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The real number \(e\) is the base of natural logarithms. It appears in certain mathematics problems involving growth or decay and is part of Stirling's formula for approximating factorials. One way to calculate \(e\) is shown below. \(e=\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\cdots\) a) Determine the approximate value of \(e\) using the first five terms of the series shown. b) How does the approximate value of \(e\) change if you use seven terms? eight terms? What do you conclude? c) What is the value of \(e\) on your calculator? d) Stirling's approximation can be expressed as \(n ! \approx\left(\frac{n}{e}\right)^{n} \sqrt{2 \pi n}\) Use Stirling's approximation to estimate \(15 !,\) and compare this result with the true value. e) A more accurate approximation uses the following variation of Stirling's formula: \(n ! \approx\left(\frac{n}{e}\right)^{n} \sqrt{2 \pi n}\left(1+\frac{1}{12 n}\right)\) Use the formula from part d) and the variation to compare estimates for \(50 !.\)

Step by step solution

01

Calculate the first five terms

To determine the approximate value of \( e \) using the first five terms of the series, we need to compute the sum of \( \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} \). Let's evaluate these terms: \( \frac{1}{0!} = 1 \) \( \frac{1}{1!} = 1 \) \( \frac{1}{2!} = \frac{1}{2} = 0.5 \) \( \frac{1}{3!} = \frac{1}{6} \approx 0.1667 \) \( \frac{1}{4!} = \frac{1}{24} \approx 0.0417 \) Summing these: \(1 + 1 + 0.5 + 0.1667 + 0.0417 \approx 2.7084 \)

02

Calculate for seven terms

Now include the next two terms, \( \frac{1}{5!} \) and \( \frac{1}{6!} \): \( \frac{1}{5!} = \frac{1}{120} \approx 0.0083 \) \( \frac{1}{6!} = \frac{1}{720} \approx 0.0014 \) Adding these to our previous sum: \(2.7084 + 0.0083 + 0.0014 \approx 2.7181 \)

03

Calculate for eight terms

Include one more term, \( \frac{1}{7!} \): \( \frac{1}{7!} = \frac{1}{5040} \approx 0.000198 \) Adding this: \(2.7181 + 0.000198 \approx 2.7183 \)

04

Analyze the approximation

As more terms are added, the value of \( e \) gets closer to its actual value. Each additional term has a smaller effect, showing convergence.

05

True value of \( e \)

Check the value of \( e \) on a calculator: \( e \approx 2.7182818 \)

06

Stirling's approximation for \(15!\)

Use Stirling's formula to estimate \(15!\):\(15! \approx \left( \frac{15}{e} \right)^{15} \sqrt{2 \pi \cdot 15} \)Calculating step-by-step: \( 15/e \approx 15/2.7183 = 5.514 \) \( 5.514^{15} \approx 4.07 \times 10^9 \) \( \sqrt{2 \pi \cdot 15} \approx 9.688 \)Thus, \(15! \approx 4.07 \times 10^9 \times 9.688 \approx 3.94 \times 10^{10} \)The true value of \(15!\) is 1,307,674,368,000. Stirling's approximation is close but not exact.

07

More accurate approximation for \(50!\)

Use the more accurate approximation:\(50! \approx \left( \frac{50}{e} \right)^{50} \sqrt{2 \pi \cdot 50} \left( 1 + \frac{1}{12 \cdot 50} \right) \)Calculate step-by-step: \( 50/e \approx 50/2.7183 = 18.39 \) \( 18.39^{50} \approx 3.04 \times 10^{63} \) \( \sqrt{2 \pi \cdot 50} \approx 17.72 \) \( 1 + \frac{1}{12 \cdot 50} \approx 1.00167 \)Thus, \( 50! \approx 3.04 \times 10^{63} \times 17.72 \times 1.00167 \approx 5.39 \times 10^{64} \)The true value of \(50!\) is approximately \(3.0414 \times 10^{64} \). The more accurate approximation provides a closer estimate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Natural Logarithms

Natural logarithms are special logarithms with the base of the mathematical constant, e. They are denoted as \( \text{ln}(x) \) and are useful in many areas of mathematics, especially in solving growth and decay problems.

The natural logarithm helps to invert exponential functions where the base is e. For example, if \( e^x = y \), then \( \text{ln}(y) = x \). This makes it easier to solve equations involving exponential growth or decay.

Natural logarithms also appear in calculus, particularly in the context of integration and differentiation of exponential functions.

Stirling's Approximation

Stirling's Approximation is a formula used to estimate large factorials. A factorial of a number n (denoted as \( n! \)) is the product of all positive integers up to n.

Stirling's Approximation can be expressed as:
\[ n ! \approx \left( \frac{n}{e} \right)^{n} \sqrt{2 \pi n} \]

This formula is used because calculating large factorials directly can be complex and computation-heavy. Stirling's Approximation simplifies these calculations by using logarithms and properties of the exponential function.

The formula provides an estimation that becomes more accurate as n increases, making it particularly useful in probability and statistical computing.

Factorials

Factorials, denoted by an exclamation mark (\( n! \)), are the product of all positive integers up to a given number n. For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

Factorials are fundamental in various fields of mathematics, including combinatorics, algebra, and calculus. They are used to calculate permutations and combinations, which are ways of arranging and selecting objects.

Understanding factorials is also crucial when working with series expansions, such as the Taylor series, and in algorithms that require factorial computations for complex problem-solving.

Growth and Decay

Growth and decay refer to changes in quantities over time, which can be modeled using exponential functions. If a quantity grows or decays at a rate proportional to its current value, it follows an exponential model.

In mathematical terms, exponential growth can be represented by \( y = y_0 e^{kt} \) where:

• \( y \) is the final amount
• \( y_0 \) is the initial amount
• \( k \) is the growth rate
• \( t \) is the time

Exponential decay follows a similar formula but with a negative growth rate.

Understanding these models is essential in fields such as biology, finance, and physics for modeling population growth, radioactive decay, and interest calculations.

Mathematical Series

A mathematical series is the summation of the terms of a sequence. For example, the series for calculating the constant e is:
\[ e = \frac{1}{0!} \ + \frac{1}{1!} \ + \frac{1}{2!} \ + \frac{1}{3!} \ + \frac{1}{4!} \ + \cdots \]

Each term in the series represents part of the total sum, and as more terms are added, the sum approaches the true value of the infinite series.

Understanding series is crucial for analyzing and approximating functions. It's especially important in calculus, where series expansions like the Taylor and Maclaurin series are used to represent complex functions as infinite sums of simpler polynomial terms.