I am attempting to compute the following discrete difference: $$\Delta[(3x+2)^2]$$ where $\Delta[f(x)] = f(x+1)-f(x)$
Method 1: \begin{align*} \Delta[(3x+2)^2] &= [3(x+1)+2]^2 - (3x+2)^2 \\ &=(3x+5)^2 - (3x+2)^2 \\ &= (9x^2+30x+25)-(9x^2+12x+4) = 18x+21 \end{align*}
Method 2:
Define $f(x)^{(k)} = f(x)f(x-1)\cdots f(x-[k-1]).$ It is known1 that $\Delta[(ax+b)^{(k)}] = ka(ax+b)^{(k-1)}.$ Using the fact that $x^2 = x^{(2)} +x$ : \begin{align*} \Delta[(3x+2)^2] &= \Delta[(3x+2)^{(2)} + (3x+2)^{(1)}] \\ &= \Delta[(3x+2)^{(2)}] + \Delta[(3x+2)^{(1)}]\\ &= 6(3x+2) + 3 = 18x+15 \end{align*}
I can't see why I'm getting two different answers. A direct expansion indicates the first calculation is correct. Can anyone help me find the flaw in the second computation?
- See e.g. Theorem 5 of Richardson's An Introduction to the Calculus of Finite Differences.
Using $$ (a x + b)^2 = (a x + b)(a (x-1) + b) + a \, (a x + b) = (a x + b)^{(2)} + a \, (a x + b)^{(1)}$$ then \begin{align} \Delta (a x + b)^{2} &= \Delta (a x + b)^{(2)} + a \, \Delta (a x + b)^{(1)} \\ &= 2 \, a \, (a x + b)^{(1)} + a \, a \, (a x + b)^{(0)} \\ &= 2 \, a \, (a x + b) + a^2 \\ &= a \, (2 a \, x + 2 b + a). \end{align} Setting $a = 3$ and $b = 2$ gives $\Delta (3 x + 2)^2 = 3 \, (6 x + 7)$.
Method 3: \begin{align} \Delta (a x + b)^{2} &= \Delta (a^2 \, x^2 + 2 a b \, x + b^2) \\ &= a^2 \, \Delta (x^2) + 2 a b \, \Delta (x) + b^2 \, \Delta (1) \\ &= a^2 \, (2 x + 1) + 2 a b \, (1) \\ &= a \, (2 a \, x + 2 b + a). \end{align} This is the same as the other results.