Question

a. Differentiate the Taylor series about 0 for the following functions. b. Identify the function represented by the differentiated series. c. Give the interval of convergence of the power series for the derivative. $$f(x)=-\ln (1-x)$$

Short answer

a) Find the Taylor series of the given function centered at x=0 and differentiate the series term by term. The Taylor series of the given function $$f(x) = -\ln{(1-x)}$$ centered at x=0 is: $$-\ln{(1-x)} = (x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots)$$ Differentiated series term by term is: $$f'(x) = 1 + x + x^2 + x^3 + \cdots = \frac{1}{1-x}$$ b) Identify the function represented by the differentiated series. The function represented by the differentiated series is: $$f'(x) = \frac{1}{1-x}$$ c) Find the interval of convergence for the power series of the derivative. The interval of convergence for the power series of the derivative is $$(-1, 1)$$.

Step-by-step solution

Find the Taylor series of f(x) centered at 0

To find the Taylor series of the function $$f(x) = -\ln (1-x)$$ at x=0, we can use the well-known geometric series $$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$$. Now, we can integrate the series with respect to the variable x to find the Taylor series of the given function: $$-\ln{(1-x)} = \int \frac{1}{1-x} dx = \int (1 + x + x^2 + x^3 + \cdots) dx$$

Integrate the geometric series term by term

Now, let's integrate the series term by term with respect to x: $$-\int \frac{1}{1-x} dx = \int (1 + x + x^2 + x^3 + \cdots) dx = (x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots) + C$$ Considering that the Taylor series is centered at 0, we have $$C=0$$, and thus the Taylor series of $$-\ln{(1-x)}$$ is: $$-\ln{(1-x)} = (x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots)$$

Differentiate the Taylor series term by term

Now, differentiate the Taylor series term by term with respect to x: $$\frac{d}{dx} \left[-\ln{(1-x)}\right]= \frac{d}{dx} \left[ x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots \right]$$ This results in the differentiated series: $$f'(x) = 1 + x + x^2 + x^3 + \cdots = \frac{1}{1-x}$$

Identify the function represented by the differentiated series

Comparing the differentiated series with the original function, we can identify the function represented by the differentiated series as: $$f'(x) = \frac{1}{1-x}$$

Find the interval of convergence for the power series of the derivative

To find the interval of convergence for the power series of the derivative, we can use the ratio test. If the absolute value of the ratio of consecutive terms approaches a meaningful constant as n approaches infinity, the series converges. Here, the ratio is $$\left|\frac{x^{n+1}}{x^n}\right| = |x|$$. Thus, the series converges if $$|x| < 1$$, so the interval of convergence for the power series of the derivative is $$(-1, 1)$$.

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a quantity changes. In the context of Taylor series, differentiation is used to find the derivative of a function represented by its series expansion. Differentiating a Taylor series term by term is often straightforward. For example, consider the derivative of the Taylor series for the function \(-\ln(1-x)\). This series starts as \(x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots\). Differentiating each term individually involves applying the power rule, which states that the derivative of \(x^n / n\) is \(x^{n-1}\). By performing term-by-term differentiation:

• The derivative of \(x\) is 1.
• The derivative of \(\frac{x^2}{2}\) is \(x\).
• The derivative of \(\frac{x^3}{3}\) is \(x^2\).
• And so on...

The resulting series \(1 + x + x^2 + \cdots\) can be recognized again as a geometric series \(\frac{1}{1-x}\). Hence, differentiation allows us to convert a series into a compact form, which is both easier to analyze and often useful in calculations and further functions.

Interval of Convergence

The interval of convergence in the context of a power series is the set of all real numbers for which the series converges, or sums to a finite value. For a Taylor series, this interval is crucial as it defines where the series accurately represents the function.To find the interval of convergence for a differentiated power series like in our example, we employ the ratio test. In our series \(1 + x + x^2 + \cdots\), we compare the absolute value of the ratio of consecutive terms:\[ \left| \frac{x^{n+1}}{x^n} \right| = |x| \]According to the ratio test, the series converges if this ratio is less than one. Thus, we find:

• The series converges for \(|x| < 1\).
• The interval of convergence is \((-1, 1)\).

If \(|x| \geq 1\), the series diverges. This interval ensures that within these bounds, the Taylor series matches the behavior of the function it’s representing.

Power Series

A power series is an infinite sum of terms in the form \(a_n x^n\), where \(a_n\) represents the coefficients of the series. Power series are incredibly useful because they can approximate functions and are used to solve differential equations, among other applications.In our exercise, the Taylor series for \(-\ln(1-x)\) is expressed as a power series:\[ x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots \]Such a series encapsulates the function's behavior around a specific point—in this case, 0. By changing values of \(x\) within its interval of convergence, different approximations of \(-\ln(1-x)\) can be calculated. Though an infinite series, for practical applications, we might truncate it after a few terms to make approximations.Power series can represent not only simple functions but also complex ones under specific conditions, like within the radius of convergence. Their flexibility makes them a powerhouse tool in calculus.

Logarithmic Functions

Logarithmic functions, such as \(\ln(x)\), are essential in many areas of mathematics. They are the inverses of exponential functions and have unique properties, such as the ability to transform multiplicative relationships into additive ones.In this exercise, we consider the function \(-\ln(1-x)\). Logarithmic transformations are often used in both pure math and applied fields like statistics and economics, due to their utility in modeling growth rates and compounding processes.The Taylor series for \(-\ln(1-x)\) illustrates the principle where complex functions can be represented as an infinity of simpler polynomial terms. Understanding this not only reinforces the relationship between different types of functions but also enhances our ability to manipulate and interpret them for practical use.