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

R7.3

Find:

(a) The scalar product ${\displaystyle \displaystyle <f,g>.}$

(b) The magnitude of ${\displaystyle \displaystyle f}$ and ${\displaystyle \displaystyle g}$.

(c) The angle between ${\displaystyle \displaystyle f}$ and ${\displaystyle \displaystyle g}$.

For:

(1) ${\displaystyle \displaystyle f(x)=cos(x),g(x)=x}$ for ${\displaystyle \displaystyle -2\leq x\leq 10}$.

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

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

Plugging in our given ${\displaystyle \displaystyle f}$ and ${\displaystyle \displaystyle g}$ functions:

${\displaystyle \displaystyle \int _{-2}^{10}xcos(x)dx}$ (7.3.1)

Integrating (7.3.1) by parts where:

${\displaystyle \displaystyle du=cos(x),v=x,u=sin(x),dv=dx}$

We find:

${\displaystyle \displaystyle \int vdu=vu-\int udv}$

${\displaystyle \displaystyle \int xcosx=xsinx-\int sinxdx}$

This leads us to the solution:

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

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

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

Plugging our given values in and using trigonometric identities:

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

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

${\displaystyle \displaystyle u=2x,du=2dx}$

We find:

${\displaystyle \displaystyle {\frac {1}{4}}\int _{-2}^{10}cos(u)du=\int _{-2}^{10}{\frac {1}{2}}dx}$

This leads to:

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

And allows us to solve for our answer:

${\displaystyle \displaystyle \parallel f(x)\parallel =2.47}$

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

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

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

Giving us our final magnitude of:

${\displaystyle \displaystyle \parallel g(x)\parallel =18.33}$

(c) We know that:

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

Plugging in solved values shows:

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

Leading us to:

${\displaystyle \displaystyle \theta =1.74radians}$ 

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

Plugging in our given ${\displaystyle \displaystyle f}$ and ${\displaystyle \displaystyle g}$ functions:

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

Integrating (7.3.5), we find:

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

This leads us to the solution:

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

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

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

Plugging our given values in and using trigonometric identities:

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

This allows us to solve for our answer:

${\displaystyle \displaystyle \parallel f\parallel =.632456}$

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

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

Which leads up to our final magnitude:

${\displaystyle \displaystyle \parallel g(x)\parallel =.5345}$  

(c) We know that:

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

Plugging in (7.3.8) and (7.3.9) shows:

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

Leading us to:

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