Question

Find the \(n^{\text {th }}\) term of the indicated Taylor polynomial. Find a formula for the \(n^{\text {th }}\) term of the Maclaurin polynomial for \(f(x)=\frac{1}{1-x}\).

Short answer

The nth term is \( x^n \).

Step-by-step solution

Understand the Maclaurin Series

A Maclaurin series is a type of Taylor series centered at 0. The general form for a function \( f(x) \) expanded in a Maclaurin series is \( f(x) = \sum_{n=0}^{warrow} \frac{f^{(n)}(0)}{n!} x^n \). For the function \( f(x) = \frac{1}{1-x} \), we need to find \( f^{(n)}(0) \).

Differentiate the Function

Calculate the derivatives of \( f(x) = \frac{1}{1-x} \). The first derivative is \( f'(x) = \frac{1}{(1-x)^2} \), the second derivative \( f''(x) = \frac{2}{(1-x)^3} \), and so on. In general, the \( n^{\text{th}} \) derivative is \( f^{(n)}(x) = \frac{n!}{(1-x)^{n+1}} \).

Evaluate at Zero

Evaluate each derivative at 0 to use for the Maclaurin series: \( f(0) = 1 \), \( f'(0) = 1 \), \( f''(0) = 2 \), and so on. This simplifies the general derivative form to \( f^{(n)}(0) = n! \).

Construct the General Formula for the n-th Term

Substitute \( f^{(n)}(0) \) into the Maclaurin series formula. The \( n^{\text{th}} \) term is given by \( \frac{f^{(n)}(0)}{n!} x^n \). Therefore, \( t_n = \frac{n!}{n!} x^n = x^n \).

Verify the Series

The result \( x^n \) aligns with the geometric series expansion \( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \), confirming the correctness of \( t_n = x^n \).

Key concepts

Taylor Series and Maclaurin Series

The Taylor series is a way to represent a function as an infinite sum of terms that are calculated from the values of its derivatives at a single point. Specifically, for a function centered at point \( a \), the Taylor series is:

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

When this formula is specifically centered around zero, it becomes the Maclaurin series.

The Maclaurin series is simply a special case of the Taylor series, which considers only \( a = 0 \). This makes it particularly handy for functions easy to evaluate at zero, like \( f(x) = \frac{1}{1-x} \), because you don’t have to worry about making calculations around another center point. Maclaurin series are frequently used in calculus to simplify complex functions, making them easier to work with in various calculations and solving differential equations. It's especially useful when a function and its derivatives are well-defined and manageable at \( x = 0 \).

Geometric Series and Relation to Maclaurin Series

A geometric series is a series with a constant ratio between successive terms. Its general form is:

• \( S = a + ar + ar^2 + ar^3 + \,\ldots\)

where \( a \) is the first term and \( r \) is the common ratio. If \(|r|<1\), the sum of an infinite geometric series is \( \frac{a}{1-r} \).

So how does this relate to the function \( f(x) = \frac{1}{1-x} \)? The series \( 1 + x + x^2 + x^3 + \ldots \) is precisely that—a geometric series where \( a = 1 \) and \( r = x \). When finding the nth term of the Maclaurin series for this function, we see it coincides with the nth term of a geometric series.

This simple form occurs because the geometric series naturally fits the form of the Maclaurin expansion of \( \frac{1}{1-x} \). So, the work done in finding the Maclaurin series illustrates a classic connection to the geometric series, confirming the validity of term \( t_n = x^n \) from both perspectives.

Understanding the nth Term in Series Expansions

In both Taylor and Maclaurin series, finding the \( n^{\text{th}} \) term involves determining how each term contributes to approximating the original function. This term acts as a building block of the entire series representation.

For the Maclaurin series of \( f(x) = \frac{1}{1-x} \), we observe that the \( n^{\text{th}} \) term is \( x^n \). Here's how we break it down:

• First, recognize that each derivative of the function taken at zero simplifies the calculation, making the result directly \( f^{(n)}(0) = n! \).
• Given this, the term in the series becomes \( \frac{n!}{n!} x^n = x^n \) because the factorial terms cancel each other.

This balanced approach ensures the nth term accurately portrays the nth part of the series which, when summed, converges to represent the function itself. The nth term thus holds pivotal importance as it reflects the ease or complexity of determining the series expansion's individual parts.