Problem 54 - A Graphical Approach to College Algebra

Question

It can be shown that $$(1+x)^{n}=1+n x+\frac{n(n-1)}{2 !} x^{2}+\frac{n(n-1)(n-2)}{3 !} x^{3}+\cdots$$ for any real number \(n\) ( not just positive integer values) and any real number \(x\), where \(|x|<1 .\) Use this result to approximate each quantity in Exercises \(53-56\) to the nearest thousandth. $$\frac{1}{1.04^{5}}$$

Step-by-Step Solution

Answer

The approximate value of \( \frac{1}{1.04^5} \) is 0.817.

Step 1: Rewrite the expression

The given expression to approximate is \( \frac{1}{1.04^5} \). To use the binomial expansion, rewrite this as \( (1 + 0.04)^{-5} \) to match the form \((1+x)^n \). This helps in identifying \( x = 0.04 \) and \( n = -5 \).

Step 2: Apply the binomial series expansion

Use the binomial series expansion formula:\[(1+x)^{n}=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots\]Substitute \( x = 0.04 \) and \( n = -5 \) into this formula. This gives:\[(1 + 0.04)^{-5} = 1 - 5 \times 0.04 + \frac{(-5)(-6)}{2} \times (0.04)^2 - \frac{(-5)(-6)(-7)}{6} \times (0.04)^3 + \cdots\]

Step 3: Calculate each term

Calculate the first few terms:1. First term: \(1\)2. Second term: \(-5 \times 0.04 = -0.20\)3. Third term: \(\frac{30}{2} \times 0.0016 = 24 \times 0.0008 = 0.0192\)4. Fourth term: \(- \frac{210}{6} \times 0.000064 = -35 \times 0.000064 = -0.00224\)Add these terms: \(1 - 0.20 + 0.0192 - 0.00224\).

Step 4: Sum the series

Add the calculated terms to approximate the series:\[1 - 0.20 + 0.0192 - 0.00224 = 0.81696\]Rounding to the nearest thousandth gives \(0.817\).

Step 5: Conclusion

Therefore, the approximate value of \( \frac{1}{1.04^5} \) is 0.817 when rounded to the nearest thousandth.

Key Concepts

Series Expansion

A series expansion is a way to express a mathematical function as the sum of terms calculated from the values of its derivatives at a single point. This concept forms the basis for many mathematical approximations and is essential in calculus for simplifying complex functions.

In our example, we use the binomial series expansion. The general formula for binomial series expansion is:

• \[(1+x)^{n} = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots\]

This formula allows us to expand expressions of the form \((1+x)^n\) where \(|x|<1\).

The series expansion is invaluable in physics and engineering by providing simpler polynomial expressions instead of more complicated functions. It makes calculations manageable and helps in understanding the behavior of functions around certain points.

Approximation

Approximation in mathematics refers to estimating a value when exact values are unnecessary or complicated to calculate. It is a practical tool that simplifies complex expressions and equations into usable results.

In the given problem, approximating \(\frac{1}{1.04^5}\) becomes easier with binomial expansion.

• We first rewrite \(\frac{1}{1.04^5}\) as \((1+0.04)^{-5}\), identifying \(x = 0.04\) and \(n = -5\).
• By applying the binomial series expansion, we calculate the first few terms and sum them to approximate the value.

These approximate values are especially useful in real-world applications where exact precision is impractical, such as in financial projections or engineering adjustments.

Calculus

Calculus is the branch of mathematics that studies continuous change. It provides the underlying principles for various concepts like series expansion and approximation.

The binomial theorem itself is a direct application of calculus, allowing us to express non-integer powers of expressions in a series form. This method showcases the power of calculus in

• breaking down complex problems,
• solving differential equations, and
• performing integrations and differentiations.

In the context of this exercise, calculus lends us the tools to approximate values that are otherwise very challenging to compute by hand. The expansion makes use of derivative concepts, reflecting the term by term behavior of the function around a center point, facilitating a deeper comprehension of the function's nature.