Question

I don't understand the example from textbook
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Starting with basic {1,x,x^2} , we can use Gram-Schmidt process to generate an orthnormal basis.

||1||^2=<1,1>=3

why is it 3?

Answer 1

Either a typo or they are using a very unusual norm function. Maybe try reading back a few pages?

Answer 2

What is their definition of <1,1>? I would expect it to = 2 using an integral over the interval -1 to 1

Answer 3

more about problem

Find orthonormal basics for P_3 if the inner product on P_3 is defined by

\[\sum _{i=1}^3 p\left(x_i\right)q\left(x_i\right)\]
where x1=-1 , x2=0, x3=1.

Answer 4

@Zarkon , help

Answer 5

sorry, not as familiar with linear algebra and working with orthonormal basis
in other words i don't remember and would have to study up on it

Answer 6

@myininaya , could you help

Answer 7

@Thomas9 , can you help?

Answer 8

I think(?) p and q are one of 1, x, x^2
they define the inner product for 1 as
1*1+1*1+1*1 = 3 (i.e. 1 is not a function)

for p=1 q=x we get
1*(-1) + 1*0+ 1*1 =0

for p=x , q=x we get
-1*-1 +0*0+1*1= 2

Answer 9

with this definition you can apply Gram-Schmidt

Answer 10

I think you get b1= 1/sqrt(3) b2= x/sqrt(2)
now you have to find b3

Answer 11

I don't understand how that innerproduct works:

You have two vectors p and q, and you multiply element wise and take the sum? Just like the regular innerproduct?

Answer 12

If that's the case, isn't {1,x,x^2} already an orthonormal basis?

Answer 13

p(x) is a polynomial that you evaluate at -1, 0, 1

Answer 14

I see, in that case you're correct so far.

Answer 15

So ||1||^2= (1,1) = 1*1+1*1+1*1=3
||1||=sqrt(3)

So first step is to divide the the first vector 1 by sqrt(3).

((1,1) is the innerproduct, not sure if that notation is common)

Answer 16

x is already orthogonal to 1, so that's easy.

||x||=sqrt(2), so you have to divide that vector by sqrt(2)

Answer 17

Thanks, I have much better understading now

Answer 18

Thanks, Phi and Thomas

Answer 19

cool

Answer 20

to finish this, x^2 is orhogonal to x, but not to 1, so you have to add something to that vector and then divide by its length.

Answer 21

that last bit looks ugly

Answer 22

isn't it x^2-(x^2,1)1/|| x^2-(x^2,1)1||

I'm not sure haven't used those formulas for a while.

Answer 23

yes. but evaluate it

Answer 24

(x^2-2)/sqrt(6)?

Answer 25

tho the 1's should be unit length x^2 dot 1/|1| time 1/|1| (if that makes sense)

Answer 26

I see

Answer 27

yes, that is good. I think that has unit length.

Answer 28

but it is not orthogonal to 1 is it?

Answer 29

I don't think so no.

Answer 30

if a=1/sqrt(3)

Then the third vector should be x^2-(x^2,a)a/|| x^2-(x^2,a)a||

Answer 31

but that's the same as x^2-(x^2,1)/3/|| x^2-(x^2,1)/3||

Answer 32

which equals (x^2-2/3)/sqrt(2/3)

I think.

Answer 33

It's orthogonal to both 1 and x and has lenght 1, so that should be it.

Answer 34

i didnt read thru everyone elses post but do u want me to solve this question?