48 -> 48x -> 24x^2 -> 8x^3
9 - 8•1^3, 73 - 8•2^3, 241 - 8•3^3
Approximately, what the difference table is doing is differentiating until the leading term is constant. If we end up with a constant k levels down, we must have had a.x^k in the polynomial, as the derivative of a.x^k = ak.x^(k-1). Repeating this k times gives constant = a.k!
So the leading coefficient of the polynomial, a = constant/k!
I guess the next step is to subtract the values generated by this term from the sequence, then repeat to find the next term in the polynomial...
Give your DT "coefficients" work your way back to the solution.
I.e. in this case: you start with
48
48
48
From which you can get
104 <-- Note: this needs to be given!
152
200
248
And work your way back to
9
73
241
561
1081
1849
So you know you want a polynomial P(x)such that P(1) = 9
p(2) = 73
...
p(3) = 1849
So now you know you need at most a 5th degree polynomial.
So, you have y = a_0 + a_1 * x + a_2 * x^2 + ... + a_5 * x^5
Plug in each value of x and you get 6 equations with 6 unknowns. Solve for a_0, ..., a_5.
Bunch of the a's might well be 0, but thats ok.
Edit: This only works if your difference table eventually reduces to a set of differences which are all the same (e.g here 48, 48, 48). Otherwise, the answer is not a polynomial.