Difference Sequences

Difference Sequences

We have already looked at The Difference Operator and More Properties of the Difference Operator. We will now look at this operator in terms of sequences.

 Definition: Consider a sequence $(a_i)_{i=o}^{\infty} = (a_0, a_1, a_2, ..., a_i, ...)$ of real numbers. The First Order Difference Sequence of $(a_i)_{i=0}^{\infty}$ denoted $(\Delta a_i)_{i=0}^{\infty}$ is a sequence of real numbers whose terms satisfy $\Delta a_i = a_{i+1} - a_i$, that is, $(\Delta a_i)_{i=0}^{\infty} = (a_{1} - a_0, a_2 - a_1, a_3 - a_2, ..., a_{i+1} - a_i, ...)$.

For example, consider the following sequence of squared nonnegative integers:

(1)
\begin{align} \quad (i^2)_{i=0}^{\infty} = (0, 1, 4, 9, 16, 25, 36, ..., i^2, ...) \end{align}

The first order different sequence of $(i^2)_{i=0}^{\infty}$ is:

(2)
\begin{align} \quad (\Delta i^2)_{i=0}^{\infty} (1, 3, 5, 7, 9, 11, ..., (i+1)^2 - i^2, ... \end{align}

Note that $( \Delta i^2 )_{i=0}^{\infty}$ appears to be the sequence of increasing positive odd integers. This can easily be proven since:

(3)
\begin{align} \quad \Delta i^2 = (i + 1)^2 - i^2 = i^2 + 2i + 1 - i^@ = 2i + 1 \end{align}

Therefore $(\Delta i^2)_{i=0}^{\infty} = (2i + 1)_{i=0}^{\infty}$ which is the sequence of increasing positive odd integers.

Of course, we can also obtain higher order difference sequences.

 Definition: Consider a sequence $(a_i)_{i=o}^{\infty} = (a_0, a_1, a_2, ..., a_i, ...)$ of real numbers. The $p^{\mathrm{th}}$ Order Difference Sequence of $(a_i)_{i=0}^{\infty}$ denoted $(\Delta^p a_i)_{i=0}^{\infty}$ is a sequence of real numbers whose terms satisfy $\Delta^p a_i = \Delta^{p-1} a_{i+1} - \Delta^{p-1} a_i$.

From our example above, we have that the second order difference sequence of $(i^2)_{i=0}^{\infty}$ is:

(4)
\begin{align} \quad (\Delta^2 i^2)_{i=0}^{\infty} = (\Delta ( \Delta i^2))_{i=0}^{\infty} = (\Delta (2i + 1))_{i=0}^{\infty} = (2, 2, 2, ... ) \end{align}

We see that $(\Delta^2 i^2)_{i=0}^{\infty}$ seems to be the infinite sequence whose terms are all $2$s. Once again, proving this is relatively easy since:

(5)
\begin{align} \quad \Delta^2 i^2 = \Delta ( \Delta i^2) = \Delta (2i + 1) = [2(i+1) + 1] - [2i + 1] = 2i + 2 + 1 - 2i - 1 = 2 \end{align}

Therefore $(\Delta^2 i^2)_{i=0}^{\infty} (2i^0)_{i=0}^{\infty} = (2, 2, 2, ...)$

Furthermore, the third order difference sequence of $(i^2)_{i=0}^{\infty}$ is:

(6)
\begin{align} \quad (\Delta^3 i^2)_{i=0}^{\infty} = (\Delta (\Delta^2 i^2))_{i=0}^{\infty} = (\Delta 2i^0)_{i=0}^{\infty} = (0, 0, 0, ...) \end{align}