# Orthogonal projecitons, minimizing difference

1. Nov 30, 2013

### usn7564

1. The problem statement, all variables and given/known data
Determine the polynomial p of degree at most 1 that minimizes

$$\int_0^2 |e^x - p(x)|^2 dx$$

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

3. The attempt at a solution

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

where the inner product is

$$<f, g> = \int_0^2 f(x)\bar{g(x)} dx$$

But this fails just for the first term, ie

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

$$p(x) = 3x + \frac{1}{2}(e^2 - 7)$$

2. Nov 30, 2013

### Office_Shredder

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

3. Nov 30, 2013

### usn7564

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.

Last edited: Nov 30, 2013
4. Nov 30, 2013

### Office_Shredder

Staff Emeritus
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 :tongue:

5. Nov 30, 2013

### usn7564

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.

Last edited: Nov 30, 2013