Minggatu-Catalan Number

Minggatu-Catalan Number

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First discovered by the Chinese Mathematician Minggatu 明安图 (清 康熙, 1730), later by the French (né Belgian) École Polytechnique mathematician Eugène Charles Catalan (1814 – 1894).

Proof: Consider ways of making sums with {0, 1, 2, 3, 4…} and +, ( , ).

0 = (0)
1 way for 0
\boxed {C_{0} = 1}

0+1 =(0+1) :
1 way for 1 \boxed {C_{1} = 1}

0 +1 +2 = ((0+(1+2)) = ((0+1)+2):
2 ways for 2 \boxed {C_{2} = 2}

0+1+2+3 = (0+(1+(2+3))) = (0+((1+2)+3))= ((0+1)+(2+3)) = ((0+(1+2))+3) = (((0+1)+2)+3) :
5 ways for 3 \boxed {C_{3} = 5}

0+1+2+3+…+n = ?
? ways for n \boxed {C_{n} = ?}

Try finding the pattern for C_{4}:

Let A represents {0, 1, 2, 3, 4}.
There are 4 cases:

Case 1:
(\underbrace{(A)}_{C_{0}} + \underbrace{(A+ A+ A+ A)} _{C_{3}})

Case 2:
( \underbrace{ (A+A ) }_{C_{1}} + \underbrace{ (A+ A+ A) }_{C_{2}} )

Case 3:
( \underbrace{ (A+A+A ) }_{C_{2}} + \underbrace{ ( A+ A) }_{C_{1}})

Case 4:
( \underbrace{ (A+A+A+A) }_{C_{3}} + \underbrace{ (A) }_{C_{0}} )

C_{4} = C_{0}.C_{3} + C_{1}.C_{2} + C_{2}.C_{1}+ C_{3}.C_{0}

Generalise:
C_{n+1} = C_{0}.C_{n-0} + C_{1}.C_{n-1} + C_{2}.C_{n-2} + ...+ C_{k}.C_{n-k} + ...+ C_{n-0}.C_{0}

Recurrence Sequence of C_{n}:
\boxed{ \begin{cases} C_{0} =1, \\ C_{n+1} = \displaystyle\sum_{k=0}^{n}C_{k}C_{n-k} , & \text{( } n \geq {0} \text {)} \end{cases} }

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Step 1: Generating Function C(x) :
Let C(x) =C_{0} + C_{1}x+ C_{2}x^2 + ...+ C_{n}x^n

\boxed {\displaystyle C(x) = \sum_{n=0}^{\infty} C_{n} x^n }

C(x)^{2} = (C_{0}C_{0})+ (C_{0}C_ {1} + C_{1}C_{0})x+ (C_{0}C_{2}+ C_{1}C_{1}+ C_{2}C_{0} ) x^2 + ...

The coefficient for x^{n}: \displaystyle \sum_{k=0}^{n}C_{k}C_{n-k}

\displaystyle C(x)^2 = \sum_{n=0}^{\infty} \underbrace{ \left (\sum_{k=0}^{n}C_{k}C_{n-k} \right)}_{\text {coeff of }x^{n}} x^n = \sum_{n=0}^{\infty} \underbrace{ C_{n+1}}_{\text {recurrence}} x^n

\displaystyle x.C(x)^{2} =x. \sum_{n=0}^{\infty} C_{n+1} x^n = \sum_{n=0}^{\infty} C_{n+1} x^{n +1}
Change (n+1) to n by adjusting initial value from n= 0 to 1:
\displaystyle x.C(x)^{2} = \sum_{n=1}^{\infty} C_{n} x^n

Reset the initial value from n=1 to 0:
\displaystyle x.C(x)^{2} = \sum_{n=0}^{\infty} C_{n} x^n - C_{0}
We have seen
C_{0}= 1 and
\displaystyle C(x) = \sum_{n=0}^{\infty} C_{n} x^n

\displaystyle x.C(x)^{2} - C(x) +1 = 0
This is the quadratic equation on C (x):

C(x)= \frac{1 \pm \sqrt{1-4x} }{2x}

2x.C(x)= 1 \pm \sqrt{1-4x}

Case: 2x.C(x)= 1 + \sqrt{1-4x}
When x \rightarrow 0
2x.C(x)= 0
1 + \sqrt{1-4x}= 1+1=2
Impossible !

Case: 2x.C(x)= 1 - \sqrt{1-4x}
When x \rightarrow 0
2x.C(x)= 0
1 - \sqrt{1-4x}= 1- 1=0

Step 2: The closed form for the generating function C(x) :

\boxed { C(x)= \frac{1 - \sqrt{1-4x} }{2x} }

Step 3: The closed form for C_{n}

We expand \sqrt{1-4x} into power series:

Let \displaystyle \sqrt{1-4x}= K(x)= \sum_{k=0}^{\infty}K_{k}x^{k}

2xC(x)= 1 - \sqrt{1-4x}
\displaystyle 2xC(x)= 1 - \sum_{k=0}^{\infty}K_{k}x^{k}

\displaystyle 2x\sum_{k=0}^ {\infty}C_{k}x^{k} = 1 - \sum_{k=0}^{\infty}K_{k}x^{k}

Move 2x inside the left sum:
\displaystyle \sum_{k=0}^ {\infty}2C_{k}x^{k+1} = 1 - \sum_{k=0}^{\infty}K_{k}x^{k}

Synchronise the indices k,
\displaystyle \sum_{k=1}^ {\infty}2C_{k-1}x^{k} = 1 - K_{0} - \sum_{k=1}^{\infty}K_{k}x^{k}

\displaystyle \sum_{k=1}^ {\infty}2C_{k-1}x^{k} + \sum_{k=1}^{\infty}K_{k}x^{k} = 1 - K_{0}

\displaystyle \sum_{k=1}^ {\infty} (2C_{k-1} + K_{k})x^{k} = 1 - K_{0}

We compare the coefficient of each term x^{k} :

x^{0}: 0 = 1- K_{0}
x^{1}: 2C_{0} + K_{1} = 0
x^{2}: 2C_{1} + K_{2} = 0

x^{n}: 2C_{n} + K_{n+1} = 0

\boxed{ \begin{cases} K_{0} =1, \\ C_{n} = \displaystyle - \frac {K_{n+1}}{2} , & \text{( } n \geq {0} \text {)} \end{cases} }

We need to express K_{n+1} in term of n only.

Go back to the definition of K(x):
\displaystyle K(x)= \sum_{k=0}^{\infty}K_{k}x^{k}

\displaystyle K(x)= K_{0}+K_{1}x + K_{2}x^{2} + ...+ K_{n}x^{n}+...

Differentiate:
\displaystyle K'(x)=1.K_{1}x + 2K_{2}x^{1} + ...+ nK_{n}x^{n-1}+...

\displaystyle K'(0)= 1K_{1}

Differentiate 2nd time:
\displaystyle K''(x)=2.1K_{2} + ...+ n. (n-1)K_{n}x^{n-2}+...

\displaystyle K''(0)= 2.1K_{2}

Continue …

K^{(n)}(x)= n (n-1)(n-2)...2.1K_{n} + ...

K^{(n)}(0)= n! K_{n}

\boxed { K_{n} = \frac { K^{(n)}(0)}{n!} = \frac { K^{(n)}(0)}{n^{\underline {n}}} }

Note: See blog on “Falling Factorial”: x^{\underline {n}}

Next step, we shall find K^{(n)}(0) in term of n
Recall
K(x) = (1-4x)^{\frac {1}{2}}

K'(x) = -2.(1-4x)^{-\frac {1}{2}}
K''(x) = -2.2(1-4x)^{-\frac {3}{2}}
K'''(x) = -2.4.3(1-4x)^{-\frac {5}{2}}
K''''(x) = -2.6.5.4(1-4x)^{-\frac {7}{2}}

K^{(n)}(x) = -2.(2n-2)^{\underline {n-1}}(1-4x)^{-\frac {2n-1}{2}}

K^{(n+1)}(x) = -2.(2n)^{\underline {n}}(1-4x)^{-\frac {2n+1}{2}}

\boxed { K^{(n+1)}(0) = -2.(2n)^{\underline {n}} }

Also from above, we have seen:
\boxed { K_{n} = \frac { K^{(n)}(0)}{n^{\underline {n}}} }

Change n to (n+1):
K_{n+1} = \frac { K^{(n+1)}(0)}{(n+1)^{\underline {n+1}}}

K_{n+1} = \frac { -2.(2n)^{\underline {n}} } {(n+1)^{\underline {n+1}}}
From step 3 above, we also know:
C_{n} = - \frac {K_{n+1}} {2}

C_{n} = \frac { (2n)^{\underline {n}} } {(n+1)^{\underline {n+1}}} = \frac {1}{n+1} \frac { (2n)^{\underline {n}} } {(n)^{\underline {n}}}
since (n+1)^{\underline {n+1}} = (n+1)!= (n+1).n! = (n+1). (n)^{\underline {n}}
and
\frac { (2n)^{\underline {n}}} {(n)^{\underline {n}}} = \binom {2n}{n}  (by definition of Falling Factorial)

\boxed {\displaystyle C_{n} = \frac {1}{n+1} \binom {2n}{n}  } [QED]

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Ref:

1. History:
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2. Proof of Minggatu Catalan numbers

3. Monoid

4. Kleene Star

5. Ref:
Math Girls by Hiroshi Yuki et al.
http://www.amazon.co.uk/dp/0983951306/ref=cm_sw_r_udp_awd_UKSvtb1AXNWCE

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