In 6th grade I was first introduced to the idea of a function in the form of tables. The input would be "n" and the output "$f_n$" would be some modification of the input. I remember finding a pattern in the function "f(n)=n^2". Here is what the table looked like:
\begin{array}{|c|c|}
\hline
n& f_n\\ \hline
1&1 \\ \hline
2&4\\ \hline
3&9\\ \hline
4&16\\ \hline
5&25\\ \hline
…&…\\ \hline
n&n^2\\ \hline
\end{array}
I would then take the outputs $f_n$ and find the differences between each one: $f_n-f_{n-1}$. This would produce:
\begin{array}{|c|c|}
\hline
n& f(n)-f(n-1)\\ \hline
1&1 \\ \hline
2&3\\ \hline
3&5\\ \hline
4&7\\ \hline
5&9\\ \hline
…&…\\ \hline
\end{array}
Repeating this process (of finding the differences) for the outputs of $f_n-f_{n-1}$ would yield a continuous string of $2$s. As a 6th grader I called this process 'breaking down the function' and at the time it was just another pattern I had found.
Looking back at my work as a freshman in high school, I realize that the end result of 'breaking down a function' corresponds to the penultimate derivative (before the derivative equals zero). For example: breaking down $y=x^3$ gives a continuous string of $6$s, and the third derivative of $x^3$ is 6 (while the 2nd derivative is 6x).
Is there any significance to this pattern found by finding the differences between each output of a function over-and-over again? Does it have anything to do with derivatives? I know my question is naive, but I'm only a high school freshman in algebra II. A non-calculus (or intuitively explained calculus concepts) answer would be very helpful [note that I used an online derivative calculator to find the derivatives of these functions and I apologize for any incorrect calculus terminology].
Best Answer
Yes, this has plenty to do with the derivative. In particular, what you describe is the backwards difference operator, which is just defined as $$\nabla f(n)=f(n)-f(n-1).$$ This is an operator of interest on its own, but the connection to calculus is that we can consider this as telling us the "average" slope between $n-1$ and $n$.
What you are doing is iterating the operator. In particular, one often writes $$\nabla^{k+1} f(n)=\nabla^k f(n)-\nabla^k f(n-1)$$ to meant that $\nabla^k f(n)$ is the result of applying this operator $k$ times. For instance, one has that $\nabla^3 n^3 = 6$, as you note. More generally $\nabla^k n^k = k!$, and this lets us recover a polynomial function from its table, which is what you were up to in sixth grade.
However, we can take things further by trying to interpret these numbers - and there is a natural interpretation. For instance, $\nabla^2 f(n)$ represents how quickly $f$ is "accelerating" over the interval $[n-2,n]$, since it tells us about how the average slope changes between the interval $[n-2,n-1]$ and the interval $[n-1,n]$. If we keep going, we get that $\nabla^3 f(n)$ tells us how the acceleration changes between an interval $[n-3,n-2]$ and $[n-2,n]$. We can keep going like this for physical interpretations.
However, this operator has a problem: We'd like to interpret the values as accelerations or as slopes, but $\nabla^k f(n)$ depends on the values of $f$ across the interval $[n-k,n]$. That is, it keeps taking up information from further and further away from the point of interest. The way one fixes this is to try to measure the slope over a smaller distance $h$ rather than measure it over a length of $1$: $$\nabla_h f(n)=\frac{f(n)-f(n-h)}h$$ which is now the average slope of $f$ between $n-h$ and $n$. So, if we make $h$ smaller, we start to need to know $f$ across a smaller range. This gives better meanings to higher order differences like $\nabla_h^k f(n)$, since now they only depend on a small portion of $f$.
The derivative is just what happens to $\nabla_h$ when you send $h$ to $0$. It captures only local information about the function - so, it captures instantaneous slope or instantaneous acceleration and so on. In particular, one can work out that $\nabla f(n)$ is just the average of the derivative over the interval $[n-1,n]$. One can also work out that $\nabla^2 f(n)$ is a weighted average* of the second derivative over the interval $[n-2,n]$ and $\nabla^3 f(n)$ is another weighted average of the third derivative over $[n-3,n]$.
In particular, if the $k^{th}$ derivative is constant, then it coincides with $\nabla^k f(n)$. One can also find results that if the $k^{th}$ derivative is linear, then $\nabla^k f(n)$ differs from it by at worst a constant. In particular, $\nabla$ is good at capture "global" effects (like the highest order term in a polynomial and its coefficient) but bad at capturing "local" effects (like instantaneous changes in the slope). So, in some sense, $\nabla$ is just a rough approximation of the derivative, and has similar interpretations, just doesn't work nearly as cleanly.
(*Unfortunately, "weighted average" here is hard to explain rigorously without calculus. For the benefit of readers with more background, I really mean "convolution" assuming that $f$ is actually differentiable enough times for any of this to make sense)