Difference of the difference of consecutive squares I was playing around with the square numbers and I noticed something:
$$\left.
\begin{array}{r}
&\left.
\begin{array}{r}
&1^2 = 1\\
&2^2=4\\
\end{array}
\right\}
\ 3 \\
&\left.
\begin{array}{r}
&2^2 = 4\\
&3^2=9\\
\end{array}
\right\}
\ 5
\end{array}
\right\}
\ 2$$
$$\left.
\begin{array}{r}
&\left.
\begin{array}{r}
&3^2 = 9\\
&4^2=16\\
\end{array}
\right\}
\ 7 \\
&\left.
\begin{array}{r}
&4^2 = 16\\
&5^2=25\\
\end{array}
\right\}
\ 9
\end{array}
\right\}
\ 2$$
The fact the difference of consecutive squares is $2n+1$ is easily justified, $(n+1)^2-n^2=2n+1$.
However, my question is about this:
$$\text{Let } D_n := |a_{n+1}-a_n| = 2n+1, \text{Where } a_n=n^2$$
$$\text{Then } |D_{n+1}-D_n|=|2(n+1)+1-(2n+1)|=2$$
And this is the same as the derivative of $D_n$, $\displaystyle \frac{d}{dn}(2n+1)=2$. This makes some intuitive sense to me, $2$ is how much the difference between squares changes. However, talking about derivatives in $\mathbb{N}$ seems fallacious to me. Is this just a happy coincidence?
 A: This is what I would call a happy coincidence, but only halfway. What you're discovering is the so-called "calculus of finite differences".
Note that any sequence $(a_n)$ of integers may be thought of as a function $a : \Bbb N \rightarrow \Bbb Z$, where $a(n) = a_n$. We may then define the operator $\Delta$ on such functions, so that if $a : \Bbb N \rightarrow \Bbb Z$, then $\Delta[a] = a(n+1) - a(n)$. This is a linear operator on such functions, just as the derivative operator $\frac{\mathrm{d}}{\mathrm{d}x}$ is linear.
Also, note that $\Delta[c] = c - c = 0$ for any constant $c$, also analogous to the derivative. Note that $\Delta[n] = n + 1 - n = 1$, still analogous. This allows us to reach our result with linearity: $\Delta[2n+1] = 2\Delta[n] + \Delta[1] = 2 \times 1 + 0 = 2$.
However, in general, the "power rule" does not apply--rather, it is the "permutation rule". Where in normal derivatives, $\frac{\mathrm{d}}{\mathrm{d}x}[x^k] = kx^{k-1}$, with finite differences, the analog is $\Delta[_{n}P_{k}] = k _{n}P_{k-1}$.
There are even slightly modified versions of the product and quotient rules for $\Delta$, but I leave these as an exercise to the reader.
