generating function for the Catalan numbers

generating function for the Catalan numbers

This article derives the formula

∑

C

n

⁢

x

n

=

1

-

1

-

4

⁢

x

2

⁢

x

for the

generating function

for the

Catalan numbers

, given in the parent (

http://planetmath.org/CatalanNumbers

) article, in two different ways.

A

Dyck path

is a

lattice path

in the

Euclidean plane

from

(

0

,

0

)

to

(

2

⁢

n

,

0

)

whose steps are either

(

1

,

1

)

or

(

1

,

-

1

)

, i.e. either upwards

diagonals

or downwards diagonals, and which never goes below the

x

-axis. Here are two Dyck paths:

If you rotate a Dyck path counterclockwise by

45

∘

and then

reflect

it in the line

x

=

y

, the result is a lattice path from

(

0

,

0

)

to

(

n

,

n

)

that never goes above the diagonal. One can also think of such a path as a path from

(

1

,

0

)

to

(

n

+

1

,

n

)

that never touches or crosses the diagonal. As the article on ballot numbers (

http://planetmath.org/LatticePathsAndBallotNumbers

) explains, these paths are counted by the Catalan numbers, and thus the number of Dyck paths of length

2

⁢

n

(or

Dyck paths of semilength

n

) is equal to

C

n

, the

n

th

Catalan number.
Note that any Dyck path has a unique decomposition as follows. Every Dyck path returns to the

x

-axis at some point (possibly at its end). Split the path at the first such point. Then the original path consists of an up step (the first step of the path), an arbitrary (perhaps empty) Dyck path, a down step returning to the

x

-axis, and then another arbitrary (perhaps empty) Dyck path.
Considering the first example above, this decomposition looks like this:

So a Dyck path is either empty, or consists of an up/down and two arbitrary Dyck paths. If one thinks about the lengths of the paths involved, it is clear that in terms of the generating function, we have

C

⁢

(

x

)

=

1

+

x

⁢

C

⁢

(

x

)

2

since a nonempty Dyck path of semilength

n

consists of two Dyck paths the sum of whose semilengths is

n

-

1

together with the unique Dyck path of semilength

1

(the up-down steps).
Solving this

quadratic equation

, we get

C

⁢

(

x

)

=

1

±

1

-

4

⁢

x

2

⁢

x

and we must decide between the plus and minus sign. However, note that

C

⁢

(

x

)

is a

formal power series

with only nonnegative powers of

x

, and the

expansion

of

1

-

4

⁢

x

starts with the

constant term

1

. So if the sign were positive, then

C

⁢

(

x

)

would have a leading term of

1

x

. So we must choose the negative sign, and we get

C

⁢

(

x

)

=

1

-

1

-

4

⁢

x

2

⁢

x

Here is a second method of getting to the same result; it is more straightforward but less informative.
By the generalized binomial formula, we have that

1

+

x

=

(

1

+

x

)

1

/

2

=

∑

k

=

0

∞

(

1

/

2

k

)

⁢

x

k

=

(

1

/

2

0

)

+

∑

k

=

1

∞

(

1

/

2

k

)

⁢

x

k

=

1

+

∑

k

=

1

∞

(

1

/

2

k

)

⁢

x

k

Now,

(

1

/

2

k

)

=

(

1

/

2

)

⁢

(

1

/

2

-

1

)

⁢

(

1

/

2

-

2

)

⁢

⋯

⁢

(

1

/

2

-

k

+

1

)

k

!

=

2

-

k

⁢

(

-

1

)

k

-

1

⁢

1

⋅

3

⋅

5

⁢

⋯

⁢

(

2

⁢

k

-

3

)

k

!

=

2

-

k

⁢

(

-

1

)

k

-

1

⁢

(

k

-

1

)

!

⋅

1

⋅

3

⋅

5

⁢

⋯

⁢

(

2

⁢

k

-

3

)

k

!

⁢

(

k

-

1

)

!

=

2

1

-

2

⁢

k

⁢

(

-

1

)

k

-

1

⁢

(

2

⁢

k

-

2

)

!

k

!

⁢

(

k

-

1

)

!

=

2

1

-

2

⁢

k

⁢

(

-

1

)

k

-

1

⁢

1

k

⋅

(

2

⁢

k

-

2

k

-

1

)

and so, substituting, we get

1

+

x

=

1

+

∑

k

=

1

∞

2

1

-

2

⁢

k

⁢

(

-

1

)

k

-

1

⁢

1

k

⋅

(

2

⁢

k

-

2

k

-

1

)

⁢

x

k

Replacing

x

by

-

4

⁢

x

, we get

1

-

4

⁢

x

-

1

=

∑

k

=

1

∞

2

1

-

2

⁢

k

⁢

(

-

1

)

k

-

1

⁢

1

k

⋅

(

2

⁢

k

-

2

k

-

1

)

⁢

(

-

1

)

k

⁢

2

2

⁢

k

⁢

x

k

=

∑

k

=

1

∞

(

-

2

)

⁢

1

k

⁢

(

2

⁢

k

-

2

k

-

1

)

⁢

x

k

=

∑

k

=

0

∞

-

2

k

+

1

⁢

(

2

⁢

k

k

)

⁢

x

k

+

1

where the last equality results from replacing

k

by

k

+

1

and adjusting the range of summation. On dividing through by

-

2

⁢

x

, we get

1

-

1

-

4

⁢

x

2

⁢

x

=

∑

k

=

0

∞

1

k

+

1

⁢

(

2

⁢

k

k

)

⁢

x

k

=

∑

k

=

0

∞

C

k

⁢

x

k

and we are done.
As a final observation, note that the recurrence

C

n

=

∑

i

=

0

n

-

1

C

i

⁢

C

n

-

1

-

i

given in the parent (

http://planetmath.org/CatalanNumbers

) article can be easily derived from the functional equation for the generating function,

C

⁢

(

x

)

=

1

+

x

⁢

C

⁢

(

x

)

2

:

∑

k

C

k

⁢

x

k

=

1

+

x

⁢

(

∑

k

C

k

⁢

x

k

)

2

and the coefficient of

x

n

on the right hand side is the coefficient of

x

n

-

1

in the sum, which is precisely

∑

k

=

0

n

-

1

C

k

⁢

C

n

-

1

-

k

as desired.

Title

generating function for the Catalan numbers

Canonical name

GeneratingFunctionForTheCatalanNumbers

Date of creation

2013-03-22 17:38:47

Last modified on

2013-03-22 17:38:47

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

9

Author

rm50 (10146)

Entry type

Derivation

Classification

msc 05A10

Related topic

DyckLanguage

Defines

Dyck paths

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