Problem 89 - Thomas Calculus

Question

The linearization of \(\ln (1+x)\) at \(x=0\) Instead of approximating ln \(x\) near \(x=1,\) we approximate \(\ln (1+x)\) near \(x=0\) . We get a simpler formula this way. a. Derive the linearization \(\ln (1+x) \approx x\) at \(x=0\) b. Estimate to five decimal places the error involved in replacing \(\ln (1+x)\) by \(x\) on the interval \([0,0.1] .\) c. Graph \(\ln (1+x)\) and \(x\) together for \(0 \leq x \leq 0.5 .\) Use different colors, if available. At what points does the approximation of \(\ln (1+x)\) seem best? Least good? By reading coordinates from the graphs, find as good an upper bound for the error as your grapher will allow.

Step-by-Step Solution

Answer

a. Linearization: \(\ln(1+x) \approx x\) at \(x=0\). b. Maximum error: \(0.0047\). c. Best around \(x=0\), least around \(x=0.5\).

Step 1: Define Linearization Concept

The linearization of a function \(f(x)\) at the point \(x=a\) is given by the formula \(f(a) + f'(a)(x-a)\). This provides an approximation of the function near \(x=a\).

Step 2: Calculate Derivative

Calculate the derivative of \(f(x) = \ln(1+x)\), which is \(f'(x) = \frac{1}{1+x}\). At \(x=0\), \(f'(0) = 1\).

Step 3: Apply Linearization Formula

Using the linearization formula: \(\ln(1+x) \approx \ln(1) + 1 \cdot (x-0) = x\). Thus, the linear approximation at \(x=0\) is \(\ln(1+x) \approx x\).

Step 4: Estimate Error for Interval [0,0.1]

Find the actual values of \(\ln(1+x)\) and compare them with \(x\) on the interval \([0, 0.1]\). Calculate the maximum error, which occurs at \(x=0.1\): \(\ln(1.1) - 0.1 \approx 0.0953 - 0.1 = -0.0047\). So, the maximum error magnitude is approximately \(0.0047\).

Step 5: Graph Functions

Graph \(y = \ln(1+x)\) and \(y = x\) from \(0 \leq x \leq 0.5\). The graphs closely match near \(x=0\) and start diverging further as \(x\) approaches 0.5. The approximation is best near \(x=0\) and least good towards \(x=0.5\).

Step 6: Determine Error from Graph

By observing the graph, the point where the approximation is least accurate (farthest apart) can be seen near \(x=0.5\). The maximum error visually appears to be around 0.1, as the two curves diverge.

Key Concepts

Error Estimation

Error estimation is all about figuring out how far off our approximation can be from the exact value. In the context of linearization, it helps us understand how accurate our linear approximation is compared to the actual function.
For the function \(ln(1+x)\), we approximate it using its linear form \(x\) when \(x\) is close to 0. This is because at \(x=0\), their values match exactly, so the error, or the difference between them, is initially 0.
However, as we move away from \(x=0\), particularly towards the edge of our given interval \[0, 0.1\], the approximation starts to deviate from the actual function. By observing the difference, \(\ln(1+x) - x\), we can estimate the error.

• Calculate the actual value of \(ln(1.1)\).
• Compare it with the linear estimate \(x = 0.1\).

For instance, at \(x=0.1\), the error is roughly \(-0.0047\), showing that the approximation skips over some fine detail of the log function for small \(x\). This error estimate tells us how reliable the approximation is over this range.

Graphing Functions

Graphing functions helps us visualize where and how much two functions such as \(\ln(1+x)\) and its linear approximation \(x\) diverge. This visual insight can be very powerful.
When graphing these functions on the interval from 0 to 0.5:

• You'll witness that near \(x=0\), the curves overlap almost exactly, indicating a great match.
• As \(x\) increases, examine how the curves drift apart.

By visually comparing the vertical gap between the curves, we can pinpoint the approximation errors more intuitively.
The graph clearly shows that the functions align best near \(x=0\) and diverge as x grows towards 0.5, where the linear approximation starts to lose its precision. This is why graphical analysis is a core tool in calculus—it allows for an understanding of behavior over a range, rather than just at isolated points.

Derivative Calculation

Calculating derivatives is key in linearizing functions. The derivative represents the rate of change, which provides the slope for our linear approximation.
For \(f(x) = \ln(1+x)\), its derivative is \(f'(x) = \frac{1}{1+x}\). Getting the derivative at a point, such as \(x=0\), helps to determine the slope of the tangent line at that point.

• At \(x=0\), the derivative works out to \(f'(0) = \frac{1}{1+0} = 1\).
• This results in a tangent line with a slope of 1 at the origin, which means it rises one unit on the y-axis for each unit it moves to the right on the x-axis.

Therefore, the tangent line—our linear approximation—passes through \(x=0\) with this slope. This simple yet profound connection between derivatives and linear approximation lets us predict the function's behavior with just a small computational adjustment, especially close to where the line touches the curve. By mastering derivatives, you can become adept at crafting linear approximations for a variety of functions.