphi }_j⟩=0$ for $i\ne j$
(5) $⟨{\varphi }_j,{\varphi }_j⟩=\frac{L}{2}$ for $i=j$

Solution

From the lecture notes on p.19-9 we know that:
(2) ${\varphi }_i=\sin ({\omega }_{i}x)$
(3) $⟨{\varphi }_i,{\varphi }_j⟩=\underset{0}{\overset{L}{\int }}{\varphi }_i(x){\varphi }_j(x)dx$

Then, we have
$⟨{\varphi }_i,{\varphi }_j⟩=\underset{0}{\overset{L}{\int }}{\varphi }_i(x){\varphi }_j(x)dx=\underset{0}{\overset{L}{\int }}sin({\omega }_{i}x)\sin ({\omega }_{j}x)dx$

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:

$\underset{0}{\overset{L}{\int }}\sin (ax)sin(bx)=\frac{b\sin (\pi a)\cos (\pi b)-a\cos (\pi a)\sin (\pi b)}{a^2-b^2}$

Substituting for ${\omega }_i$ and ${\omega }_j$:
$\underset{0}{\overset{L}{\int }}\sin (ax)sin(bx)=\frac{{\omega }_j\sin (\pi {\omega }_i)\cos (\pi {\omega }_j)-{\omega }_i\cos (\pi {\omega }_i)\sin (\pi {\omega }_j)}{{\omega }_i^2-{\omega }_j^2}$
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)  $⟨{\varphi }_i,{\varphi }_j⟩=0$ for $i\ne j$

Similarly we can verify (5).
We have
$⟨{\varphi }_j,{\varphi }_j⟩=\underset{0}{\overset{L}{\int }}{\varphi }_j(x){\varphi }_j(x)dx=\underset{0}{\overset{L}{\int }}sin({\omega }_{j}x)\sin ({\omega }_{j}x)dx$

Here, we will keep the integration boundaries as $0\to L$ 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:

$\underset{0}{\overset{L}{\int }}\sin (ax{)}^2=\frac{L}{2}-\frac{\sin (2aL)}{4a}$

Substituting for ${\omega }_j$:
$\underset{0}{\overset{L}{\int }}\sin (ax{)}^2=\frac{L}{2}-\frac{\sin (2L{\omega }_j)}{4{\omega }_j}$

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)  $⟨{\varphi }_j,{\varphi }_j⟩=\frac{L}{2}$ 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:

$\{\begin{array}{c}f(x)=x(x-2)\\ g(x)=0\\ C=3\\ L=4\end{array}$

Then:

$a_j=\frac{2}{L}\underset{0}{\overset{L}{\int }}f(x)\sin {\omega }_{j}xdx$

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

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

Plugging back to the truncated series:

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

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

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 $\alpha =0.5$ :

File:Ahalf.jpg

When $\alpha =1$ :

File:Aone.jpg

When $\alpha =1.5$ :

File:Aonehalf.jpg

When $\alpha =2$ :

File:Atwo.jpg

--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) $f(x)=cos(x),g(x)=xfor-2\le x\le 10$

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

$<f,g>=\underset{a}{\overset{b}{\int }}f(x)g(x)dx$

$<f,g>=\underset{-2}{\overset{10}{\int }}x\cos (x)dx$

Using integration by parts;

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

                      $<f,g>=-7.68$

Magnitude

$\Vert f\Vert =<f,f{>}^{1/2}=\underset{a}{\overset{b}{\int }}{f}^2(x)dx$

$=\underset{-2}{\overset{10}{\int }}[\cos (x){]}^{2}dx$

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

                       $\Vert f\Vert =2.457$

$\Vert g\Vert =\underset{a}{\overset{b}{\int }}{g}^2(x)dx$

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

                       $\Vert g\Vert =\frac{1008}{3}$

Angle Between Functions

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

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

                      $\theta =89.47$

                      The two functions are nearly orthogonal.

Part 2

solved by Luca Imponenti

Scalar Product

$<f,g>=\underset{a}{\overset{b}{\int }}f(x)g(x)dx$

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

$=\underset{-1}{\overset{1}{\int }}\frac{1}{4}(15x^5-4x^3+3x)dx$

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

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

Since all exponents are even, everything in brackets cancels out

                      $<f,g>=0$

Magnitude

$\Vert f\Vert =<f,f{>}^{1/2}=\underset{a}{\overset{b}{\int }}{f}^2(x)dx$

$=\underset{-1}{\overset{1}{\int }}[\frac{1}{2}(3x^2-1){]}^{2}dx$

$=\underset{-1}{\overset{1}{\int }}\frac{1}{4}(9x^4-6x^2+1)dx$

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

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

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

                       $\Vert f\Vert =\frac{2}{5}$

$\Vert g\Vert =\underset{a}{\overset{b}{\int }}{g}^2(x)dx$

$=\underset{-1}{\overset{1}{\int }}[\frac{1}{2}(5x^3-3x){]}^{2}dx$

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

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

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

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

                       $\Vert g\Vert =\frac{2}{7}$

Angle Between Functions

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

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

File:Problem6.jpg

File:Problem6graph.jpg

K 2011 pg.482 pb. 9

Problem Statement

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

$f(x)=\{\begin{array}{c}x,if:-\pi <x<0\\ \pi -x,if:0<x<\pi \end{array}$

Solution

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

File:Problem9.jpg

File:Problem9graph.jpg

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:

$f(x)=a_0+\sum _{n=1}^{\mathrm{\infty }}(a_{n}cos(nx)+b_{n}sin(nx))$

Where:

$a_0=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)dx$
$a_n=\frac{1}{\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)cos(nx)dx$
$b_n=\frac{1}{\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)sin(nx)dx$

This particular function given in the problem can be also defined in a piecewise manner over this interval, namely:

$f(x)=\{\begin{array}{c}-x\text{ if }-\pi \le x\le 0\\ x\text{ if }0\le x\le \pi \end{array}$

Calculating the first term $a_0$:

$a_0=\frac{1}{2\pi }(\underset{-\pi }{\overset{0}{\int }}-xdx+\underset{0}{\overset{\pi }{\int }}xdx)$
$a_0=\frac{1}{2\pi }({\left[\frac{-x^2}{2}\right]}_{-\pi }^0+{\left[\frac{x^2}{2}\right]}_0^{\pi })$
$a_0=\frac{1}{2\pi }(-\left(\frac{-{\pi }^2}{2}\right)+\left(\frac{{\pi }^2}{2}\right))$
$a_0=\frac{1}{2\pi }({\pi }^2)$

                                                         $a_0=\frac{\pi }{2}$

Calculating the coefficient $a_n$:

$a_n=\frac{1}{\pi }(\underset{-\pi }{\overset{0}{\int }}-xcos(nx)dx+\underset{0}{\overset{\pi }{\int }}xcos(nx)dx)$
$a_n=\frac{1}{\pi }(-\underset{-\pi }{\overset{0}{\int }}xcos(nx)dx+\underset{0}{\overset{\pi }{\int }}xcos(nx)dx)$

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:

$a_n=\frac{1}{\pi }(-{[\frac{1}{n}xsin(nx)-\int \frac{1}{n}sin(nx)dx]}_{-\pi }^0+{[\frac{1}{n}xsin(nx)-\int \frac{1}{n}sin(nx)dx]}_0^{\pi })$
$a_n=\frac{1}{\pi }(-{[\frac{1}{n}xsin(nx)+\frac{1}{n^2}cos(nx)]}_{-\pi }^0+{[\frac{1}{n}xsin(nx)+\frac{1}{n^2}cos(nx)]}_0^{\pi })$
$a_n=\frac{1}{\pi }(-[\frac{1}{n^2}-(\frac{1}{n}(-\pi )sin(-n\pi )+\frac{1}{n^2}cos(-n\pi ))]+[\frac{1}{n}\pi sin(n\pi )+\frac{1}{n^2}cos(n\pi )-\frac{1}{n^2}])$

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:

$a_n=\frac{1}{\pi }(-[\frac{1}{n^2}-(0+\frac{1}{n^2}cos(-n\pi ))]+[0+\frac{1}{n^2}cos(n\pi )-\frac{1}{n^2}])$
$a_n=\frac{1}{\pi }(-\frac{1}{n^2}+\frac{1}{n^2}cos(-n\pi )+\frac{1}{n^2}cos(n\pi )-\frac{1}{n^2})$

Note that $cos(-x)=cos(x)$ therefore:

$a_n=\frac{1}{\pi }(-\frac{1}{n^2}+\frac{1}{n^2}cos(n\pi )+\frac{1}{n^2}cos(n\pi )-\frac{1}{n^2})$
$a_n=\frac{1}{\pi }(-\frac{2}{n^2}+\frac{2}{n^2}cos(n\pi ))$
$a_n=\frac{1}{\pi }((-\frac{2}{n^2})(1-cos(n\pi ))$
$a_n=-\frac{2}{n^2\pi }(1-cos(n\pi ))$

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 $b_n$:

$b_n=\frac{1}{\pi }(\underset{-\pi }{\overset{0}{\int }}-xsin(nx)dx+\underset{0}{\overset{\pi }{\int }}xsin(nx)dx)$
$b_n=\frac{1}{\pi }(-\underset{-\pi }{\overset{0}{\int }}xsin(nx)dx+\underset{0}{\overset{\pi }{\int }}xsin(nx)dx)$

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:

$b_n=\frac{1}{\pi }(-{[\frac{-1}{n}xcos(nx)-\int \frac{-1}{n}cos(nx)dx]}_{-\pi }^0+{[\frac{-1}{n}xcos(nx)-\int \frac{-1}{n}cos(nx)dx]}_0^{\pi })$
$b_n=\frac{1}{\pi }(-{[\frac{-1}{n}xcos(nx)+\frac{1}{n^2}sin(nx)]}_{-\pi }^0+{[\frac{-1}{n}xcos(nx)+\frac{1}{n^2}sin(nx)]}_0^{\pi })$
$b_n=\frac{1}{\pi }(-[0-(\frac{-1}{n}(-\pi )cos(-n\pi )+\frac{1}{n^2}sin(-n\pi ))]+[\frac{-1}{n}\pi cos(n\pi )+\frac{1}{n^2}sin(n\pi )-0])$

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:

$b_n=\frac{1}{\pi }(-[0-(\frac{-1}{n}(-\pi )cos(-n\pi )+0)]+[\frac{-1}{n}\pi cos(n\pi )+0])$
$b_n=\frac{1}{\pi }([\frac{1}{n}(\pi )cos(-n\pi )]+[\frac{-1}{n}\pi cos(n\pi )])$

Note that $cos(-x)=cos(x)$ therefore:

$b_n=\frac{1}{\pi }(\frac{1}{n}\pi cos(n\pi )+\frac{-1}{n}\pi cos(n\pi ))$
$b_n=\frac{1}{\pi }\left(0\right)$

                                                       $b_n=0$

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:

$f(x)=\frac{\pi }{2}-\frac{4}{\pi }cos(x)-\frac{4}{3^2\pi }cos(3x)-\frac{4}{5^2\pi }cos(5x)+...$

                                    $f(x)=\frac{\pi }{2}-\frac{4}{\pi }(cos(x)+\frac{1}{9}cos(3x)+\frac{1}{25}cos(5x)+...)$

A graph of the function, and the Fourier series for $n=1,3,5$ is shown below:

File:R7.4.12.jpg
--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:

$f(x)=\{\begin{array}{c}-x\text{ if }-\pi <x<0\\ \pi -x\text{ if }0<x<\pi \end{array}$

Solution

The Fourier series of a function with a period of $p=2\pi$ is defined:

$f(x)=a_0+\sum _{n=1}^{\mathrm{\infty }}(a_{n}cos(nx)+b_{n}sin(nx))$

Where:

$a_0=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)dx$
$a_n=\frac{1}{\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)cos(nx)dx$
$b_n=\frac{1}{\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)sin(nx)dx$

Calculating the first term $a_0$:

$a_0=\frac{1}{2\pi }(\underset{-\pi }{\overset{0}{\int }}xdx+\underset{0}{\overset{\pi }{\int }}(\pi -x)dx)$
$a_0=\frac{1}{2\pi }({\left[\frac{x^2}{2}\right]}_{-\pi }^0+{[\pi x-\frac{x^2}{2}]}_0^{\pi })$
$a_0=\frac{1}{2\pi }(-\left(\frac{{\pi }^2}{2}\right)+({\pi }^2-\frac{{\pi }^2}{2}))$
$a_0=\frac{1}{2\pi }(-\frac{{\pi }^2}{2}+{\pi }^2-\frac{{\pi }^2}{2})$
$a_0=\frac{1}{2\pi }({\pi }^2-{\pi }^2)$

                                                         $a_0=0$

Calculating the coefficient $a_n$:

$a_n=\frac{1}{\pi }(\underset{-\pi }{\overset{0}{\int }}xcos(nx)dx+\underset{0}{\overset{\pi }{\int }}(\pi -x)cos(nx)dx)$

Using integration by parts with the following substitutions for the integral $\underset{-\pi }{\overset{0}{\int }}xcos(nx)dx$:

$u=x$ 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 $\underset{0}{\overset{\pi }{\int }}(\pi -x)cos(nx)dx$:

$u=\pi -x$ and therefore $du=-dx$
$dv=cos(nx)dx$ and therefore $v=\frac{1}{n}sin(nx)$

This yields for the overall expression:

$a_n=\frac{1}{\pi }({[\frac{1}{n}xsin(nx)-\int \frac{1}{n}sin(nx)dx]}_{-\pi }^0+{[\frac{1}{n}(\pi -x)sin(nx)-\int -\frac{1}{n}sin(nx)dx]}_0^{\pi })$
$a_n=\frac{1}{\pi }({[\frac{1}{n}xsin(nx)+\frac{1}{n^2}cos(nx)]}_{-\pi }^0+{[\frac{1}{n}(\pi -x)sin(nx)-\frac{1}{n^2}cos(nx)]}_0^{\pi })$
$a_n=\frac{1}{\pi }([\frac{1}{n^2}-(\frac{1}{n}(-\pi )sin(-n\pi )+\frac{1}{n^2}cos(-n\pi ))]+[-\frac{1}{n^2}cos(n\pi )+\frac{1}{n^2}])$

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:

$a_n=\frac{1}{\pi }([\frac{1}{n^2}-0-\frac{1}{n^2}cos(-n\pi )]+[-\frac{1}{n^2}cos(n\pi )+\frac{1}{n^2}])$

$a_n=\frac{1}{\pi }(\frac{1}{n^2}-\frac{1}{n^2}cos(-n\pi )-\frac{1}{n^2}cos(n\pi )+\frac{1}{n^2})$

Note that $cos(-x)=cos(x)$ therefore:

$a_n=\frac{1}{\pi }(\frac{1}{n^2}-\frac{1}{n^2}cos(n\pi )-\frac{1}{n^2}cos(n\pi )+\frac{1}{n^2})$

$a_n=\frac{1}{\pi }(\frac{2}{n^2}-\frac{2}{n^2}cos(n\pi ))$
$a_n=\frac{1}{\pi }(\left(\frac{2}{n^2}\right)(1-cos(n\pi )))$
$a_n=\frac{2}{n^2\pi }(1-cos(n\pi ))$

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 $b_n$:

$b_n=\frac{1}{\pi }(\underset{-\pi }{\overset{0}{\int }}xsin(nx)dx+\underset{0}{\overset{\pi }{\int }}(\pi -x)sin(nx)dx)$

Using integration by parts with the following substitutions for the integral $\underset{-\pi }{\overset{0}{\int }}xsin(nx)dx$:

$u=x$ 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 $\underset{0}{\overset{\pi }{\int }}(\pi -x)sin(nx)dx$:

$u=\pi -x$ and therefore $du=-dx$
$dv=sin(nx)dx$ and therefore $v=\frac{-1}{n}cos(nx)$

This yields for the overall expression:

$b_n=\frac{1}{\pi }({[\frac{-1}{n}xcos(nx)-\int \frac{-1}{n}cos(nx)dx]}_{-\pi }^0+{[\frac{-1}{n}(\pi -x)cos(nx)-\int -\frac{-1}{n}cos(nx)dx]}_0^{\pi })$
$b_n=\frac{1}{\pi }({[\frac{-1}{n}xcos(nx)+\frac{1}{n^2}sin(nx)]}_{-\pi }^0+{[\frac{-1}{n}(\pi -x)cos(nx)-\frac{1}{n^2}sin(nx)]}_0^{\pi })$
$b_n=\frac{1}{\pi }([0-(\frac{-1}{n}(-\pi )cos(-n\pi )+\frac{1}{n^2}sin(-n\pi ))]+[-\frac{1}{n^2}sin(n\pi )-\left(\frac{-1}{n}\pi \right)])$
$b_n=\frac{1}{\pi }(-\frac{1}{n}\pi cos(-n\pi )-\frac{1}{n^2}sin(-n\pi )-\frac{1}{n^2}sin(n\pi )+\frac{1}{n}\pi )$

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:

$b_n=\frac{1}{\pi }(-\frac{1}{n}\pi cos(-n\pi )-0-0+\frac{1}{n}\pi )$
$b_n=\frac{1}{\pi }(\frac{1}{n}\pi -\frac{1}{n}\pi cos(-n\pi ))$

Note that $cos(-x)=cos(x)$ therefore:

$b_n=\frac{1}{\pi }(\frac{1}{n}\pi -\frac{1}{n}\pi cos(n\pi ))$
$b_n=\frac{1}{\pi }(\left(\frac{\pi }{n}\right)(1-cos(n\pi )))$
$b_n=\frac{1}{n}(1-cos(n\pi ))$

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:

$f(x)=0+(\frac{4}{\pi }cos(x)+\frac{4}{3^2\pi }cos(3x)+\frac{4}{5^2\pi }cos(5x)+...)+(2sin(x)+\frac{2}{3}sin(3x)+\frac{2}{5}sin(5x)+...)$

                   $f(x)=\frac{4}{\pi }(cos(x)+\frac{1}{9}cos(3x)+\frac{1}{25}cos(5x)+...)+2(sin(x)+\frac{1}{3}sin(3x)+\frac{1}{5}sin(5x)+...)$

A graph of the function, and the Fourier series for $n=1,3,5$ is shown below:

File:R7.4.13.jpg
--Egm4313.s12.team11.sheider (talk) 06:00, 22 April 2012 (UTC)

R7.5

Solved by Daniel Suh

Problem Statement

Consider the following,
$⟨{\varphi }_{2j-1},{\varphi }_{2k-1}⟩=\underset{0}{\overset{p}{\int }}{\varphi }_{2j-1}(x)\cdot {\varphi }_{2k-1}(x)dx$

$⟨{\varphi }_{2j-1},{\varphi }_{2k-1}⟩=\underset{0}{\overset{p}{\int }}\sin j\omega x\cdot \sin k\omega xdx$

with $j\ne kandj,k=1,2,...$, 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

$⟨{\varphi }_{2j-1},{\varphi }_{2k-1}⟩=\underset{0}{\overset{p}{\int }}\sin jwx\cdot \sin kwxdx$

$⟨{\varphi }_{2j-1},{\varphi }_{2k-1}⟩=\underset{0}{\overset{2\pi }{\int }}\sin 2\omega x\cdot \sin 3\omega xdx$

$⟨{\varphi }_{2j-1},{\varphi }_{2k-1}⟩=\underset{0}{\overset{2\pi }{\int }}\frac{1}{2}\cos (2\omega x-3\omega x)-\frac{1}{2}\cos (2\omega x+3\omega x)dx$

$⟨{\varphi }_{2j-1},{\varphi }_{2k-1}⟩=\frac{1}{2}\underset{0}{\overset{2\pi }{\int }}\cos (-\omega x)dx-\frac{1}{2}\underset{0}{\overset{2\pi }{\int }}\cos (5\omega x)dx$

$⟨{\varphi }_{2j-1},{\varphi }_{2k-1}⟩=\frac{1}{2}\sin (-\omega x){|}_0^{2\pi }-\frac{1}{2}\sin (5\omega x){|}_0^{2\pi }$

$⟨{\varphi }_{2j-1},{\varphi }_{2k-1}⟩=(0-0)-(0-0)$

                      $⟨{\varphi }_{2j-1},{\varphi }_{2k-1}⟩=0$

Part 2

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

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

>> Z = trapz(X,Y)

Z =

 2.9490e-017