Problem 13 - Essential Calculus Early Transcendentals

Question

Find the Taylor series for \(f(x)\) centered at the given value of \(a\) . [Assume that \(f\) has a power series expansion. Do not show that \(R_{n}(x) \rightarrow 0.1]\) $$f(x)=\ln x, \quad a=2$$

Step-by-Step Solution

Answer

The Taylor series for \(\ln x\) at \(a=2\) is: \(\ln 2 + \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(n-1)!}{2^n n!} (x-2)^n\).

Step 1: Identify Taylor Series Formula

The Taylor series of a function \( f(x) \) centered at \( a \) is given by \( f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n \). This is the formula we'll use to compute the Taylor series.

Step 2: Compute the Function and Its Derivatives

First, calculate \( f(x) = \ln x \) at \( a=2 \), which gives \( f(2) = \ln 2 \). Then compute derivatives of \( f(x) \):1. \( f'(x) = \frac{1}{x} \)2. \( f''(x) = -\frac{1}{x^2} \)3. \( f^{(3)}(x) = \frac{2}{x^3} \), etc.Evaluate these derivatives at \( x = 2 \):- \( f'(2) = \frac{1}{2} \)- \( f''(2) = -\frac{1}{4} \)- \( f^{(3)}(2) = \frac{2}{8} = \frac{1}{4} \) and so on.

Step 3: Substitute Values into the Taylor Series Formula

Substitute the evaluations from Step 2 into the Taylor series formula:\[ f(x) = \ln 2 + \frac{1}{2}(x-2) - \frac{1}{4}\frac{(x-2)^2}{2!} + \frac{1}{4}\frac{(x-2)^3}{3!} - \ldots \]

Step 4: Write the General Form of the Series Terms

For the general term in the series, note the pattern and write:\[ \frac{f^{(n)}(2)}{n!} = (-1)^{n-1} \frac{(n-1)!}{2^n n!} \] for derivatives, where every odd derivative follows the series. Consequently, the Taylor series is:\[ f(x) \approx \ln 2 + \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(n-1)!}{2^n n!} (x-2)^n \]

Key Concepts

Power Series Expansion

A power series expansion allows us to express functions as infinite sums of terms based on powers of a variable. In essence, we can express a function in a similar way to polynomials. This is particularly useful for approximating complex functions and performing analysis in calculus.

A Taylor series is a form of power series expansion that approximates a function near a specific point, known as the center, typically denoted by \(a\). The formula for a Taylor series centered at \(a\) is:

• \( f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n \)

Each term in the series involves the derivatives of the function at the center \(a\), which makes it suitable for calculating function values very close to \(a\).

Using power series helps simplify complex functions such as \(e^x\), trigonometric functions, and the natural logarithm, which might otherwise be challenging to deal with directly.

Natural Logarithm

The natural logarithm, denoted as \( \ln x \), is a fundamental mathematical function linking various areas of mathematics, from algebra to calculus. It is the logarithm to the base \(e\), where \(e\) is an irrational constant approximately equal to 2.718.

The natural logarithm comes into play extensively in calculus for its simple derivative and integral properties:

• The derivative of \( \ln x \) is \( \frac{1}{x} \).
• Its integral leads to natural growth functions seen in many scientific fields.

When examining functions like \( \ln x \), power series like Taylor series become an essential tool. They allow us to expand \( \ln x \) around specific points, like \( a = 2 \) in the exercise, to make computation simpler and to analyze behavior around that point.

Understanding the natural logarithm deeply enhances comprehension of exponential growth and decay, making it vital for fields like economics, physics, and biology.

Derivatives of Functions

Derivatives measure the rate at which a function changes as its input changes. They're a cornerstone of calculus and are essential in forming a Taylor series.

To find a Taylor series for a function, we need to compute its derivatives at a specific point \(a\). For our function \(f(x) = \ln x \), we compute several derivatives:

• The first derivative: \( f'(x) = \frac{1}{x} \)
• The second derivative: \( f''(x) = -\frac{1}{x^2} \)
• The third derivative: \( f^{(3)}(x) = \frac{2}{x^3} \), and so on.

Each derivative tells us something about the function's behavior. For instance, the first derivative provides information on the slope of the tangent line to the function at any point.

These derivatives are evaluated at our chosen center \(a = 2\) in this problem. They are then used in the Taylor series formula to approximate the function's value at points near \(x = 2\). Understanding derivatives is crucial for reasoning about changes in function behavior, optimization problems, and many applications beyond mathematics.