Approximating gamma ratios

Ratios of gamma functions come up often in applications. If the two gamma function arguments differ by an integer, then it’s easy to calculate their ratio exactly by using (repeatedly if necessary) the fact at Γ(x + 1) = x Γ(x).

If the arguments differ by 1/2, there is no closed formula, but the there are useful approximations. I’ve needed something like this a few times lately.

The simplest approximation is

\frac{\Gamma\left(x + 1 \right) }{\Gamma\left(x + \frac{1}{2} \right) } \sim x^{1/2}

You could motivate or interpret this as saying Γ(x + 1/2) is approximately the geometric mean between Γ(x + 1) and Γ(x). As we’ll see in the plot below, this approximation is good to a couple significant figures for moderate values of x.

There is another approximation that is a little more complicated but much more accurate.

\frac{\Gamma\left(x + 1 \right) }{\Gamma\left(x + \frac{1}{2} \right) } \sim \left(x^2 + \frac{x}{2} + \frac{1}{8}\right)^{1/4}

The following plot shows the relative error in both approximations.

gamma ratio approximation errors

By the way, the first approximation above is a special case of the more general approximation

\frac{\Gamma(x+a)}{\Gamma(x)} \sim x^a

Source:  J. S. Frame. An Approximation to the Quotient of Gamma Function. The American Mathematical Monthly, Vol. 56, No. 8 (Oct., 1949), pp. 529-535


One thought on “Approximating gamma ratios

  1. The next one in that sequence is a bit more accurate still for large x, but gives a poorer approximation near x=0
    (x^4 + x^3 + (1/2)*x^2 + (1/8)*x)^(1/8).
    The function
    is an overestimate with a relative error at most 0.023 for all x >= 0. The relative error decreases for large x and for x=100 it is about 10^(-6).
    The function
    has a relative error at most 0.002 for all x>=0. The relative error decreases for large x and for x=100 is about 10^(-11).

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