Problem 61 - Thomas Calculus

Question

The linearization is the best linear approximation This is why we use the linearization.) Suppose that \(y=f(x)\) is differentiable at \(x=a\) and that \(g(x)=m(x-a)+c\) is a linear function in which \(m\) and \(c\) are constants. If the erroo \(E(x)=f(x)-g(x)\) were small enough near \(x=a,\) we might think of using \(g\) as a linear approximation of \(f\) instead of the linearization \(L(x)=\) \(f(a)+f^{\prime}(a)(x-a) .\) Show that if we impose on \(g\) the conditions 1\. \(E(a)=0\) 2\. \(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\) then \(g(x)=f(a)+f^{\prime}(a)(x-a) .\) Thus, the linearization \(L(x)\) gives the only linear approximation whose error is both zero at \(x=a\) and negligible in comparison with \(x-a\)

Step-by-Step Solution

Answer

The function \( g(x) = f(a) + f'(a)(x-a) \) matches the linearization, ensuring zero error at \( x = a \) and negligible error near \( x = a \).

Step 1: Understanding the Error Conditions

Let's express the error function as \( E(x) = f(x) - g(x) \). According to condition 1, the error \( E(x) \) at \( x = a \) must be zero, so \( E(a) = f(a) - g(a) = 0 \). This implies that \( g(a) = f(a) \).

Step 2: Expressing the Linear Function g(x)

The function \( g(x) \) is expressed as \( g(x) = m(x-a) + c \). From Step 1, we found \( g(a) = f(a) \), so substituting \( x = a \) gives:\[ g(a) = m(a-a) + c = c \]Thus, \( c = f(a) \). Then, \( g(x) = m(x-a) + f(a) \).

Step 3: Working with the Limit Condition

Condition 2 states that \( \lim_{x \to a} \frac{E(x)}{x-a} = 0 \). Substituting \( E(x) = f(x) - g(x) \), we have:\[ \lim_{x \to a} \frac{f(x) - g(x)}{x-a} = 0 \]This simplifies to \( \lim_{x \to a} \frac{f(x) - (m(x-a) + f(a))}{x-a} = 0 \).

Step 4: Applying the Limit Expression

Rearranging the expression in Step 3, we get: \( \frac{f(x) - f(a) - m(x-a)}{x-a} = 0 \). This expression is the definition of the derivative \( f'(a) \), therefore, we equate it:\[ \lim_{x \to a} \frac{f(x) - f(a)}{x-a} - m = f'(a) - m = 0 \]Thus, \( m = f'(a) \).

Step 5: Conclusion of the Linear Function

From Steps 2 and 4, we found that \( g(x) = f(a) + f'(a)(x-a) \). This matches the linearization \( L(x) = f(a) + f'(a)(x-a) \), proving that these are the conditions for \( g(x) \) to be a linear approximation of \( f(x) \).

Key Concepts

Linear Approximation

Linear approximation is a fundamental tool in calculus for finding the best straight-line representation of a differentiable function near a certain point. It helps us simplify complex functions into easier linear forms, especially useful in calculations and predictions.
The core idea is to use the tangent line at a specific point to approximate the behavior of the function nearby. This is precisely what the linearization process does, providing a linear model:

• Defined as the function: \[ L(x) = f(a) + f^{\prime}(a)(x-a) \],
  where \( f(a) \) is the function value and \( f^{\prime}(a) \) is the slope of the tangent line at \( x = a \).
• The slope \( f^{\prime}(a) \) represents the rate of change of the function at that point, indicating how steep the tangent line is.

Linear approximation simplifies calculations by allowing a complex function to be approximated linearly over small intervals, leading to easier analytical manipulations.

Error Function

In the context of linear approximation, the error function quantifies the difference between the actual function \( f(x) \) and its linear approximation \( g(x) \). Mathematically, it's expressed as:

• \( E(x) = f(x) - g(x) \).

Understanding this error is crucial as it highlights how accurate the linear approximation is near the point \( x = a \). Several factors aid in minimizing this error:

• Condition 1 states \( E(a) = 0 \), ensuring the error is zero at the point of tangency.
• Condition 2 involves the limit \( \lim_{x \to a} \frac{E(x)}{x-a} = 0 \), indicating the error becomes negligible as \( x \) approaches \( a \).

These conditions ensure that the linear approximation \( g(x) \) closely mirrors the original function's behavior around \( x = a \). The error function's properties make it indispensable for accurate linear approximations.

Derivative

Derivatives play a vital role in calculus as they define the slope of a function at a given point, essentially capturing how the function changes. For linear approximation, understanding derivatives is key:

• The derivative \( f^{\prime}(a) \) represents the slope of the tangent line to the function at point \( x = a \).
• This slope is crucial to forming the linear function \( L(x) = f(a) + f^{\prime}(a)(x-a) \).
• It tells how aggressive or mild the rise of the curve is at the given point \( a \).

In working through the exercise, the derivative helped find the correct slope \( m \) for the approximation \( g(x) \). Calculating the derivative ensures we precisely align our linear model with the behavior of the original function around the point of interest.

Differentiability

Differentiability is an essential property of functions that forms the basis for linear approximation. For a function to be differentiable at a point \( x = a \), it must possess a derivative at that point, indicating a well-defined tangent line. Key aspects include:

• A differentiable function is smooth and continuous around the point \( x = a \), without any sharp corners or breaks.
• If \( y = f(x) \) is differentiable at \( x = a \), it means the derivative \( f^{\prime}(a) \) exists and the linear approximation \( L(x) = f(a) + f^{\prime}(a)(x-a) \) is valid.
• Differentiability guarantees that as \( x \) approaches \( a \), the linearization accurately reflects the function's local behavior.

This property ensures that our linear model is not only accurate but also optimal around \( x = a \), minimizing the approximation error efficiently.