Question

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

Short answer

The \( n^{\text{th}} \) term is \( \frac{(-1)^{n+1}}{n}(x-1)^n \).

Step-by-step solution

Understand Taylor Series

The Taylor series for a function \( f(x) \) around a point \( a \) is given by \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f^{(n)}(a)}{n!}(x-a)^n + \cdots \). Each term is derived from the derivative of the function evaluated at the point \( a \).

Calculate Function Derivatives

Calculate the first few derivatives of \( f(x) = \ln x \):- \( f(x) = \ln x \) implies \( f'(x) = \frac{1}{x} \).- \( f''(x) = -\frac{1}{x^2} \).- \( f'''(x) = \frac{2}{x^3} \).- \( f^{(n)}(x) = (-1)^{n+1} \frac{(n-1)!}{x^n} \).

Evaluate Derivatives at Point

Evaluate each derivative at \( x = 1 \):- \( f(1) = \ln(1) = 0 \).- \( f'(1) = 1 \).- \( f''(1) = -1 \).- Continue this pattern to find \( f^{(n)}(1) = (-1)^{n+1}(n-1)! \).

Determine Taylor Polynomial Terms

Using the derivatives evaluated at \( x = 1 \), the \( n^{\text{th}} \) term of the Taylor series is \( \frac{f^{(n)}(1)}{n!}(x-1)^n \). Substituting \( f^{(n)}(1) \), this becomes:\[ \frac{(-1)^{n+1}(n-1)!}{n!}(x-1)^n \]

Simplify the Expression

Simplify \( \frac{(-1)^{n+1}(n-1)!}{n!} \) as follows:- \( \frac{(n-1)!}{n!} = \frac{1}{n} \) as \( n! = n \times (n-1)! \).- Thus, the \( n^{\text{th}} \) term becomes \( \frac{(-1)^{n+1}}{n}(x-1)^n \).

Key concepts

Taylor Series

The Taylor series is a powerful mathematical tool that allows us to approximate complex functions using an infinite sum of polynomial terms. Think of it like zooming in on a function at a particular point and representing it with polynomials. This helps simplify calculations and understand function behavior in detail.

Here's a simple way to understand it: Given a function \( f(x) \), its Taylor series expansion around a point \( a \) is:

• \( f(a) \) - This is the value of the function at the point \( a \).
• \( f'(a)(x-a) \) - This represents the linear term using the first derivative.
• \( \frac{f''(a)}{2!}(x-a)^2 \) - Here comes the quadratic term with the second derivative.
• Higher order terms, etc.

By adding these together, we approximate the shape of \( f(x) \) around \( a \). The more terms you include, the more accurate your approximation becomes.

Function Derivatives

Function derivatives tell us how a function changes as we adjust \( x \). They're crucial for building each term of a Taylor series. For example, if you're dealing with \( f(x) = \ln x \), you need to compute its derivatives:

• First derivative: \( f'(x) = \frac{1}{x} \)
• Second derivative: \( f''(x) = -\frac{1}{x^2} \)
• Third derivative: \( f'''(x) = \frac{2}{x^3} \)
• General derivative: \( f^{(n)}(x) = (-1)^{n+1} \frac{(n-1)!}{x^n} \)

Each derivative represents how the logarithmic function changes, which contributes to building the terms of the Taylor series.

Logarithmic Function

The logarithmic function \( f(x) = \ln x \) is an essential concept that emerges in various fields such as mathematics, engineering, and even computer science. It is often used to model phenomena that change logarithmically, like sound intensity or pH values.

In our exercise, we find the Taylor polynomial for \( \ln x \) around \( x = 1 \). This is particularly useful because \( \ln x \) itself can be complex to work with directly. By translating it into a polynomial approximation, we simplify many calculations. Remember, logarithms embody the inverse of exponentiation, providing a perfect venue for using derivatives to express change efficiently.

Polynomial Terms

In a Taylor polynomial, each term is crafted from the base function and its derivatives, multiplied by powers of \((x-a)\). The general term for a Taylor polynomial is \( \frac{f^{(n)}(a)}{n!}(x-a)^n \). This shows how the derivatives directly contribute to forming each polynomial term.

In the solution provided, for \( f(x) = \ln x \), the task is simplified to:

• Find the derivative at \( x = 1 \): \( f^{(n)}(1) = (-1)^{n+1}(n-1)! \)
• Plug into the formula: \( \frac{(-1)^{n+1}}{n}(x-1)^n \)

This final expression for each term neatly encapsulates all the derivatives' effects, while \( \frac{1}{n} \) adjusts the factorial growth, ensuring the polynomial mimics the natural log function very closely around \( x = 1 \).