Question

I have a note from my prof, however, I don't understand what he was talking about. Let me copy it V = P_n define (x,y)= ∫bax(t)y¯(t)dt an inner product on P_n Let f in P'_n given by [x,f]= x(t0) in Complex By Riesz Representation Theorem, there exists unique P in P_n such that ∫bax(t)P¯(t)dt=x(t0) for every x in P_n P depends on n, a, b, t0 end note.

Answer 1

Can you please explain me and how to apply to the problem Convert P2 ont0 an inner product space by writing (x,y) =\(\int_0^1x(t)\bar y(t)dt\) whenever x, y are in P2 and find a complete orthonormal set in that space

Answer 2

@KingGeorge or I apply Gram-schmidt?
if so, Let say the required basis is {u1,u2,u3}
the basis of P2 = {1,t,t^2}
u1= \(\dfrac{1}{\sqrt{2}}\)
u'2 = t -<t, u1>u1 = t-\(\int_0^1 t*\frac{1}{\sqrt{2}}dt *\dfrac{1}{\sqrt{2}}\) = \(t-\dfrac{t^2}{4}\)
so, u2=\(\dfrac{u'2}{<u'2, u'2>}\) =\(\dfrac{(4t-t^2)/4}{\int_0^1(t-t^2/4)^2dt}\)= \(\dfrac{60}{43}(4t-t^2)\)
I need check, please

Answer 3

I'm not sure that's the correct process for gram-schmidt. If I were to do it for the first two vectors \(u_1,u_2\), I would just set\[u_1=1\]and then\[u_2=t-\frac{\langle1,t\rangle}{\langle1,1\rangle}\cdot1\]

Answer 4

Hold on, friend, the inner product is defined by (x,y) = \(\int_0^1 x(t)y(t) dt\)
so that, ||u1|| is not 1

Answer 5

Why can't \(u_1=1\)? And we can modify the magnitude of the vectors later anyways.

A couple quick computations reveals\[\langle1,t\rangle=\int_0^1t\,dt=\frac{1}{2}\]and\[\langle1,1\rangle=\int_0^11\,dt=1.\]So\[u_2=t-\frac{1}{2}.\]

Answer 6

oh yes, you are right. I am sorry, I confused. :)

Answer 7

let me redo and ask for check. hihihi... I am sorry for my slowness.

Answer 8

No worries.

Answer 9

u1 =1
u'2 = t - <t,1>1= t- 1/2
--> u2 = \(\dfrac{u'2}{||u2||}\) = \(\dfrac{t-1/2}{\int_0^1(t-1/2)^2dt}\)= 12t -6

Answer 10

*||u'2|| at denominator

Answer 11

That seems good to me.

Answer 12

and u'3 = t^2 -<t^2,t>t - <t^2,1>1, right?

Answer 13

u'3 = t^2 -7/12
u3 = \(\dfrac{u'3}{||u'3||}\)

Answer 14

You need a couple extra factors in there. It should be\[u_3=t^2-\frac{\langle t,t^2\rangle}{\langle t,t\rangle}t-\frac{\langle 1,t^2\rangle}{\langle 1,1\rangle}1\]

Answer 15

I get\[u_3=t^2-\frac{1/4}{1/3}t-\frac{1}{3}=t^2-\frac{3}{4}t-\frac13\]

Answer 16

but that is orthonomal basis on regular space, not inner product space.

Answer 17

The gram-Schmidt process only works on inner product spaces, so what we get at the end should be an orthonormal basis. Well, we need to modify \(u_3\) a bit by dividing by \(||u_3||\), but after that it should all work.

Answer 18

If I check orthogonal property of this basis by <u1,u3 > =\(\int_0^1(t^2-3/4t-1/3)*1dt = t^3/3-3/8t^2-t/3 \) from 0 to 1 ( I mean replace 1 there) we don't have 0

Answer 19

Hm. I'm stumped as to what's going wrong actually. I guess I'll have to think about it.

Answer 20

me too. I need understanding what is wrong. Thanks for help. If you figure out, please, let me know

Answer 21

Oh. I think I see it. We should have\[u_3=t^2-\frac{\langle 12t-6,t^2\rangle}{\langle 12t-6,12t-6\rangle}(12t-6)-\frac{\langle 1,t^2\rangle}{\langle 1,1\rangle}1\]

Answer 22

Using that instead, we have\[u_3=t^2-\frac{1}{12}(12t-6)-\frac{1}{3}=t^2-t+\frac{1}{6}\]

Answer 23

Normalizing it, we get\[u_3=\frac{t^2-t+\frac16}{\int_0^1(t^2-t+\frac16)^2dt}=180t^2-180t+30.\]

Answer 24

And this vector is actually orthogonal.
http://www.wolframalpha.com/input/?i=int_0%5E1+%28180t%5E2-180t%2B30%29dt
http://www.wolframalpha.com/input/?i=int_0%5E1+%28180t%5E2-180t%2B30%29%2812t-6%29dt

Answer 25

yes!! thank you very much.

Answer 26

You're welcome.