Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the Maclaurin polynomial of order 4 for \(f(x)\) and use it to approximate \(f(0.12) .\) $$ f(x)=\ln (1+x) $$

Step-by-Step Solution

Answer

The approximation of \( f(0.12) \) is approximately 0.1134.

Step 1: Compute the derivatives of f(x)

Start by computing the first few derivatives of the function \( f(x) = \ln(1+x) \). You will need enough derivatives to construct a 4th-degree polynomial.1. \( f(x) = \ln(1+x) \)2. \( f'(x) = \frac{1}{1+x} \)3. \( f''(x) = -\frac{1}{(1+x)^2} \)4. \( f'''(x) = \frac{2}{(1+x)^3} \)5. \( f^{(4)}(x) = -\frac{6}{(1+x)^4} \)

Step 2: Evaluate the derivatives at x=0

Evaluate each derivative at \( x = 0 \) as we need the coefficients for the Maclaurin series.1. \( f(0) = \ln(1+0) = 0 \)2. \( f'(0) = \frac{1}{1+0} = 1 \)3. \( f''(0) = -\frac{1}{(1+0)^2} = -1 \)4. \( f'''(0) = \frac{2}{(1+0)^3} = 2 \)5. \( f^{(4)}(0) = -\frac{6}{(1+0)^4} = -6 \)

Step 3: Construct the Maclaurin Polynomial

Using the Maclaurin series expansion, we write the polynomial:\[P_4(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \frac{f^{(4)}(0)x^4}{4!}\]Substitute the derivative evaluations:\[P_4(x) = 0 + 1\cdot x + \frac{-1 \cdot x^2}{2} + \frac{2 \cdot x^3}{6} + \frac{-6 \cdot x^4}{24}\]Simplify:\[P_4(x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}\]

Step 4: Approximate f(0.12) using the polynomial

Now, substitute \( x = 0.12 \) into the polynomial \( P_4(x) \):\[P_4(0.12) = 0.12 - \frac{0.12^2}{2} + \frac{0.12^3}{3} - \frac{0.12^4}{4}\]Calculate each term:- First term: \( 0.12 \)- Second term: \( - \frac{0.12^2}{2} = -0.0072 \)- Third term: \( \frac{0.12^3}{3} \approx 0.000576 \)- Fourth term: \( -\frac{0.12^4}{4} \approx -0.000020736 \)Add them up:\[P_4(0.12) \approx 0.12 - 0.0072 + 0.000576 - 0.000020736 = 0.113355264\]

Step 5: Final Result

The Maclaurin polynomial of order 4 for \( f(x) = \ln(1+x) \) is \( P_4(x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} \). The approximation for \( f(0.12) \) using this polynomial is approximately 0.1134.

Key Concepts

Taylor Series

The Taylor series is a way of expressing a function as an infinite sum of terms. Each term is derived from the derivatives of the function at a single point. This approach allows us to approximate complex functions using polynomials, which are much simpler to work with. It’s particularly useful for calculating function values around points where direct computation might be challenging.
In a Taylor series centered at 0, which we call a Maclaurin series, we express a function in terms of its derivatives evaluated at zero. For a function \( f(x) \), the Maclaurin series is given as:

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

This notation captures the essence of how a function evolves around the point \( x = 0 \). It provides a polynomial approximation that becomes more accurate as we include more terms.

Derivative Calculation

Calculating derivatives is a crucial step when constructing a Taylor series. Each derivative provides the coefficient for the corresponding term in the polynomial expansion. For the function \( f(x) = \ln(1+x) \), we start by finding its derivatives up to the fourth order.
Here's how it unfolds:

• The zeroth derivative, \( f(x) = \ln(1+x) \)
• First derivative, \( f'(x) = \frac{1}{1+x} \)
• Second derivative, \( f''(x) = -\frac{1}{(1+x)^2} \)
• Third derivative, \( f'''(x) = \frac{2}{(1+x)^3} \)
• Fourth derivative, \( f^{(4)}(x) = -\frac{6}{(1+x)^4} \)

Evaluating these derivatives at \( x = 0 \) gives us the coefficients: \( f(0)=0 \), \( f'(0)=1 \), \( f''(0)=-1 \), \( f'''(0)=2 \), and \( f^{(4)}(0)=-6 \). Each of these coefficients plays a vital role in constructing the Maclaurin polynomial.

Polynomial Approximation

Polynomial approximation allows us to estimate complex function values through simpler polynomial expressions. By using a finite number of terms from a series, we can achieve reasonable approximations. For the function \( f(x) = \ln(1+x) \), the 4th-order Maclaurin polynomial is:

• \( P_4(x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} \)

Each term is calculated from the coefficients obtained from the derivatives. The degree of the polynomial, in this case, four, determines the number of terms and thereby affects the accuracy of the approximation.
When approximating \( f(0.12) \), we substitute \( x = 0.12 \) into \( P_4(x) \). The calculation steps involve multiplying and summing the terms:

• First term: \( 0.12 \)
• Second term: \( - \frac{0.12^2}{2} = -0.0072 \)
• Third term: \( \frac{0.12^3}{3} = 0.000576 \)
• Fourth term: \( -\frac{0.12^4}{4} = -0.000020736 \)

Adding up these terms gives an approximation: \( \approx 0.1134 \). This shows how Taylor series can be used to simplify computation.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is only defined for \( x > 0 \). It’s an essential function in mathematics with applications ranging from calculus to complex analysis. The natural logarithm of 1 plus a number \( x \), especially when \( x \) is small as in \( \ln(1+x) \), is frequently used in problem-solving.
The function \( \ln(1+x) \) can be rewritten using its Taylor series at \( x = 0 \) to approximate values for \( x \) around zero. This is particularly useful because directly computing logarithms for arbitrary values can be cumbersome without a calculator.
By understanding the derivatives and series expansion, one can construct a Maclaurin series to approximate these values accurately. This makes the complex function \( \ln(1+x) \) more accessible and simpler to handle mathematically. The approximation becomes more effective as more terms are used, providing a balance between simplicity and accuracy.