Problem 60 - Calculus Early Transcendentals

Question

Use the remainder term to estimate the absolute error in approximating the following quantities with the nth-order Taylor polynomial centered at \(0 .\) Estimates are not unique. $$\ln 1.04 ; n=3$$

Step-by-Step Solution

Answer

Estimate the absolute error when approximating the value of ln(1.04) using a third-order Taylor polynomial centered at 0. To estimate the absolute error, we calculated the remainder term and took its absolute value: Absolute Error ≈ |R₃(x)| = |-6(0.04)⁴| Now we can compute the approximate absolute error: Absolute Error ≈ 6 * (0.04)⁴ ≈ 9.83 * 10⁻⁶ The approximate absolute error of the given approximation is 9.83 * 10⁻⁶.

Step 1: Find the Taylor expansion of \(\ln(x)\) centered at \(0\)

We want to expand \(\ln(x)\) around \(x=1\) to avoid the singularity at \(x=0\). This will give a series in \(x-1\) instead of \(x\). For this, we'll use the Maclaurin series for \(\ln(1+x)\): $$\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{(n+1)}}{n}(x-1)^n$$

Step 2: Truncate the Taylor expansion to the third order

We only need the first three terms to find the third-order Taylor polynomial: $$\ln(1+x) \approx x - \frac{x^2}{2} + \frac{x^3}{3}$$

Step 3: Evaluate the polynomial at \(x=0.04\)

Since we want to evaluate \(\ln(1.04)\), substitute \(0.04\) for \(x\): $$\ln(1.04) \approx 0.04 - \frac{0.04^2}{2} + \frac{0.04^3}{3}$$

Step 4: Find the fourth derivative of \(\ln(1+x)\)

The Lagrange form of the remainder term requires the \((n+1)\)th derivative, so we need the fourth derivative of \(\ln(1+x)\): $$f'(x)=\frac{1}{1+x},f''(x)=\frac{-1}{(1+x)^2},f'''(x)=\frac{2}{(1+x)^3},f^{(4)}(x)=\frac{-6}{(1+x)^4}$$

Step 5: Estimate the remainder term

Now, we can put everything together to estimate the remainder term: $$R_3(x) \approx \frac{-6}{(1+c)^4}(0.04)^4$$ Estimates are not unique because the \(c\) value isn't known. But it is reasonable to use \(0 \leq c \leq 0.04\). In this case, take the largest value possible for the error: $$R_3(x) \approx \frac{-6}{(1+0)^4}(0.04)^4 = -6 (0.04)^4$$

Step 6: Compute the absolute error

Since the remainder term is an approximation of the error, we can use its absolute value to determine the absolute error: Absolute Error \(\approx |R_3(x)| = |-6(0.04)^4|\) Now you can calculate this value to find the approximate absolute error of the given approximation.

Key Concepts

Remainder Term

The remainder term plays a crucial role in estimating the accuracy of a Taylor polynomial approximation. When we use a Taylor polynomial to approximate a function, the remainder term represents the difference between the actual value of the function and the value given by the polynomial.

For a Taylor series centered at 0 (a Maclaurin series), the remainder term, specifically the Lagrange form, is often expressed as:

• \( R_n(x) \approx \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{(n+1)} \)

In this context, \( f^{(n+1)}(c) \) is the \((n+1)\)th derivative of the function evaluated at some point \(c\) in the interval between \( a \) and \( x \).
This term helps in understanding how the approximation behaves as we use higher degree polynomials.

In our problem of estimating \( \ln(1.04) \), using a third-order polynomial for approximation means we consider up to the fourth derivative in the remainder term. By evaluating the fourth derivative of \( \ln(1+x) \), we can estimate how much we might be over or under by in our approximation — knowing the behavior of the derivatives can help us gauge the polynomial's reliability in that region.

Absolute Error Estimation

Absolute error estimation provides a measure of how far off our polynomial approximation is from the true value of the function.

This is crucial because it tells us the "worst-case" difference between the estimated value from the polynomial and the actual value of the function. It's computed as the absolute value of the remainder term:

• \[ \text{Absolute Error} \approx |R_n(x)| = \left|\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{(n+1)}\right| \]

To apply this, you need the n+1th derivative that was calculated at some point within your range.
By evaluating this term, you can say with certain confidence that your approximation will not deviate by more than the calculated error.

For instance, in approximating \( \ln(1.04) \), we compute this using the fourth derivative of \( \ln(1+x) \) to give us an understanding of how good our third-degree polynomial approximation is.

Maclaurin Series

The Maclaurin series, a special type of Taylor series, provides a way to approximate functions about zero. It’s a valuable tool in calculus for deriving polynomial expressions for functions. Particularly, any function that can be expressed smoothly (without sharp points or discontinuities) around zero can have its Maclaurin series written as:

• \( f(x) =f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \)

In the context of logarithmic functions like \( \ln(1+x) \), the Maclaurin series helps express this function as a polynomial, often for practical evaluations where exact values might be costly or difficult to compute.

For example, the expansion of \( \ln(1+x) \) is used to derive its polynomial form, allowing for estimates of \( \ln(1.04) \), thus offering an algebraic expression to facilitate further analysis or computation.

Derivatives of Logarithmic Functions

Understanding derivatives is fundamental in building Taylor and Maclaurin series, especially when dealing with logarithmic functions.

The derivatives of the logarithmic function \( \ln(1+x) \) are important in constructing its series:

• The first derivative: \( f'(x) = \frac{1}{1+x} \)
• The second derivative: \( f''(x) = \frac{-1}{(1+x)^2} \)
• The third derivative: \( f'''(x) = \frac{2}{(1+x)^3} \)
• The fourth derivative: \( f^{(4)}(x) = \frac{-6}{(1+x)^4} \)

These derivatives help in determining coefficients of the series expansion terms and are essential in approximating and estimating real-world phenomena using mathematical models.
This act of differentiation elaborates how each subsequent polynomial term impacts the overall sum and accuracy of the approximation.