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Here's how I think about this. Suppose you're given n terms a\sb 1, ..., a\sb n. http://latex.mathoverflow.net/png?a%5F1%2C%20%2E%2E%2E%2C%20a%5Fn%2E Define b\sb i = a\sb i / (i-1)\cdots(i-n) http://latex.mathoverflow.net/png?b%5Fi%20%3D%20a%5Fi%20%2F%20%28i%2D1%29%5Ccdots%28i%2Dn%29, where the denominator skips the factor (i-i) http://latex.mathoverflow.net/png?%28i%2Di%29. Consider the function f(x) = b\sb 1(x-2)(x-3)\cdots(x-n) + (x-1)b\sb 2(x-3)\cdots(x-n) + \cdots + (x-1)\cdots(x-n+1)b\sb n http://latex.mathoverflow.net/png?b%5F1%28x%2D2%29%28x%2D3%29%5Ccdots%28x%2Dn%29%20%2B%20%28x%2D1%29b%5F2%28x%2D3%29%5Ccdots%28x%2Dn%29%20%2B%20%5Ccdots%20%2B%20%28x%2D1%29%5Ccdots%28x%2Dn%2B1%29b%5Fn + (x-1)\cdots(x-n)c http://latex.mathoverflow.net/png?%2B%20%28x%2D1%29%5Ccdots%28x%2Dn%29c. Then for i an integer between 1 and n, f(i) = b\sb i \ast (i-1) \ast \cdots \ast (i-n) http://latex.mathoverflow.net/png?f%28i%29%20%3D%20b%5Fi%20%2A%20%28i%2D1%29%20%2A%20%5Ccdots%20%2A%20%28i%2Dn%29, except that you skip the factor (i-i). Thus, f(i) = a_i. But by changing c, you can make the next term f(n+1) whatever you want.