derivation of generating function for the reciprocal central binomial coefficients

derivation of generating function for the reciprocal central binomial coefficients

According to the article, the ordinary generating function for

(

2

⁢

n

n

)

-

1

is

4

⁢

(

4

-

x

+

x

⁢

arcsin

⁡

(

x

2

)

)

(

4

-

x

)

3

/

2

To see this, let

C

n

=

(

2

⁢

n

n

)

-

1

, and

C

⁢

(

x

)

=

∑

n

≥

0

C

n

⁢

x

n

its ordinary generating function. Then

C

n

+

1

=

(

2

⁢

n

+

2

n

+

1

)

-

1

=

(

n

+

1

)

!

⁢

(

n

+

1

)

!

(

2

⁢

n

+

2

)

!

=

(

n

+

1

)

⁢

(

n

+

1

)

(

2

⁢

n

+

2

)

⁢

(

2

⁢

n

+

1

)

⋅

n

!

⁢

n

!

(

2

⁢

n

)

!

=

n

+

1

2

⁢

(

2

⁢

n

+

1

)

⋅

C

n

Thus

(

4

⁢

n

+

2

)

⁢

C

n

+

1

=

(

n

+

1

)

⁢

C

n

so that

∑

n

≥

0

(

4

⁢

n

+

2

)

⁢

C

n

+

1

⁢

x

n

=

∑

n

≥

0

(

n

+

1

)

⁢

C

n

⁢

x

n

A little

algebra

gives

4

⁢

∑

n

≥

0

(

n

+

1

)

⁢

C

n

+

1

⁢

x

n

-

2

⁢

∑

n

≥

0

C

n

+

1

⁢

x

n

=

∑

n

≥

0

n

⁢

C

n

⁢

x

n

+

∑

n

≥

0

C

n

⁢

x

n

so that

4

⁢

C

′

⁢

(

x

)

-

2

x

⁢

(

C

⁢

(

x

)

-

1

)

=

x

⁢

C

′

⁢

(

x

)

+

C

⁢

(

x

)

and, collecting terms,

(

4

⁢

x

-

x

2

)

⁢

C

′

⁢

(

x

)

=

(

x

+

2

)

⁢

C

⁢

(

x

)

-

2

We now have a first-order linear ODE to solve. Put it in the form

C

′

⁢

(

x

)

+

-

x

-

2

x

⁢

(

4

-

x

)

⁢

C

⁢

(

x

)

=

-

2

4

⁢

x

-

x

2

and we must now integrate the

coefficient

of

C

⁢

(

x

)

. Expand by

partial fractions

and integrate to get

∫

-

x

-

2

x

⁢

(

4

-

x

)

⁢

𝑑

x

=

ln

⁡

(

(

4

-

x

)

3

/

2

x

)

Thus the solution to the equation is

C

⁢

(

x

)

=

x

(

4

-

x

)

3

/

2

⁢

(

k

+

∫

(

4

-

x

)

3

/

2

x

⋅

-

2

x

⁢

(

4

-

x

)

⁢

𝑑

x

)

=

k

⁢

x

(

4

-

x

)

3

/

2

-

2

⁢

x

(

4

-

x

)

3

/

2

⁢

∫

4

-

x

x

3

/

2

⁢

𝑑

x

=

k

⁢

x

(

4

-

x

)

3

/

2

-

2

⁢

x

(

4

-

x

)

3

/

2

⁢

(

-

2

⁢

(

4

-

x

)

x

⁢

(

4

-

x

)

-

arcsin

⁡

(

x

2

-

1

)

)

=

4

4

-

x

+

x

(

4

-

x

)

3

/

2

⁢

(

k

+

2

⁢

arcsin

⁡

(

x

2

-

1

)

)

To determine the constant

k

, note that we should have

C

′

(

x

)

|

x

=

0

=

1

2

; looking at

lim

x

→

0

⁡

C

′

⁢

(

x

)

we see that for

k

=

π

this equation holds. Thus

C

⁢

(

x

)

=

4

4

-

x

+

x

(

4

-

x

)

3

/

2

⁢

(

π

+

2

⁢

arcsin

⁡

(

x

2

-

1

)

)

We show below that the following is an identity:

z

+

1

2

=

sin

⁡

(

π

4

+

1

2

⁢

arcsin

⁡

(

z

)

)

Assuming that result, substitute

x

2

-

1

for

z

and simplify to get

x

2

=

sin

⁡

(

π

4

+

1

2

⁢

arcsin

⁡

(

x

2

-

1

)

)

so that

4

⁢

arcsin

⁡

(

x

2

)

=

π

+

2

⁢

arcsin

⁡

(

x

2

-

1

)

and then

C

⁢

(

x

)

=

4

4

-

x

+

x

(

4

-

x

)

3

/

2

⁢

(

4

⁢

arcsin

⁡

(

x

2

)

)

=

4

⁢

(

4

-

x

+

x

⁢

arcsin

⁡

(

x

2

)

)

(

4

-

x

)

3

/

2

as desired.

Finally, to prove the identity, first expand the right-hand using the formula for

sin

⁡

(

a

+

b

)

, and then apply the half-angle formulas:

sin

⁡

(

π

4

+

1

2

⁢

arcsin

⁡

(

z

)

)

=

2

2

⁢

(

cos

⁡

(

1

2

⁢

arcsin

⁡

(

z

)

)

+

sin

⁡

(

1

2

⁢

arcsin

⁡

(

z

)

)

)

=

2

2

⁢

(

1

+

cos

⁡

(

arcsin

⁡

(

z

)

)

2

+

1

-

cos

⁡

(

arcsin

⁡

(

z

)

)

2

)

=

2

2

⁢

(

1

+

1

-

z

2

2

+

1

-

1

-

z

2

2

)

=

1

2

⁢

(

1

+

1

-

z

2

+

1

-

1

-

z

2

)

Now square this expression to get

1

4

⁢

(

2

+

2

⁢

1

-

1

+

z

2

)

=

|

z

|

+

1

2

Thus the identity holds for

0

≤

z

≤

1

; an almost identical computation using

-

z

in of

z

shows that it also holds for

-

1

≤

z

≤

0

.

Title

derivation

of

generating function

for the

reciprocal

central binomial coefficients

Canonical name

DerivationOfGeneratingFunctionForTheReciprocalCentralBinomialCoefficients

Date of creation

2013-03-22 19:04:58

Last modified on

2013-03-22 19:04:58

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

4

Author

rm50 (10146)

Entry type

Result

Classification

msc 05A10

Classification

msc 05A15

Classification

msc 05A19

Classification

msc 11B65

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