User:Beneficii/differences

For complex numbers $$h$$ and $$x$$, $$\Delta^0_h[f](x) = f(x)$$.

For a non-negative integer $$n \ge 0$$, $$\Delta^{n+1}_h[f](x) = \Delta^n_h[f](x + h) - \Delta^n_h[f](x)$$.

$$\Delta_h[f](x) = \Delta^1_h[f](x)$$.

=Formulas=

Definition
For integers $$n$$ and $$i$$ where $$n \ge i \ge 0$$, $$\binom{n}{i} = \frac{n!}{(n-i)!i!}$$.

Identities
$$\binom{0}{0} = \binom{n}{0} = \binom{n}{n} = 1$$.

$$\binom{n+1}{i+1} = \binom{n}{i+1} + \binom{n}{i}$$.

General formula
$$\Delta^n_h[f](x) = \sum^n_{i=0} (-1)^i \binom{n}{i} f(x + (n - i)h)$$.

Base case
For $$n = 0$$,

$$\begin{align} \Delta^0_h[f](x) & = \sum^0_{i=0} (-1)^i \binom{0}{i} f(x + (0 - i)h) \\ & = (-1)^0 \binom{0}{0} f(x + (0 - 0)h) \\ & = f(x) \end{align}$$.

Induction step
For $$n = a$$,

$$\Delta^a_h[f](x) = \sum^a_{i=0} (-1)^i \binom{a}{i} f(x + (a - i)h)$$.

For $$n = a+1$$,

$$\begin{align} \Delta^{a+1}_h[f](x) & = \Delta^{a}_h[f](x + h) - \Delta^{a}_h[f](x) \\ & = \sum^a_{i=0} (-1)^i \binom{a}{i} f((x + h) + (a - i)h) - \sum^a_{i=0} (-1)^i \binom{a}{i} f(x + (a - i)h) \\ & = \sum^a_{i=0} (-1)^i \binom{a}{i} f(x + (a + 1 - i)h) - \sum^a_{i=0} (-1)^i \binom{a}{i} f(x + (a - i)h) \\ & = \sum^a_{i=0} (-1)^i \binom{a}{i} f(x + (a + 1 - i)h) - \sum^{a+1}_{i=1} (-1)^{i-1} \binom{a}{i-1} f(x + (a - (i-1))h) \\ & = \sum^a_{i=1} (-1)^i \binom{a}{i} f(x + (a + 1 - i)h) + \sum^{a}_{i=1} (-1)^{i} \binom{a}{i-1} f(x + (a + 1 - i)h) \\ & \qquad + \left [ (-1)^0 \binom{a}{0} f(x + (a+1 - 0)h) \right ] + \left [ (-1)^{a+1} \binom{a}{(a+1)-1} f(x + (a + 1 - (a+1))h) \right ] \\ & = \sum^a_{i=1} (-1)^i f(x + (a + 1 - i)h) \left [\binom{a}{i} + \binom{a}{i-1} \right ] \\ & \qquad + \left [ (-1)^0 \binom{a+1}{0} f(x + (a+1 - 0)h) \right ] + \left [ (-1)^{a+1} \binom{a}{a} f(x + (a + 1 - (a+1))h) \right ] \\ & = \sum^a_{i=1} (-1)^i \binom{a+1}{i} f(x + (a + 1 - i)h) \\ & \qquad + \left [ (-1)^0 \binom{a+1}{0} f(x + (a+1 - 0)h) \right ] + \left [ (-1)^{a+1} \binom{a+1}{a+1} f(x + (a + 1 - (a+1))h) \right ] \\ & = \sum^{a+1}_{i=0} (-1)^i \binom{a+1}{i} f(x + (a + 1 - i)h). \end{align}$$

Q.E.D.

Axiom
If for any independent complex variable $$x$$, $$\Delta^a_h[f](x) = C$$, where $$C$$ is a constant complex number, then $$\Delta^{a+1}_h[f](x) = 0$$ and, where $$y$$ is an independent complex variable, $$\Delta^a_h[f](x) = \Delta^a_h[f](y)$$.