Mathematics 52 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

well it is of higher level maths . is it ?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

oh good one feldy90

OpenStudy (anonymous):

ignore the last >**

OpenStudy (anonymous):

ok but what about the first

OpenStudy (anonymous):

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

OpenStudy (anonymous):

oh is this it?

OpenStudy (anonymous):

Yep! :D

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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>

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Wow this is way longer than I expected!!

OpenStudy (anonymous):

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

OpenStudy (akshay_budhkar):

wow joe!!!

OpenStudy (anonymous):

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

OpenStudy (anonymous):

and hi akshay!

OpenStudy (akshay_budhkar):

hey joe! you got my invite?

OpenStudy (anonymous):

no i didnt o.O send again?

OpenStudy (akshay_budhkar):

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (akshay_budhkar):

and sorry feldy to interrupt!! extremely sorry

OpenStudy (anonymous):

Thats OK :) thanks guys!!

OpenStudy (anonymous):

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.