Problem 54 - A Graphical Approach to Precalculus with Limits

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 to the nearest thousandth. $$\frac{1}{1.04^{5}}$$

Step-by-Step Solution

Answer

The approximated value of \(\frac{1}{1.04^5}\) is approximately 0.822.

Step 1: Identify the given expression

The formula provided is an expansion for \((1+x)^n\) known as the Binomial Theorem for any real number \(n\). We are asked to approximate \(\frac{1}{1.04^5}\), which can be rewritten as \((1.04)^{-5}\).

Step 2: Rewrite using the expansion formula

Using the identity \((1+x)^n = (1+x)^{-5}\), let \(n = -5\) and \(x = 0.04\) because \(1 + 0.04 = 1.04\). This gives us: \((1+0.04)^{-5} = 1 + (-5)(0.04) + \frac{(-5)(-6)}{2!}(0.04)^2 + \frac{(-5)(-6)(-7)}{3!}(0.04)^3 + \cdots\).

Step 3: Compute the first few terms

Calculate the first few terms of the expansion for better accuracy: - First Term: \(1\)- Second Term: \(-5 \times 0.04 = -0.2\)- Third Term: \(\frac{(-5)(-6)}{2!}(0.04)^2 = \frac{30}{2} \times 0.0016 = 0.024\)- Fourth Term: \(\frac{(-5)(-6)(-7)}{3!}(0.04)^3 = \frac{-210}{6} \times 0.000064 = -0.00224\)

Step 4: Approximate the value

Sum the calculated terms to approximate \((1.04)^{-5}\):\[1 - 0.2 + 0.024 - 0.00224 = 0.82176\]Rounded to the nearest thousandth, this becomes \(0.822\).

Step 5: Verify conditions and conclude

Ensure \(|x| = |0.04| < 1\) and that the series converges faster for smaller \(x\), which holds true here. Thus, we can reasonably approximate \((1.04)^{-5}\) using the first few terms.

Key Concepts

Binomial Expansion

The binomial expansion is a powerful tool in mathematics, especially when dealing with expressions of the form \((1+x)^n\). The classic binomial theorem is typically applied to positive integer exponents, but it can be extended to real exponents as well. This generalized expansion is defined as:

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

This formula becomes particularly useful for approximating values when \(|x| < 1\).
In this exercise, the expansion helps approximate expressions like \((1.04)^{-5}\), using negative exponents.

Approximations

Approximations are essential in mathematics, especially when exact calculations are cumbersome. By using only the first few terms of a series, we can get a value close enough to the desired result for practical purposes.
For instance, in the problem at hand, using the series expansion for \((1.04)^{-5}\), we selected the first few terms to determine a reasonable estimate. Typically:

• The first term gives a rough estimate.
• Additional terms increase the accuracy.

Approximating leads to solutions such as \(0.822\), rounded to the nearest thousandth, which is an acceptable result for most applied scenarios.

Negative Exponents

Negative exponents indicate reciprocal values. For example, \(a^{-n} = \frac{1}{a^n}\).
This concept is critical when using binomial expansions for values like \((1+x)^{-n}\). Here, the exercise utilized a negative exponent by converting \(\frac{1}{1.04^{5}}\) into \((1.04)^{-5}\).This allows:

• Applying binomial expansion to approximate reciprocal expressions.
• Expressing real-world problems, such as discount rates in financial models.

Understanding negative exponents helps in tackling problems across different domains.

Series Convergence

Series convergence is crucial when using expansions. A series converges if the sum approaches a finite number as more terms are added. This property is vital in approximations to ensure reliability.
In our binomial expansion example:

• Convergence improves approximation accuracy.
• Smaller \(|x|\) makes series converge faster, ensuring minimal error with fewer terms.

This is why initial check \(|x| < 1\) is critical for applying the expansion method. A convergent series allows us to rely on partial sums to represent complex functions approximately.