Question

Decide whether the following product is an inner product on P_2... Thanks!

Answer 1

well it is of higher level maths . is it ?

Answer 2

\[<a_o+a_1x+a_2x^2, b_o+b_1x+b_2x^2> = a_ob_o+a_1b_1+a_2b_2>\]

Answer 3

oh good one feldy90

Answer 4

ignore the last >**

Answer 5

ok but what about the first

Answer 6

Sorry? I just meant the last > after a_2b_2 was a typo - the question is still the same :) can you help? :)

Answer 7

oh is this it?

Answer 8

Yep! :D

Answer 9

hmm...im a little rusty on this. I believe there are 4 axioms you must check to see if an operation qualifies as an Inner Product.

Answer 10

That would make sense - it's a question estimated to be worth 4 marks haha

Answer 11

Lets call the inner product of f and g<f,g>.

First, the operation must be symmetric. In ohter words, <f,g> = <g,f>

So lets check this. \[f= a_0+a_1x+a_2x^2, g = b_0+b_1x+b_2x^2\]\[<f,g> = a_0b_0+a_1b_1+a_2b_2 = b_0a_0+b_1a_1+b_2a_2 = <g,f>\]

Answer 12

So its symmetric, woot. Now the second axiom is that if f, g, and h are polynomials in the space, then:
<f+g,h> = <f,h>+<g,h>

Answer 13

So lets see:\[f = a_0+a_1x+a_2x^2, g = b_0+b_1x+b_2x^2, h = c_0+c_1x+c_2x^2\]\[f+g = (a_0+b_0)+(a_1+b_1)x+(a_2+b_2)x^2\]
\[<f+g,h> = (a_0+b_0)c_0+(a_1+b_1)c_1+(a_2+b_2)c_2\]\[= a_0c_0+b_0c_0+a_1c_1+b_1c_1+a_2c_2+b_2c_2\]\[=(a_0c_0+a_1c_1+a_2c_2)+(b_0c_0+b_1c_1+b_2c_2)\]\[<f,h>+<g,h>\]

Answer 14

so we are good on that one too. Two more axioms to go lol >.<

Answer 15

Wow this is way longer than I expected!!

Answer 16

The next one is if f and g are polynomials, and c is a scalar, then <cf,g> = c<f,g>

So lets test and see:\[f = a_0+a_1x+a_2x^2 \iff cf = ca_0+ca_1x+ca_2x^2\]\[g = b_0+b_1x+b_2x^2\]\[<cf,g> = ca_0b_0+ca_1b_1+ca_2b_2 = c(a_0b_0+a_1b_1+a_2b_2) \]\[ = c<f,g>\]

Answer 17

wow joe!!!

Answer 18

if you are writing it on paper, it doesnt take too long, its typing it out thats making it seem long lol

Answer 19

and hi akshay!

Answer 20

hey joe! you got my invite?

Answer 21

no i didnt o.O send again?

Answer 22

well i will give you a link now sign up there.. ok?

Answer 23

Still, I appreciate it so much!

The open study slogan of "Be a hero to your peers" doesn't even begin to cut it, Joe! Haha

Answer 24

The last axiom is pretty easy to check. its that:
\[<f,f> \space \geq 0, <f,f> = 0 \iff f = 0\]

Answer 25

im on the page, you can delete the stuff here.

Answer 26

and sorry feldy to interrupt!! extremely sorry

Answer 27

Thats OK :) thanks guys!!

Answer 28

so lets check:\[f = a_0+a_1x+a_2x^2\]\[<f,f> = a_0^2+a_1^2+a_2^2\]
This will always be positive because of the squares, and likewise, will only be equal to 0 if each a was equal to 0.
So that checks the fourth axiom, and because all 4 axioms check out, it qualifies as an Inner Product.