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Let $q\in \prod_{n-1}$ be interpolation polynomial for $x_1,...,x_n$ and $p\in \prod_{n-1}$ be interpolation polynomial for $x_0,...,x_{n-1}$. Show that function $$N(x)=q(x)\left( 1+\frac{x_n-x}{x_0-x_n}\right) - \frac{x_n-x}{x_0-x_n}p(x)$$ is interpolation polynomial for $x_0,...,x_n$.

My idea was to represent $p(x)$ and $q(x)$ as Lagrange polynomials. Hence $$q(x)=\sum_{k=1}^nf(x_k)\prod_{i=1 \\ i \neq k}^n\frac{x-x_i}{x_k-x_i}$$ $$p(x)=\sum_{k=0}^{n-1}f(x_k)\prod_{i=0 \\ i \neq k}^{n-1}\frac{x-x_i}{x_k-x_i}$$ and just write it in $$N(x)=q(x) + \frac{x_n-x}{x_0-x_n}(q(x)-p(x))$$ but I'm stuck and feel like this not the right way to do it. Any ideas?

I found this although I'm curious if one can do this with Lagrange.

Awerde
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1 Answers1

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I think you are making this harder than it needs to be.

What you need to show, if I am interpreting your statements correctly, is that $N(x) =p(x) $ for $x=x_0,...,x_{n-1}$ and $N(x) =q(x) $ for $x=x_1,...,x_{n}$.

If $p$ and $q$ interpolate to the same values, you need to have $p(x) =q(x) $ for $x=x_1,...,x_{n-1}$.

This should be enough.

marty cohen
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  • $x_0,...,x_n$ are the same for $p(x)$ and $q(x)$ so I don't see how $N(x)=p(x)$ and $N(x)=q(x)$ – Awerde Jan 29 '22 at 16:26
  • I assume you want to do something similar to the answer in the question I linked - for now, it seems the most reasonable way to do it – Awerde Jan 29 '22 at 16:29