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

well it is of higher level maths . is it ?

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

oh good one feldy90

ignore the last >**

ok but what about the first

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

oh is this it?

Yep! :D

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.

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

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

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>

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

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

Wow this is way longer than I expected!!

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

wow joe!!!

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

and hi akshay!

hey joe! you got my invite?

no i didnt o.O send again?

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

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

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

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

and sorry feldy to interrupt!! extremely sorry

Thats OK :) thanks guys!!

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.

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