Find (a) the scalar product, (b) the magnitude of f {\displaystyle f\!} and g {\displaystyle g\!} ,(c) the angle between f {\displaystyle f\!} and g {\displaystyle g\!} for:
1) f ( x ) = c o s ( x ) , g ( x ) = x f o r − 2 ≤ x ≤ 10 {\displaystyle f(x)=cos(x),\ g(x)=x\ for-2\leq x\leq 10\!}
2) f ( x ) = 1 2 ( 3 x 2 − 1 ) , g ( x ) = 1 2 ( 5 x 3 − 3 x ) f o r − 1 ≤ x ≤ 1 {\displaystyle f(x)={\frac {1}{2}}(3x^{2}-1),\ g(x)={\frac {1}{2}}(5x^{3}-3x)\ for-1\leq x\leq 1\!}
solved by Luca Imponenti
< f , g >= ∫ a b f ( x ) g ( x ) d x {\displaystyle <f,g>=\int _{a}^{b}f(x)g(x)\ dx\!}
< f , g >= ∫ − 1 1 [ 1 2 ( 3 x 2 − 1 ) ] [ 1 2 ( 5 x 3 − 3 x ) ] d x {\displaystyle <f,g>=\int _{-1}^{1}[{\frac {1}{2}}(3x^{2}-1)][{\frac {1}{2}}(5x^{3}-3x)]\ dx\!}
Since all exponents are even, everything in brackets cancels out
< f , g >= 0 {\displaystyle <f,g>=0\!}
‖ f ‖ =< f , f > 1 / 2 = ∫ a b f 2 ( x ) d x {\displaystyle \|f\|=<f,f>^{1/2}=\int _{a}^{b}f^{2}(x)\ dx\!}
‖ f ‖ = 2 5 {\displaystyle \|f\|={\frac {2}{5}}\!}
‖ g ‖ = ∫ a b g 2 ( x ) d x {\displaystyle \|g\|=\int _{a}^{b}g^{2}(x)\ dx\!}
‖ g ‖ = 2 7 {\displaystyle \|g\|={\frac {2}{7}}\!}
c o s ( θ ) = < f , g > ‖ f ‖ ‖ g ‖ {\displaystyle cos(\theta )={\frac {<f,g>}{\|f\|\|g\|}}\!}
Since < f , g >= 0 {\displaystyle <f,g>=0\!} the two functions are orthogonal
θ = 90 {\displaystyle \theta =90\!}
and 
,(c) the angle between and for:
![{\displaystyle <f,g>=\int _{-1}^{1}[{\frac {1}{2}}(3x^{2}-1)][{\frac {1}{2}}(5x^{3}-3x)]\ dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dfc6e327a734ba8f03bbf1230e50e5e98223f28f)
![{\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})]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53cff8357d3b5f93f2fbaae853728c278b35d84e)
![{\displaystyle =\int _{-1}^{1}[{\frac {1}{2}}(3x^{2}-1)]^{2}\ dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96ccb23d62629ff7917a2a6e03e2a6750c6a1cb2)
![{\displaystyle ={\frac {1}{4}}[({\frac {9}{5}}1^{5}-2(1)^{3}+1)-({\frac {9}{5}}(-1)^{5}-2(-1)^{3}+(-1))]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a67a46b3168d49f86e6b9d5193940336b666e72)
![{\displaystyle ={\frac {1}{4}}[{\frac {4}{5}}-(-{\frac {4}{5}})]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c25eee2b658f4822793325fe55dbe628ec494f4e)
![{\displaystyle =\int _{-1}^{1}[{\frac {1}{2}}(5x^{3}-3x)]^{2}\ dx\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7dab4c84a1b03ac40fba4b50302dfb4d1a45262)
![{\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})]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4531b4d40107736928a68a1c13e726c81505e0f)
![{\displaystyle ={\frac {1}{4}}[{\frac {4}{7}}-(-{\frac {4}{7}})]\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3dc29b7811deeaa641b8e5ecbb01cb934edd33a2)
the two functions are orthogonal
