What is the function if the second differences have a common multiplier?

Question

Ok, so I know that if for a constant difference in x values, there is a constant different in y values (ie the 1st differences are all the same) then the data is linear. If the second differences differences are the same, then the data is quadratic. Third differences, cubic, etc.

I also know that if for a constant difference in x values there is a constant multiplier for y values, the relationship is y=AB^x. Further, if instead there is a constant multiplier for 1st differences, then the relationship is y=AB^x+k.

My question is: What if the second differences have a constant multiplier? What relationship would the original x and y have?

I made a data set that fits the criteria, and have done some simple statistical regressions on it, but nothing good is coming up. I bet that it's going to end up being a combined function that has polynomial and exponential parts, but I'm just not sure how to approach it.here is a data set I made

I'm really curious. It came up while discussing exponentials with one of my students.

Answer 1

You're exactly right. The general solution is $y = m x + c + A B^x$. You can use your proof for the two cases you've already done to get to this! Hint: For a sequence with constant multipliers for the second differences, you know the general form of the first difference sequence.

Answer 2

Let $w_n$ be the second order difference.

The solution of

$$w_{n+1}=Rw_n$$ is $w_n=w_0R^n=w_0R^n$.

Now with $z_n$ the first order difference,

$$z_{n+1}-z_n=w_0R^n$$ and by induction

$$z_n=w_0R^{n-1}+w_0R^{n-2}+\cdots w_0R+z_0=w_0\frac{R^n-1}{R-1}+z_0.$$

Finally,

$$y_{n+1}-y_n=\frac{w_0}{R-1}R^n-\frac{w_0}{R-1}+z_0$$

and by induction,

$$y_n=\frac{w_0}{(R-1)^2}R^n-\frac{w_0}{(R-1)^2}+\left(-\frac{w_0}{R-1}+z_0\right)n+y_0.$$

More generally, the $k^{th}$ order differences will yield the exponential and a polynomial in $n$ of degree $k-1$.