*Difficulty:*★★★☆☆ undergraduate

Consider the function:

where the independent variable *x* ranges between 0 and *X*, and the exponents are large: . [We could call it a “polynomial”, though the exponents need not be integers. Specifically it is the product of “monomials” in *x* and *X* – *x*, so might possibly be called a “sparse” polynomial in this sense.] Surprisingly, it closely resembles a gaussian curve, over our specified domain .

The turning point is where the derivative equals zero. This occurs when *x* is the surprisingly simple expression:

at which the function has value:

An arbitrary gaussian, not necessarily normalised, has form: . This has centre *D* which we equate with , and maximum height *C* which we set to the above expression. We can fix the final parameter, the standard deviation, by matching the second derivatives at the turning point. Hence the variance is:

Hence our gaussian approximation may be expressed:

The integral of the original curve turns out to be:

This uses the binomial coefficient , which is extended to non-integer values by replacing the factorials with Gamma functions. We could then apply Stirling’s approximation to each factorial, to obtain:

though this is more messy to write out. On the other hand, the integral of the gaussian approximation is:

We evaluated this integral over all real numbers, because the expression is simpler and still approximately the same. The ratio of the above two expressions is .