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Generalization of Binomial Coefficients: Beta Function 📂Functions

Generalization of Binomial Coefficients: Beta Function

Theorem: Binomial Coefficients Expressed Through the Beta Function

0≤k≤n0 \le k\le n is satisfied for two natural numbers k,nk,n, the following equation holds. (nk)=nCk=C(n,k)=1(n+1)B(n−k+1,k+1) \binom{n}{k}={}_{n}C_{k}=C(n,k)=\frac{1}{(n+1)B(n-k+1,k+1)} For two natural numbers m,nm,n, the following equation holds. B(m,n)=[mnm+n(m+nn)]−1 B(m,n)=\left[ \frac{mn}{m+n} \begin{pmatrix} m+n \\ n \end{pmatrix}\right]^{-1}

Description

The beta function, defined by B(p,q):=∫01tp−1(1−t)q−1dtB(p,q):=\displaystyle \int_{0}^{1}t^{p-1}(1-t)^{q-1}dt, can also be seen as a generalization of binomial coefficients. The proof is not difficult, but a lemma has to be proved first.

Proof

Lemma 1

B(p,q)=B(p+1,q)+B(p,q+1) B(p,q)=B(p+1,q) +B(p,q+1)

Proof of Lemma 1

B(p+1,q)+B(p,q+1)=∫01tp−1(1−t)q−1dt+∫01tp−1(1−t)p−1dt=∫01tp−1(1−t)q−1[t+(1−t)]dt=∫01tp−1(1−t)q−1dt=B(p,q) \begin{align*} B(p+1,q) + B(p,q+1) =& \int_{0}^{1} t^{p-1} (1-t)^{q-1}dt + \int_{0}^{1}t^{p-1}(1-t)^{p-1}dt \\ =& \int_{0}^{1}t^{p-1}(1-t)^{q-1}\big[t+(1-t) \big]dt \\ =& \int_{0}^{1}t^{p-1}(1-t)^{q-1}dt \\ =& B(p,q) \end{align*}

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Lemma 2

  • (a): B(p+1,q)=pp+qB(p,q)B(p+1,q)=\dfrac{p}{p+q}B(p,q)
  • (b): B(p,q+1)=qp+qB(p,q)B(p,q+1)=\dfrac{q}{p+q}B(p,q)

Proof of Lemma 2

B(p+1,q)=∫01tp(1−t)q−1dt=[−1qtp(1−t)q]01+∫01pqtp−1(1−t)qdt=0+pq∫01tp−1(1−t)qdt=pqB(p,q+1) \begin{align*} B(p+1,q) =& \int_{0}^{1}t^{p}(1-t)^{q-1}dt \\ =& \left[ -\frac{1}{q}t^{p}(1-t)^{q}\right]_{0}^{1} + \int_{0}^{1} \frac{p}{q}t^{p-1}(1-t)^{q}dt \\ =& 0 + \frac{p}{q}\int_{0}^{1} t^{p-1}(1-t)^{q}dt \\ =& \frac{p}{q} B(p,q+1) \end{align*} Integration by parts was used for the second equality. Since by Lemma 1, (a)(a), substituting this into the equation yields: B(p+1,q)=pqB(p,q)−pqB(p+1,q)⇒q+pqB(p+1,q)=pqB(p,q)⇒B(p+1,q)=pp+qB(p,q) B(p+1,q)=\frac{p}{q}B(p,q) -\frac{p}{q}B(p+1,q) \\ \Rightarrow \frac{q+p}{q}B(p+1,q)=\frac{p}{q}B(p,q) \\ \Rightarrow B(p+1,q)=\frac{p}{p+q}B(p,q) Since by Lemma 1, (b)(b), substituting this into the equation yields: B(p,q)−B(p,q+1)=pqB(p,q+1)⇒B(p,q)=p+qqB(p,q+1)⇒B(p,q+1)=qp+qB(p,q) B(p,q)-B(p,q+1)=\frac{p}{q}B(p,q+1) \\ \Rightarrow B(p,q)=\frac{p+q}{q}B(p,q+1) \\ \Rightarrow B(p,q+1)=\frac{q}{p+q}B(p,q)

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Main Proof

First, let’s show that B(1,1)=1B(1,1)=1. This can be directly seen from the definition. B(1,1)=∫01t0(1−t)0dt=1−0=1 B(1,1)=\int_{0}^{1}t^{0}(1-t)^{0}dt=1-0=1 Let’s say m,n∈Nm,n \in \mathbb{N}. Applying Lemma 2’s (a)(a) repeatedly to B(m,n)B(m,n) yields: B(m,n)=m−1m+n−1B(m−1,n)=m−1m+n−1⋅m−2m+n−2B(m−2,n)=m−1m+n−1⋅m−2m+n−2⋅⋯1m+n−(m−1)B(1,n)=(m−1)!(m+n−1)(m+n−2)⋯(n+1)B(1,n) \begin{align*} B(m,n) =& \frac{m-1}{m+n-1}B(m-1,n) \\ =& \frac{m-1}{m+n-1}\cdot \frac{m-2}{m+n-2}B(m-2,n) \\ =& \frac{m-1}{m+n-1} \cdot \frac{ m-2 }{m+n-2}\cdot \cdots \frac{ 1 }{m+n-(m-1) }B(1,n) \\ =& \frac{ (m-1)! }{ (m+n-1)(m+n-2)\cdots(n+1) }B(1,n) \end{align*} Applying Lemma 2’s (b)(b) repeatedly results in: B(m,n)=(m−1)!(m+n−1)(m+n−2)⋯(n+1)B(1,n)=(m−1)!(m+n−1)(m+n−2)⋯(n+1)n−1nB(1,n−1)=(m−1)!(m+n−1)(m+n−2)⋯(n+1)n−1n⋅n−2n−1B(1,n−2)=(m−1)!(m+n−1)(m+n−2)⋯(n+1)n−1n⋅n−2n−1⋯1n+1−(n−1)B(1,1)=(m−1)!(n−1)!(m+n−1)(m+n−2)⋯(n+1)n(n−1)⋯2B(1,1)=(m−1)!(n−1)!(m+n−1)!=m+nmnm!n!(m+n)!=[mnm+n(m+nn)]−1 \begin{align*} B(m,n) =& \frac{ (m-1)! }{ (m+n-1)(m+n-2)\cdots(n+1) }B(1,n) \\ =& \frac{ (m-1)! }{ (m+n-1)(m+n-2)\cdots(n+1) } \frac{ n-1 }{ n }B(1,n-1) \\ =& \frac{ (m-1)! }{ (m+n-1)(m+n-2)\cdots(n+1) } \frac{ n-1 }{ n }\cdot \frac{ n-2 }{ n-1 }B(1,n-2) \\ =& \frac{ (m-1)! }{ (m+n-1)(m+n-2)\cdots(n+1) } \frac{ n-1 }{ n }\cdot \frac{ n-2 }{ n-1 }\cdots \frac{ 1 }{ n+1-(n-1) }B(1,1) \\ =& \frac{ (m-1)!(n-1)! }{ (m+n-1)(m+n-2)\cdots(n+1)n(n-1)\cdots2 } B(1,1) \\ =& \frac{ (m-1)!(n-1)! }{ (m+n-1)! } \\ =& \frac{ m+n }{ mn }\frac{ m!n! }{ (m+n)! } \\ =& \left[ \frac{ mn }{ m+n } \begin{pmatrix} m+n \\ n \end{pmatrix} \right]^{-1} \end{align*} Substituting m=n−k+1m=n-k+1, n=k+1n=k+1 into the third equation from the bottom of the above sequence gives: B(n−k+1,k+1)=(n−k)!k!(n+1)!=(n−k)!k!(n+1)n! B(n-k+1,k+1)=\frac{(n-k)!k!}{ (n+1)! }=\frac{ (n-k)!k! }{(n+1)n! } Therefore: n!(n−k)!k!=(nk)=1(n+1)B(n−k+1,k+1) \frac{ n! }{(n-k)!k! }=\begin{pmatrix} n \\ k \end{pmatrix}=\frac{ 1 }{ (n+1)B(n-k+1,k+1) }

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See Also