Problem 56 - 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. $$(1.03)^{0.2}$$

Step-by-Step Solution

Answer

The approximation of \\( (1.03)^{0.2} \\\) is approximately \\( 1.006 \\\).

Step 1: Identify the given expression

We are asked to approximate \( (1.03)^{0.2} \). This expression can be compared to the form \( (1+x)^n \) given in the problem, where \( n = 0.2 \) and \( 1+x = 1.03 \). Therefore, \( x = 0.03 \).

Step 2: Apply the binomial series expansion

According to the series expansion \( (1+x)^n = 1 + n x + \frac{n(n-1)}{2!} x^2 + \frac{n(n-1)(n-2)}{3!} x^3 + \cdots \), substitute \( n = 0.2 \) and \( x = 0.03 \) to get initial terms: 1. \( 1^{st} \ text{ term} : 1 \)2. \( 2^{nd} \ text{ term} : 0.2 \cdot 0.03 \)3. \( 3^{rd} \ text{ term} : \frac{0.2 \cdot (0.2-1)}{2!} \cdot (0.03)^2 \) 4. Additional terms involve higher powers of \( x \) and can be neglected due to smallness.

Step 3: Calculate each term

Now compute the relevant terms: 1. \( 1^{st} \ text{ term} : 1 \)2. \( 2^{nd} \text{ term} : 0.2 \times 0.03 = 0.006 \)3. \( 3^{rd} \text{ term} : \frac{0.2 \times (-0.8)}{2} \times (0.03)^2 = -0.000072 \)

Step 4: Sum the calculated terms

To approximate \( (1.03)^{0.2} \), add the significant computed terms:\[ 1 + 0.006 - 0.000072 \approx 1.005928 \]

Step 5: Round the approximation

Round \( 1.005928 \) to the nearest thousandth, resulting in \( 1.006 \).

Key Concepts

Series Expansion

The concept of series expansion is a powerful mathematical tool that allows us to represent functions that are otherwise complex to handle. The binomial theorem in particular provides a formula to expand expressions of the form \((1+x)^n\), even when \(n\) is not a positive integer. It does so by expressing \((1+x)^n\) as an infinite series:

• 1st term: 1
• 2nd term: \(n \times x\)
• 3rd term: \(\frac{n(n-1)}{2!} \times x^2\)

The pattern continues, adding more terms with increasing powers of \(x\). Adding up these terms gives us an approximation of the function. By focusing on the significant terms, due to the small value of \(x\), we've found it's often sufficient to approximate with just a few terms. This approach is particularly useful for practical calculations.

Real Numbers

In mathematical terms, real numbers include all the numbers that can be found on the number line. This includes both rational numbers (like integers and fractions) and irrational numbers (like \(\sqrt{2}\) and \(\pi\)).
The binomial theorem's series expansion can be applied to any real number \(n\), offering a more flexible way to handle these numbers. This is different from the traditional use of integers and shows the theorem's strength.
In the given exercise, we considered \(n=0.2\), a real number, to show that we can approximate expressions involving non-integer exponents using series expansion. This capability broadens the theorem's application beyond simple expressions.

Approximation

Approximation is a critical concept in mathematics, allowing us to make complex calculations more manageable. In our exercise, we approximated \((1.03)^{0.2}\) to obtain a readily useful result.
Using series expansion, we calculate only the first few terms of the series. This is because smaller terms contribute less to the total value and can be ignored for a quick approximate solution.

• 1st term: 1
• 2nd term: 0.006
• 3rd term: -0.000072 (very small)

By summing these terms, we get an excellent approximation of \(1.03^{0.2}\), which simplifies our task of manual computation. This is where the art of approximation shows its value - simplifying complex problems into simpler, solvable ones.

Exponents

Exponents denote the power to which a number is raised. They are symbols of compact notation for expressing repeated multiplication, like \(x^n\).
In the exercise, we encounter \(n=0.2\), showing that the concept of exponents goes beyond integers. We can use real numbers as exponents, leading to fractional powers and roots.

• For \((1.03)^{0.2}\), 0.2 indicates a 'fifth' root
• This is not easily calculated by hand

That's why using binomial series expansion becomes a practical approach. It allows us to approximate complex exponential expressions like \((1+x)^n\) even when \(n\) is fractional. The flexibility of handling different kinds of exponents shows the expansive applicability of the binomial theorem.