R7.3
Find:
(a) The scalar product ![{\displaystyle \displaystyle <f,g>.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4d62788b222db9d03b6743e73c1c3c4ea618bef2)
(b) The magnitude of
and
.
(c) The angle between
and
.
For:
(1)
for
.
(2)
for
.
(a) We know that ![{\displaystyle \displaystyle <f,g>=\int _{a}^{b}f(x)g(x)dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1fae0f595b3e0489888c976daa917a6630f11909)
Plugging in our given
and
functions:
(7.3.1)
Integrating (7.3.1) by parts where:
![{\displaystyle \displaystyle du=cos(x),v=x,u=sin(x),dv=dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec0f40e5bff4aa9efad92568547712098e60a882)
We find:
![{\displaystyle \displaystyle \int vdu=vu-\int udv}](https://wikimedia.org/api/rest_v1/media/math/render/svg/df649fab0b12c50a1b542923f71e4cdc336703d3)
![{\displaystyle \displaystyle \int xcosx=xsinx-\int sinxdx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a7d70c09bf0001dbab8979fcf9cb588370edc9d)
This leads us to the solution:
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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d4899aa794044bf3a3a32f76daf5ce2261fc214)
Plugging our given values in and using trigonometric identities:
(7.3.3)
Using u substitution on (7.3.3) to integrate, where:
![{\displaystyle \displaystyle u=2x,du=2dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69797ec9d231e2e96fb6c198d1becf72959cc245)
We find:
![{\displaystyle \displaystyle {\frac {1}{4}}\int _{-2}^{10}cos(u)du=\int _{-2}^{10}{\frac {1}{2}}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/55381c545a85ad6e12b3d9d0385175cac7b206a5)
This leads to:
![{\displaystyle \displaystyle [{\frac {1}{2}}(sinxcosx+x)\mid _{-2}^{10}]^{1/2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/da08a0ab29ffca597b29f0b649e6b4daa1825fa1)
And allows us to solve for our answer:
![{\displaystyle \displaystyle \parallel f(x)\parallel =2.47}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b272ca8e1db2e0d78abf528e023e43f442ca4c57)
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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b624a4f5efffdc3b0564303284abf70a3de97f73)
![{\displaystyle \displaystyle \parallel g(x)\parallel =[{\frac {x^{3}}{3}}\mid _{-2}^{10}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c54914378e6f54a1ff41db8fa0dfec72b896384b)
Giving us our final magnitude of:
![{\displaystyle \displaystyle \parallel g(x)\parallel =18.33}](https://wikimedia.org/api/rest_v1/media/math/render/svg/223240070cec6b403cd435babbada7fc69b8b032)
(c) We know that:
![{\displaystyle \displaystyle cos(\theta )={\frac {<f,g>}{\parallel f\parallel \parallel g\parallel }}(7.3.4)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0545a6aa484eda5c9d0d76f34dad5ae8d2540d0)
Plugging in solved values shows:
![{\displaystyle \displaystyle cos(\theta )={\frac {-7.68}{(2.478)(18.33)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c9ad1608a1990beefea8759462586c1e65179cbe)
Leading us to:
(a) We know that ![{\displaystyle \displaystyle <f,g>=\int _{a}^{b}f(x)g(x)dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1fae0f595b3e0489888c976daa917a6630f11909)
Plugging in our given
and
functions:
(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}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b6ea337af3e279088721f88979d31832cdc0e4f2)
This leads us to the solution:
![{\displaystyle \displaystyle <f,g>=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0b18d4ecaa626d04b1f1730e2ad7623a7a6b133)
(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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d4899aa794044bf3a3a32f76daf5ce2261fc214)
Plugging our given values in and using trigonometric identities:
(7.3.7)
This allows us to solve for our answer:
![{\displaystyle \displaystyle \parallel f\parallel =.632456}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe6b558e1bdc7d11967f88043d2acddb7ed9ba4b)
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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d3310be0dd25e6bc99f0d6a59c054242c9fb8e7)
Which leads up to our final magnitude:
(c) We know that:
![{\displaystyle \displaystyle cos(\theta )={\frac {<f,g>}{\parallel f\parallel \parallel g\parallel }}(7.3.4)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0545a6aa484eda5c9d0d76f34dad5ae8d2540d0)
Plugging in (7.3.8) and (7.3.9) shows:
![{\displaystyle \displaystyle cos(\theta )={\frac {0}{(.632456)(.5345)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0adb61c6cfa459c8ab8494926a4c164d5db5ecc)
Leading us to:
radians