Chapter 10: Problem 9
9-12. For each function: a. Find the fourth Taylor polynomial at \(x=0\). b. Graph the original function and the Taylor polynomial together on the indicated window. $$ \begin{aligned} &f(x)=\ln (x+1)\\\ &\text { (for (b), use the window }[-2,2] \text { by }[-2,2]) \end{aligned} $$
Short Answer
Expert verified
The fourth Taylor polynomial at \(x=0\) is \(P_4(x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4\).
Step by step solution
01
Understand the Problem
We need to find the fourth Taylor polynomial for the function \( f(x) = \ln(x+1) \) centered at \( x = 0 \). Then, we will graph both the original function and the Taylor polynomial over the window \([-2, 2]\) for both x and y axes.
02
Calculate the Derivatives
To construct the Taylor polynomial, start by finding the derivatives of \( f(x) \) at \( x = 0 \). The Taylor polynomial will use the first four non-zero derivatives. - \( f(x) = \ln(x+1) \)- \( f'(x) = \frac{1}{x+1} \)- \( f''(x) = -\frac{1}{(x+1)^2} \)- \( f'''(x) = \frac{2}{(x+1)^3} \)- \( f^{(4)}(x) = -\frac{6}{(x+1)^4} \)Evaluate each of these derivatives at \( x = 0 \).
03
Evaluate Derivatives at x = 0
Plug \( x = 0 \) into each derivative to find the coefficients for the Taylor polynomial:- \( f(0) = \ln(1) = 0 \)- \( f'(0) = \frac{1}{1} = 1 \)- \( f''(0) = -\frac{1}{1^2} = -1 \)- \( f'''(0) = \frac{2}{1^3} = 2 \)- \( f^{(4)}(0) = -\frac{6}{1^4} = -6 \)
04
Construct the Fourth Taylor Polynomial
Use the formula for the Taylor polynomial: \[ P_n(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 \]Substitute the derivatives evaluated at \( x = 0 \):\[ P_4(x) = 0 + 1 \cdot x - \frac{1}{2} \cdot x^2 + \frac{2}{6} \cdot x^3 - \frac{6}{24} \cdot x^4 \]Simplify:\[ P_4(x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 \]
05
Graph the Original Function and Taylor Polynomial
Graph \( f(x) = \ln(x+1) \) and \( P_4(x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 \) over the window \([-2, 2]\) for both x and y.Note: The Taylor polynomial provides a good approximation near \( x = 0 \). As you move further from the center, the approximation may become less accurate. Ensure the graph captures the difference in behavior, especially near the x-intercept and closely around zero.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivatives
In the calculus world, derivatives play a crucial role in understanding how a function behaves. They show the rate of change of a function concerning a variable. Let’s break down the idea with an example. Given a function, \(f(x) = \ln(x+1)\), we can find its derivatives to get more insight into its behavior.
Derivatives evaluated at a particular point are crucial for constructing Taylor polynomials, which we will explore next.
- First, calculate the first derivative, \(f'(x) = \frac{1}{x+1}\). This tells us the slope of the tangent line to the graph at any point, showing how steep or flat the curve is at that point.
- Next, the second derivative, \(f''(x) = -\frac{1}{(x+1)^2}\), gives us information about the concavity of the function. If \(f''(x)\) is positive, the function is concave up, and if it's negative, the function is concave down.
- We also have higher-order derivatives like \(f'''(x)\) and \(f^{(4)}(x)\), which carry additional information about the function's behavior and contribute to forming a Taylor polynomial.
Derivatives evaluated at a particular point are crucial for constructing Taylor polynomials, which we will explore next.
Function Graphing
Graphing functions gives us a visual insight and understanding of their behavior. For our example, we have two functions to graph: the original function \(f(x) = \ln(x+1)\) and its fourth Taylor polynomial \(P_4(x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4\). Graphing might seem daunting without a calculator or software, but here's how you can do it manually.
When graphing, pay special attention:
- First, create a coordinate system over the window \[[-2, 2]\].
- Graph the original function by plotting points derived from substituting values into \(f(x)\) like \( x = -1, 0, 1, \) and observe the \`ln\` behavior which rises slowly as \(x\) increases.
- Graph the fourth Taylor polynomial using its simpler polynomial expression to see how it approximates the curve of \(f(x)\) near the center, \(x = 0\).
When graphing, pay special attention:
- Both graphs should intersect at \(x = 0\) because we constructed the Taylor polynomial to match \(f(x)\) here.
- Notice how the Taylor polynomial diverges from \(f(x)\) as \(x\) moves further from zero, illustrating how approximations can lose accuracy over larger intervals.
Polynomial Approximation
Polynomial approximation is a mathematical technique used to represent complex functions as polynomials. A Taylor polynomial is one of these approximations. It captures the behavior of a function near a given point, making calculations more manageable. Here’s how Taylor's approximation works with our example:
This method works best near the point of expansion (here \(x = 0\)). That's because the polynomial captures the essence of the function's tangents up to the fourth degree. However, as you move away from \(x = 0\), the approximation's accuracy diminishes. This happens because the higher-degree terms in the function that we ignored become significant, causing divergence from the polynomial.
- We expressed \(f(x) = \ln(x+1)\) using a degree-4 Taylor polynomial centered at \(x = 0\).
- The polynomial, \(P_4(x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4\), was constructed using derivatives calculated at \(x = 0\).
- For each term \(f^{(n)}(0) \frac{x^n}{n!}\), the coefficient is the \(n\)th derivative evaluated at \(x = 0\), which defines the polynomial's behavior around the center point.
This method works best near the point of expansion (here \(x = 0\)). That's because the polynomial captures the essence of the function's tangents up to the fourth degree. However, as you move away from \(x = 0\), the approximation's accuracy diminishes. This happens because the higher-degree terms in the function that we ignored become significant, causing divergence from the polynomial.