Chapter 9: Problem 15
Find the Taylor polynomial of order 3 based at 1 for \(f(x)=x^{3}-2 x^{2}+3 x+5\) and show that it is an exact represen tation of \(f(x)\).
Short Answer
Expert verified
The Taylor polynomial is \( P_3(x) = x^3 - 2x^2 + 3x + 5 \), exactly matching \( f(x) \).
Step by step solution
01
Write Down the Function and Its Derivatives
The given function is \( f(x) = x^3 - 2x^2 + 3x + 5 \). We need its derivatives up to the third order: 1. \( f(x) = x^3 - 2x^2 + 3x + 5 \) 2. First Derivative: \( f'(x) = 3x^2 - 4x + 3 \) 3. Second Derivative: \( f''(x) = 6x - 4 \) 4. Third Derivative: \( f'''(x) = 6 \) 5. Fourth Derivative: \( f^{(4)}(x) = 0 \) (and for higher derivatives too).
02
Calculate Function Values at x = 1
Plug \( x = 1 \) into the function and its derivatives:1. \( f(1) = 1^3 - 2(1)^2 + 3(1) + 5 = 1 - 2 + 3 + 5 = 7 \) 2. \( f'(1) = 3(1)^2 - 4(1) + 3 = 3 - 4 + 3 = 2 \) 3. \( f''(1) = 6(1) - 4 = 2 \) 4. \( f'''(1) = 6 \)
03
Construct the Taylor Polynomial
The Taylor polynomial of order 3 based at \( x = 1 \) is given by:\[ P_3(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3 \]Substituting the evaluated derivatives:\[ P_3(x) = 7 + 2(x-1) + \frac{2}{2}(x-1)^2 + \frac{6}{6}(x-1)^3 \]Simplifying the terms gives:\[ P_3(x) = 7 + 2(x-1) + (x-1)^2 + (x-1)^3 \]
04
Verify the Polynomial is the Exact Representation of f(x)
Since the third derivative is a constant (6) and all higher derivatives are zero, the Taylor polynomial up to the third degree provides the exact form of \( f(x) \). Rewriting \( P_3(x) = 7 + 2(x-1) + (x-1)^2 + (x-1)^3 \) confirms that it simplifies to the original function \( f(x) = x^3 - 2x^2 + 3x + 5 \) when expanded:- Expansion yields: \( (x-1)^3 = x^3 - 3x^2 + 3x - 1 \) - \( (x-1)^2 = x^2 - 2x + 1 \) - Combining with existing terms transforms back to \( x^3 - 2x^2 + 3x + 5 \), proving the match.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivatives
When working with Taylor polynomials, derivatives of the function play a crucial role. A derivative represents the rate at which a function's value changes as its input changes. For most functions, evaluating derivatives is a step-by-step process that involves reducing the original function to simpler expressions that reflect its rate of change.
- The first derivative, denoted as \(f'(x)\), gives the slope of the function, showing how the function changes at any point \(x\).
- Higher-order derivatives, such as the second derivative \(f''(x)\), indicate the rate of change of the previous derivative, providing insight into the concavity of the function.
Function Evaluation
Evaluating a function means calculating its value at specific points. For Taylor polynomials, this means substituting a particular base point, such as \(x = 1\) in our example, into both the original function and its derivatives. This isn't just a routine calculation; it's about understanding how the function 'sits' at this base point.
- We start by evaluating the function itself, \(f(1)\), which results in a numerical value representing its state at that point.
- Following this, we calculate the first, second, and third derivatives at \(x = 1\). These evaluated derivatives not just shed light on the behavior of the function in the vicinity of \(x = 1\), but they also enable us to build the Taylor polynomial accurately.
Polynomial Expansion
Polynomial expansion is applying algebraic techniques to express a polynomial in a different form or combine terms. When constructing a Taylor polynomial, starting with an expression like \(P_3(x) = 7 + 2(x-1) + (x-1)^2 + (x-1)^3\), the aim is to simplify it until it directly corresponds to the original function.
- Each term in this polynomial reflects the behavior of \(f(x)\) at \(x=1\). For instance, \((x-1)^3\) represents part of the cubic nature of \(f(x)\). The task is about rewriting this polynomial in standard form.
- This transformation requires expanding powers of \((x-1)\) to their full algebraic expressions, like \((x-1)^2 = x^2 - 2x + 1\), and then methodically combining them.