Question

How do I verify an inner product?

Answer 1

What do you mean?

Answer 2

You need to check if it satisfies the inner product axioms:
1) conjugate symmetry
2) linearity in the first argument
3) positive-definiteness

Answer 3

here:

Answer 4

V is a vector space of polynomials of degree less than or equal to 3.

Answer 5

You need to show that it satisfies each of those axioms.

Answer 6

I never heard of any of those..

Answer 7

The inner product is defined by those axioms.

Answer 8

1)\[\langle x, y \rangle =\overline{\langle y, x \rangle}\]
2)\[\langle x + c, y \rangle = \langle x , y \rangle + \langle c, y \rangle\]\[\langle ax, y \rangle = a\langle x , y \rangle \]
3)\[\langle x, x \rangle \geq 0\] and 0 only when x=0

Answer 9

what does <y,x> with a line on top mean?

Answer 10

conjugate. if your course doesn't deal with the complex field just ignore it

Answer 11

oh we don't deal with complex numbers.

Answer 12

anyways show that the inner product satisfies the axioms to "verify" it

Answer 13

okay thanks ill just go over some problems and attempt this.

Answer 14

Tell me what the problem is.

Answer 15

okay hold on..

Answer 16

Here.. part C

Answer 17

Yes so have you tried showing that it satisfies each of the axioms?

Answer 18

<x+c,y> .. but what are my xs and cs and ys?

Answer 19

Its about showing linearity. Have you ever been asked to prove a transformation is linear?

Answer 20

yes.

Answer 21

Its the same concept.

Answer 22

okay let me try..

Answer 23

\[\langle f(x), g(x) \rangle = \int_{\pi}^{\pi}f(x)g(x) = \int_{\pi}^{\pi}g(x)f(x) = \langle g(x)f(x) \rangle\]
\[\langle f(x) + h(x), g(x) \rangle\]\[= \int_{\pi}^{\pi}(f(x) + h(x))g(x) = \int_{\pi}^{\pi}f(x)g(x) + \int_{\pi}^{\pi}h(x)g(x)
= \langle g(x)f(x) \rangle + \langle h(x)f(x) \rangle\]

Answer 24

alchemista i was doing it this way.. hold on ima show u.. dunno if that's the same thing..

Answer 25

err sorry small mistake

Answer 26

\[ = \langle f(x), g(x) \rangle + \langle h(x), g(x) \rangle\]

Answer 27

here is this the same thing?

Answer 28

yeah

Answer 29

okay thanks and then I do basically alpha(f(x)) = f(alphax)

Answer 30

also prove conjugate symmetry and positive definiteness.

basically show that the integral of f(x)^2 is positive and 0 only when you have the 0 function which is easy.

Answer 31

that's for the positive definiteness part

Answer 32

Hey Alchemista i'll post the continuation as a new thread i was working on the 2nd part of the proof..