Problem 40 - Thomas Calculus

Question

The estimate \(\sqrt{1+x}=1+(x / 2)\) is used when \(x\) is small. Estimate the error when \(|x|<0.01.\)

Step-by-Step Solution

Answer

The maximum error is approximately -0.0000125.

Step 1: Understand the Approximation Formula

We are given the approximation formula \( \sqrt{1+x} \approx 1 + \frac{x}{2} \) for small values of \(x\). This means we can use this formula to approximate the square root of \(1+x\) when \(x\) is a small number close to zero.

Step 2: Identify True Value and Error

For error estimation, we need to compare the approximate value \(1 + \frac{x}{2}\) against the true value \(\sqrt{1+x}\). The error can be expressed as the difference between these two expressions.

Step 3: Define the Error Expression

The error in this approximation is given by the formula: \( \text{Error} = \sqrt{1+x} - \left( 1 + \frac{x}{2} \right) \). This expression represents how far off our approximation is from the actual value.

Step 4: Use a Taylor Expansion to Estimate the Error

For small values of \(x\), the first non-zero term after the linear term in the Taylor expansion of \(\sqrt{1+x}\) is \(-\frac{{x^2}}{8}\). Therefore, the error becomes approximately \(-\frac{{x^2}}{8}\) for small \(x\).

Step 5: Evaluate Error for Maximum |x| Condition

We are given that \(|x| < 0.01\), which implies the magnitude of \(x\) is less than 0.01. To estimate the maximum possible error, substitute \(x = 0.01\) into the error expression: \( \text{Error} \approx -\frac{{(0.01)^2}}{8} \).

Step 6: Calculate the Error

Calculate the error using the derived formula: \( \text{Error} \approx -\frac{{0.0001}}{8} = -0.0000125 \). The error is approximately \(-0.0000125\), meaning the approximation is slightly lesser than the true value by this amount.

Key Concepts

Understanding the Approximation Formula

When dealing with small values of a variable, an approximation formula can simplify complex expressions. In the exercise provided, we use the approximation formula:

• \( \sqrt{1+x} \approx 1 + \frac{x}{2} \)

This formula is particularly useful when \(x\) is very small, as it allows us to estimate \(\sqrt{1+x}\) without needing to compute the square root directly.
This simplification is achieved by focusing on the initial linear term of the formula. Why does this work? Because when \(x\) is close to zero, higher order terms in expansion make less impact on the result.
Hence, this approximation becomes practical and efficient for calculations involving small deviations in mathematical expressions.

The Concept of Taylor Expansion

Taylor expansion provides a powerful tool for creating approximations of functions. When we want to approximate a function around a specific point, we expand it into a series:

• The series contains terms of increasing powers.
• Each term's contribution decreases as you move further away from the expansion point.

For this exercise, the Taylor series expansion of \(\sqrt{1+x}\) is used to derive the approximation, beginning from:

• The initial term \(1\).
• The first derivative term \(\frac{x}{2}\).
• And the quadratic term \(-\frac{x^2}{8}\).

The Taylor expansion helps in understanding how our approximation formula is crafted. By ignoring higher order terms like \(-\frac{x^2}{8}\), we get an approximate yet useful version of the original function for small values of \(x\).
This reduction of complexity, retaining the most significant components, is key in making calculations feasible with minor errors.

Importance of Small Values Approximation

The effectiveness of an approximation formula often hinges on the value of the variable being approximated. For small values of \(x\), the impact of higher degree terms in the expansion is minimal. This assumption simplifies the expression considerably:
- Imagine holding a magnifying glass. When focused on smaller sections, details of the larger picture become less relevant.
In practical scenarios, measuring or computing very precise values might be unnecessary. Instead, a close enough value derived swiftly through approximations can suffice.
Using small values approximation, we estimate the error in the exercise:

• By substituting \(|x|<0.01\), we calculated the error to be \(-0.0000125\).
• This tiny error confirms the efficacy of small value assumptions in yielding accurate approximations without complex calculations.

Thus, approximations for small values are not just computational tricks, but intelligent strategies for tackling complex mathematical problems efficiently.