Problem 49 - A Graphical Approach to College Algebra

Question

The factorial of a positive integer \(n\) can be computed as a product: \(n !=1 \cdot 2 \cdot 3 \cdot \cdots \cdot n\) Calculators and computers can evaluate factorials quickly. Before the days of technology, mathematicians developed a formula, called Stirling's formula, for approximating large factorials. Interestingly enough, it involves the irrational numbers \(\pi\) and \(e\). $$n ! \approx \sqrt{2 \pi n} \cdot n^{n} \cdot e^{-n}$$ As an example, the exact value of \(5 !\) is \(120,\) and Stirling's formula gives the approximation as 118.019168 with a graphing calculator. This is "off" by less than \(2,\) an error of only \(1.65 \% .\) Use a calculator and Stirling's formula to find the exact value of \(10 !\) and its approximation.

Step-by-Step Solution

Answer

Exact: 3,628,800; Approximation: 3,598,695.6 (small error).

Step 1: Understanding Factorial Calculation

Factorial of a positive integer \( n \), denoted as \( n! \), is the product of all positive integers from \( 1 \) to \( n \). Therefore, for \( 10! \), we calculate \( 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \).

Step 2: Exact Calculation of 10!

Calculate \( 10! \) as follows: \\( 10 \times 9 = 90 \90 \times 8 = 720 \720 \times 7 = 5040 \5040 \times 6 = 30240 \30240 \times 5 = 151200 \151200 \times 4 = 604800 \604800 \times 3 = 1814400 \1814400 \times 2 = 3628800 \). \Thus, \( 10! = 3628800 \).

Step 3: Applying Stirling's Approximation

Use Stirling's formula, \( n! \approx \sqrt{2 \pi n} \cdot n^n \cdot e^{-n} \), to approximate \( 10! \).Substitute \( n = 10 \): \[10! \approx \sqrt{2 \pi \times 10} \cdot 10^{10} \cdot e^{-10}\]Calculate each component:1. \( \sqrt{20\pi} \approx 7.989\)2. \( 10^{10} = 10000000000 \)3. \( e^{-10} \approx 0.0000453999 \)Thus, combine these: \[10! \approx 7.989 \times 10000000000 \times 0.0000453999 \approx 3598695.6\]

Step 4: Comparing Results

The exact value of \(10!\) is 3,628,800.The Stirling's approximation is approximately 3,598,695.6.Comparing these, the approximation is close to the exact value with a small error.

Key Concepts

Stirling's formula

Stirling's formula provides a handy way to approximate the factorial of a large number. This formula becomes especially useful when dealing with big numbers because computing large factorials directly can be very challenging. The essence of Stirling's formula is to offer a simple way to get a good estimate of a factorial using mathematics involving irrational numbers like \(\pi\) and \(e\). It is represented mathematically as:\[ n! \approx \sqrt{2 \pi n} \cdot n^{n} \cdot e^{-n} \]The beauty lies in its efficiency for large numbers and its surprisingly close estimates. A practical example of this is using Stirling's formula for 5! which results in an approximation that is very close to the exact factorial value with a minimal error.

Approximation

An approximation is a useful mathematical concept that gives us a value close to the real one, without the need for exact computation. This approach is incredibly helpful, especially when calculating large numbers or when the exact computation is either complex or unnecessary. In the context of factorials, exact calculation for large numbers can become overwhelming and time-consuming. So, approximations like Stirling's provide a quick and fairly accurate way to estimate values. For example, while the exact value of 10! is 3,628,800, using Stirling's formula gives an approximation of about 3,598,695.6. This close-to-reality value offers enough precision for many practical purposes without the hassle of full computation.

Irrational Numbers

Irrational numbers, such as \(\pi\) and \(e\), play a fundamental role in advanced mathematical calculations, including approximations like Stirling's formula. These numbers cannot be expressed as simple fractions and have non-repeating, infinite decimal expansions.In the context of Stirling's formula, \(\pi\) is used to help in approximating the factorial value, and \(e\) appears in the exponential part of the formula. Their inclusion is essential because they help closely mimic the natural behavior of growth and decay processes in calculations, leading to a better approximation balance between what’s practical and what’s accurate.Utilizing irrational numbers allows mathematicians to deal with infinite precision and make complex calculations easier to handle and implement.

Mathematical Calculation

Mathematical calculations encompass various operations and techniques used to solve problems. When dealing with factorials, the calculation can be intensive, involving multiplication of sequential numbers.For instance, to find 10!, each number from 1 to 10 is sequentially multiplied, resulting in 3,628,800. This straightforward calculation can be demanding as numbers grow larger.Using Stirling’s formula simplifies this by reducing complex operations into more manageable ones, involving multiplication, power, and exponentiation with a mix of constants like \(\pi\) and \(e\). This makes it a pivotal tool for mathematicians, allowing them to estimate results without doing enormous calculations, while still maintaining a high level of accuracy. The combination of mathematical constants and operations showcases the power of mathematical calculation in streamlining tasks efficiently.