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Technoguyrob comments on Beautiful Math - Less Wrong
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If you get a non constant, yes. For a linear function, f(a+1) - f(a) = f'(a). Inductively you can then show that the nth one-step difference of a degree n polynomial f at a point a is f^(n)(a). But this doesn't work for anything but n. Thanks for pointing that out!
Ah, yes, that's a good point, because the leading coefficient be the same whether you use the x^k basis or the falling factorial basis.