Problem 9 - Calculus Early Transcendentals

Question

Confirm that the stated formula is the local linear approximation at \(x_{0}=0\) $$ e^{x} \approx 1+x $$

Step-by-Step Solution

Answer

The local linear approximation of \( e^x \) at \( x_0 = 0 \) is \( 1 + x \).

Step 1: State the General Formula for Linear Approximation

The formula for the linear approximation of a function \( f(x) \) near \( x = x_0 \) is given by the first order Taylor expansion: \( f(x) \approx f(x_0) + f'(x_0)(x - x_0) \).

Step 2: Identify the Function and Point of Interest

We are given \( f(x) = e^x \) and we need to find the linear approximation at \( x_0 = 0 \).

Step 3: Calculate the Function Value at \(x_0\)

Calculate \( f(x_0) \) at \( x_0 = 0 \):\[ f(0) = e^0 = 1. \]

Step 4: Compute the Derivative of the Function

The derivative of \( f(x) = e^x \) is \( f'(x) = e^x \). Thus, at \( x_0 = 0 \), \( f'(0) = e^0 = 1 \).

Step 5: Substitute Values into the Linear Approximation Formula

Substitute the values we've calculated into the general approximation formula: \( f(x) \approx f(x_0) + f'(x_0)(x - x_0) \). This gives us: \[ e^x \approx 1 + 1 \cdot x = 1 + x. \]

Step 6: Verify the Result

Confirm that the calculated approximation formula matches the given approximation. Indeed, \( e^x \approx 1 + x \), which confirms that the given formula is the correct local linear approximation at \( x_0 = 0 \).

Key Concepts

Taylor Expansion

Taylor expansion is a way to approximate functions using infinite series. However, sometimes we only need a quick, simple approximation. This is where the first few terms of the Taylor series come in handy.

For many functions, calculating each term of an entire infinite series isn’t practical. Instead, we pick a starting point, like a point of expansion, to simplify our calculations.

• The Taylor expansion begins with the function’s value at some point, say x = x_0.
• Next, it includes terms involving the function’s derivatives at that point.

The beauty of Taylor expansion is its flexibility. You can expand any smooth function around a point, such as x_0 = 0 for our e^x function. In this specific case, the Taylor expansion for small values of x simplifies to the first two terms, neatly capturing the behavior of e^x near the point of interest.

Derivative of Exponential Function

The exponential function, e^x, is unique because its derivative is the same as the function itself. This property makes calculations easier when tackling problems like linear approximation.

Let's break it down:

• The derivative of f(x) = e^x is f'(x) = e^x.
• At any point, even at x_0 = 0, this derivative doesn’t change complexity: f'(0) = e⁰ = 1.

This is a simple yet powerful concept because it instantly gives us the rate of change of e^x at any point. This property simplifies expressions for linear approximations, helping you quickly write down a formula without computing awkward derivatives.

First Order Approximation

First order approximation is a practical technique used in calculus to ESTIMATE function values with a linear expression. Often, especially near points like x_0 = 0, this provides a good balance between simplicity and accuracy.

For our exercise with e^x, we explored this type of approximation using the Taylor expansion's first order terms:

• For a function f(x), the approximation formula is:
  f(x) ≈ f(x_0) + f'(x_0)(x - x_0).
• This formula captures the function's value and the slope (rate of change) at x_0.

By using only these first terms, e^x becomes f(x) ≈ 1 + x when centered at x_0 = 0. Thus, the first order approximation paints an accurate picture using just a linear perspective, making it incredibly intuitive and quick for smaller x values.