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For n + 1 supporting points and values (x$_{0}$,y$_{0}$),...,(x$_{n}$,y$_{n}$) and 0 ≤ j ≤ n as well as 0 ≤ i ≤ n-j let p$_{i,j}\in$ P$_{j}$ defined by p$_{i,j}$,(x$_{k}$)=y$_k$, k = i, i + 1,...,i+j. Let the numbers y$_{i,j}$, be defined by y$_{i,0}$,0 = yi , i = 0,1,...,n, and y$_{i,j}$=$\frac{y_{i+1,j-1}-y_{i,j-1}}{x_{i+j}-x_{i}}$ for 1≤j≤n and 0≤i≤n-j.

Show that p$_{i,j}$(x)=y$_{i,j}$,jx$^{j}$+r$_{i,j}$(x) with a polynomial r$_{i,j}\in$ P$_{j-1}$ for j≥1 and i=0,1,...,n-j holds.

My fellow students and I really don't have a clue about this assignment. We are totally at a loss. We would be very happy to get some tips.

Melila
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1 Answers1

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Let $f$ be some function with $f(x_i)=y_i$. Then it should be easy to confirm that $$ y_{i,j}=f[x_i,x_{i+1},...,x_{i+j}] $$ In consequence the polynomials can be written as Newton polynomials $$\begin{align} p_{i,j}(x)&=f(x_i)+f[x_i,x_{i+1}](x-x_i)+...+f[x_i,x_{i+1},...,x_{i+j}](x-x_i)(x-x_{i+1})···(x-x_{i+j-1}) \\ &=y_{i,0}+y_{i,1}(x-x_i)+...+y_{i,j}(x-x_i)(x-x_{i+1})···(x-x_{i+j-1}) \end{align}$$ so that indeed the monomial coefficient of degree $j$ is $y_{i,j}$.

Lutz Lehmann
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