Problem 145 - Thomas Calculus

Question

The linearization of \(e^{x}\) at \(x=0\) \begin{equation}\begin{array}{l}{\text { a. Derive the linear approximation } e^{x} \approx 1+x \text { at } x=0 \text { . }} \\ {\text { b. Estimate to five decimal places the magnitude of the error }} \\ \quad {\text { involved in replacing } e^{x} \text { by } 1+x \text { on the interval }[0,0.2].} \\\ {\text { c. Graph } e^{x} \text { and } 1+x \text { together for }-2 \leq x \leq 2 . \text { Use different }} \\ \quad {\text { colors, if available. On what intervals does the approximation }} \\ \quad {\text { appear to overestimate } e^{x} ? \text { Underestimate } e^{x} ?} \end{array}\end{equation}

Step-by-Step Solution

Answer

a. Linear approximation is \(1 + x\). b. Error magnitude is about 0.0214. c. \(1+x\) underestimates \(e^x\) on most intervals.

Step 1: Linearization Formula

The linearization of a function at a point is given by the formula: \[ L(x) = f(a) + f'(a)(x-a) \]where \(f(x) = e^x\) and \(a = 0\).

Step 2: Calculate Function Value and Derivative at 0

First, compute \(f(0)\) and \(f'(0)\):- \(f(0) = e^0 = 1\)- The derivative is \(f'(x) = e^x\), so \(f'(0) = e^0 = 1\).

Step 3: Derive Linear Approximation

Substitute these values into the linearization formula:\[L(x) = 1 + 1 imes (x - 0) = 1 + x\]Thus, the linear approximation of \(e^x\) at \(x = 0\) is \(1 + x\).

Step 4: Estimate Error Magnitude

The error of the approximation on the interval \([0, 0.2]\) is given by the Taylor Remainder Theorem:The error is \(|R(x)| = |e^x - (1+x)|\).We approximate \(e^{0.2} \approx 1.2214\), so the error at \(x = 0.2\) is:\[|1.2214 - 1.2| = 0.0214\].Therefore, the magnitude of the error is approximately 0.0214.

Step 5: Graph and Intervals of Approximations

Graph \(e^x\) and \(1+x\) together over the interval \([-2, 2]\). In this graph, \(e^x\) always curves upwards whereas \(1+x\) is a straight line. From \(-2\) to \(0\), the line \(1+x\) underestimates \(e^x\), while from \(0\) to around \(2\), \(1+x\) also underestimates \(e^x\) but is very close around \(x=0\). The overestimation happens for very small intervals just above 0.

Key Concepts

Derivative

Understanding derivatives is fundamental in calculus, as they tell us how a function changes at any given point. In simple terms, the derivative represents the slope of the tangent line to the function at a specific point. For the function \(f(x) = e^x\), the derivative is particularly straightforward because the derivative of \(e^x\) remains \(e^x\) itself. This property makes \(e^x\) unique among functions. To find its derivative at any point, including \(x = 0\), you simply use the formula \(f'(x) = e^x\). Hence, at \(x = 0\), \(f'(0) = e^0 = 1\), indicating that the function has a slope of 1 at this point.

Taylor Remainder Theorem

The Taylor Remainder Theorem helps us understand how well a Taylor polynomial approximates a function within a specific interval. This is especially useful in linearization, where we use a linear polynomial to approximate a curve near a point. The theorem lets us estimate how much error exists between the actual function and the polynomial approximation.For a linear approximation of \(e^x\) around \(x = 0\), the Taylor series only goes up to the first derivative term. Therefore, the polynomial is \(1 + x\). The error, known as the remainder, can be thought of as the difference between the true value of \(e^x\) and the approximation provided by \(1 + x\). In this context, if you consider the linearization at \(x = 0\) up to \(x = 0.2\), the remainder tells you how far from \(e^x\) your approximation \(1 + x\) deviates, like knowing the error \(|R(0.2)| = |e^{0.2} - 1.2|\).

Error Estimation

Estimating the error is crucial for knowing how accurate our linear approximation is. When replacing \(e^x\) with \(1 + x\) over a small interval, such as \[0, 0.2\], it's vital to compute how much this approximation differs from reality. Let's say you estimate \(e^{0.2} \approx 1.2214\), and using \(1 + 0.2 = 1.2\), your estimation error becomes \(|1.2214 - 1.2| = 0.0214\). This result shows that, within the given interval, our approximation has a margin of error of about 0.0214. This calculation helps ensure that the linear model fits closely within the region near the selected linearization point, providing confidence in using the approximation for quick calculations.

Graphing Functions

Graphing functions such as \(e^x\) and its linear approximation \(1 + x\) can provide intuitive insights into the approximation's accuracy. When graphed between \[-2, 2\], the exponential function \(e^x\) shows an upward curve due to its continually increasing value.Alternatively, the linear approximation \(1 + x\) appears as a straight line. In the negative interval, \([-2, 0]\), \(1+x\) clearly underestimates \(e^x\). When crossing to the positive side of \(x=0\), \(1+x\) still tends to underestimate \(e^x\), particularly as \(x\) grows larger. Near the origin, the two graphs are extremely close, validating the approximation's effectiveness at values close to \(x = 0\). Graphs visually demonstrate where the approximation is better or worse on a given interval.