Question

Use the inner product
=f(−3)g(−3)+f(0)g(0)+f(3)g(3) in P2 to find the orthogonal projection of f(x)=2x^2+4x+2 onto the line L spanned by g(x)=3x^2−5x+8.

Answer 1

If you were asked to do the same thing with vectors, do you know where you would start?

Answer 2

yep..don't understand the format

Answer 3

and my teacher leaves it to us to teach ourselves

Answer 4

Well, think about it in terms of vectors in the following way. Let
\[<1, 4, 5> \text{ and } <0, 2,0> \]
be two vectors, A and B respectively. If I want the orthogonal projection of A onto B, I first turn B into a unit vector by dividing by the magnitude, so it becomes
\[<0,1,0>\]
Next, I take the inner product of this with the vector A, yielding
\[<0,1,0> \cdot <1,4,5> = 4\]
Finally I multiply this scalar by my old unit vector, yielding an orthogonal projection of
\[<0,1,0> \cdot 4 = <0,4,0>\]

Do you follow this?

Answer 5

yeah i do, any idea on how to apply it to this question?

Answer 6

Yes. Do exactly the same thing. You must abstract away from Euclidean space, but the concepts are the same. You are given the definition of some inner product above. It's not a dot product, like you have with vectors, but it's a perfectly valid inner product. So working through it just like we did with the vectors:

Answer 7

I have two objects, f and g. I want the orthogonal projection of f onto g. First, I make g a unit object by dividing by its magnitude (what do we mean by magnitude? Simply the object divided by the square root of its inner product with itself). Next, I take the inner product of this object with my other object f. Finally, I multiply this scalar by the unit object g.

Answer 8

Ill work the problem out on here, and let me know if you have a problem with it. ill show you what iv done. my hw is online so it gives instant feedback

Answer 9

if you dont mind

Answer 10

((<8,2,32><50,8,20>)/sqrt(2964))dot <50,8,20>

Answer 11

thats essentially what i did after just plugging in f(-3),f(0),f(3),etc

Answer 12

i come up with 19.3966...which is incorrect

Answer 13

Sure, I'll work it out here anyway. The magnitude of g is <g,g> which is apparently
\[|g| = \sqrt{<g,g>} = \sqrt{g(-3)^2 + g(0)^2 + g(3)^2} = \sqrt{50^2 + 8^2 + 20^2}\approx 54.44\]
so I define the "unit function"
\[g_u = \frac{g}{|g|} \]
and take the inner product of this with f. That is,
\[<f,g_u> = \frac{f(-3)g(-3) + f(0)g(0) + f(3)g(3)}{|g|}= \frac{400 + 16 + 640}{54.44} \approx 19.4 \]
Finally, I multiply this scalar by my unit object to find my projection P:
\[P = 19.4\cdot g_u = 58.19x^2 - 96.99 x + 155.18\]

Answer 14

Oops. I meant to say "the magnitude of g is sqrt(<g,g>)... " the calculation is right though. Unless I made a typo :)

Answer 15

nope pellet. hahah I did

Answer 16

Just a sec

Answer 17

on my web homework they are looking for a single magnitude....lol thanks for the explanation. i might have to ask my teacher about it, not seeing what i'm doing incorrect