Problem 61 - Thomas Calculus

Question

Obtain the Taylor series for 1\(/(1+x)^{2}\) from the series for \(-1 /(1+x) .\)

Step-by-Step Solution

Answer

The Taylor series for \(\frac{1}{(1+x)^2}\) is \(-1 + 2x - 3x^2 + 4x^3 - \cdots\).

Step 1: Recognize the Given Function

The initial given function is \(-\frac{1}{1+x}\). This is the derivative of the geometric series \(\frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots\).

Step 2: Find the Derivative

The derivative of \(-\frac{1}{1+x}\) with respect to \(x\) is expressed as \(f'(x) = \frac{1}{(1+x)^2}\). This function's Taylor series representation is what we're trying to find.

Step 3: Differentiate the Geometric Series

Take the derivative of the geometric series term by term: \((-1)' = 0\), \((-x)' = -1\), \((x^2)' = 2x\), \((-x^3)' = -3x^2\), and so on. Differentiating the series \(-1 + x - x^2 + x^3 - \cdots\), we get the series: \(0 - 1 + 2x - 3x^2 + 4x^3 - \cdots\).

Step 4: Reinterpret the Differentiated Series

The resulting series \(-1 + 2x - 3x^2 + 4x^3 - \cdots\) is the Taylor series for the derivative \(\frac{1}{(1+x)^2}\), which is our desired result.

Key Concepts

geometric series

A geometric series is a type of sequence where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In our exercise, the series starts with

• \(\frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots\).

This series has a common ratio of

• \(-x\).

Each term is obtained by multiplying the previous term by

• \(-x\).

This power of the geometric series helps express functions as a sum of infinite terms which can be differentiated or integrated term by term.
This makes geometric series often serve as a foundation for more complex functions like the Taylor series, which we are examining here.

derivatives

Derivatives measure how a function changes as its input changes. For example, the derivative of a function \(f(x)\) gives us a function \(f'(x)\) that describes the rate of change or the slope of \(f(x)\) at any given point. In this exercise, we differentiated

• \(-\frac{1}{1+x}\)

by finding its derivative, resulting in

• \(\frac{1}{(1+x)^2}\).

When dealing with series, taking derivatives of each term in a series like

• \(-1 + x - x^2 + x^3 - \cdots\)

can help us find a Taylor series for more complex functions. Term by term differentiation is a powerful technique that simplifies the manipulation of series using calculus rules, providing a bridge from simple geometric series to more generalized forms like power series.

power series expansion

A power series expansion is a way to express functions as an infinite sum of terms involving powers of the variable. This is crucial when constructing Taylor series, which is a specific type of power series centered at a particular point. The exercise provided

• a power series expansion for the function \(\frac{1}{(1+x)^2}\)

by differentiating a geometric series. Taylor series expansions allow us to approximate functions using polynomials, making them useful for calculations and solving differential equations.
The mechanism showcased here involved expanding

• a simpler geometric series

, then differentiating term by term to find the expansion for the derived function, illustrating the flexibility and broad applications of power series in mathematical problem-solving.