I found an approximation in falling factorial. It says that:
(a)n-1=a(a-1)(a-2)...(a-n+2) approximate to [a-(n-2)/2]n-1.
Can someone tell me how to derive the approximation?
The falling factorial is a product of a-0, a-1, a-2,...,a-(n-2). The median of 0, 1, 2,...,n-2 is (n-2)/2, and it is a median of n-1 numbers. The approximation comes by replacing each of the n-1 numbers subtracted off by their median.
Replace n with n+1 to get a(a-1)...(a-n+1) ~ (a - (n-1)/2)^n.
Assume that a > n-1,take the natural logarithm, and divide by n to get
[ ln(a) + ln(a-1) + ... + ln(a-n+1) ] / n ~ ln(a - (n-1)/2).
For n odd, this just says that the arithmetic mean is approximated by the median, while for n even, it says that the arithmetic mean is approximated by something close to the median. Reversing these steps provides a simple derivation of your approximation.
Another way to view this that avoids taking logarithms is simply to observe that the geometric mean [a(a-1)...(a-n+1)]^[1/n] is approximated by the median a - (n-1)/2.
Establishing the error in this approximation would be a logical next step. Intuitively, the approximation should improve as the argument a get larger.
The approximation is valid for large a, in fact for a >> n^(3/2), and the relative error behaves like -n^3/(24a^2), as can be estimated byTaylor-expanding the error in the logarithm for large a.
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