Orthogonal projecitons, minimizing difference

[usn7564]

Homework Statement

Determine the polynomial p of degree at most 1 that minimizes

[tex]\int_0^2 |e^x - p(x)|^2 dx[/tex]

Hint: First find an orthogonal basis for a suitably chosen space of polynomials of degree at most 1

The Attempt at a Solution

I assumed what I wanted was a p(x) of the form
[tex]
p(x) = \frac{<e^x, 1>}{<1,1>} + \frac{<e^x, x>}{<x,x>}x
[/tex]

where the inner product is

[tex]<f, g> = \int_0^2 f(x)\bar{g(x)} dx[/tex]

But this fails just for the first term, ie

[tex]\frac{<e^x, 1>}{<1,1>} = \frac{e^2-1}{2}[/tex] does not coincide with the correct answer


Correct answer:
[tex]p(x) = 3x + \frac{1}{2}(e^2 - 7)[/tex]

 

[Office_Shredder]

Do you understand why the hint says you should find orthogonal polynomials first? And do you know if 1 and x are orthogonal?

 

[usn7564]

Do you understand why the hint says you should find orthogonal polynomials first? And do you know if 1 and x are orthogonal?

For the first question, yeah I believe so. Thinking in 'normal' linear algebra with a vector u in R^3 the best approximation of that vector in any plane in R^3 will be the orthogonal projection of u onto that plane, and you need an orthogonal basis to find that. Applying the same principle here, or that's what I think anyway.
As for the second, err, I automatically assumed so for whatever reason. Checking now I see that's clearly not the case. I suppose I have to tinker a bit to find a basis that's actually an orthogonal set. Perhaps the inner product shouldn't be what it is too.

 

[Office_Shredder]

As for the second, err, I automatically assumed so for whatever reason. Checking now I see that's clearly not the case. I suppose I have to tinker a bit to find a basis that's actually an orthogonal set. That or my inner product is way off.

A good way to find an orthogonal set is to use Gram Schmidt orthogonalization. Admittedly in the two dimensional case you can just figure it out by staring for a while/solving explicitly the equation for two polynomials to be orthogonal, but it's good practice anyway, and you'll feel smarter for doing it

 

[usn7564]

A good way to find an orthogonal set is to use Gram Schmidt orthogonalization. Admittedly in the two dimensional case you can just figure it out by staring for a while/solving explicitly the equation for two polynomials to be orthogonal, but it's good practice anyway, and you'll feel smarter for doing it

Obviously, Christ, should be the same as always except it's not a dot product anymore. Didn't feel like it was even part of my toolbox for some inexplicable reason.

Should be able to figure out the rest now, thank you.