Given a series how do I create a function to describe that series

Question

Given a series of numbers how do I find the equation that describes the series?

For example given the following series of numbers...

$$352, 1424, 2528, 3664, 4832, 6032, 7264, 8528, 9824, 11152, 12512, 13904$$

The first differences are

$$1072, 1104, 1136, 1168, 1200, 1232, 1264, 1296, 1328, 1360, 1392$$ The second differences are all $32$

How can I find the function that describes this series.

Ultimately I am trying to determine which number in the series is a perfect square without doing trial division, There is a maximum of one perfect square in the series. My intention was to find the intersection of the set of perfect squares and the series above (and any other series that follows the same shape).

But before I worry about finding the intersection I am trying to understand how to convert a series of points into a function.

I hope that makes sense.

Answer 1

The second differences are all $32$.
The k^th first difference is thus $1072+(k-1)32$
The terms of the series are thus: $$352+\sum_{k=1}^{n-1}(1072+(k-1)32)=352+1072(n-1)+32\frac{(n-1)(n-2)}2=16(n^2+64n-43)$$ Another way to look at it is: $$t_n={}^{n-1}{\rm C}_0\Delta_0+{}^{n-1}{\rm C}_1\Delta_1+{}^{n-1}{\rm C}_2\Delta_2+...$$ where $\Delta_0=32,\Delta_1=1072,\Delta_2=32,\Delta_{r,r\ge3}=0$ I hope you know about binomials for the C's.

Answer 2

Hint

If you plot the numbers, they are along a parabola the equation of which being (easy to establish; just pick three data points) $$y(n)=16 n^2 + 1024 n − 688$$ In the post general case, try to run polynomial regression.

Edit

You noticed that the second difference is a constant. Make a link between the second difference and the second derivative; if the function is polynomial, then it must be quadratic. You can extend that to higher orders.

Edit

Let us suppose that the numbers can be represented by a polynomial, say $$a_n=\alpha +\beta n+\gamma n^2+\delta n^3$$ then the numbers of the first difference are given by $$b_n=a_{n+1}-a_n=(\beta +\gamma +\delta )+n (2 \gamma +3 \delta )+3 \delta n^2$$ the numbers of the second difference are given by $$c_n=b_{n+1}-b_n=(2 \gamma +6 \delta )+6 \delta n$$ the numbers of the third difference are given by $$d_n=c_{n+1}-c_n=6 \delta$$ So, as in the case of your post, the second differences are constant then $\delta=0$ and $2\gamma=32$ so $\gamma=16$.

Now, go backward as ADG showed.