Download 5 generating functions

Generating Functions
A robust tool to solve a wide variety of problems that deal
with sequences, series, permutations, combinations, and
more.
A generating function gives us a way to mathematically
represent or encode such problems.
– Definition
– Purpose
– Examples
Generating function
• A polynomial with either
– Interesting choices for its coefficients
(useful for recurrence relations)
– Interesting choices for its exponents
(useful for permutations/combinations)
• Customary to write in ascending powers of x
– Coefficients can correspond to terms of a sequence
Sequences have no mathematical operators. Writing as a
polynomial provides a way to manipulate sequences.
– Convenient if number of terms is infinite
• We almost never substitute a value for x
Purpose
• Our textbook shows how to solve simple:
– Permutation & combinations
– Recurrence relations
Using techniques that are narrowly defined.
• Generating functions allow you to solve a much
wider variety of problems in a systematic, less ad-hoc
way.
• Disadvantage: not very easy to use
– Can be tedious
– Requires techniques such as binomial expansion or partial
fractions along the way
Preliminary remarks
• How many terms do you get if you multiply…?
(a + b)(c + d)
(a + b)(c + d)(e + f)
(a + b + c)(d + e + f + g)
• Notice that each term of the product tells you which
number you selected inside each ( ).
– If you replace each letter with “1”, you get # of terms!
• Analogous concept: how many factors does a
number have, e.g. 1200?
Combinations
• Consider the multiplication of (1+ax)(1+bx)(1+cx)
• We need to create terms, where we select something
from each factor:
– Select the 1 or ax from (1+ax)
– Select the 1 or bx from (1+bx)
– Select the 1 or cx from (1+cx)
– And you sum all 8 possibilities to get:
1 + (a+b+c)x + (ab + bc + ac)x2 + (abc)x3
• What are the coefficients saying? 
• If you take away the letters a,b,c, you get the answer
to a different question: (1+x)3 = 1 + 3x + 3x2 + x3
continued
• A simple generating function for combinations is
(1+x)(1+x)(1+x)…(1+x) = (1+x)n
– The coefficient of xr is the number of ways to choose r
objects from a set of n distinct objects.
– You are basically deciding which x’s to multiply to produce
the xr.
• What if the collection from which we are selecting
has duplicates? Notice the contrast:
– Select 2 objects from { A, B, C, D }
– Select 2 objects from ( A, A, B, B ).
Duplicates
• To select from ( A, A, B, B ) is like selecting from a set
{ A, B} with the ability to choose up to two of each!
• Try this generating function:
(1 + ax + a2x2) (1 + bx + b2x2)
= 1 + (a+b)x + (a2 + ab + b2)x2 + (ab2 + a2b)x3 + a2b2x4.
• If we just want the number of ways, we don’t need
to specify “a” and “b”:
(1+x+x2) (1+x+x2) = 1 + 2x + 3x2 + 2x3 + x4.
• What do the coefficients 1, 2, 3, 2, 1 tell us?
• Try another problem like this.
Duplicates (2)
• I have 10 A’s and 3 B’s. How many ways are there to
select r objects, given that the number of A’s I can
select must be a multiple of 5?
(1 + x5 + x10)(1 + x + x2 + x3)
= 1 + x + x2 + x3 + x5 + x6 + x7 + x8 + x10 + x11 + x12 + x13
• What answers do you see in this polynomial?
• Restriction on A’s: this problem can be stated as a
coin problem with pennies and nickels!
Exponent on x = total number of cents
• We can easily generalize to different numbers plus
more categories such as C’s, D’s, etc.
Formulating
• We need a separate factor for each distinct value we are
selecting, or for each distinct category, die roll, digit in a
number, etc.
• Decide on what the exponent of x means. Possibilities:
– Number of objects we select
– The sum of the numbers selected
– How many odd / even / multiple of 3 …
• When done, coefficient of some term gives us our answer.
– You want to know just how many, or a full enumeration?
• More than one variable can be used if you need to keep track
of more than one quantity about the selection.
Applications
• How to select r objects from a set
– Allowing for repetition of the same element
– Restrictions
– Answer = coefficient of xr in some ( ) ( ) generating
function.
• Similar problems that are only superficially different
– Coin problems
– Dice problems
Review
• What combination question would this generating
function help solve?
(1 + ax + a2x2) (1 + bx) (cx + c2x2)
– We can elegantly encode restrictions: maximum,
minimum, even number, etc.
– What if we just wanted the # of possibilities?
– What if the a’s were unlimited?
• The number of ways to choose coins to obtain c
cents from pennies and nickels.
– Think of the set { P, N } but the number of N’s you choose
must be a multiple of 5.
Dice application
• How many ways can we roll a sum of 7 with 2 dice?
• Formulation:
(x + x2 + x3 + x4 + x5 + x6) (x + x2 + x3 + x4 + x5 + x6)
= (x + x2 + x3 + x4 + x5 + x6)2
And we want the coefficient of x7.
• We can match x with x6; x2 with x5; etc.
– Thought process is parallel to how you might do with
without generating functions.
– The answer is 6.
• Note that this question is equivalent to asking for
coefficient of x5 in (1 + x + x2 + x3 + x4 + x5)2.
3 dice
• But, what if we had more than 2 dice? Generating
functions “scale” very well.
• Sum of 7 for 3 dice? No problem:
• We want coefficient of x7 in (x + x2 + x3 + x4 + x5 + x6)3
• Equivalent to asking for coefficient of x4 in
(1 + x + x2 + x3 + x4 + x5)3.
• Inside ( ) we have a geometric series = (1 – x6)/(1 – x)
• So (1 + x + x2 + x3 + x4 + x5)3 = (1 – x6)3(1 – x)–3
• Let’s work out (1 – x6)3 and (1 – x)–3….
continued
We need to use the fact that the coefficients of (a+b)n are 1, n,
n(n – 1)/2!, n(n – 1)(n – 2)/3!, n(n – 1)(n – 2)(n – 3)/4!, etc.
(1 – x6)3 = 1 – 3x6 + 3x12 – x18
(1 – x)–3 = 1 + 3x + (3)(4)/2! x2 + (3)(4)(5)/3! x3 + …
• If we multiply these polynomials, how do we find a
term containing x4?
– x18, x12 and x6 can’t help us, because 2nd polynomial has
only positive powers of x.
– The only possibility is to multiply “1” from first polynomial
by the x4 term from the 2nd polynomial.
– Its coefficient is (3)(4)(5)(6)/4! = (5)(6) / 2 = 15. 
Alternate calculation
• Determining the coefficient of a single term can be
phrased as a combination question.
• The coefficient of x4 in (1 + x + x2 + x3 + x4 + x5)3
means we want 3 exponents to sum to 4, where each
exponent ranges from 0 to 5.
–
–
–
–
–
In other words, e1 + e2 + e3 = 4 where each ei = 0,1,…,5.
How can 3 numbers sum to 4?
0+0+4, 0+1+3, 0+2+2, 1+1+2
In each case, decide which ei represents each number.
Total number of ways = 3 + 6 + 3 + 3 = 15.
Ball in urn
• Many combination questions are of the ball-in-urn type. We
have n identical balls to place in r urns. The balls are identical,
so we simply need to know how many wind up in each urn.
• Basic approach: the generating function is of the form
(1 + x + x2 + x3 + …)r
and we want the coefficient of xn.
Highest term in each factor can simply be xn.
• Distribute 12 bottles of wine among 4 tables…
– What if each table must get 1+ bottle?
– What if no table may get more than 4?
– The first table must have at most 1 bottle, the 2nd must receive at least
1, the third must receive 0, 2 or 4; and the last must receive 2 or 3?
Practice formulating
• A combination problem can be encoded as a
generating function. Try these…
• How many ways are there to select 4 objects from
[ 5 A’s, 5 B’s, 5 C’s ]?
– Now, what if we wanted an equal number of A’s and C’s?
• How many ways can you make $1 change using
pennies, nickels, dimes and/or quarters?
• How many integers from 0 to 999 have a sum of
digits of 12?
Two variables
• Yes, it’s possible for a generating function to have 2 variables.
• Why? Maybe you want to keep track of two quantities at the
same time.
• Example: (1, 1, 2, 3, 3, 4, 4, 5, 5, 6). How many ways to…
– select r different numbers
– select numbers that have a sum of r
– select r numbers that have a sum of s.
• Example: (0, 1, 2, 2, 3, 4)
(1 + x)(1 + xy)(1 + xy2 + x2y4)(1 + xy3)(1 + xy4)
Some of the terms include 3x3y5, 2x3y4, 3x2y4, x2y7
Review
• The coefficient of x4 in (1 + x + x2)5 can answer what
combinatorical question?
• What if we had several dice, and…
–
–
–
–
You don’t have to roll all of them.
The dice do not all have the same number of sides.
You want to achieve a certain sum with a certain number of dice.
You want a sum of 31 with 4 6-sided, 2 8-sided, 2 12-sided and you
must roll at least 1 of each.
• Can you think of a question that would need 3 variables?
• Using the Coefficient command in Wolfram Alpha
– Example: Coefficient[(1+x+x^2)^10 (1+y)^4,x^5 y]