Problem 47 - Algebra and Trigonometry with Analytic Geometry

Question

Approximate (1.2) \({ }^{10}\) by using the first three terms in the expansion of \((1+0.2)^{10}\), and compare your answer with that obtained using a calculator.

Step-by-Step Solution

Answer

Approximation: 3.72; Exact value: 6.1917. The approximation is not accurate.

Step 1: Understand the Binomial Expansion

The binomial theorem states that for any integer \( n \), the expression \((1+x)^n\) can be expanded as: \((1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k\), where \(\binom{n}{k}\) is the binomial coefficient, representing \(n!/(k!(n-k)!\)." In this problem, \(x = 0.2\) and \(n = 10\)."

Step 2: Calculate First Three Terms of Expansion

We wish to approximate \((1.2)^{10}\) using the first three terms of the expansion of \((1+0.2)^{10}\). So we calculate:1. The zeroth term: \(\binom{10}{0} (0.2)^0 = 1\).2. The first term: \(\binom{10}{1} (0.2)^1 = 10 \times 0.2 = 2\).3. The second term: \(\binom{10}{2} (0.2)^2 = \frac{10 \times 9}{2} \times 0.04 = 18 \times 0.04 = 0.72\).

Step 3: Sum the Terms

Add the first three terms obtained from the expansion to approximate \((1.2)^{10}\): \[1 + 2 + 0.72 = 3.72\] So, the approximate value of \((1.2)^{10}\) is 3.72 using the first three terms of the expansion.

Step 4: Use a Calculator for Exact Calculation

Now, compute \((1.2)^{10}\) using a calculator. Doing this gives a value of approximately 6.191736422. Compare this with our approximate value of 3.72.

Step 5: Compare and Analyze the Results

The approximation using the first three terms of the binomial expansion (3.72) is significantly lower than the exact value (6.191736422). The approximation is not very accurate due to neglecting the higher-order terms of the expansion, which are significant in this case.

Key Concepts

Binomial Coefficient

In mathematics, understanding binomial coefficients is crucial when studying expansions like \( (1+x)^n \). The binomial coefficient is denoted as \( \binom{n}{k} \), which represents the number of ways to choose \( k \) elements from a set of \( n \) elements. The formula to calculate the binomial coefficient is given by:

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

where \( n! \) (n factorial) is the product of all positive integers up to \( n \). For example, if you wanted to compute \( \binom{10}{2} \), it would be calculated as \( \frac{10 \times 9}{2 \times 1} = 45 \). This formula enables us to find the coefficients for each term in the expansion.
In the case of approximating \( (1.2)^{10} \), we use binomial coefficients to calculate the terms of the series. Understanding the role of binomial coefficients helps greatly when approximating powers due to their integral role in forming each component of the expansion.

Binomial Expansion

The Binomial Expansion is a powerful method to expand expressions like \( (1+x)^n \). According to the Binomial Theorem, such expressions can be rewritten as a sum of terms using the binomial coefficients. The formula is:

• \( (1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \)

This formula tells us how to expand \( (1+x)^n \) into a polynomial where \( x \) is raised to successive powers. Each term combines a coefficient \( \binom{n}{k} \), with a power of \( x \).
In our example of \( (1.2)^{10} \), we express it as \( (1+0.2)^{10} \). By inserting values into the expansion formula for the first three terms, we obtain:

• The zeroth term: \( \binom{10}{0} \times (0.2)^0 = 1 \)
• The first term: \( \binom{10}{1} \times (0.2)^1 = 2 \)
• The second term: \( \binom{10}{2} \times (0.2)^2 = 0.72 \)

By adding these terms, we approximate \( (1.2)^{10} \) to get a value of 3.72, though more terms would be needed for greater accuracy.

Approximation Methods

Approximation methods are techniques used to estimate values or functions, especially when exact calculations are complex or impractical. One common method is the truncation of series expansions. By taking only the first few terms of an infinite series, we can approximate a function.
For calculating \( (1.2)^{10} \), we used the binomial expansion and kept only the first three terms to simplify the calculation. The rationale behind this is that early terms generally carry the most weight, and as you add higher powers, their contribution diminishes. However, in this case, stopping at three terms resulted in an approximation of 3.72, far from the actual computed value of approximately 6.191.

• This occurs because the dropped terms in the series actually become significant in determining the overall accuracy.

To improve approximations for functions like powers of numbers slightly above 1, including more terms in the series typically yields a more precise result.