asymptotics of central binomial coefficient

asymptotics of central binomial coefficient

By making use of the expression

4

-

n

⁢

(

2

⁢

n

n

)

=

∏

m

=

1

n

2

⁢

m

-

1

2

⁢

m

,

we may obtain estimates of the

central binomial
coefficient

(

2

⁢

n

n

)

for large values of

n

.
We begin by making some elementary

algebraic

manipulations of this product:

4

-

n

⁢

(

2

⁢

n

+

1

)

⁢

(

2

⁢

n

n

)

=

(

2

⁢

n

+

1

)

⁢

∏

m

=

1

n

(

2

⁢

m

-

1

)

∏

m

=

1

n

2

⁢

m

=

∏

m

=

1

n

(

2

⁢

m

+

1

)

∏

m

=

1

n

2

⁢

m

=

∏

m

=

1

n

2

⁢

m

+

1

2

⁢

m

Then we multiply this by our previous expression, factor-by-factor:

16

-

n

⁢

(

2

⁢

n

+

1

)

⁢

(

2

⁢

n

n

)

2

=

∏

m

=

1

n

(

2

⁢

m

+

1

)

⁢

(

2

⁢

m

-

1

)

(

2

⁢

m

)

2

.

With a bit of rearrangement and manipulation, this gives
us the following formula for the central binomial coefficient,

(

2

⁢

n

n

)

=

4

n

2

⁢

n

+

1

⁢

∏

m

=

1

n

(

1

-

1

4

⁢

m

2

)

,

which we shall examine to obtain estimates of the central
binomial coefficient for large

n

.

To make use of this formula, we make two key observations
about the product which appears on the right-hand side.
First, since each of the terms in the product lies between

0

and

1

, the product is an

decreasing function

of

n

,
Thus, we have

4

-

a

⁢

2

⁢

a

+

1

⁢

(

2

⁢

a

a

)

>

4

-

b

⁢

2

⁢

b

+

1

⁢

(

2

⁢

b

b

)

when

a

<

b

. Secondly, the product converges in the limit

n

→

∞

. In fact, it turns out to be Wallis’ product
for

π

, so we have

lim

n

→

∞

⁡

4

-

n

⁢

2

⁢

n

+

1

⁢

(

2

⁢

n

n

)

=

2

π

.

This limit may be reread as an approximate formula when

n

is large:

(

2

⁢

n

n

)

≈

2

π

⁢

4

n

2

⁢

n

+

1

As an example, let us consider the case

n

=

10

. The
exact answer is

(

20

10

)

=

184756

. The approximate
answer is

(

4

10

⁢

2

/

21

⁢

π

)

=

182570.38

⁢

…

,
which agrees with the exact answer to a percent. Also
note that the estimate is smaller than the exact answer —
this is a general feature which is due to the observation
made above that the product is an decreasing function of

n

.
Moreover, this observation also implies that the percentage
error decreases as

n

increases; in particular, the approximation
is good within a percent when

n

>

10

.

It is worth noting that same result can be obtained from Stirling’s
formula. In fact, one can deduce Stirling’s formula by a similar
line of reasoning.

With a little more work, wee can improve our approximation.
We begin by considering the product

∏

m

=

1

n

64

⁢

m

2

-

9

64

⁢

m

2

-

25

.

On the one hand, we can evaluate this product by

factoring

the

numerator

and denominator and cancelling terms:

∏

m

=

1

n

64

⁢

m

2

-

9

64

⁢

m

2

-

25

=

∏

m

=

1

n

(

8

⁢

m

-

3

)

⁢

(

8

⁢

m

+

3

)

(

8

⁢

m

-

5

)

⁢

(

8

⁢

m

+

5

)

=

5

⋅

11

3

⋅

13

⋅

13

⋅

19

11

⋅

21

⋅

21

⋅

27

19

⋅

29

⁢

⋯

⁢

⋯

⁢

(

8

⁢

n

-

11

)

⁢

(

8

⁢

n

+

5

)

(

8

⁢

n

-

13

)

⁢

(

8

⁢

n

-

3

)

⋅

(

8

⁢

n

-

3

)

⁢

(

8

⁢

n

+

3

)

(

n

⁢

m

-

5

)

⁢

(

8

⁢

n

+

5

)

=

5

3

⋅

8

⁢

n

+

3

8

⁢

n

+

5

From this expression, it follows that the product converges
to

5

/

3

as

n

→

∞

.
On the other hand, we can multiply this product by the product
we obtained earlier termwise:

(

∏

m

=

1

n

64

⁢

m

2

-

9

64

⁢

m

2

-

25

)

⁢

(

∏

m

=

1

n

(

1

-

1

4

⁢

m

2

)

)

=

∏

m

=

1

n

64

⁢

m

2

-

9

64

⁢

m

2

-

25

⋅

4

⁢

m

2

-

1

4

⁢

m

2

=

∏

m

=

1

n

256

⁢

m

4

-

100

⁢

m

2

+

9

256

⁢

m

4

-

100

⁢

m

2

=

∏

m

=

1

n

(

1

+

9

256

⁢

m

4

-

100

⁢

m

2

)

By taking limits on both sides, we conclude that

∏

m

=

1

∞

(

1

+

9

256

⁢

m

4

-

100

⁢

m

2

)

=

10

3

⁢

π

.

Embracing our earlier formula for the central binomial
coefficient with both hands, we obtain

16

-

n

⋅

80

⁢

n

2

+

70

⁢

n

+

15

24

⁢

n

+

15

⁢

(

2

⁢

n

n

)

2

=

∏

m

=

1

n

(

1

+

9

256

⁢

m

4

-

100

⁢

m

2

)

.

Juggling

terms from one hand to the other, we obtain
a new formula for the central binomial coefficient:

(

2

⁢

n

n

)

=

4

n

⁢

24

⁢

n

+

15

80

⁢

n

2

+

70

⁢

n

+

15

⁢

∏

m

=

1

n

(

1

+

9

256

⁢

m

4

-

100

⁢

m

2

)

Despite the increase in complexity, this formula is
actually an improvement over the old formula. The reason
for this is that the product converges more rapidly
because the

polynomial

in the denominator is now
of the fourth order.

As before, we may take the limit

n

→

∞

:

lim

n

→

∞

⁡

4

-

n

⁢

16

⁢

n

2

+

14

⁢

n

+

3

8

⁢

n

+

5

⁢

(

2

⁢

n

n

)

=

2

π

This gives us a new, improved asymptotic formula:

(

2

⁢

n

n

)

≈

2

π

⁢

4

n

⁢

8

⁢

n

+

5

16

⁢

n

2

+

14

⁢

n

+

3

To see just how good this formula is, let us revisit
the case

n

=

10

. Now, we get the approximation

(

4

10

⁢

170

/

1743

⁢

π

)

=

184756.93

⁢

…

.
Because the terms in the new product are greater
than unity, the approximation is an overestimate,
so we round it down to

184756

, which just so
happens to be the exact answer. The approximation
is that good!

Title

asymptotics of central binomial coefficient

Canonical name

AsymptoticsOfCentralBinomialCoefficient

Date of creation

2013-03-22 17:40:50

Last modified on

2013-03-22 17:40:50

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

18

Author

rspuzio (6075)

Entry type

Result

Classification

msc 11B65

Classification

msc 05A10

Related topic

CatalanNumbers

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