DHARMDISCRETE NUMERIC FUNCTIONS AND GENERATING FUNCTIONS 33Thus,
S–1 (∇an) = S–1 (an – an–1) = S–1 an – S–1 an–1 for n ≥ 1.
∴ S–1 an = an+1 for n ≥ 1 and S–1 an–1 = an–1+1 for n ≥ 1
Hence,
S–1 an – S–1 an–1 = an+1 – an for n ≥ 1 = ∆an RHS
Example 2.12. Let a numeric function an, i.e.,
an = n^3 – 2n^2 + 3n + 2 for n ≥ 0
Then determine ∆an, ∆^2 an, ∆^3 an, ......., ∆nan.
Sol. Remember the asked terms ∆an, ∆^2 an, ∆^3 an, ......., ∆nan are known as first order, second
order, third order and so on the nth order forward difference.
Since, ∆an = an+1 – an, for n ≥ 0, then we can determine an+1 from an as,
an = n^3 – 2n^2 + 3n + 2 for n ≥ 0
so an+1 = (n + 1)^3 – 2(n + 1)^2 + 3(n + 1) + 2 for (n + 1) ≥ 0 or n ≥ 0.
Therefore,
∆an = an+1 – an = {(n + 1)^3 – n^3 } – 2{(n + 1)^2 – n^2 } + 3{(n + 1) – n}
∆an = 3n^2 – n + 2 for n ≥≥≥≥≥ 0
Now find ∆^2 an = ∆(∆an), let ∆an = bn.
So ∆^2 an = ∆(bn) = bn+1 – bn for n ≥ 0
Since, bn = 3n^2 – n + 2, so bn+1 = 3(n + 1)^2 – (n + 1) + 2, for n ≥ 0
Therefore, ∆(bn) = bn+1 – bn = 3(2n + 1) – 1 = 6n + 2 for n ≥ 0.
Hence, ∆^2 an = 6n + 2 for n ≥≥≥≥≥ 0
Further assume that cn = ∆^2 an = 6n + 2 so cn+1 = 6(n + 1) + 2 for n ≥ 0.
Then
∆^3 an = ∆(∆^2 an) = ∆(cn) = cn+1 – cn = 6(n + 1) + 2 – 6n – 2
∆^3 an = 6 for n ≥ 0.
Similarly we can determine ∆ian = 0 (where i ≥ 4) for n ≥≥≥≥≥ 0.
Example 2.13. Let a numeric function an be a polynomial of the form α 0 + α 1 n + α 2 n^2 + ...... +
αk nk. Show that ∆k+1an = 0.
Sol. From the previous example we have seen that increase in the degree of the forward differ-
ence then the degree of the polynomial is reduced by one on successive steps, i.e., if
an = α fk(n) where α is an scalar, then
∆an = α fk–1(n)
∆^2 an = α fk–2(n)
...
∆kan = α f^0 (n) = α
Therefore if we go further and determine the forward difference of k + 1th order then we
have,
∆k+1an = ∆(∆kan) = ∆(α) = α – α = 0. [∴ an+1 = α and so an = α]
Hence, proved.
Example 2.14 Let an and bn are two numeric functions such that
an = n + 1 and bn = an for n ≥ 0.
Determine ∆(anbn).