Question

Use the definition to find the Taylor series (centered at \(c\) ) for the function. $$ f(x)=\ln x, \quad c=1 $$

Short answer

The Taylor series of \( f(x) = \ln x \) centered at \( c = 1 \) is \( f(x) = (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3} - ... = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x-1)^n}{n} \)

Step-by-step solution

Find the Derivative of the Function at the Point c

The first few derivatives of \( f(x) = \ln x \) at \( c = 1 \) are found as follows: \n\( f'(x) = \frac{1}{x} \) then evaluating at \( c = 1 \) gives \( f'(1) = 1 \)\n\( f''(x) = -\frac{1}{x^2} \) then evaluating at \( c = 1 \) gives \( f''(1) = -1 \)\nNotice the pattern. All \( n^{th} \) derivatives of \( ln(x) \) at \( x = 1 \) can be written as \( (-1)^{n+1} (n-1)! \) for \( n \geq 1 \) and \( f(1) = 0 \)

Substituting into the Taylor Series Formula

Substitute the value of the derivatives into the Taylor series formula, we will have \n\( f(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n+1} (n-1)!(x-1)^n}{n!} \)

Simplify the Result

The \( n! \) in the denominator and \( (n-1)! \) in the numerator can simplify a bit: \n\( f(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x-1)^n}{n} \) \n Note that the lower limit of the sum is now 1 since the term for \( n = 0 \) is zero.

Write Out the Final Series

Replace the summation notation by the first few terms of the series to get a clear picture of the series: \n\( f(x) = (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3} - ... = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x-1)^n}{n} \)

Key concepts

Taylor Series Expansion

The Taylor Series expansion is an essential concept in calculus, allowing us to express complex functions as an infinite sum of polynomials, making them much easier to work with. In essence, it's a way to approximate any function by a polynomial series, particularly useful in calculations and theoretical assessments.

The Taylor series for a function is centered around a point, usually denoted as 'c'. The series is composed by iteratively taking the function's derivatives at that point and then creating terms in the series accordingly. Each term is then multiplied by a power of \(x - c\), with the exponent corresponding to the order of the derivative, and divided by the factorial of that order.

To build the series, you begin by evaluating the function and its derivatives at the point 'c'. These values then act as coefficients for the polynomial terms in the sum. The formula for a Taylor series expansion about the point 'c' is: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^{n} \.\]
Using the Taylor series, we can create a powerful tool for estimating values of functions near the point 'c', understand their behavior, and even solve certain equations that would otherwise be complex or impossible to manage directly.

Derivatives in Calculus

A cornerstone of calculus, derivatives represent the rate at which one quantity changes with respect to another. They are fundamental in depicting the 'slope' or 'gradient' at a particular point on a curve. The act of finding a derivative is called differentiation.

In the context of Taylor series, derivatives play a pivotal role as they form the coefficients of the polynomial terms in the series expansion. When calculating a Taylor series, finding the derivatives at the center point \(c\) is essential. You start with the first derivative, which gives information about the slope of the function at that point, and proceed to higher-order derivatives, which reflect more nuanced aspects of the function's behavior, such as curvature and concavity.

The pattern of derivatives, as seen in the exercise step 1 with the function \(f(x) = \ln x\), show a systematic relationship which can be exploited to form a general expression for the \(n^{th}\) term. This relationship between successive derivatives is what allows us to craft a Taylor series expansion for functions where the pattern of differentiation can be discerned and expressed neatly.

Natural Logarithm Properties

The natural logarithm, denoted as \( \ln(x) \), is a logarithmic function to the base \(e\), where \(e\) is an irrational and transcendental constant approximately equal to 2.71828. It is a crucial function in calculus because of its unique properties, which are fundamental to differential and integral calculus.

One key property of the natural logarithm is that the derivative of \( \ln(x) \) with respect to \(x\) is \(1/x\). This derivative is also instrumental in the solution of the original exercise, as it forms the basis for the successive derivatives that were calculated. Another important property is the fact that \( \ln(e) = 1 \), and the logarithm of 1 for any base is always 0, hence \( \ln(1) = 0 \). These properties are often used in solving integration and differentiation problems involving logarithmic functions.

Understanding the behavior of the natural logarithm, especially its derivatives, is essential for exercises like finding the Taylor series of \( \ln(x) \) at \(c = 1\). As illustrated in the exercise, recognizing these properties and the pattern they produce can significantly simplify the process of series expansion, helping students grasp concepts in calculus with greater ease.