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I need some hint on this particular homework problem.

Show the following two polynomials $R(x)$ and $L(x)$ both interpolate the given points $$\left\{(x_1,\ y_1),\ (x_2,\ y_2),\ (x_3,\ y_3),\ (x_4,\ y_4)\right\}$$

I want to ask the community how I go about verifying they are interpolating the same data points? I am generalizing my question. I hope this is not too localized.

Is there some properties involved? My first thought was plugged in the points but $R(x)$ and $L(x)$ shouldn't give equal values... somehow I think I need to reconstruct the actual $f(x)$ function?

$$R(x) = 3 - 2(x+1) + 0 (x+1)(x) + (x+1)(x)(x-1)$$ $$L(x) = -1 +4(x+2) - 3(x+2)(x+1)+(x+2)(x+1)(x)$$

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It'd be nice to know what $R(x)$ and $L(x)$ are. –  EuYu Oct 10 '12 at 23:59
    
@EuYu Thanks. I just edited, although I feel a bit bad revealing the actual question. But I think it's also good for learning and demonstration –  User007 Oct 11 '12 at 0:03
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Don't try reconstructing the function $f(x)$ that determines the four points (whatever they are), since you can't. There are infinitely many different possible functions (even if you restrict them to be polynomials) that pass through a set of four points. –  Rick Decker Oct 11 '12 at 0:22
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2 Answers

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The point of a polynomial interpolating a set of points is that they must intercept those points. An easy example would be two points and a line. What does it mean for the line to interpolate the two points? It would mean that the line passes through both of the points. In particular, if $f$ is your interpolating polynomial, then you must show that $$f(x_i) = y_i$$ for each $(x_i,\ y_i)$. There is perhaps a short cut in checking for small polynomials. If you have a set of $n$ points, then you can always find a polynomial of degree at most $n-1$ which interpolates the set of points, moreover this polynomial is unique. If you have two polynomials of degree less than or equal to $n-1$, then at least one of the two cannot interpolate the set of points.

You have four points and two different degree three polynomials. Only one of these can actually interpolate the points.

Edit: Ah, I see the polynomials are actually the same. That explains a bit.

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two quick questions. $f(x_i) = y_i$ does it have to exact? usually the interpolation will give an error (and we have an upper bound). And, is there a different way than combining terms in the polynomials to show they are actually the same? like using those data points. –  User007 Oct 11 '12 at 0:17
    
Normally interpolation is considered exact. I would probably use the term regression for non-exact curve fitting. As for comparing polynomials, I really don't know any easier way than simply writing them in standard form and comparing them. You might use a probabilistic argument. Try the polynomials for a few simple values ($x=0,\ 1, -1,\ \text{etc}$). There's a good chance if the polynomials are different that these values won't match. But if you want to prove that they're equal, expanding and comparing would be needed. –  EuYu Oct 11 '12 at 0:18
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Expand the two polynomials and collect terms. They're both equal (to $x^3-3x+1$) so of course they'll both interpolate (or not) the set of points.

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thanks. I was thinking along numerically. or some kind of verification instead of combining, which will work. But thanks. –  User007 Oct 11 '12 at 0:15
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