Problem 53 - 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. $$(1.02)^{-3}$$

Step-by-Step Solution

Answer

The approximate value of \((1.02)^{-3}\) to the nearest thousandth is 0.942.

Step 1: Identify the formula components

We are given the binomial expansion formula: \[(1+x)^{n} = 1 + nx + \frac{n(n-1)}{2!} x^{2} + \frac{n(n-1)(n-2)}{3!} x^{3} + \cdots\]Here, our task is to approximate \((1.02)^{-3}\). We identify that \((1.02)^{-3}\) can be written as \((1 + 0.02)^{-3}\), where \(x = 0.02\) and \(n = -3\). This fits the formula since \(|x| < 1\).

Step 2: Substitute values in the formula

Substitute \(n = -3\) and \(x = 0.02\) into the formula:\[(1 + 0.02)^{-3} = 1 - 3(0.02) + \frac{-3(-3-1)}{2!} (0.02)^2 - \frac{-3(-3-1)(-3-2)}{3!} (0.02)^3 + \cdots\]

Step 3: Calculate the first term

The first term of the expansion is 1. Thus, it is:\[1\]

Step 4: Calculate the second term

The second term is calculated using \(-3(0.02)\):\[-3 \times 0.02 = -0.06\]

Step 5: Calculate the third term

The third term is calculated using \(\frac{-3(-4)}{2!} (0.02)^2\):\[\frac{-3(-4)}{2} \times (0.02)^2 = 6 \times 0.0004 = 0.0024\]

Step 6: Calculate the fourth term

The fourth term is calculated using \(\frac{-3(-4)(-5)}{3!} (0.02)^3\):\[\frac{-3(-4)(-5)}{6} \times (0.02)^3 = -10 \times 0.000008 = -0.00008\]

Step 7: Sum the terms for approximation

Now sum the first four terms to approximate the value:\[1 - 0.06 + 0.0024 - 0.00008 = 0.94232\]

Step 8: Round to the nearest thousandth

Round 0.94232 to the nearest thousandth to get the final approximation:\[0.942\]

Key Concepts

Binomial Expansion

The Binomial Theorem is a fundamental concept in algebra. It provides a way to expand expressions that are raised to a power, like \((1+x)^n\). For any real number \(n\), not just integers, the binomial expansion is a series:

• The series starts with 1, then terms are added which involve powers of \(x\).
• The coefficients of each term are calculated using combinations of \(n\).
• This means for \((1+x)^n\), the expansion is: \[1 + nx + \frac{n(n-1)}{2!} x^2 + \frac{n(n-1)(n-2)}{3!} x^3 + \cdots\]

This expansion can be used for any real \(n\) provided that \(|x| < 1\). For calculations like \((1.02)^{-3}\), this generalized approach is more applicable than direct calculation, simplifying the process through a series of manageable steps.

Approximation Methods

When calculating expressions like \((1.02)^{-3}\) using binomial expansion, approximation helps simplify calculations. By truncating after several terms, the series provides an estimated value instead of an exact result.This process involves:

• Calculating about three to four terms in the expansion:
• As more terms are added, the approximation becomes more accurate.
• Ignoring terms with negligible impact due to small powers of \(x\) such as higher powers beyond \(x^3\).
• Rounding is then done to ensure the approximation meets precision requirements like the nearest thousandth.

This method balances between computational effort and precision, especially useful when \(x\) is small as it makes higher powers of \(x\) even smaller.

Negative Exponents

A negative exponent, as seen in \((1.02)^{-3}\), indicates an inverse power. It can be rewritten as \(\frac{1}{{(1.02)^3}}\). However, by using binomial expansion, dealing with negatives becomes easier.Handling negative exponents includes:

• Using the expansion formula: turning inverse values into a sum of terms.
• The formula transforms the original power problem into a series of additions and subtractions.
• Negative integer exponents become easier to manage as each term involves completing simple arithmetic operations.

Thus, what may seem complex due to negative powers is simplified through systematic expansion, turning larger computational problems into smaller, manageable parts.