Use Stirling's formula to estimate C(100, 50) to three significant figures, and give the relative error of the estimate as a function of n for C(2n, n).

Use Stirling's formula to estimate C(100, 50) to three significant figures, and give the relative error of the estimate as a function of n for C(2n, n).

Approach: Apply Stirling to each of the three factorials in the central binomial coefficient and cancel, then use the next order term of the expansion to size the error.

1.01e29. Stirling gives m! approximately sqrt(2 pi m) (m/e)^m, so the central binomial coefficient C(2n, n) = (2n)!/(n!)^2 becomes sqrt(4 pi n) (2n/e)^{2n} / (2 pi n (n/e)^{2n}) = 4^n / sqrt(pi n). At n = 50 that is 4^50/sqrt(50 pi) = 1.2677e30 / 12.533 = 1.0115e29, against the exact value 1.0089e29. The next term of the asymptotic expansion is C(2n, n) = 4^n/sqrt(pi n) * (1 - 1/(8n) + ...), so the relative error of the leading form is about 1/(8n), which at n = 50 is 1/400 = 0.25%, matching the observed gap. The same 4^n/sqrt(pi n) form is what makes a symmetric random walk return to the origin with probability of order 1/sqrt(n) at step 2n.

Follow-up: How many decimal digits does 100! have, and what does Stirling give for the number of trailing zeros compared with the exact count?

Key concepts: stirling approximation, central binomial coefficient, relative error, asymptotic expansion.