# Finding the function for a sequence using differences

I am trying to find a function generating/suitable for the following sequence:
$${2,6,12,20,30,42,56...}$$

The first order differences are:
$$4, 6, 8, 10, 12, 14...$$

and the second order differences are:
$$2, 2, 2, 2, 2, 2...$$

this means that:
$$y''= 2 \implies y' = 2\cdot x + c$$ and we can see that for $$c = 2$$ we can get the correct numbers i.e. the first order differences (replacing starting from $$x = 1$$). So $$y' = 2\cdot x + 2$$
This means that $$y = x^2 + 2\cdot x + c$$
But I can't find any value of $$c$$ that would give the original sequence.

Now I see that $$y'$$ could also be $$y' = 2\cdot x + 4$$ as it gives the correct numbers if we replace starting from $$x = 0$$ but then we have $$y = x^2 + 4\cdot x + c$$ and I again can't find any $$c$$ that matches the original sequence.

What am I doing wrong here?

• Well, inspection gives $a_n=n(n+1)$
– lulu
Apr 6, 2022 at 17:00
• You've figured out $x_{n+1}-x_n = 2n+2$, $x_1 = 2$, so try working recursively. (wolframalpha.com/…) Apr 6, 2022 at 17:06
• @User203940: Isn't that is $y'$ though? That is why I go towards $n^2 + 2$. Otherwise I am confused here
– Jim
Apr 6, 2022 at 17:07
• When you say "function generating" this sequence, do you mean an explicit formula for the $n$-th term, or do you mean the literal generating function? Apr 6, 2022 at 17:16
• @user170231: the generating function. Is there a difference with the formula for the $n-th$ term?
– Jim
Apr 6, 2022 at 18:21

You start with the sequence $$y=2,6,12,20,30,42,56$$ when $$x=1,2,3,\dots$$.

You have found that $$y(2)-y(1)=4, y(3)-y(2)=6, y(4)-y(3)=8,\dots$$, but this is not the same as finding $$y'(1), y'(2),\dots$$, because the value of $$y'$$ is not constant over these intervals.

In fact, since $$y'$$ changes linearly (for $$y$$ is a quadratic), you have actually found $$y'(1.5),y'(2.5),y'(3.5),\dots=4,6,8,\dots$$ i.e. these numbers $$4,6,8,\dots$$ are the average rate of change across the intervals $$[1,2],[2,3],\dots$$, which happens to be the exact rate of change at the midpoint of the interval.

For $$y''$$, which is constant everywhere, this nuance does not matter: $$y''=2$$ and you have correctly concluded that $$y$$ will have a term of $$x^2$$ (with coefficient $$1$$).

However, for the coefficient of $$x$$, you need to note that $$y'(1.5)=4$$ so if $$y=x^2+bx+c$$ then $$y'=2x+b$$, so $$y'(1.5)=2\times 1.5+b=4\implies b=1$$.

Thus, we have that $$y=x^2+x+c$$ and by comparing sequences, we see $$c=0$$.

Sequences are not generally thought of in these terms of raw calculus, which studies continuous change over $$\mathbb{R}$$ rather than discrete changes over $$\mathbb{N}$$. Instead, we might study recurrence relations.

A method often used for identifying quadratics would be to just find the $$x^2$$ coefficient by halving the difference of differences (here, $$2$$) and then subtract that off the original sequence to see what you're left with: $$2,6,12,20,30,\dots-1,4,9,16,25,\dots=1,2,3,4,5,\dots$$. In general, you know this sequence will be linear; here, you can just see outright it is the identity sequence $$x$$, so the full sequence is $$x^2+x$$.

In this case, as a commenter said, you could possibly spot the solution outright: if you notice the numbers are all times tables values you might see $$56=7\times 8,42=6\times 7,30=5\times 6$$ and conclude $$y=x\times (x+1)$$.

• "are the average rate of change across the intervals $[1,2],[2,3],\dots$, which happens to be the exact rate of change at the midpoint of the interval." I can understand that because the change is linear and we have the change across $1$ unit, we have the rate of change from $[2,3]$ to be $\frac{12-6}{3-2}=\frac{6}{1}=6$. But how do we infer that it is the exact change at the midpoint? I.e. $2.5$?
– Jim
Apr 7, 2022 at 8:46
• @Jim we know the gradient from $2$ to $3$ is a straight line and the average value on this line segment is $6$, because $y(3)-y(2)=\int_2^3 y'(x) dx=6$. The average value (by thinking about integration as area) also happens to be the area underneath the line segment. This area - a triangle plus a rectangle - is the same area as underneath a horizontal line that passes through the midpoint i.e. $\int_2^3 \frac{1}{2}(y(3)-y(2)) dx$. Sketch it out with some different lines and values.
– A.M.
Apr 7, 2022 at 9:33
• "we know the gradient from 2 to 3 is a straight line" we know this because of the second differences which are constant right? Otherwise I am not sure how do we infer this
– Jim
Apr 7, 2022 at 10:02
• @Jim yes, correct.
– A.M.
Apr 7, 2022 at 10:27

Let $$\{a_n\}_{n\in\mathbb N}=\{2,6,12,20,30,42,56,\ldots\}$$. Let $$\{b_n\}_{n\in\mathbb N}$$ be the sequence of the first-order forward differences of $$\{a_n\}$$, and $$\{c_n\}_{n\in\mathbb N}$$ the sequence of second-order differences. So for $$n\ge1$$,

$$b_n = a_{n+1} - a_n \\ c_n = b_{n+1} - b_n$$

As you observed, $$c_n=2$$ for all $$n$$. Then

\begin{align*}b_{n+1} &= b_n + 2 \\ &= b_{n-1}+2\times2 \\ &= b_{n-2}+3\times2 \\ &~~\vdots \\ &= b_1 + 2n \end{align*}

which means $$b_n = b_1+2(n-1)=2(n+1)$$.

Now solve for $$a_n$$:

\begin{align*} a_{n+1} &= a_n + 2(n+1) \\ &= a_{n-1} + 2((n+1) + n) \\ &= a_{n-2} + 2((n+1) + n + (n-1)) \\ &~~\vdots \\ &= a_1 + 2 \sum_{k=0}^{n-1} (n+1-k)\\ &= a_1 + n^2 + 3n \end{align*}

so that $$a_n = 2 + (n-1)^2 + 3(n-1) = n^2 + n$$.

In case you are interested in the generating function, it takes the form

$$f(x) = \sum_{n=0}^\infty a_n x^n$$

Recall that if $$|x|<1$$, we have

$$\frac1{1-x} = \sum_{n=0}^\infty x^n$$

and finding $$f(x)$$ is just a matter of differentiating both sides twice.

$$\frac1{(1-x)^2} = \sum_{n=0}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^{n-1} = \sum_{n=0}^\infty (n+1)x^n$$

$$\frac2{1-x^3} = \sum_{n=0}^\infty (n^2+n) x^{n-1}$$

$$\implies f(x) = \frac{2x}{1-x^3} = \sum_{n=0}^\infty (n^2+n) x^n$$