Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For any function $f$ and distinct reals $x_1,\ldots,x_n$, denote by $f[x_0,\ldots,x_n]$ the coefficient of $x^n$ of the minimal polynomial interpolating $f$ at $x_0,\ldots,x_n$.

Let $f$ and $g$ be real-valued functions defined on the real line and let $x_0,x_1,\ldots,x_n$ be $n+1$ distinct real numbers. Derive a formula for the $n$-th order divided difference $(fg)[x_0,x_1,\ldots,x_n]$ of the product function $fg$ at $x_0,x_1,\ldots,x_n$ in terms of the divided differences $f[x_0,x_1,\ldots,x_k]$ and $g[x_k,x_{k+1},\ldots,x_n]$ of $f$ and $g$, for $k=0,1,\ldots,n$.

For $n=1$, I've found that $(fg)[x_0,x_1]=f[x_0]g[x_0,x_1]+f[x_0,x_1]g[x_1]$. But I didn't find any nice form for $n=2$. If I can guess the correct formula, I'm quite sure it can be proved by induction.

share|cite|improve this question
You'll want to look up the Leibniz formula. See this as well. – J. M. Apr 18 '12 at 16:14
Thank you very much! – matt Apr 19 '12 at 4:33

the answer is $ \sum_{j=0}^{n}f[x_{0}, x_{1},..., x_{j}]g[x_{j}, x_{j+1},..., x_{n}] $

share|cite|improve this answer

Your Answer


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.