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

R7.3

Problem Statement

Find:

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

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

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

For:

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

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

Solution (1)

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

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

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

Integrating (7.3.1) by parts where:

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

We find:

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

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

This leads us to the solution:

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

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

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

Plugging our given values in and using trigonometric identities:

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

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

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

We find:

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

This leads to:

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

And allows us to solve for our answer:

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

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

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

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

Giving us our final magnitude of:

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

(c) We know that:

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

Plugging in solved values shows:

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

Leading us to:

$\displaystyle \theta =1.74radians$ 

Solution (2)

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

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

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

Integrating (7.3.5), we find:

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

This leads us to the solution:

$\displaystyle <f,g>=0$

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

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

Plugging our given values in and using trigonometric identities:

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

This allows us to solve for our answer:

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

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

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

Which leads up to our final magnitude:

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

(c) We know that:

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

Plugging in (7.3.8) and (7.3.9) shows:

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

Leading us to:

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