Problem 34 - Calculus: One and Several Variables

Question

Find the Lagrange form of the remainder \(R_{n}(x)\). $$f(x)=\ln (1+x) ; \quad n=5$$

Step-by-Step Solution

Answer

The Lagrange form of the remainder for the given function and degree is: \[R_{5}(x)= \frac{-x^6}{(1+c)^6 \cdot 6}\] where \(c\) is some value between 0 and x.

Step 1: Compute the (n+1)-th derivative of f(x)

Given \(f(x) = \ln(1+x)\), we need to find the 6th derivative, since \(n = 5\). To do this, we'll take one derivative at a time and observe any patterns that emerge: 1st derivative: \(f'(x) = \frac{1}{1+x}\) 2nd derivative: \(f''(x) = -\frac{1}{(1+x)^2}\) 3rd derivative: \(f'''(x) = \frac{2}{(1+x)^3}\) 4th derivative: \(f^{(4)}(x) = -\frac{2 \cdot 3}{(1+x)^4}\) 5th derivative: \(f^{(5)}(x) = \frac{2 \cdot 3 \cdot 4}{(1+x)^5}\) Observing the pattern, we can easily determine the 6th derivative: 6th derivative: \(f^{(6)}(x) = -\frac{2 \cdot 3 \cdot 4 \cdot 5}{(1+x)^6}\)

Step 2: Plug the 6th derivative into the formula for the Lagrange remainder

Using the formula for the Lagrange remainder and plugging in the 6th derivative, we get: \[R_{5}(x)= \frac{f^{(6)}(c)}{6!}(x-0)^6\] Now substitute the expression for the 6th derivative: \[R_{5}(x)= \frac{-\frac{2 \cdot 3 \cdot 4 \cdot 5}{(1+c)^6}}{6!}x^6\] To simplify, we can cancel out some terms. We have a \(5!\) in the denominator and \(2 \cdot 3 \cdot 4 \cdot 5\) in the numerator, so we can cancel them: \[R_{5}(x)= \frac{-1}{(1+c)^6 \cdot 6}x^6\] Thus, the Lagrange form of the remainder for the given function and degree is: \[R_{5}(x)= \frac{-x^6}{(1+c)^6 \cdot 6}\] where \(c\) is some value between 0 and x.

Key Concepts

Understanding Taylor Series

The Taylor series is a powerful mathematical tool used to approximate functions. It expands a function into an infinite sum of terms calculated from the values of its derivatives at a single point. This approach allows us to represent complex functions with simpler polynomial expressions, making them easier to work with. The general form of a Taylor series for a function \( f(x) \) about the point \( a \) is given by:

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

Each term involves a derivative of the function, evaluated at \( a \), and is scaled by a factorial term and a power of \( (x-a) \). This allows the series to closely match the function near \( a \), but as you move further away, the series may become less accurate. In practice, we commonly truncate the series after a finite number of terms. However, when we do this, we must take into account the error introduced, known as the Lagrange remainder.

Delving into Derivatives

Derivatives are at the heart of calculus, representing the rate at which a function changes. For any function \( f(x) \), the first derivative, \( f'(x) \), provides a measure of the slope at any point \( x \), indicating how the function behaves at that point. Higher derivatives, such as \( f''(x) \), \( f'''(x) \), and so on, offer insights into the function's curvature and changes in curvature.When dealing with Taylor series and their approximations, derivatives are used to construct each term in the series. For instance, as seen in the given exercise, the function \( \ln(1+x) \) was differentiated multiple times to form various components of its Taylor series:

• 1st derivative: \( f'(x) = \frac{1}{1+x} \)
• 2nd derivative: \( f''(x) = -\frac{1}{(1+x)^2} \)
• 3rd derivative: \( f'''(x) = \frac{2}{(1+x)^3} \)
• Continuing this pattern helps determine the required term for the Lagrange remainder.

Derivatives help us understand and utilize the Taylor series effectively, especially in calculating the error or the Lagrange remainder when truncating the series.

Exploring Logarithmic Functions

Logarithmic functions, like \( \ln(1+x) \), are crucial in many areas of mathematics and science. The function \( \ln(1+x) \) is particularly interesting because it can be expanded using the Taylor series around \( x=0 \), often referred to as the Maclaurin series:

• The expansion begins with \( x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \)

This series helps approximate the logarithmic function using a polynomial when \( x \) is close to 0. Notably, the series requires the computation of derivatives to generate each component, as highlighted in the exercise.Logarithmic functions have unique properties that make them suitable for modeling many real-world scenarios, such as exponential growth and decay, and they are often transformed into polynomials for simpler computation. Knowing how to expand these functions using Taylor series can greatly simplify analysis and problem solving.