DHARMDISCRETE NUMERIC FUNCTIONS AND GENERATING FUNCTIONS 51Differentiate w.r.t z
1.5/(1 – 5z)^2 = 0 + 5.1 + 2.5^2 z + 3.5^3 z^2 + .......... + n.5n zn–1 + ..................
or 5/(1 – 5z)^2 = 0.5^0 + 1.5^1 + 2.5^2 z + 3.5^3 z^2 + .......... + n.5n zn–1 + .............
Multiply both sides with z, so we have
5 z/(1 – 5z)^2 = 0.5^0 z + 1.5z + 2.5^2 z^2 + 3.5^3 z^3 + .......... + n.5n zn + ..............
or 5 z/(1 – 5z)^2 = 0.5^0 + 1.5z + 2.5^2 z^2 + 3.5^3 z^3 + ..........+ n.5n zn + ................
Therefore, 5z/(1 – 5z)^2 is the generating function for (0.5^0 , 1.5^1 , 2.5^2 , ........, n.5n, .............).
Example 2.35 Determine generating function and discrete numeric function of series
- 3/2 + 2 + 0 + 8 + ........
Sol. Let scale of relation of the given series be 1 – pz – qz^2. Obtain p and q from the equations,
0 – 2p + 3/2q = 0 and 8 – 0 – 2q = 0
So we obtain q = 4 and p = 3. Whence, scale of relation is 1 – 3z – 4z^2.
Hence, the generating function A(z) for series, – 3/2 + 2.z + 0.z^2 + 8.z^3 + ........ is
= [– 3/2 + (2 + 3.3/2)z]/(1 – 3z – 4z^2 ) (using equation (2.7))
= (– 3/2 + 13/2z )/ (1 – 3z – 4z^2 )
So, A(z) = (– 3/2 + 13/2z)/(1 – 3z – 4z^2 )
Since A(z) can be written as (using partial fraction)
= (8/5)/(1 + z) – (31/10)/(1 – 4z)
So numeric function will be the coefficient of zn that will be,
an = (8/5).(– 1)n.1 – (31/10).4n
or an = (– 1)n .(8/5) – (31/10).4n for n ≥ 0.
2.5 Application of Generating Function to Solve Combinatorial Problems.........................
The important application of generating function representation of numeric function is the
solving of combinatorial problems. Consider the numeric function an, i.e.,an =
m
nF
HGI
KJ for a fixed value of nSince, am
n n
m
n nmn=F
HGI
KJ==
F
HGI
KJ=>R
S||
T
|
|100forforwe recall following definitions that are frequently used in solving of combinatorial problems,- The expression n(n – 1) ..... 2 .1 is called n – factorial, and denoted by n!, with the
convention 0! = 1.
m
n
F
HGI
KJ is called binomial coefficient, i.e., (number of n – subsets of an m – set)
m
nF
HGI
KJ
= m!/n! (m – n)!- Binomial coefficients m
n
F
HGI
KJ
have no combinatorial meaning for negative values of m.