University of Florida/Egm4313/s12.team8.dupre/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: $$\displaystyle =xsinx+cosx+C     (7.3.2)$$ where C=0.

(b) We know that the magnitude of a function is obtained as follows: $$\displaystyle \parallel f\parallel ==[\int_{a}^b{f^2(x)dx}]^{1/2} $$ 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: $$\displaystyle \parallel g(x)\parallel =[\int_{-2}^{10}x^2dx]^{1/2} $$ $$\displaystyle \parallel g(x)\parallel =[\frac{x^3}{3}\mid _{-2}^{10}]$$ Giving us our final magnitude of: $$\displaystyle \parallel g(x)\parallel =18.33 $$

(c) We know that: $$\displaystyle cos(\theta )=\frac{}{\parallel f\parallel \parallel g\parallel }    (7.3.4)$$ Plugging in solved values shows: $$\displaystyle cos(\theta )=\frac{-7.68}{(2.478)(18.33)} $$ Leading us to: $$\displaystyle \theta =1.74 radians $$

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: $$\displaystyle =0 $$

(b) We know that the magnitude of a function is obtained as follows: $$\displaystyle \parallel f\parallel ==[\int_{a}^b{f^2(x)dx}]^{1/2} $$ Plugging our given values in and using trigonometric identities: $$\displaystyle \parallel f(x)\parallel =[\int_{-1}^{1}(\frac{3}{2}x^2-\frac{1}{2})^2dx]^{1/2}=[\frac{9}{20}x^5-\frac{x^3}{2}+\frac{x}{4}\mid _{-1}^{1}]^{1/2}$$    (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: $$\displaystyle \parallel g(x)\parallel =[\int_{-1}^{1}(\frac{5}{2}x^3-\frac{3}{2}x)^2dx]^{1/2}=[\frac{25}{28}x^7-\frac{3}{2}x^5+\frac{3}{4}x^3]\mid _{-1}^{1}]^{1/2}$$ Which leads up to our final magnitude: $$\displaystyle \parallel g(x)\parallel = .5345 $$

(c) We know that: $$\displaystyle cos(\theta )=\frac{}{\parallel f\parallel \parallel g\parallel }    (7.3.4)$$ 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