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}