Orthogonal projecitons, minimizing difference

usn7564
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Homework Statement


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


The Attempt at a Solution



I assumed what I wanted was a p(x) of the form
<br /> p(x) = \frac{&lt;e^x, 1&gt;}{&lt;1,1&gt;} + \frac{&lt;e^x, x&gt;}{&lt;x,x&gt;}x<br />

where the inner product is

&lt;f, g&gt; = \int_0^2 f(x)\bar{g(x)} dx

But this fails just for the first term, ie

\frac{&lt;e^x, 1&gt;}{&lt;1,1&gt;} = \frac{e^2-1}{2} does not coincide with the correct answer


Correct answer:
p(x) = 3x + \frac{1}{2}(e^2 - 7)
 
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Do you understand why the hint says you should find orthogonal polynomials first? And do you know if 1 and x are orthogonal?
 
Office_Shredder said:
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.
 
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usn7564 said:
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 :-p
 
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Office_Shredder said:
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 :-p
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.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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