Chapter 11: Problem 9
Find the Taylor polynomial \(T_{3}(x)\) for the function \(f\) centered at the number \(a\). Graph \(f\) and \(T_{3}\) on the same screen. $$ f(x)=x e^{-2 x}, \quad a=0 $$
Short Answer
Expert verified
The third Taylor polynomial is \(T_3(x) = x - 2x^2 + \frac{4}{3}x^3\). Graph \(f(x)=xe^{-2x}\) and \(T_3(x)\) to compare.
Step by step solution
01
Understand a Taylor Polynomial
A Taylor polynomial is an approximation of a function around a point \(a\). The \(n\)-th Taylor polynomial \(T_n(x)\) for a function \(f(x)\) is given by: \[ T_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k \] where \(f^{(k)}(a)\) represents the \(k\)-th derivative of \(f\) evaluated at \(x=a\). We need to find \(T_3(x)\), the third-degree polynomial.
02
Calculate Derivatives of the Function
First, calculate the derivatives of \(f(x) = x e^{-2x}\): - \(f(x) = x e^{-2x}\) - First derivative: \(f'(x) = e^{-2x} - 2x e^{-2x}\) - Second derivative: \(f''(x) = -4e^{-2x} + 4x e^{-2x}\) - Third derivative: \(f'''(x) = 8e^{-2x} - 8x e^{-2x}\)
03
Evaluate Derivatives at a=0
Substitute \(x = 0\) into each derivative to evaluate them: - \(f(0) = 0 \times e^{0} = 0\) - \(f'(0) = e^{0} - 2 \times 0 \times e^{0} = 1\) - \(f''(0) = -4 \times e^{0} + 4 \times 0 \times e^{0} = -4\) - \(f'''(0) = 8 \times e^{0} - 8 \times 0 \times e^{0} = 8\)
04
Construct the Third-Degree Taylor Polynomial
Using the formula for the Taylor polynomial and values of derivatives at \(a = 0\): - \(T_3(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3\) - Substitute values: \[ T_3(x) = 0 + 1 \cdot x - 4 \cdot \frac{x^2}{2} + 8 \cdot \frac{x^3}{6} \] - Simplify: \[ T_3(x) = x - 2x^2 + \frac{4}{3}x^3 \]
05
Graph the Function and Polynomial
Using any graphing tool: - Plot \(f(x) = x e^{-2x}\), which is an exponentially decaying function. - Plot \(T_3(x) = x - 2x^2 + \frac{4}{3}x^3\) on the same graph. This shows how \(T_3(x)\) approximates \(f(x)\) near \(x = 0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Function Derivatives
To understand Taylor polynomials, a good starting point is the derivative of a function, which measures how a function changes as its input changes. Derivatives give us insight into a function's behavior by telling us the slope of the function at any given point. For a function like \(f(x) = x e^{-2x}\), derivatives help us explore its tangent lines and curvature.
Let's break it down further:
Let's break it down further:
- Zeroth Derivative: This simply refers to the function itself, \(f(x) = x e^{-2x}\). At \(a = 0\), we find \(f(0) = 0\).
- First Derivative \(f'(x):\) This shows how \(f\) is changing at each point. For this function, \(f'(x)\) is calculated as \(e^{-2x} - 2x e^{-2x}\), with \(f'(0) = 1\).
- Second Derivative \(f''(x):\) This reflects the curvature of \(f(x)\). For \(f(x) = x e^{-2x}\), \(f''(x) = -4e^{-2x} + 4x e^{-2x}\), resulting in \(f''(0) = -4\).
- Third Derivative \(f'''(x):\) The third derivative provides information on how the curvature itself changes, calculated as \(8e^{-2x} - 8x e^{-2x}\), leading to \(f'''(0) = 8\).
Polynomial Approximation
Polynomial approximation is the idea of using a polynomial to represent the behavior of a function around a particular point. This concept is particularly useful when dealing with functions that are complex or transcendental, like \(f(x) = x e^{-2x}\), because it allows us to approximate the function's value and behavior with a simpler, more understandable expression.
In this context, the Taylor polynomial is a key tool as it allows us to create a polynomial approximation by utilizing the function's derivatives.
The Taylor polynomial \(T_3(x)\) for \(f(x) = x e^{-2x}\) about \(a = 0\) is constructed using the derivatives we calculated:
In this context, the Taylor polynomial is a key tool as it allows us to create a polynomial approximation by utilizing the function's derivatives.
The Taylor polynomial \(T_3(x)\) for \(f(x) = x e^{-2x}\) about \(a = 0\) is constructed using the derivatives we calculated:
- The constant term is 0 because \(f(0) = 0\).
- The linear term is \(x\), coming from \(f'(0) = 1\).
- The quadratic term is \(-2x^2\), derived from the formula \(\frac{-4}{2!}x^2\).
- The cubic term is \(\frac{4}{3}x^3\), based on \(\frac{8}{3!}x^3\).
Graphing Functions
When it comes to visualizing Taylor polynomials and the functions they approximate, graphing is an essential tool. Graphing provides a visual representation of how the polynomial approximates the function around the center point, in this case, \(x = 0\).
To graph \(f(x) = x e^{-2x}\) alongside its Taylor polynomial \(T_3(x) = x - 2x^2 + \frac{4}{3}x^3\), it's beneficial to use graphing software or graphing calculators. This comparison helps reveal the polynomial’s effectiveness as an approximation around the center.
To graph \(f(x) = x e^{-2x}\) alongside its Taylor polynomial \(T_3(x) = x - 2x^2 + \frac{4}{3}x^3\), it's beneficial to use graphing software or graphing calculators. This comparison helps reveal the polynomial’s effectiveness as an approximation around the center.
- Exponential Decay of \(f(x):\) \(f(x) = x e^{-2x}\) rapidly decreases as \(x\) increases due to the exponential decay factor \(e^{-2x}\).
- Comparing Curves: Plot \(f(x)\) and \(T_3(x)\) on the same graph to observe how well \(T_3(x)\) follows \(f(x)\) around \(x = 0\). This shows where the approximation is most accurate.
- Closeness to \(f(x):\) Near \(x = 0\), \(T_3(x)\) and \(f(x)\) should lie almost atop one another, indicating a tight approximation at this point.
As \(x\) moves further from \(0\), the deviation between \(f(x)\) and \(T_3(x)\) will typically become more pronounced.