University of Florida/Egm4313/s12.team11.imponenti/R7.3

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

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

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

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

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

Since all exponents are even, everything in brackets cancels out

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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