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

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 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

                         $\theta =90$