University of Florida/Egm4313/s12.team8.dupre/R7.3

Problem Statement

Find:

(a) The scalar product $<f,g>.$

(b) The magnitude of $f$ and $g$.

(c) The angle between $f$ and $g$.

For:

(1) $f(x)=cos(x),g(x)=x$ for $-2\le x\le 10$.

(2) $f(x)=\frac{1}{2}(3x^2-1),g(x)=\frac{1}{2}(5x^2-3x)$ for $-1\le x\le 1$.

Solution (1)

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

Plugging in our given $f$ and $g$ functions:

$\underset{-2}{\overset{10}{\int }}xcos(x)dx$ (7.3.1)

Integrating (7.3.1) by parts where:

$du=cos(x),v=x,u=sin(x),dv=dx$

We find:

$\int vdu=vu-\int udv$

$\int xcosx=xsinx-\int sinxdx$

This leads us to the solution:

$<f,g>=xsinx+cosx+C(7.3.2)$ where C=0.

(b) We know that the magnitude of a function is obtained as follows:

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

Plugging our given values in and using trigonometric identities:

$\parallel f(x)\parallel =[\underset{-2}{\overset{10}{\int }}co{s}^2(x)dx{]}^{1/2}=[\underset{-2}{\overset{10}{\int }}(\frac{1}{2}(cos(2x))+\frac{1}{2})dx{]}^{1/2}$ (7.3.3)

Using u substitution on (7.3.3) to integrate, where:

$u=2x,du=2dx$

We find:

$\frac{1}{4}\underset{-2}{\overset{10}{\int }}cos(u)du=\underset{-2}{\overset{10}{\int }}\frac{1}{2}dx$

This leads to:

$[\frac{1}{2}(sinxcosx+x)\underset{-2}{\overset{10}{\mid }}{]}^{1/2}$

And allows us to solve for our answer:

$\parallel f(x)\parallel =2.47$

Doing the same process (without needing to integrate by parts) for our given g function:

$\parallel g(x)\parallel =[\underset{-2}{\overset{10}{\int }}{x}^{2}dx{]}^{1/2}$

$\parallel g(x)\parallel =[\frac{x^3}{3}\underset{-2}{\overset{10}{\mid }}]$

Giving us our final magnitude of:

$\parallel g(x)\parallel =18.33$

(c) We know that:

$cos(\theta )=\frac{<f,g>}{\parallel f\parallel \parallel g\parallel }(7.3.4)$

Plugging in solved values shows:

$cos(\theta )=\frac{-7.68}{(2.478)(18.33)}$

Leading us to:

$\theta =1.74radians$ 

Solution (2)

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

Plugging in our given $f$ and $g$ functions:

$\frac{1}{2}\underset{-1}{\overset{1}{\int }}(3x^2-1)(5x^3-3x)dx$ (7.3.5)

Integrating (7.3.5), we find:

$\frac{1}{2}[\frac{15}{6}{x}^6-\frac{14}{4}{x}^4+\frac{3}{2}{x}^2\underset{-1}{\overset{1}{\mid }}]$

This leads us to the solution:

$<f,g>=0$

(b) We know that the magnitude of a function is obtained as follows:

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

Plugging our given values in and using trigonometric identities:

$\parallel f(x)\parallel =[\underset{-1}{\overset{1}{\int }}(\frac{3}{2}{x}^2-\frac{1}{2}{)}^{2}dx{]}^{1/2}=[\frac{9}{20}{x}^5-\frac{x^3}{2}+\frac{x}{4}\underset{-1}{\overset{1}{\mid }}{]}^{1/2}$ (7.3.7)

This allows us to solve for our answer:

$\parallel f\parallel =.632456$

We now obtain the magnitude of the g function in the same way, as follows:

$\parallel g(x)\parallel =[\underset{-1}{\overset{1}{\int }}(\frac{5}{2}{x}^3-\frac{3}{2}x{)}^{2}dx{]}^{1/2}=[\frac{25}{28}{x}^7-\frac{3}{2}{x}^5+\frac{3}{4}{x}^3]\underset{-1}{\overset{1}{\mid }}{]}^{1/2}$

Which leads up to our final magnitude:

$\parallel g(x)\parallel =.5345$  

(c) We know that:

$cos(\theta )=\frac{<f,g>}{\parallel f\parallel \parallel g\parallel }(7.3.4)$

Plugging in (7.3.8) and (7.3.9) shows:

$cos(\theta )=\frac{0}{(.632456)(.5345)}$

Leading us to:

$\theta =\frac{\pi }{2}$ radians