Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer

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!

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.