University of Florida/Egm4313/s12.team11.R7

Scalar Product

${\displaystyle <f,g>=\int _{a}^{b}f(x)g(x)\ dx\!}$

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

Using integration by parts;

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

                      ${\displaystyle <f,g>=-7.68\!}$

Magnitude

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

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

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

                       ${\displaystyle \|f\|=2.457\!}$

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

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

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

Angle Between Functions

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

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

                      ${\displaystyle \theta =89.47}$

                      The two functions are nearly orthogonal.

Scalar Product

${\displaystyle <f,g>=\int _{a}^{b}f(x)g(x)\ dx\!}$

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

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

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

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

Since all exponents are even, everything in brackets cancels out

                      ${\displaystyle <f,g>=0\!}$

Magnitude

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Angle Between Functions

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

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

                         ${\displaystyle \theta =90\!}$

Problem Statement

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

${\displaystyle f(x)=\left|x\right|\!}$

Solution

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

Problem Statement

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

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

Solution

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

Solved by Jonathan Sheider

Problem Statement

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

Given:

${\displaystyle f(x)=|x|\!}$

Solution

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

${\displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }(a_{n}cos(nx)+b_{n}sin(nx))\!}$

Where:

${\displaystyle a_{0}={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x)dx\!}$
${\displaystyle a_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)cos(nx)dx\!}$
${\displaystyle b_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)sin(nx)dx\!}$

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

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

Calculating the first term ${\displaystyle a_{0}\!}$:

${\displaystyle a_{0}={\frac {1}{2\pi }}\left(\int _{-\pi }^{0}-xdx+\int _{0}^{\pi }xdx\right)\!}$
${\displaystyle a_{0}={\frac {1}{2\pi }}\left(\left[{\frac {-x^{2}}{2}}\right]_{-\pi }^{0}+\left[{\frac {x^{2}}{2}}\right]_{0}^{\pi }\right)\!}$
${\displaystyle a_{0}={\frac {1}{2\pi }}\left(-\left({\frac {-\pi ^{2}}{2}}\right)+\left({\frac {\pi ^{2}}{2}}\right)\right)\!}$
${\displaystyle a_{0}={\frac {1}{2\pi }}(\pi ^{2})\!}$

                                                         ${\displaystyle a_{0}={\frac {\pi }{2}}\!}$

Calculating the coefficient ${\displaystyle a_{n}\!}$:

${\displaystyle a_{n}={\frac {1}{\pi }}\left(\int _{-\pi }^{0}-xcos(nx)dx+\int _{0}^{\pi }xcos(nx)dx\right)\!}$
${\displaystyle a_{n}={\frac {1}{\pi }}\left(-\int _{-\pi }^{0}xcos(nx)dx+\int _{0}^{\pi }xcos(nx)dx\right)\!}$

Using integration by parts with the following substitutions:

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

This yields for the integral:

${\displaystyle a_{n}={\frac {1}{\pi }}\left(-\left[{\frac {1}{n}}xsin(nx)-\int {\frac {1}{n}}sin(nx)dx\right]_{-\pi }^{0}+\left[{\frac {1}{n}}xsin(nx)-\int {\frac {1}{n}}sin(nx)dx\right]_{0}^{\pi }\right)\!}$
${\displaystyle a_{n}={\frac {1}{\pi }}\left(-\left[{\frac {1}{n}}xsin(nx)+{\frac {1}{n^{2}}}cos(nx)\right]_{-\pi }^{0}+\left[{\frac {1}{n}}xsin(nx)+{\frac {1}{n^{2}}}cos(nx)\right]_{0}^{\pi }\right)\!}$
${\displaystyle a_{n}={\frac {1}{\pi }}\left(-\left[{\frac {1}{n^{2}}}-\left({\frac {1}{n}}(-\pi )sin(-n\pi )+{\frac {1}{n^{2}}}cos(-n\pi )\right)\right]+\left[{\frac {1}{n}}\pi sin(n\pi )+{\frac {1}{n^{2}}}cos(n\pi )-{\frac {1}{n^{2}}}\right]\right)\!}$

Note that for all n = 1,2,3... : ${\displaystyle sin(n\pi )=0\!}$ as ${\displaystyle sin(\pi )=sin(2\pi )=sin(3\pi )...=0\!}$ therefore these terms are evaluated as zero, which yields:

${\displaystyle a_{n}={\frac {1}{\pi }}\left(-\left[{\frac {1}{n^{2}}}-\left(0+{\frac {1}{n^{2}}}cos(-n\pi )\right)\right]+\left[0+{\frac {1}{n^{2}}}cos(n\pi )-{\frac {1}{n^{2}}}\right]\right)\!}$
${\displaystyle a_{n}={\frac {1}{\pi }}\left(-{\frac {1}{n^{2}}}+{\frac {1}{n^{2}}}cos(-n\pi )+{\frac {1}{n^{2}}}cos(n\pi )-{\frac {1}{n^{2}}}\right)\!}$

Note that ${\displaystyle cos(-x)=cos(x)\!}$ therefore:

${\displaystyle a_{n}={\frac {1}{\pi }}\left(-{\frac {1}{n^{2}}}+{\frac {1}{n^{2}}}cos(n\pi )+{\frac {1}{n^{2}}}cos(n\pi )-{\frac {1}{n^{2}}}\right)\!}$
${\displaystyle a_{n}={\frac {1}{\pi }}\left(-{\frac {2}{n^{2}}}+{\frac {2}{n^{2}}}cos(n\pi )\right)\!}$
${\displaystyle a_{n}={\frac {1}{\pi }}\left(\left(-{\frac {2}{n^{2}}}\right)(1-cos(n\pi )\right)\!}$
${\displaystyle a_{n}=-{\frac {2}{n^{2}\pi }}\left(1-cos(n\pi )\right)\!}$

To evaluate the term ${\displaystyle (1-cos(n\pi ))\!}$:

Note that ${\displaystyle cos(n\pi )=-1\!}$ for odd n values as ${\displaystyle cos(\pi )=cos(3\pi )=cos(5\pi )...=-1\!}$

And that ${\displaystyle cos(n\pi )=1\!}$ for even n values as ${\displaystyle cos(2\pi )=cos(4\pi )=cos(6\pi )...=1\!}$.

Therefore, it can be concluded that for odd n values:

${\displaystyle (1-cos(n\pi ))=1-(-1)=2\!}$

And for even n values:

${\displaystyle (1-cos(n\pi )=1-(1)=0\!}$

Therefore, for the coefficient ${\displaystyle a_{n}\!}$, all even terms will equal zero while all odd terms will have a multiplier of 2, this yields (for all odd n values):

                                            ${\displaystyle a_{n}=-{\frac {4}{n^{2}\pi }}{\text{  for  }}n=1,3,5...\!}$

Calculating the coefficient ${\displaystyle b_{n}\!}$:

${\displaystyle b_{n}={\frac {1}{\pi }}\left(\int _{-\pi }^{0}-xsin(nx)dx+\int _{0}^{\pi }xsin(nx)dx\right)\!}$
${\displaystyle b_{n}={\frac {1}{\pi }}\left(-\int _{-\pi }^{0}xsin(nx)dx+\int _{0}^{\pi }xsin(nx)dx\right)\!}$

Using integration by parts with the following substitutions:

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

This yields for the integral:

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

Note that for all n = 1,2,3... : ${\displaystyle sin(n\pi )=0\!}$ as ${\displaystyle sin(\pi )=sin(2\pi )=sin(3\pi )...=0\!}$ therefore these terms are evaluated as zero, which yields:

${\displaystyle b_{n}={\frac {1}{\pi }}\left(-\left[0-\left({\frac {-1}{n}}(-\pi )cos(-n\pi )+0\right)\right]+\left[{\frac {-1}{n}}\pi cos(n\pi )+0\right]\right)\!}$
${\displaystyle b_{n}={\frac {1}{\pi }}\left(\left[{\frac {1}{n}}(\pi )cos(-n\pi )\right]+\left[{\frac {-1}{n}}\pi cos(n\pi )\right]\right)\!}$

Note that ${\displaystyle cos(-x)=cos(x)\!}$ therefore:

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

                                                       ${\displaystyle b_{n}=0\!}$

Therefore there will be no ${\displaystyle sin(nx)\!}$ terms in the Fourier representation. In conclusion, the Fourier series representation for the given function is as follows:

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

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

A graph of the function, and the Fourier series for ${\displaystyle n=1,3,5\!}$ 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 ${\displaystyle 2\pi \!}$. Show the details of your work.
Sketch or graph the partial sums up to that including ${\displaystyle cos(5x)\!}$ and ${\displaystyle sin(5x)\!}$

Given:

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

Solution

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

${\displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }(a_{n}cos(nx)+b_{n}sin(nx))\!}$

Where:

${\displaystyle a_{0}={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x)dx\!}$
${\displaystyle a_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)cos(nx)dx\!}$
${\displaystyle b_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)sin(nx)dx\!}$

Calculating the first term ${\displaystyle a_{0}\!}$:

${\displaystyle a_{0}={\frac {1}{2\pi }}\left(\int _{-\pi }^{0}xdx+\int _{0}^{\pi }(\pi -x)dx\right)\!}$
${\displaystyle a_{0}={\frac {1}{2\pi }}\left(\left[{\frac {x^{2}}{2}}\right]_{-\pi }^{0}+\left[\pi x-{\frac {x^{2}}{2}}\right]_{0}^{\pi }\right)\!}$
${\displaystyle a_{0}={\frac {1}{2\pi }}\left(-\left({\frac {\pi ^{2}}{2}}\right)+\left(\pi ^{2}-{\frac {\pi ^{2}}{2}}\right)\right)\!}$
${\displaystyle a_{0}={\frac {1}{2\pi }}\left(-{\frac {\pi ^{2}}{2}}+\pi ^{2}-{\frac {\pi ^{2}}{2}}\right)\!}$
${\displaystyle a_{0}={\frac {1}{2\pi }}\left(\pi ^{2}-\pi ^{2}\right)\!}$

                                                         ${\displaystyle a_{0}=0\!}$

Calculating the coefficient ${\displaystyle a_{n}\!}$:

${\displaystyle a_{n}={\frac {1}{\pi }}\left(\int _{-\pi }^{0}xcos(nx)dx+\int _{0}^{\pi }(\pi -x)cos(nx)dx\right)\!}$

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

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

Using integration by parts with the following substitutions for the integral ${\displaystyle \int _{0}^{\pi }(\pi -x)cos(nx)dx\!}$:

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

This yields for the overall expression:

${\displaystyle a_{n}={\frac {1}{\pi }}\left(\left[{\frac {1}{n}}xsin(nx)-\int {\frac {1}{n}}sin(nx)dx\right]_{-\pi }^{0}+\left[{\frac {1}{n}}(\pi -x)sin(nx)-\int -{\frac {1}{n}}sin(nx)dx\right]_{0}^{\pi }\right)\!}$
${\displaystyle a_{n}={\frac {1}{\pi }}\left(\left[{\frac {1}{n}}xsin(nx)+{\frac {1}{n^{2}}}cos(nx)\right]_{-\pi }^{0}+\left[{\frac {1}{n}}(\pi -x)sin(nx)-{\frac {1}{n^{2}}}cos(nx)\right]_{0}^{\pi }\right)\!}$
${\displaystyle a_{n}={\frac {1}{\pi }}\left(\left[{\frac {1}{n^{2}}}-\left({\frac {1}{n}}(-\pi )sin(-n\pi )+{\frac {1}{n^{2}}}cos(-n\pi )\right)\right]+\left[-{\frac {1}{n^{2}}}cos(n\pi )+{\frac {1}{n^{2}}}\right]\right)\!}$

Note that for all n = 1,2,3... : ${\displaystyle sin(n\pi )=0\!}$ as ${\displaystyle sin(\pi )=sin(2\pi )=sin(3\pi )...=0\!}$ therefore these terms are evaluated as zero, which yields:

${\displaystyle a_{n}={\frac {1}{\pi }}\left(\left[{\frac {1}{n^{2}}}-0-{\frac {1}{n^{2}}}cos(-n\pi )\right]+\left[-{\frac {1}{n^{2}}}cos(n\pi )+{\frac {1}{n^{2}}}\right]\right)\!}$

${\displaystyle a_{n}={\frac {1}{\pi }}\left({\frac {1}{n^{2}}}-{\frac {1}{n^{2}}}cos(-n\pi )-{\frac {1}{n^{2}}}cos(n\pi )+{\frac {1}{n^{2}}}\right)\!}$

Note that ${\displaystyle cos(-x)=cos(x)\!}$ therefore:

${\displaystyle a_{n}={\frac {1}{\pi }}\left({\frac {1}{n^{2}}}-{\frac {1}{n^{2}}}cos(n\pi )-{\frac {1}{n^{2}}}cos(n\pi )+{\frac {1}{n^{2}}}\right)\!}$

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

To evaluate the term ${\displaystyle (1-cos(n\pi ))\!}$:

Note that ${\displaystyle cos(n\pi )=-1\!}$ for odd n values as ${\displaystyle cos(\pi )=cos(3\pi )=cos(5\pi )...=-1\!}$

And that ${\displaystyle cos(n\pi )=1\!}$ for even n values as ${\displaystyle cos(2\pi )=cos(4\pi )=cos(6\pi )...=1\!}$.

Therefore, it can be concluded that for odd n values:

${\displaystyle (1-cos(n\pi ))=1-(-1)=2\!}$

And for even n values:

${\displaystyle (1-cos(n\pi ))=1-(1)=0\!}$

Therefore, for the coefficient ${\displaystyle a_{n}\!}$, all even terms will equal zero while all odd terms will have a multiplier of 2, this yields (for all odd n values):

                                            ${\displaystyle a_{n}={\frac {4}{n^{2}\pi }}{\text{  for  }}n=1,3,5...\!}$

Calculating the coefficient ${\displaystyle b_{n}\!}$:

${\displaystyle b_{n}={\frac {1}{\pi }}\left(\int _{-\pi }^{0}xsin(nx)dx+\int _{0}^{\pi }(\pi -x)sin(nx)dx\right)\!}$

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

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

Using integration by parts with the following substitutions for the integral ${\displaystyle \int _{0}^{\pi }(\pi -x)sin(nx)dx\!}$:

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

This yields for the overall expression:

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

Note that for all n = 1,2,3... : ${\displaystyle sin(n\pi )=0\!}$ as ${\displaystyle sin(\pi )=sin(2\pi )=sin(3\pi )...=0\!}$ therefore these terms are evaluated as zero, which yields:

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

Note that ${\displaystyle cos(-x)=cos(x)\!}$ therefore:

${\displaystyle b_{n}={\frac {1}{\pi }}\left({\frac {1}{n}}\pi -{\frac {1}{n}}\pi cos(n\pi )\right)\!}$
${\displaystyle b_{n}={\frac {1}{\pi }}\left(\left({\frac {\pi }{n}}\right)(1-cos(n\pi ))\right)\!}$
${\displaystyle b_{n}={\frac {1}{n}}\left(1-cos(n\pi )\right)\!}$

To evaluate the term ${\displaystyle (1-cos(n\pi ))\!}$:

Note that ${\displaystyle cos(n\pi )=-1\!}$ for odd n values as ${\displaystyle cos(\pi )=cos(3\pi )=cos(5\pi )...=-1\!}$

And that ${\displaystyle cos(n\pi )=1\!}$ for even n values as ${\displaystyle cos(2\pi )=cos(4\pi )=cos(6\pi )...=1\!}$.

Therefore, it can be concluded that for odd n values:

${\displaystyle (1-cos(n\pi ))=1-(-1)=2\!}$

And for even n values:

${\displaystyle (1-cos(n\pi ))=1-(1)=0\!}$

Therefore, for the coefficient ${\displaystyle b_{n}\!}$, all even terms will equal zero while all odd terms will have a multiplier of 2, this yields (for all odd n values):