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Consider the function $\displaystyle f(x) = \frac{1}{\alpha (x-\beta)^2 + 1}$ in the interval $I = [-1,1]$. Set $\beta = 0$.

How do I get the expression for the least-squares polynomial, say $\tilde f(x)$ of degree $m = 0$. In terms of the parameter $\alpha$ which could be any. By using the nodes $x_0=-1$, $x_1=0$, $x_2=1$

And how can I deduce which kind of interpolant does a least-squares approximating polynomial of degree $m=1$ corresponds to, when using 2 distinct nodes?

I'm going through some exercises for my exam next week, but I couldn't understand this.

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up vote 1 down vote accepted

What's the point of mentioning $\beta$ if it's zero ??

It's not very clear what "using the nodes" means. I'll assume it means that error is supposed to be measured only at these points.

A polynomial of degree $m=0$ is a constant; let's call it $k$. Then the squared error is $\left(k-f(-1)\right)^2 + \left(k-f(0)\right)^2 + \left(k-f(1)\right)^2$.

Substitute for $f(-1), f(0), f(1)$. This will give you an expression involving $\alpha$ and $k$. Differentiate wrt to $k$, and find places where this derivative is zero.

The second part of the question is even more fuzzy, but I expect similar techniques will work.

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Well, I don't know, but the exercise says that. "Using the nodes" I suppose it means that the points $(x_i, f(x_i))$, are the points being approximated by the least-squares. For the second question, well that's why I decided to ask here, because I don't know. Could you give the general formula for the least-squares of m-degree. – BRabbit27 Jan 20 '13 at 7:09
Well, the problem is ... if $m \ge 1$, then the least squares polynomial will actually be the interpolating polynomial (i.e. the one that passes through the given two points). And, a polynomial of degree 1 is a straight line, of course. Maybe that was the point of the question.+ – bubba Jan 21 '13 at 4:16

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