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Technoguyrob comments on Beautiful Math - Less Wrong
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This is obvious after you learn calculus. The "nth difference" corresponds to nth derivative (a sequence just looks at integer points of a real-valued function), so clearly a polynomial of degree n has constant nth derivative. It would be even more accurate to say that an nth antiderivative of a constant is precisely a degree n polynomial.
Iterated finite differences correspond to derivatives in some non-obvious way I can't remember (and can't be bothered to find out).
Notice that the result doesn't hold if the points aren't evenly spaced, so the solution must use this fact.
Differences and derivatives are not the same, though there is the obvious analogy. If you want to take derivatives and antiderivatives, you want to write in the x^k basis or the x^k/k! basis. If you want to take differences and sums, you want to write in the falling factorial basis or the x choose k basis.
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.
Neither finite differences nor calculus are new to me, but I didn't pick up the correlation between the two until now, and it really is obvious.
This is why I love mathematics - there's always a trick hidden up the sleeve!