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102 LearnersLast updated on October 7, 2025
What is Stirling's Formula
Stirling's formula is an approximation used in mathematics to estimate factorials, particularly useful for large numbers. In this topic, we will explore the derivation, applications, and uses of Stirling's formula.
Understanding Stirling's Formula
Stirling's formula is a powerful tool used to approximate factorials. Let’s delve into the derivation, application, and accuracy of Stirling's formula.
Derivation of Stirling's Formula
Stirling's formula provides an approximation for n!, which is the factorial of n. It is derived using methods from calculus and asymptotic analysis.
The formula is expressed as:\( [ n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n ] \) Where \((\pi)\) is the constant pi, and\( (e)\) is the base of natural logarithms.
Applications of Stirling's Formula
Stirling's formula is widely used in various fields of mathematics and science for simplifying calculations involving large factorials. It is particularly useful in probability theory, statistical mechanics, and combinatorics where exact computation of factorials becomes cumbersome.
Accuracy of Stirling's Formula
Stirling's approximation is most accurate for large values of n. For small values, the error can be significant. However, the relative error decreases as n increases, making it an invaluable approximation for large-scale problems.
Importance of Stirling's Formula
In mathematics and science, Stirling's formula is crucial for simplifying complex factorial computations. Here are some reasons why it's important:
- It allows the calculation of large factorials without computational overload.
- It aids in understanding the behavior of functions involving factorials.
- Stirling's formula is foundational in asymptotic analysis and big O notation.
Tips and Tricks to Remember Stirling's Formula
Stirling's formula can seem complex, but with some tips, it can be easier to remember:
- Recall that the formula involves \((\sqrt{2\pi n}) \)and \((\left(\frac{n}{e}\right)^n). \)
- Visualize the formula as a product of a "scale factor"\( (\sqrt{2\pi n})\) and an "exponential term" \((\left(\frac{n}{e}\right)^n). \)
- Practice using the formula in different contexts to reinforce memory.
Common Mistakes and How to Avoid Them While Using Stirling's Formula
Even though Stirling's formula is a powerful tool, users may encounter some common errors. Here are some pitfalls and how to avoid them.
Mistake 1
Misapplying Stirling's Formula for Small n
Stirling's formula is less accurate for small values of n. To avoid errors, ensure that n is large enough for the approximation to be valid. For small n, compute factorials directly if possible.
Mistake 2
Ignoring the Factor of \(( \sqrt{2\pi n} )\)
A common mistake is neglecting the factor of\((\sqrt{2\pi n})\). This can lead to significant errors. Always include this factor when using the formula.
Mistake 3
Confusing e with Euler's Constant
Sometimes, students confuse the base of natural logarithms, e, with Euler's constant, \((\gamma)\). Remember that in Stirling's formula, e is the base of the natural logarithm.
Mistake 4
Overestimating the Accuracy for Moderate n
While Stirling's formula is effective for large n, it may not be precise for moderate values. Always consider the context and the necessary precision before applying the formula.
Mistake 5
Forgetting to Multiply by n
When using the exponential component \((\left(\frac{n}{e}\right)^n),\) do not forget to multiply by n when setting up the expression. This misstep can lead to incorrect calculations.
Examples of Problems Using Stirling's Formula
Problem 1
Estimate 10! using Stirling's formula.
Approximately 3.6 million
Explanation
Using Stirling's formula:\( [ 10! \approx \sqrt{2\pi \times 10} \left(\frac{10}{e}\right)^{10} ]\)
\([ \approx \sqrt{20\pi} \times \left(\frac{10}{2.718}\right)^{10} ]\)
\([ \approx 3.6 \times 10^6 ]\)
Problem 2
How would you apply Stirling's formula to approximate 15!?
Approximately 1.3 trillion
Explanation
Using Stirling's formula: \([ 15! \approx \sqrt{2\pi \times 15} \left(\frac{15}{e}\right)^{15} ]\)
\([ \approx \sqrt{30\pi} \times \left(\frac{15}{2.718}\right)^{15} ]
\)
\([ \approx 1.3 \times 10^{12} ]\)
Problem 3
Estimate 20! with Stirling's formula.
Approximately 2.4 quintillion
Explanation
Using Stirling's formula: \([ 20! \approx \sqrt{2\pi \times 20} \left(\frac{20}{e}\right)^{20} ] \)
\([ \approx \sqrt{40\pi} \times \left(\frac{20}{2.718}\right)^{20} ]
\)
\([ \approx 2.4 \times 10^{18} ]\)
Problem 4
Using Stirling's formula, approximate 25!.
Approximately 1.5 septillion
Explanation
Using Stirling's formula: \([ 25! \approx \sqrt{2\pi \times 25} \left(\frac{25}{e}\right)^{25} ] \)
\([ \approx \sqrt{50\pi} \times \left(\frac{25}{2.718}\right)^{25} ]
\)
\([ \approx 1.5 \times 10^{24} ]\)
Problem 5
How accurate is Stirling's approximation for 30!?
Very accurate with a small relative error
Explanation
For large n such as 30, Stirling's formula provides a very close approximation to the exact value of 30!, with the relative error being minimal compared to the factorial's magnitude.
FAQs on Stirling's Formula
1.What is Stirling's formula?
2.What is the basic form of Stirling's formula?
3.Why use Stirling's formula?
4.Is Stirling's formula accurate for small n?
5.What fields use Stirling's formula?
Glossary for Stirling's Formula
- Stirling's Formula: An approximation for estimating factorials, particularly useful for large numbers.
- Factorial: A product of all positive integers up to a given number n, denoted as n!.
- Asymptotic Analysis: A method of describing limiting behavior and approximations for large numbers.
- Natural Logarithm: A logarithm to the base e, where e is approximately equal to 2.718.
- Relative Error: The absolute error divided by the true value, indicating the accuracy of an approximation.
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Jaskaran Singh Saluja
About the Author
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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: He loves to play the quiz with kids through algebra to make kids love it.
