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Chapter 3: Problem 15

Faster than a calculator Use the approximation \((1+x)^{k} \approx\) \(1+k x\) to estimate the following. $$ \text { a. }(1.0002)^{50} \quad \text { b. } \sqrt[3]{1.009} $$

Short Answer

Expert verified
a. Approximately 1.01. b. Approximately 1.003.

Step by step solution

01

Identify the expression structure for part (a)

For part (a), the expression \((1.0002)^{50}\) fits the structure \((1 + x)^k\) where \(x = 0.0002\) and \(k = 50\).
02

Apply the approximation for part (a)

Using the approximation \((1+x)^k \approx 1+kx\), apply it to the expression identified:\[ (1.0002)^{50} \approx 1 + 50 \times 0.0002 = 1 + 0.01 = 1.01. \]
03

Identify the expression structure for part (b)

For part (b), the expression \(\sqrt[3]{1.009}\) can be written as \((1 + x)^{1/3}\) with \(x = 0.009\) and \(k = \frac{1}{3}\).
04

Apply the approximation for part (b)

Similarly, apply the approximation formula: \[ \sqrt[3]{1.009} \approx 1 + \frac{1}{3} \times 0.009 = 1 + 0.003 = 1.003. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Approximation
The binomial approximation is a mathematical technique that helps to simplify the computation of expressions raised to a power. When you encounter expressions like \((1+x)^k\), where \(x\) is a small number, determining the exact value can be cumbersome.
This approximation states that \((1+x)^{k} \approx 1+kx\) when \(x\) is much smaller than 1. This is useful when quick estimates are needed.
  • For example, in our exercise, when estimating \((1.0002)^{50}\), we identified that \(x = 0.0002\) and \(k = 50\).
  • Since \(x\) is very small, using \(1+kx\) provides an easier computation with reasonably accurate results.
This approximation is derived from the binomial theorem, which expands \((1+x)^k\) into an infinite series, but only its first few terms are used when \(x\) is small.
Exponents
Exponents are fundamental in algebra and mathematics as a whole. They represent repeated multiplication of a number or expression. For instance, \(a^n\) means \(a\) multiplied by itself \(n\) times.
In our approximation exercise, the expressions \((1.0002)^{50}\) and \(\sqrt[3]{1.009}\) are both examples exploiting the concept of exponents.
  • When you see an exponent, think about how many times you multiply the base number by itself.
  • In expressions where exponents seem complicated, approximation techniques can simplify the process.
By converting complex powers into simpler forms using approximation techniques, we create more accessible calculations, especially when precise computing tools are not available.
Roots
Roots are closely related to exponents, often being the inverses of exponentiation. For instance, while the square root \(\sqrt{a}\) finds a number that, when squared, returns \(a\), cube root operations seek a number that results in \(a\) when cubed.
The exercise illustrates this concept with the cube root approximation of \(1.009\), i.e., \((1+x)^{1/3}\).
  • Approximating roots often involves rewriting them in exponential form. In this case, \(\sqrt[3]{1.009}\) is equivalent to \((1 + 0.009)^{1/3}\).
  • Such transformations make it possible to apply binomial approximations effectively, helping to estimate complex roots.
These strategies are particularly useful when numbers are slightly above 1, allowing for straightforward calculations without complex tools.
Algebraic Expressions
Algebraic expressions are combinations of numbers, variables, and operations that define mathematical relationships. The ability to simplify and manipulate these expressions is key to solving equations and inequalities efficiently.
In our approximation task, both parts (a) and (b) involve simplifying expressions of the form \((1+x)^k\).
  • The process involves recognizing the structure and making strategic approximations.
  • This often requires rewriting expressions efficiently, as seen with \(\sqrt[3]{1.009}\) expressed as \((1+0.009)^{1/3}\).
Understanding how to handle algebraic expressions helps in interpreting and solving diverse mathematical problems efficiently, laying a groundwork for more complex calculations.

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