Editing Finite difference (section)

===Inductive proof===
====Base case====
Let {{math|''Q''(''x'')}} be a polynomial of degree {{math|1}}:
<math display=block>\Delta_h [Q](x) = Q(x + h) - Q(x) = [a(x + h) + b] - [ax + b] = ah = ah^11!</math>

This proves it for the base case.

====Inductive step====
Let {{math|''R''(''x'')}} be a polynomial of degree {{math|''m''&nbsp;−&nbsp;1}} where {{math|''m'' &ge; 2}} and the coefficient of the highest-order term be {{math|''a'' &ne; 0}}. Assuming the following holds true for all polynomials of degree {{math|''m''&nbsp;−&nbsp;1}}:
<math display="block">\Delta_h^{m-1} [R](x) = ah^{m-1}(m-1)!</math>

Let {{math|''S''(''x'')}} be a polynomial of degree {{math|''m''}}. With one pairwise difference:
<math display=block>\Delta_h [S](x) = [a(x+h)^{m} + b(x+h)^{m-1} + \text{l.o.t.}] - [ax^m + bx^{m-1} + \text{l.o.t.}] = ahmx^{m-1} + \text{l.o.t.} = T(x)</math>

As {{math|''ahm'' &ne; 0}}, this results in a polynomial {{math|''T''(''x'')}} of degree {{math|''m'' − 1}}, with {{math|''ahm''}} as the coefficient of the highest-order term. Given the assumption above and {{math|''m'' − 1}} pairwise differences (resulting in a total of {{math|''m''}} pairwise differences for {{math|''S''(''x'')}}), it can be found that:
<math display="block">\Delta_h^{m-1} [T](x) = ahm \cdot h^{m-1}(m-1)! = a h^m m!</math>

This completes the proof.