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# asymptotics of central binomial coefficient

By making use of the expression

$4^{{-n}}{2n\choose n}=\prod_{{m=1}}^{n}{2m-1\over 2m},$ |

we may obtain estimates of the central binomial coefficient ${2n\choose n}$ for large values of $n$. We begin by making some elementary algebraic manipulations of this product:

$4^{{-n}}(2n+1){2n\choose n}={(2n+1)\prod_{{m=1}}^{n}(2m-1)\over\prod_{{m=1}}^{% n}2m}={\prod_{{m=1}}^{n}(2m+1)\over\prod_{{m=1}}^{n}2m}=\prod_{{m=1}}^{n}{2m+1% \over 2m}$ |

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

$16^{{-n}}(2n+1){2n\choose n}^{2}=\prod_{{m=1}}^{n}{(2m+1)(2m-1)\over(2m)^{2}}.$ |

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

${2n\choose n}={4^{n}\over\sqrt{2n+1}}\sqrt{\prod_{{m=1}}^{n}\left(1-{1\over 4m% ^{2}}\right)},$ |

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

$4^{{-a}}\sqrt{2a+1}{2a\choose a}>4^{{-b}}\sqrt{2b+1}{2b\choose b}$ |

when $a<b$. Secondly, the product converges in the limit $n\to\infty$. In fact, it turns out to be Wallis’ product for $\pi$, so we have

$\lim_{{n\to\infty}}4^{{-n}}\sqrt{2n+1}{2n\choose n}=\sqrt{2\over\pi}.$ |

This limit may be reread as an approximate formula when $n$ is large:

${2n\choose n}\approx\sqrt{2\over\pi}{4^{n}\over\sqrt{2n+1}}$ |

As an example, let us consider the case $n=10$. The exact answer is ${20\choose 10}=184756$. The approximate answer is $(4^{{10}}\sqrt{2}/\sqrt{21\pi})=182570.38\ldots$, 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

$\prod_{{m=1}}^{n}{64m^{2}-9\over 64m^{2}-25}.$ |

On the one hand, we can evaluate this product by factoring the numerator and denominator and cancelling terms:

$\displaystyle\prod_{{m=1}}^{n}{64m^{2}-9\over 64m^{2}-25}$ | $\displaystyle=\prod_{{m=1}}^{n}{(8m-3)(8m+3)\over(8m-5)(8m+5)}$ | ||

$\displaystyle={5\cdot 11\over 3\cdot 13}\cdot{13\cdot 19\over 11\cdot 21}\cdot% {21\cdot 27\over 19\cdot 29}\cdots\cdots{(8n-11)(8n+5)\over(8n-13)(8n-3)}\cdot% {(8n-3)(8n+3)\over(nm-5)(8n+5)}$ | |||

$\displaystyle={5\over 3}\cdot{8n+3\over 8n+5}$ |

From this expression, it follows that the product converges to $5/3$ as $n\to\infty$. On the other hand, we can multiply this product by the product we obtained earlier termwise:

$\displaystyle\left(\prod_{{m=1}}^{n}{64m^{2}-9\over 64m^{2}-25}\right)\left(% \prod_{{m=1}}^{n}\left(1-{1\over 4m^{2}}\right)\right)$ | $\displaystyle=\prod_{{m=1}}^{n}{64m^{2}-9\over 64m^{2}-25}\cdot{4m^{2}-1\over 4% m^{2}}$ | ||

$\displaystyle=\prod_{{m=1}}^{n}{256m^{4}-100m^{2}+9\over 256m^{4}-100m^{2}}$ | |||

$\displaystyle=\prod_{{m=1}}^{n}\left(1+{9\over 256m^{4}-100m^{2}}\right)$ |

By taking limits on both sides, we conclude that

$\prod_{{m=1}}^{\infty}\left(1+{9\over 256m^{4}-100m^{2}}\right)={10\over 3\pi}.$ |

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

$16^{{-n}}\cdot{80n^{2}+70n+15\over 24n+15}{2n\choose n}^{2}=\prod_{{m=1}}^{n}% \left(1+{9\over 256m^{4}-100m^{2}}\right).$ |

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

${2n\choose n}=4^{n}\sqrt{24n+15\over 80n^{2}+70n+15}\,\sqrt{\prod_{{m=1}}^{n}% \left(1+{9\over 256m^{4}-100m^{2}}\right)}$ |

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 $n\to\infty$:

$\lim_{{n\to\infty}}4^{{-n}}\,\sqrt{16n^{2}+14n+3\over 8n+5}{2n\choose n}=\sqrt% {2\over\pi}$ |

This gives us a new, improved asymptotic formula:

${2n\choose n}\approx\sqrt{2\over\pi}\,4^{n}\,\sqrt{8n+5\over 16n^{2}+14n+3}$ |

To see just how good this formula is, let us revisit the case $n=10$. Now, we get the approximation $(4^{{10}}\sqrt{170/1743\pi})=184756.93\ldots$. 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!

## Mathematics Subject Classification

11B65*no label found*05A10

*no label found*

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