Product of r Choose k with r Minus Half Choose k/Formulation 1
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Theorem
Let $k \in \Z$, $r \in \R$.
- $\dbinom r k \dbinom {r - \frac 1 2} k = \dfrac {\dbinom {2 r} k \dbinom {2 r - k} k} {4^k}$
where $\dbinom r k$ denotes a binomial coefficient.
Proof 1
First we establish the following:
\(\ds \paren {r - \frac 1 2}^{\underline k}\) | \(=\) | \(\ds \paren {r - \frac 1 2} \paren {r - \frac 3 2} \paren {r - \frac 5 2} \dotsm \paren {r - \frac 1 2 - k + 1}\) | Definition of Falling Factorial | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {2^k \paren {r - \frac 1 2} \paren {r - \frac 3 2} \paren {r - \frac 5 2} \cdots \paren {r - \frac 1 2 - k + 1} } {2^k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {2 r - 1} \paren {2 r - 3} \paren {2 r - 5} \dotsm \paren {2 r - 2 k + 1} } {2^k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {2 r \paren {2 r - 1} \paren {2 r - 2} \paren {2 r - 3} \dotsm \paren {2 r - 2 k + 2} \paren {2 r - 2 k + 1} } {2^k \paren {2 r \paren {2 r - 2} \paren {2 r - 4} \dotsm \paren {2 r - 2 k + 2} } }\) | ||||||||||||
\(\text {(1)}: \quad\) | \(\ds \) | \(=\) | \(\ds \dfrac {\paren {2 r}^{\underline{2 k} } } {2^k \times 2^k r^{\underline k} }\) | Definition of Falling Factorial |
Then:
\(\ds \dbinom r k \dbinom {r - \frac 1 2} k\) | \(=\) | \(\ds \dfrac {r^{\underline k} } {k!} \dfrac {\paren {r - \frac 1 2}^{\underline k} } {k!}\) | Definition of Binomial Coefficient | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {r^{\underline k} } {k!} \dfrac {\paren {2 r}^{\underline {2 k} } } {k! \, 2^{2 k} r^{\underline k} }\) | from $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {2 r}^{\underline {2 k} } } {\paren {k!}^2 \, 4^k}\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {2 r}^{\underline k} \paren {2 r - k}^{\underline k} } {\paren {k!}^2 \, 4^k}\) | Falling Factorial of Sum of Integers | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {4^k} \dfrac {\paren {2 r}^{\underline k} } {k!} \dfrac {\paren {2 r - k}^{\underline k} } {k!}\) | separating out | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {4^k} \dbinom {2 r} k \dbinom {2 r - k} k\) | Definition of Binomial Coefficient |
Proof 2
From Binomial Coefficient expressed using Beta Function:
- $(1): \quad \dbinom r k \dbinom {r - \frac 1 2} k = \dfrac 1 {\paren {r + 1} \map \Beta {k + 1, r - k + 1} \paren {r + \frac 1 2} \map \Beta {k + 1, r - k + \frac 1 2} }$
Then:
\(\ds \dbinom r {k + 1} \dbinom {r - \frac 1 2} {k + 1}\) | \(=\) | \(\ds \dfrac 1 {\paren {r + 1} \map \Beta {k + 2, r - k} \paren {r + \frac 1 2} \map \Beta {k + 2, r - k - \frac 1 2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {\paren {r + 1} \frac {k + 1} {r + 1} \map \Beta {k + 1, r - k} \paren {r + \frac 1 2} \frac {k + 1} {r + \frac 1 2} \map \Beta {k + 1, r - k - \frac 1 2} }\) | Beta Function of $x$ with $y+1$ by $\dfrac {x+y} y$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {\paren {r + 1} \frac {k + 1} {r + 1} \frac {r + 1} {r - k} \map \Beta {k + 1, r - k + 1} \paren {r + \frac 1 2} \frac {k + 1} {r + \frac 1 2} \frac {r + \frac 1 2} {r - k - \frac 1 2} \map \Beta {k + 1, r - k + \frac 1 2} }\) | Beta Function of $x$ with $y+1$ by $\dfrac {x+y} y$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {r - k} \paren {r - k - \frac 1 2} } {\paren {k + 1}^2} \times \frac 1 {\paren {r + 1} \map \Beta {k + 1, r - k + 1} \paren {r + \frac 1 2} \map \Beta {k + 1, r - k + \frac 1 2} }\) | simplifying | |||||||||||
\(\text {(2)}: \quad\) | \(\ds \) | \(=\) | \(\ds \dfrac {\paren {r - k} \paren {r - k - \frac 1 2} } {\paren {k + 1}^2} \dbinom r k \dbinom {r - \frac 1 2} k\) | from $(1)$ |
Again from Binomial Coefficient expressed using Beta Function:
- $(3): \quad \dbinom {2 r} k \dbinom {2 r - k} k = \dfrac 1 {\paren {2 r + 1} \map \Beta {k + 1, 2 r - k + 1} \paren {2 r - k + 1} \map \Beta {k + 1, 2 r - 2 k + 1} }$
Then:
\(\ds \dbinom {2 r} {k + 1} \dbinom {2 r - k - 1} {k + 1}\) | \(=\) | \(\ds \dfrac 1 {\paren {2 r + 1} \map \Beta {k + 2, 2 r - k} \paren {2 r - k} \map \Beta {k + 2, 2 r - 2 k - 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {\paren {2 r + 1} \frac {k + 1} {2 r + 1} \map \Beta {k + 1, 2 r - k} \paren {2 r - k} \frac {k + 1} {2 r - k} \map \Beta {k + 1, 2 r - 2 k - 1} }\) | Beta Function of $x$ with $y+1$ by $\dfrac {x+y} y$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {\paren {k + 1}^2 \map \Beta {k + 1, 2 r - k} \map \Beta {k + 1, 2 r - 2 k - 1} }\) | simplification | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {\paren {k + 1}^2 \frac {2 r + 1} {2 r - k} \map \Beta {k + 1, 2 r - k + 1} \frac {2 r - k} {2 r - 2 k - 1} \map \Beta {k + 1, 2 r - 2 k} }\) | Beta Function of $x$ with $y+1$ by $\dfrac {x+y} y$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {2 r - 2 k - 1} } {\paren {k + 1}^2 \paren {2 r + 1} \map \Beta {k + 1, 2 r - k + 1} \map \Beta {k + 1, 2 r - 2 k} }\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {2 r - 2 k - 1} } {\paren {k + 1}^2 \paren {2 r + 1} \map \Beta {k + 1, 2 r - k + 1} \frac {2 r - k + 1} {2 r - 2 k} \map \Beta {k + 1, 2 r - 2 k + 1} }\) | Beta Function of $x$ with $y+1$ by $\dfrac {x+y} y$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {2 r - 2 k - 1} \paren {2 r - 2 k} } {\paren {k + 1}^2} \times \frac 1 {\paren {2 r + 1} \map \Beta {k + 1, 2 r - k + 1} \paren {2 r - k + 1} \map \Beta {k + 1, 2 r - 2 k + 1} }\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {2 r - 2 k} \paren {2 r - 2 k - 1} } {\paren {k + 1}^2} \dbinom {2 r} k \dbinom {2 r - k} k\) | from $(3)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {4 \paren {r - k} \paren {r - k - \frac 1 2} } {\paren {k + 1}^2} \dbinom {2 r} k \dbinom {2 r - k} k\) |
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Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $47$