Please do not judge me too harshly for my lack of knowledge, but at school we have gone over Quadratic functions recently. Now, these types of functions are not new to me, however when we viewed a table of one of these functions I found something interesting:
$$
\begin{array}{c|lcr}
x & f(x) & \text{1st Diff.} & \text{2nd Diff} \\
\hline
-2 & 9 & – & – \\
-1 & 6 & -3 &- \\
0 & 5 & -1 & 2 \\
1 & 6 & 1 & 2 \\
2 & 9 & 3 & 2
\end{array}
$$
Now, as you might guess, the explicit function of this quadratic is
$$
f(x) = x^2 + 5
$$
But what was interesting was that the 2nd difference is constantly $+2$, which would be the slope of the derivative function, since $a = 1$:
$$
f'(x) = 2(1)x
$$
We've seen a few more and every time the $a$ coefficient of the quadratic was $1/2$ that of the second difference, which only added to the connection.
So, what is the relationship between the 2nd difference and the derivative of this function? I looked online but nothing had the answer that I was looking for.
Thank you!
Gil Keidar
Best Answer
For a quadratic $f(x) = a x^2 + b x + c$, the derivative is $f'(x) = 2 a x + b$, and the slope of the derivative, i.e. the second derivative, is $2 a$.
The second difference is $f(x+2) - 2 f(x+1) + f(x)$, which is also $2a$.
This is not really an accident: it turns out that in general the second difference of a function can be written as a kind of weighted average of the second derivative (as long as that second derivative exists and is continuous). The formula (which you probably won't understand at this level) is
$$ f(x+2) - 2 f(x+1) + f(x) = \int_0^2 (1 - |1-t|) f''(t)\; dt $$
Quadratics are exactly the functions whose second derivatives are constant, and the weighted average of a constant is that constant. That's why for a quadratic the second difference is equal to the second derivative.