Question

\[=\int\limits_{-1}^{1}f(x)g(x)dx\] \[f(x)=x\] \[g(x)=x ^{2}-x+2\] find a. b. ||f|| c. ||g|| d. d(f,g)

Answer 1

where are you stuck?

Answer 2

Actually, I think I'm on a line it solving it now. I'll put my results in in a few minutes if you wouldn't mind verifying it for me.

Answer 3

ok

Answer 4

Here's my a and b... if I have those right, I assume I've got this figured out...

\[a. \int\limits_{-1}^{1}-x ^{3}+x ^{2}-2x\]
\[b. \frac{2}{3}\]

Answer 5

I don't get that

Answer 6

\(<f,g>=-\frac{2}{3}\)

Answer 7

\[\|f\|=\frac{\sqrt{6}}{3}\]

Answer 8

\[<f,g>=\int\limits_{-1}^{1}x\times(x^2-x+2)dx\]

Answer 9

so for a. <f,g> since it's just defined as integral(f(x)(g(x))dx don't I just multiply my two formulas to get my integrand?

Answer 10

\[\|f\|=\sqrt{<f,f>}=\sqrt{\int\limits_{-1}^{1}x\times x\,dx}\]

Answer 11

except when I integrated that, I got 2/3 not -2/3

Answer 12

\[<f,g>=\int\limits_{-1}^{1}x\times(x^2-x+2)dx=\int\limits_{-1}^{1}(x^3-x^2+2x)dx\]

Answer 13

oh... I see had a typo... f(x) is -x not x... my bad

Answer 14

oh...ok

Answer 15

so I get 2/3 for a, now with the typo corrected, is my answer for b correct?

Answer 16

no...my answer is still valid since -x*-x=x*x=x^2

Answer 17

yeah, I gotta recalculate b... I left out entire steps that I see now...
lemme see if I can get what you got.

Answer 18

\[\|f\|=\sqrt{<f,f>}=\sqrt{\int\limits_{-1}^{1}(-x)\times (-x)\,dx}\]

\[\sqrt{\int\limits_{-1}^{1}x^2\,dx}=\sqrt{\left.\frac{x^3}{3}\right|_{-1}^1}=\sqrt{\frac{1}{3}+\frac{1}{3}}=\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}\]

Answer 19

ok... that's what I get now too. thanks!

Answer 20

good

Answer 21

I need to go...but here is what I got for c and d

c)\[\frac{4\sqrt{165}}{15}\]

d)\[\frac{\sqrt{2490}}{15}\]

Answer 22

great. thanks again!