GENERATING FUNCTIONS

Before going to start the exact discussion on generating functions let us begin our tour from the origin of generating functions. Assume a series a₀ + a₁ + a₂ + …….. + a_n in which, from and after a certain term is equal to the sum of a fixed number of preceding terms multiplied respectively by certain constant is called recurring series. For example in the series 1 + 2z + 3z² + 4z³ + 5z⁴ + …. , where each term after the second is equal to the sum of the two preceding terms multiplied respectively by the constants 2z and – z². These quantities being constants because they are remain same for all values of n. Thus,

5z⁴ = (2z). 4z³ + (– z²). 3z² Therefore we assume that

a₄ = 2z. a₃ – z². a₂

In general when n is greater then 1, each term is represented by its two immediately preceded terms through the equation,

a_n = 2z. a_n–1 – z². a_n–2

or a_n – 2z. a_n–1 + z². a_n–2 = 0 (2.1)

Equation 2.1 is called scale of relation in which coefficients a_n, a_n–1, and a_n–2 are taken with their proper signs. Thus the series 1 + 2z + 3z² + 4z³ + 5z⁴ + …… has scale of relation 1 – 2z + z². Consequently, if the scale of relation of a recurring series is given, then we could find any term of the series, from sufficient number of known preceding terms.

For example, let 1 – pz – qz² – wz³ (2.2)

is the scale of relation of the series a₀ + a₁ z + a₂ z² + …….. + a_n zⁿ (2.3) Then we have, a_n zⁿ ≥ p_z. a_n–1 z^n–1 + qz². a_n–2 z^n–2 + wz³. a_n–3 z^n–3;

or a_n = p. a_n–1 + q. a_n–2 + w. a_n–3; (2.4)

Thus, any coefficient can be found when coefficients of the three preceding terms are known.

Let us take a series shown in (2.3) and extended up to infinite terms, i.e.,

a₀ + a₁ z + a₂ z² + …….. + a_n zⁿ + ………… (2.5)

and assume the scale of relation is 1 – pz – qz². So we have a_n – p a_n–1 – q a_n–2 = 0

Assume, A(z) = a₀ + a₁ z + a₂ z² + …….. …….. + a_n–1 z^n–1

– pz.A(z) = – pa₀ z – pa₁ z² – ……….. – p_an–2 z^n–1 – p a_n–1 zⁿ – qz².A(z) = – qa₀ z² – ……….. – qa_n–3 z^n–1 – q a_n–2 zⁿ – qa_n–1 zⁿ⁺¹ (1 – pz – qz²) A(z) = a₀ + (a₁ – pa₀) z + …………. – (pa_n–1 + qa_n–2) zⁿ – qa_n–1.zⁿ⁺¹ For the coefficient of every other power of z is zero in consequence of the relation

a_n – p a_n–1 – q a_n–2 = 0 i.e., A(z) = [a₀ + (a₁ – pa₀) z]/(1 – pz – qz²)

– [(pa_n–1 + qa_n–2) zⁿ – qa_n–1 zⁿ⁺¹]/(1 – pz – qz²) (2.6) For large value of n series (2.5) is infinite and so second fraction of the equation (2.6) decreases indefinitely. Thus we have the sum

A(z) = [a₀ + (a₁ – pa₀) z]/(1 – pz – qz²) (2.7) If we develop this fraction in ascending powers of z, we shall obtain as many terms of the original series as we please. Therefore, the expression (2.7) is called generating function of the infinite series (2.5).

As we mention in the previous section that for a numeric function a_n we uses a₀, a₁, a₂,

…, a_n, …..to denote its values at 0, 1, 2, ……, n, …… Here we introduce an alternative way to represent numeric functions, such as for a numeric function (a₀, a₁, a₂, …, a_n, …..) we define an infinite series,

a₀ + a₁ z + a₂ z² + …….. + a_n zⁿ + …………

Here z is a formal variable and powers of z is used as generator in the infinite series such that the coefficient of zⁿ returns the value of the numeric function at n. For a numeric function a_n we write A(z) to denote its generating function and so the above infinite series can be written in closed form as shown in equation (2.7), i.e.,

A(z) = [a₀ + (a₁ – pa₀)z]/(1 – pz – qz²) (where (1 – pz – qz²) is the scale of relation)

For example, the numeric function (5⁰, 5¹, 5², …, 5ⁿ, …..) we define an infinite series, 5⁰ + 5¹ z + 5² z² + …….. + 5ⁿ zⁿ + …………

Therefore we obtain the summation of infinite series A(z) = 1/(1 – 5z). A(z) is the gener-ating function that represent the numeric function in a compact way. Hence for a numeric function we can easily obtain its generating function and vise - versa. In fact an alternate way to representing numeric function leads to efficiency and easiness in some context that we wish to carry out.

Consider a numeric function a_n = 1 for n ≥ 0, then its generating function A(z) = 1/(1 – z). Similarly the generating function of the numeric function b_n = kⁿ for n ≥ 0 will be B(z) = 1/(1 – kz). It is quite often that generating function can be expressed as a group of equivalent partial fractions and the general term of series may be easily found where the coefficient of zⁿ represent the corresponding numeric function. Thus, suppose generating function can be decomposed into partial fractions,

P/(1 – αz) + Q/(1 + βz) + R/(1 – γz) Then coefficient of the zⁿ in the general term is

{P. αⁿ + (– 1)ⁿ B. βⁿ + (n + 1) R. γⁿ} for n ≥ 0 that will be equivalent to numeric function.

Example 2.27 Find the numeric function for the generating function A(z) = (1 – 8z)/(1 – z – 6z²)

Sol. Since A(z) can be expressed in partial fraction i.e., A(z) = 2/(1 + 2z) – 1/(1 – 3z)

Then obtain the coefficient of zⁿ in the general term and it will come out to be (– 1)ⁿ 2ⁿ⁺¹ – 3ⁿ for n ≥ 0

That is the required numeric function of A(z). Alternatively, we say a_n = (– 1)ⁿ 2ⁿ⁺¹ – 3ⁿ for n ≥ 0.

Example 2.28 Find the generating function and the numeric function for the series 1 + 6 + 24 + 84 + …………..

Sol. The given series is transformed into another infinite series by introducing a formal pa-rameter z and using power of z as generator in it, i.e.

1 + 6z + 24z² + 84z³ + ………….. (2.8)

The scale of relation of the above series is (1 – 5z + 6z²). To explain the method of determining the scale of relation, let’s assume (1 – pz – qz²) is the scale of relation for the series (2.8). Therefore to obtain p and q we have the equations,

24 – 6 p – q = 0 and 84 – 24 p – 6q = 0 whence, p = 5, and q = – 6, So the scale of relation is (1 – 5z + 6z²).

Hence, using equation (2.7) we obtain the generating function A(z) = (1 + z)/(1 – 5z + 6z²).

Since A(z) can be written as,

A(z) = 4/(1 – 3z) – 3/(1 – 2z)

Now obtain the coefficient of zⁿ in the general term of A(z) that will return the numeric function a_n i.e.

a_n = 4.3ⁿ – 3.2ⁿ for n ≥ 0 Example 2.29 Find numeric function for generating function

A(z) = 2/(1 – 4z²) Sol. Since A(z) can be written as

A(z) = 1/(1 – 2z) + 1/(1 + 2z) Thus we obtain the numeric function a_n as,

a_n = 2ⁿ + (– 1)ⁿ . 2ⁿ for n ≥ 0

Now we shall discuss the common features of generating function corresponding to the features of numeric functions. Let a_n, b_n, and c_n are numeric function and A(z), B(z), and C(z) are their corresponding generating functions then,

• If c_n = a_n + b_n then corresponding generating function C(z) = A(z) + B(z). For example, Let a_n = 2ⁿ and b_n = 3ⁿ for n ≥ 0 then c_n = 2ⁿ + 3ⁿ for n ≥ 0. So C(z) = 1/(1 – 2z) + 1/(1 – 3z) = (2 + 3z – 6z²)/(1 – 2z).

• If b_n = ka_n, where k is any arbitrary constant then corresponding generating func-tion is B(z) = kA(z). For example let a_n = 5.2ⁿ for n ≥ 0 then A(z) = 5/(1 – 2z) is the numeric function.

• If b_n = kⁿa_n, where k is any arbitrary constant then corresponding generating func-tion is B(z) = A(kz) = 1/(1 – kz). Since numeric funcfunc-tion b_n is expressed in infinite series as

a₀ + k.a₁z + k² a₂z² + ………. + kⁿ a_nzⁿ + ……

= a₀ + a₁(kz) + a₂(kz)² + ………. + a_n (kz)ⁿ + ……

Therefore, generating function B(z) = A(kz) = 1/(1 – kz).

• Since A(z) is the generating function of function a_n so generating function of S^I a_n will be z^I A(z), for any positive integer I. For example, let generating function B(z) = z⁸/(1 – 3z) then it can be written as,

B(z) = z⁸. 1/(1 – 3z) ≡ z^I A(z)

Then its corresponding numeric function will be S⁸ a_n where a_n is the numeric func-tion for the funcfunc-tion 1/(1 – 3z) that is equal to 3ⁿ.

Therefore, we obtain numeric function S⁸ 3ⁿ, i.e.,

S for 0

• Likewise, generating function for the numeric function S^–I an for any integer I, will be

z^–I [A(z) – a₀ – a₁z – a₂z² – …….. – a_I–1 z^I–1]

Since A(z) = a₀ + a₁z + a₂z² + …….. + a_I–1 z^I–1 + a_I z^I ……. + a_n zⁿ + ….

or, A(z) – a₀ – a₁z – a₂z² – …….. – a_I–1 z^I–1 = a_I z^I + a_I+1 z^I+1 + ….. + a_n zⁿ + ….

Multiplied both side by z^–I, hence we obtain

z^–I [A(z) – a₀ – a₁z – a₂z² – …….. – a_I–1 z^I–1] = a_I + a_I+1 z + ….. + a_n z^n–I + ….

= a_n+I zⁿ for n ≥ 0 That is equivalent to S^–I a_n.

For example, let a_n = 3ⁿ⁺⁵ for n ≥ 0 which is equivalent to numeric function S^–5.3ⁿ and its corresponding generating function can be computed by using the expression,

A(z) = z^–I [A(z) – a₀ – a₁z – a₂z² – …….. – a_I–1 z^I–1] i.e., A(z) = z^–5 [1/(1 – 3z) – 3⁰ – 3¹z – 3²z² – 3³ z³ – 3⁴ z⁴]

= z^–5 [3⁵.z⁵/(1 – 3z)] = 3⁵/(1 – 3z).

Example 2.30 Find the generating function for the numeric function a_n such that

a

Sol. Let A(z) is the generating function for a_n where,

A(z) = a₀ + a₁z + a₂z² + …….. + a_n zⁿ + ….

Since values of a₀ to a₅ and a₁₀ onwards are equal to zero. So A(z) = a₆ z⁶ + a₇ z⁷ + a₈ z⁸ + a₉ z⁹ + a₁₀ z¹⁰ = 1.z⁶ + 2.z⁷ + 3.z⁸ + 4.z⁹ + 0.z¹⁰ = z⁶.(1 + 2z + 3z² + 4z³)

Note that Generating function to the convolution of numeric functions i.e. c_n = a_n*b_n will be A(z) . B(z). Since

c_n = a_n*b_n = a₀b_n + a₁b_n–1 + ……… + a_n–1b₁ + a_nb₀ which is the coefficient of zⁿ in the product of series

(a₀ + a₁z + a₂z² + …… + a_n zⁿ + ….) . (b₀ + b₁z + b₂z² + ….+ a_n zⁿ + ….) And it’s generating function is equal to the product of the generating functions of both A(z) and B(z), i.e.

C(z) = A(z) . B(z)

Example 2.31 Let a_n = 3ⁿ and b_n = 5ⁿ for n ≥ 0, then determine the generating function for the convolution of numeric functions a_n and b_n.

Sol. Since generating function corresponding to numeric function a_n and b_n are A(z) = 1/(1 – 3z) and 1/(1 – 5z) respectively. Let c_n = a_n*b_n then its generating function is given by C(z) = A(z) . B(z). Therefore

C(z) = 1/(1 – 3z) . 1/(1 – 5z)

Alternatively, C(z) can be written as using partial fraction C(z) = – 3/2(1 – 3z) + 5/2(1 – 5z) Therefore its numeric function will be,

c_n = – 3/2. 3ⁿ + 5/2 . 5ⁿ or c_n = (– 1/2). 3ⁿ⁺¹ + (1/2) .5ⁿ⁺¹ for n ≥ 0

• If c_n = Σ

I=0

n a_I then its generating is C(z) = A(z) . 1/(1 – z) where A(z) is the generating function for function a_I for I = 0 to n. To prove this fact we recall that if the convolu-tion of numeric funcconvolu-tion a_n and b_n is c_n, then

c_n = _I=0Σⁿ a_I . b_n–I

And its generating function is given as C(z) = A(z) . B(z) Assume b_n–I = 1 then

c_n = _I=0Σⁿ a_I . 1

Since generating function for function b_n–I = 1 for n ≥ 0 or (1, 1, 1, …….. 1, ….) is B(z) = 1/(1 – z). Therefore,

C(z) = A(z) . 1/(1 – z)

where A(z) is the generating function of the numeric function Σ

I=0 n a_I

Example 2.32 Determine the generating function for the numeric function (1, 2, 3, ……., n,

….).

Sol. Since 1/(1 – z) = 1 + z + z² + z³ …………. + zⁿ + zⁿ⁺¹ + ….

Differentiate both side w.r.t.z, i.e.,

1/(1 – z)² = 1 + 2z + 3z² + …………. + nz^n–1 + (n + 1)zⁿ + ….

Thus, we obtain the generating function 1/(1 – z)² for the numeric function (1, 2, 3, ….., n, …).

Conversely, for generating function 1/(1 – z)², the coefficient of zⁿ (general term) will be,

= (– 1)ⁿ (– 2) (– 2 – 1) …….. (– 2 – n + 1)/n!

= 2 .3 ……. (n + 1)/n!

= (n + 1)

Therefore we obtain the numeric function a_n = n + 1 for n ≥ 0 for it values (1, 2, …, n ,..) Example 2.33 Determine the generating function for the numeric function (0²,1², 2², ……., n²,

….).

Sol. Since 1/(1 – z) = 1 + z + z² + z³ …………. + zⁿ + ….

Differentiate both sides w.r.t.z, i.e.,

1/(1 – z)² = 1 + 2z + 3z² + …………. + nz^n–1 + ….

Multiply both sides by z so we obtain

z.1/(1 – z)² = 1.z + 2z² + 3z³ + …………. + nzⁿ + ….

Differentiate again w.r.t.z and multiply both sides with z, i.e.,

z. (1 + z)/(1 – z)³ = 0² + 1².z + 2².z² + 3².z³ + ………….+ n².zⁿ + ….

Thus, we obtain the generating function z. (1 + z)/(1 – z)³ for the numeric function (0²,1², 2², ……., n², ….).

Conversely, the coefficient of z_n in z. (1 + z)/(1 – z)³ is n²†, therefore a_n = n² for n ≥ 0.

Example 2.34 Find generating functions of the following discrete numeric functions (i) 1, 2/3, 3/9, 4/27, ……..(n + 1)/3ⁿ, ……

(ii) 0.5⁰, 1.5¹, 2.5², …….., n.5ⁿ, ………..

Sol. (i) Let A(z) is the generating function of the series (1, 2/3, 3/9, 4/27, …….. (n + 1)/3ⁿ, ……), i.e.,

A(z) = 1 + z.2/3 + z².3/9 + z³.4/27 + …….. + zⁿ.(n + 1)/3ⁿ + ………

or A(z) = 1 + 2.(z/3) + 3.(z/3)² + 4.(z/3)³ + ……..+ (n + 1).(z/3)ⁿ +…...

Since, 1/(1 – z)² = 1 + 2z + 3z² + …………. + (n + 1)zⁿ + ….

Replace z by z/3 so above equation becomes

1/(1 – z/3)² = 1 + 2.(z/3) + 3.(z/3)² + …………. + (n + 1).(z/3)ⁿ + ….

Therefore, 1/(1 – z/3) is the required generating function.

(ii) Let B(z) is the generating function of the series (0.5⁰, 1.5¹, 2.5², …….., n.5ⁿ, ………..), i.e.,

B(z) = 0.5⁰ + 1.5¹ z + 2.5² z² + …….. + n.5ⁿ zⁿ + ………..

Since, 1/(1 – z) = 1 + z + z² + z³ …………. + zⁿ + ………….

Replace z by 5z thus we obtain the equation

1/(1 – 5z) = 1 + 5z + (5z)² + (5z)³ …………. + (5z)ⁿ + ………….

† Since general term of 1/(1 – z)³, is given as

= (– 1)ⁿ. zⁿ .(– 3). (– 4). ………(– 3 – n + 1)/n!

Thus, the coefficient of zⁿ in 1/(1 – z)³ is

= (3). (4). ……… (2 + n)/n!

= (n + 2) (n + 1)/1. 2

Therefore the coefficient of zⁿ in the expansion of z.(1 + z)1/(1 – z)³ is

= n .(n + 1)/2 + (n – 1) .n/2

= n .2n/2

= n²

Differentiate w.r.t z

1.5/(1 – 5z)² = 0 + 5.1 + 2.5²z + 3.5³z² + ………. + n.5ⁿ z^n–1 + …………...

or 5/(1 – 5z)² = 0.5⁰ + 1.5¹ + 2.5²z + 3.5³z² + ………. + n.5ⁿ z^n–1 + ………….

Multiply both sides with z, so we have

5z/(1 – 5z)² = 0.5⁰ z + 1.5z + 2.5²z² + 3.5³z³ + ………. + n.5ⁿ zⁿ + …...

or 5z/(1 – 5z)² = 0.5⁰ + 1.5z + 2.5²z² + 3.5³z³ + ……….+ n.5ⁿ zⁿ + …………....

Therefore, 5z/(1 – 5z)² is the generating function for (0.5⁰, 1.5¹, 2.5², …….., 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 – qz². Obtain p and q from the equations, 0 – 2p + 3/2q = 0 and 8 – 0 – 2q = 0 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 z_n that will be, a_n = (8/5).(– 1)ⁿ.1 – (31/10).4ⁿ

or a_n = (– 1)ⁿ .(8/5) – (31/10).4ⁿ for n ≥ 0.

2.5 APPLICATION OF GENERATING FUNCTION TO SOLVE COMBINATORIAL

In document Fundamentals of Mathematics for Computer Science (Page 62-68)