Question

Prove that f_1 (x)=1,f_2 (x)=x,f_3 (x)=((3x^2-1))/2 are orthogonal over (-1,1)

I know the method, Just need a start
general form for
\(f_m (x) ~~ and~~ f_n(x)\)

Answer 1

\( f_1 (x)=1, \\ f_2 (x)=x,\\ f_3 (x)=((3x^2-1))/2 \)

Answer 2

Isn't this how you do it ? \[\int\limits_{-1}^{1}x.\frac{ 3x ^{2}-1 }{ 2 }=\int\limits_{-1}^{1}(3/2 .x ^{3}-x/2)dx = 0\]

Answer 3

for m\(\ne\) n,
\( ∫_a^b [f_n (x)] [ f_m (x)] dx=0\)

for \(m= n\)
\(∫_a^b [f_n (x)]^2 dx \ne 0\)

this is how i do it

Answer 4

So what's the general form ?

Answer 5

just need to work 3 integrals, right ?

m,n :
1,2
1,3
2,3

Answer 6

how do you put them in matrix @OOOPS ?

Answer 7

need to prove
1,1
2,2
3,3
not =0 also, right ? or not required ?

Answer 8

A true scale plot is attached.

Answer 9

1,3 is not an odd function, i doubt it would be 0 ...

1,2 and 2,3 are odd, directly 0

Answer 10

oh yeah, all 3 are 0

Answer 11

discrete :

\(\large a.b = 0\)

means \(a\) and \(b\) are orthogonal

so i think proving first condition is sufficient, not sure though.

Answer 12

ok, i am quite sure i need to prove 1,1 ...2,2...3,3 not =0 too
will do it

Answer 13

continuous :

\(\large \int a.b = 0 \implies a, b \) are orthogonal

and \(0\) vector is orthogonal to every vector, so im not sure why you're quite sure haha :3

Answer 14

wait, why ?

Answer 15

lol, i didn't get ganeshie's point :P

Answer 16

and 1,1 will be 0
:O

Answer 17

Actually 1,1 can't be =0

Answer 18

oh yes, its not odd function
ok
so last doubt, why its not requied to show 1,1 2,2 3,3 not =0

Answer 19

And you're right 2.2 and 3.3 shouldn't =0 either.