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Let $f(x)=a_mx^{m}$ + (lower degree terms) be a polynomial. Show that

$$f[x_0,...,x_n,x] = \begin{cases} {degree}[m-n-1], & n < m-1 & \\ a_m, &n =m-1 \\ 0 & n>m-1\end{cases} $$where $f[x_0,...,x_n,x]$ is Newton divided difference.

I am having difficulty proving this question. I don't know how to prove it. What I know up to this point is that from writing the general polynomial of degree $m$, I see that there are $m+1$ independent parameters $a_0,...,a_m$. From a function $p(x_i)$, $i = 0,...,n$ then it imposes $n+1$ conditions on $p(x)$. From that and the conditions in my question how can I prove this?

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up vote 1 down vote accepted

Only 45 minutes to go before class. I'm in the room to the left of our classroom if you still need help. The general direction you should go in is that the forward difference from [x0, ..., x[n], x] is equal to f to the n+1th derivative of some number between the maximal range of [x0, ..., x[n]] divided by (n+1)!... I have no idea how to use latex or I'd type it out for you!

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If you have the book, it's on page 154, f^(n+1)(ξ(x))/(n+1)!, where f^(n+1) is f to the n+1th derivative, sorry about not knowing latex! :( – user60977 Feb 4 '13 at 19:49
I was just looking at that. Thanks a lot! – user60514 Feb 4 '13 at 19:50

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