Stirling's Approximation Calculator
Written by Thierno Sadou Diallo, formula verified per our methodology • Last checked on 9/10/2026
Stirling's approximation estimates the factorial with n! ≈ √(2πn) × (n/e)ⁿ. For n=10, this approximation gives about 3,598,696, versus the exact value of 3,628,800 (10!) — a relative error of only 0.83%, which shrinks further as n increases.
Explanation
The factorial calculator already published computes n! EXACTLY, by successively multiplying all integers from 1 to n — a perfectly precise method but one that quickly becomes impractical for very large values of n, both in computation time and in the size of the numbers handled (100! already has 158 digits). Stirling's approximation, discovered by the Scottish mathematician James Stirling in the 18th century, offers a CLOSED-FORM alternative: a direct formula requiring no repeated multiplication, and whose precision continuously improves as n increases, becoming practically exact for large values. This asymptotic convergence property — the relative error between the approximation and the exact value steadily decreases as n grows, never completely disappearing but quickly becoming negligible — is what makes this approximation so valuable in applied mathematics: from statistical physics (where it appears directly in calculating the entropy of systems containing a very large number of particles) to algorithm analysis in computer science (to estimate the complexity of procedures involving factorials, such as permutation sort), through probability and information theory, where it allows manipulating expressions containing factorials of very large numbers without ever having to calculate them explicitly.
Example: approximation of 10!
Inputs
n = 10.
Calculation
Stirling(10) = √(2π×10) × (10/e)¹⁰ ≈ √(62.83) × (3.6788)¹⁰ ≈ 3,598,696.
Result
Stirling's approximation gives about 3,598,696, compared to the exact value of 10! (3,628,800) — a relative error of only 0.83%.
Frequently asked questions
Why does the approximation become more precise as n increases?
Because Stirling's approximation is ASYMPTOTIC in nature: it is mathematically built to get closer and closer to the exact value as n tends to infinity, even though it never becomes perfectly exact for a finite n. This property was verified directly in this calculator: the relative error goes from about 0.83% at n=10 to about 0.42% at n=20, a reduction consistent with the expected asymptotic behavior of the formula.
In what contexts is this approximation really useful in practice?
It is particularly valuable as soon as n becomes too large for a practical exact calculation (beyond a few hundred, the exact factorial becomes a number with hundreds of digits, expensive to handle), or when you only need an ORDER OF MAGNITUDE rather than a digit-exact value — for example to quickly estimate the number of possible permutations of a large set, or to simplify mathematical expressions in statistical physics where only the overall behavior matters, not the exact precision of the last digit.
Is there an even more precise version of this approximation?
Yes, the formula presented here is the simplest version of Stirling's approximation; more elaborate versions add extra correction terms (a series of corrections in decreasing powers of 1/n) that further reduce the gap with the exact value for a given n. The simple version used here nonetheless remains the most commonly cited and is more than enough for most practical and teaching uses.