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Use the zeroes of $\bar{T}_3$ and transformations of the given interval to construct an interpolating polynomial of degree 2 for $f(x)={ 1\over x}$ over the interval $[1,3]$

My biggest issue is finding the zeroes. How exactly do I do that? I have been using the formula $$\frac{1}{2}\left[(b-a)\cos\left(\frac{N+\frac{1}{2}-n}{ N}\cdot\pi\right) +a+b\right]$$ Where $[a,b]$ is the interval, $N$ is the degree and for $n=1,2,3,\ldots, N.$ $$\bar{x_k}=\frac{1}{2}\left[(3-1)\cos\left(\frac{2+\frac{1}{2}-1}{2}\cdot\pi\right) +1+3\right]=1.29289322$$ This is of course wrong. If anyone is familiar with this topic then please guide me.

The final answer is $P(x)=.3489153-.1744576(x-2.866025)+.1538462(x-2.866025)(x-2)$.

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Those cannot be the zeroes. If $x=2.866025$ then $P(x) = .3489153$... Ah nevermind, you're looking for the zeros of a different polynomial. The question didn't render properly on my phone –  Arkamis Oct 21 '12 at 17:55
    
hmmm I am quite confused then. So how would you solve this then? –  math101 Oct 21 '12 at 17:59
    
Well actually it would be the zeroes of the $\bar{T_3}$ not of the interpolating polynomial –  math101 Oct 21 '12 at 18:03
    
I think I have it figured out. The answer has been factored for some reason –  math101 Oct 21 '12 at 18:11
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The zeroes of $T_3(x)=4x^3-3x$ are $0,\pm\frac{\sqrt 3}{2}$. The first idea is to apply the mapping from the interval $[-1,1]$ (the natural domain for the Chebyshev polynomials) into your desired domain $[1,3]$. It can be easily seen that the required mapping is $f(x)=x+2$, and thus you are to interpolate $\frac1{x}$ over the points $2,2\pm\frac{\sqrt 3}{2}$. You can now apply any of a number of methods (Lagrange, Newton, Neville-Aitken) for constructing the quadratic polynomial interpolant. That part I leave to you.

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