Define the "forward difference operator" $\Delta$ acting on a function $f(x)$ by $$\Delta f(x) = f(x+1)-f(x).$$ 1. Let $f(x)=x+3$. Compute $\Delta f(x)$.
Solution: By definition, $$\begin{array}{ll} \Delta f(x) &= f(x+1)-f(x) \\ &= [(x+1)+3]-[x+3] \\ &= (x+4)-(x+3) \\ &= 1. \end{array}$$ 2. Let $g(x)=x^2+2x-4$. Compute $\Delta g(x)$.
Solution: By definition, $$\begin{array}{ll} \Delta g(x) &= g(x+1)-g(x) \\ &=[(x+1)^2+2(x+1)-4] - [x^2+2x-4] \\ &= [x^2+2x+1+2x+2-4] - [x^2+2x-4] \\ &= 2x+3 \end{array}$$ 3. Consider the function $h(x)=a(x)b(x)$, where $a(x)$ and $b(x)$ are some functions. The well-known product rule for differentiation says $h'(x)=a'(x)b(x)+a(x)b'(x)$. Find a product rule for the difference operator by computing $\Delta h(x)$ and expressing the result in terms of $a(x)$, $\Delta a(x)$, $b(x)$, and $\Delta b(x)$.
Solution: Compute $$\begin{array}{ll} \Delta h(x) &= a(x+1)b(x+1) - a(x)b(x) \\ &= a(x+1)b(x+1) - a(x)b(x) + 0 \\ &= a(x+1)b(x+1) - a(x)b(x) + a(x+1)b(x) - a(x+1)b(x) \\ &= a(x+1)[b(x+1)-b(x)] + b(x)[a(x+1)-a(x)] \\ &= a(x+1) \Delta b(x) + b(x) \Delta a(x). \end{array}$$
Let $a$ be a number and let $k=0,1,2,\ldots$. Define the "rising factorial" $(a)_k$ by $$(a)_k = a(a+1)(a+2)(a+3)\ldots (a+k-1).$$
4. Compute $(3)_2$.
Solution: $(3)_2 = 3(4) = 12$
5. Compute $(2)_3$.
Solution: $(2)_3 = 2(3)(4) = 24$