Problem 24 - Thomas Calculus

Question

Find the Maclaurin series for the functions. \((x+1) \ln (x+1)\)

Step-by-Step Solution

Answer

Maclaurin series: \( (x+1)\ln(x+1) = x + \frac{1}{2}x^2 - \frac{1}{6}x^3 + \ldots \)

Step 1: Maclaurin Series Definition

The Maclaurin series of a function \(f(x)\) is a Taylor series expansion of \(f(x)\) about 0. It is given by \(f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots\) We need to compute the derivatives of \((x+1)\ln(x+1)\) at \(x=0\).

Step 2: Zeroth Derivative Evaluation

To find the zeroth derivative, evaluate \(f(x) = (x+1) \ln(x+1)\) at \(x = 0\). We have: \(f(0) = (0+1) \ln(0+1) = \ln(1) = 0\).

Step 3: First Derivative and Evaluation

Differentiate \(f(x) = (x+1)\ln(x+1)\) to get \(f'(x) = \ln(x+1) + 1\). Evaluate at \(x=0\): \(f'(0) = \ln(1) + 1 = 1\).

Step 4: Second Derivative and Evaluation

Differentiate \(f'(x) = \ln(x+1) + 1\) to find \(f''(x) = \frac{1}{x+1}\). Evaluate at \(x=0\): \(f''(0) = \frac{1}{1} = 1\).

Step 5: Third Derivative and Evaluation

Differentiate \(f''(x) = \frac{1}{x+1}\) to get \(f'''(x) = -\frac{1}{(x+1)^2}\). Evaluate at \(x=0\): \(f'''(0) = -1\).

Step 6: Assemble the Series

Using the Maclaurin series formula, substitute the derivatives evaluated at \(x=0\): \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots \]This yields:\[ (x+1)\ln(x+1) = 0 + x + \frac{1}{2}x^2 - \frac{1}{6}x^3 + \ldots \]

Key Concepts

Taylor Series

The Taylor series is a fundamental concept in calculus that is used to express functions as infinite sums of terms calculated from the values of their derivatives at a single point. It is a way to approximate more complex functions with polynomial expressions. The general formula for a Taylor series is:

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

Here, \(a\) is the point about which the series is expanded, \(f(a)\) is the function value, and \(f'(a), f''(a),\) etc., are its derivatives evaluated at \(a\). In a Maclaurin series, \(a\) is specifically zero.

Derivatives

Derivatives are a core pillar of calculus, representing the rate of change of a function with respect to one of its variables. When working with Taylor or Maclaurin series, derivatives are crucial because they provide the coefficients for each term in the series.
To find a derivative, you differentiate the function, progressively reducing it each time. This involves applying rules such as the power rule, product rule, and chain rule. In the context of the given exercise:

• 0th derivative: Evaluate the value of the function directly, which is \( \ln(1) = 0 \).
• 1st derivative: Differentiate to get \( \ln(x+1) + 1 \) and evaluate to \( 1 \).
• 2nd derivative: Differentiate again to get \( \frac{1}{x+1} \) and evaluate to \( 1 \).
• 3rd derivative: Further differentiation yields \( -\frac{1}{(x+1)^2} \) which evaluates to \( -1 \).

These derivatives reveal the behavior of the function around zero, forming the basis for calculating the series coefficients.

Series Expansion

In mathematics, a series expansion is a method of expressing a function as a sum of more straightforward components, often involving powers of variables. The Maclaurin and Taylor series are examples of such expansions, particularly useful for simplifying calculations and solving complex mathematical problems.
The goal of series expansion is to approximate a function by a polynomial. For the function \((x+1)\ln(x+1)\), using the derivatives found, the Maclaurin series can be assembled as:

• Starts with the function value at zero: \(f(0) = 0\).
• Add the first derivative term: \(f'(0)x = x\).
• Include the second derivative term: \(\frac{f''(0)}{2!}x^2 = \frac{1}{2}x^2\).
• Finally, add the third derivative term: \(-\frac{f'''(0)}{3!}x^3 = -\frac{1}{6}x^3\).

These components form the series: \(0 + x + \frac{1}{2}x^2 - \frac{1}{6}x^3 + \ldots\). This series approximation makes dealing with the function much simpler, especially for small values of \(x\).

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is the logarithm to the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. It is a fundamental mathematical function with unique properties that make it invaluable in calculus and other branches of mathematics.
Some key properties of the natural logarithm include:

• \(\ln(1) = 0\): This property was essential when evaluating the function at zero in the problem.
• \(\ln(ab) = \ln(a) + \ln(b)\): Useful in understanding the multiplication of terms inside logarithms.
• \(\frac{d}{dx}\ln(x) = \frac{1}{x}\): This derivative rule shows how the logarithm function changes, central to the derivative calculations in series expansions.

By using the properties of the natural logarithm and integrating them with derivative operations, you can systemically break down complex expressions, forming the backbone of series calculations like Maclaurin expansions.