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This is a question on my homework.

Let $P_n$ be the interpolation polynomial of f at the nodes $x_0, \ldots , x_n$. Suppose we know $P_n(x) = 3x^n + Q(x)$, where $Q$ is a polynomial of degree at most $n − 1$. Find $f[x_0,\ldots , x_n]$.

I am unsure how to solve this. Basically what I think is that we know our nodes $x_0, \ldots , x_n$, and we know $P(x_j)=3x_j^n + Q(x_j)$, so we have the interpolation polynomial's value at each point.

And we have the nondescript function $Q(x)$, so we have some value $Q(x_j)$ at some point $j$. And I think $f[x_0,\ldots , x_n]$ would have to be something similar to the $x_1,x_0$ where you take $(Q(1)-Q(0))/(1-0)$, but more generalized, which I do not know how to find.

I am also unsure of how $P(x)$ is relevant in this case. Would we need to do some subtraction in this case, or is this on the wrong track? I am just very unsure how to go about doing this and would appreciate any help possible.

Masacroso
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girlwithadragontattoo
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1 Answers1

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The Newton interpolation polynomial for nodes $x_0,\ldots,x_n$ and a function $f$ is defined by

$$N[f;x_0,\ldots,x_n]:=\sum_{k=0}^na_k\omega_k[x_0,\ldots,x_{k-1}]$$

where $$\omega_k[x_0,\ldots,x_{k-1}]:=\prod_{j=0}^{k-1}(x-x_j)$$

are called the Newton's polynomials, and the $a_k$ are coefficients defined recursively by

$$\begin{align}&a_0=f(x_0)\\&a_0+a_1\omega_1[x_0](x_1)=f(x_1)\\&a_0+a_1\omega_1[x_0](x_2)+a_2\omega_2[x_0,x_1](x_2)=f(x_2)\\&\ldots\ldots\ldots\\&\sum_{j=0}^na_j\omega_j[x_0,\ldots,x_{j-1}](x_n)=f(x_n)\end{align}$$

where we had used the standard convention of the empty product. Thus the leading coefficient of $N$ is defined by the maximum $k$ such that $a_k\neq 0$, because by definition $$\deg(\omega_k[x_0,\ldots,x_{k-1}])<\deg(\omega_{k+1}[x_0,\ldots,x_k])$$

Using your notation $a_k=f[x_0,\ldots,x_k]$.

Masacroso
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