Chapter 9: Problem 30
Find the first four distinct Taylor polynomials about \(x=x_{0},\) and use a graphing utility to graph the given function and the Taylor polynomials on the same screen. $$ \ln (x+1) ; x_{0}=0 $$
Short Answer
Expert verified
The first four Taylor polynomials are: 0, \(x\), \(x - \frac{x^2}{2}\), and \(x - \frac{x^2}{2} + \frac{x^3}{3}\).
Step by step solution
01
Identify the Function and Center
We are given the function \(f(x) = \ln(x+1)\) and the center \(x_0 = 0\). We will expand this function into its Taylor series around \(x_0 = 0\).
02
Compute the Derivatives
The first few derivatives of \(f(x) = \ln(x+1)\) are needed. - \(f'(x) = \frac{1}{x+1}\)- \(f''(x) = -\frac{1}{(x+1)^2}\)- \(f'''(x) = \frac{2}{(x+1)^3}\) - \(f''''(x) = -\frac{6}{(x+1)^4}\)Evaluate these at \(x = 0\). So, \(f'(0) = 1\), \(f''(0) = -1\), \(f'''(0) = 2\), \(f''''(0) = -6\).
03
Formulate the Taylor Polynomials
Based on the derivatives, the first four distinct Taylor polynomials are: - \(P_0(x) = f(0) = 0\) - \(P_1(x) = 0 + 1 \cdot x = x\)- \(P_2(x) = x - \frac{x^2}{2}\)- \(P_3(x) = x - \frac{x^2}{2} + \frac{x^3}{3}\) \(\text{and} \) - \(P_4(x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}\)
04
Graph the Function and Polynomials
Using a graphing utility, plot \(f(x) = \ln(x+1)\) and the Taylor polynomials \(P_0(x), P_1(x), P_2(x), P_3(x), P_4(x)\). Observe how with each polynomial, the approximation improves around \(x = 0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Taylor Series
The Taylor series is a powerful tool in calculus for approximating functions using an infinite sum of terms. The series is created by evaluating the derivatives of a function at a single point, called the center. For any function \(f(x)\), the Taylor series expands the function around a central point \(x_0\) using the formula:
- \( f(x) = f(x_0) + f'(x_0)(x-x_0) + \frac{f''(x_0)}{2!}(x-x_0)^2 + \frac{f'''(x_0)}{3!}(x-x_0)^3 + \, \ldots \)
Derivative Calculation
The calculation of derivatives is critical in forming Taylor polynomials. To build a Taylor series, you need to know the first few derivatives of the function at the point \(x_0\). In our example with \(f(x) = \ln(x+1)\), derivatives are computed as follows:
- First derivative: \(f'(x) = \frac{1}{x+1}\)
- Second derivative: \(f''(x) = -\frac{1}{(x+1)^2}\)
- Third derivative: \(f'''(x) = \frac{2}{(x+1)^3}\)
- Fourth derivative: \(f''''(x) = -\frac{6}{(x+1)^4}\)
Graphing Utility
Graphing utilities are modern tools used to visualize mathematical functions and their approximations. When you graph the original function \(f(x) = \ln(x+1)\) alongside its Taylor polynomials, it's clear how each polynomial provides a different level of approximation.Steps to Graph Using a Utility:
1. Input the original function \( \ln(x+1)\).2. Add each Taylor polynomial, e.g., \(P_1(x) = x,\) \(P_2(x) = x - \frac{x^2}{2},\) etc.3. Overlay these on the same graph.4. Observe proximity of the polynomials to the original curve near \(x = 0\).These plots illustrate the convergence of the Taylor series to the function. As more terms are added (higher polynomials), the fit becomes closer to \(\ln(x+1)\) around the center.
1. Input the original function \( \ln(x+1)\).2. Add each Taylor polynomial, e.g., \(P_1(x) = x,\) \(P_2(x) = x - \frac{x^2}{2},\) etc.3. Overlay these on the same graph.4. Observe proximity of the polynomials to the original curve near \(x = 0\).These plots illustrate the convergence of the Taylor series to the function. As more terms are added (higher polynomials), the fit becomes closer to \(\ln(x+1)\) around the center.
Function Approximation
Function approximation is at the heart of using Taylor polynomials. It involves representing a function with a simpler form, typically a polynomial, which is easier for calculations. The closer the approximation matches the original function, the more accurate it generally is.
Why Use Taylor Polynomials?
- They provide insight into a function's behavior near a point.
- Useful in calculations where exact function evaluation is complex.
- Helps in estimating values for integration or differential equations.