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Consider the function

$$f(x) = \frac{1}{\alpha (x-\beta)^2 + 1}$$

in the interval $I = [-1,1]$. Set $\beta = 0$. The Lagrange interpolating polynomial of $f(x)$ with degree $n=2$ for equally spaced nodes in $I$.

I tried several times and I want to know if I got the right interpolant for different values of $\alpha$. What I got is

$$P(x) = \frac{-x^2 + x + 4}{4\alpha + 4} - \frac{\alpha x^2 - \alpha}{\alpha + 1}$$

BRabbit27
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  • Did you interpolate at $x_0=-1, x_1=0, x_2=1$? – Git Gud Jan 19 '13 at 10:05
  • Yes, at least it is what I understood by the "equally spaced nodes", am I right? If degree = 2 then I need 3 nodes, therefore, the interval divided in 3 nodes gives me [-1, 0, 1] isn't it? – BRabbit27 Jan 19 '13 at 10:36
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    I agree that you're asked to interpolate at those points, however that doesn't guarantee that you'll get a polynomial of degree $2$. You can only be sure sure the degree will be at most $2$, I believe. – Git Gud Jan 19 '13 at 10:40
  • If $\beta =0$, then $$f(x) = \frac{1}{\alpha x^2 + 1}$$ Is $\alpha\geq 0$ in the hypothesis? Notice that if $x=1$ and $\alpha=-1$, then the function isn't defined. Furthermore $f(-1)=\frac{1}{\alpha+1}$, but $P(-1)=\frac{1}{2\alpha +2}$, so that can't be the correct polynomial. The Lagrange polynomial must coincide with the function at the interpolating points. – Git Gud Jan 19 '13 at 10:42
  • It's better that you type your work in the question so people can point out where you went wrong. – Git Gud Jan 19 '13 at 10:48
  • There's nothing supposed about $\alpha$, it's like for any value of it, of course there will be some cases where the polynomial isn't defined, but I'm not interested in that, for the moment. What I've just written is the only thing I have. @GitGud you are completely right, the polynomial will be at most of degree 2. – BRabbit27 Jan 19 '13 at 19:15
  • Can anyone develop the equations to get the polynomial as a function of x and $\alpha$ ? – BRabbit27 Jan 19 '13 at 19:17
  • Earlier I got $P(x)=\frac{x(x-1)}{2\alpha +2} - (x+1)(x-1) + \frac{x(x+1)}{2\alpha +2}$, but I could very easily have made a miscalculation too. – Git Gud Jan 19 '13 at 19:40

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In the following I'll use the notation of the wikipedia link provided below.

Suppose $\beta=0$. We get $\displaystyle f(x) = \frac{1}{\alpha x^2 + 1}$.

Let $x_0=-1, x_1=0$ and $x_2=1$. We get $\displaystyle f(x_0)=\frac{1}{\alpha +1}=f(x_2)$ and $\displaystyle f(x_1)=1$.

We'll interpolate at $(x_0,y_0), (x_1,y_1), (x_2,y_2)$, where $y_0=f(x_0), y_1=f(x_1)$ and $y_2=f(x_2)$, hoping that the Lagrange Polynomial we find is of the degree 2.

The Lagrange Polynomial is given by $\displaystyle L(x)=y_0l_0(x)+y_1l_1(x)+y_2l_2(x)$, where

$$l_0(x)=\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}=\frac{(x-0)(x-1)}{(-1-0)(-1-1)}=\frac{x(x-1)}{2}$$ $$l_1(x)=\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}=\frac{(x-(-1))(x-1)}{(0-(-1))(0-1)}=\frac{(x+1)(x-1)}{-1}=-(x^2-1)$$ $$l_2(x)=\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}=\frac{(x-(-1))(x-0)}{(1-(-1))(1-0)}=\frac{x(x+1)}{2}$$

So it comes $\displaystyle L(x)=f(x_0)\frac{x(x-1)}{2}-f(x_1)(x^2-1)+f(x_2)\frac{x(x+1)}{2}=\\ \displaystyle =\frac{1}{\alpha +1}\frac{x(x-1)}{2}-(x^2-1)+\frac{1}{\alpha +1}\frac{x(x+1)}{2}$

which, fortunately, is a polynomial of degree 2.

Git Gud
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  • Perfect, my problem was at evaluating the first term $l_0$ I got confused with the x's, but now I got it right. – BRabbit27 Jan 19 '13 at 20:53