Chapter 7: Problem 4
Let \(f(x)=\frac{1}{x+2}, c=-1\), and \(n=2 .\) Show that $$ \frac{1}{x+2}=1-(x+1)+(x+1)^{2}+R_{3} $$ where, for some \(z\) between \(x\) and \(-1\), $$ R_{3}=-\frac{(x+1)^{3}}{(2+z)^{4}} $$
Short Answer
Expert verified
The function is expressed correctly as a Taylor polynomial plus a remainder, confirming the identity.
Step by step solution
01
Identify Function and Derive Taylor Polynomial Terms
We have the function \(f(x) = \frac{1}{x+2}\) and a center \(c = -1\). We need to calculate the Taylor polynomial terms \(T_2(x)\) up to degree 2 around \(c = -1\). The Taylor series expansion for a function \(f\) around \(c\) is given by:\[f(x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \dots + R_n\] where \(R_n\) is the remainder term. Here, we're focusing on \(n=2\).
02
Calculate Function Values and Derivatives at c
Calculate \(f(-1)\), \(f'(-1)\), and \(f''(-1)\):- \(f(-1) = \frac{1}{-1+2} = 1\)- \(f'(x) = -\frac{1}{(x+2)^2}\). So, \(f'(-1) = -\frac{1}{1^2} = -1\).- \(f''(x) = \frac{2}{(x+2)^3}\). Hence, \(f''(-1) = \frac{2}{1^3} = 2\).
03
Write Taylor Polynomial T_2(x)
Now substitute the values into the Taylor series formula:\[T_2(x) = f(-1) + f'(-1)(x + 1) + \frac{f''(-1)}{2!}(x + 1)^2\]\[= 1 - (x + 1) + \frac{2}{2}(x + 1)^2\]\[= 1 - (x + 1) + (x + 1)^2\]
04
Evaluate Remainder Term R_3
The remainder term \(R_3\) for \(n=2\) in Taylor's theorem is given by:\[R_3 = \frac{f'''(z)}{3!}(x+1)^3\] where \(z\) is some value between \(x\) and \(-1\). First, calculate \(f'''(x)\):\[f'''(x) = -\frac{6}{(x+2)^4}\]. Therefore, the remainder is:\[R_3 = -\frac{6}{(2+z)^4} \cdot \frac{(x+1)^3}{3!} \]\[= -\frac{(x+1)^3}{(2+z)^4}\]This matches the given remainder expression.
05
Combine the Terms
Combine the polynomial \(T_2(x)\) and the remainder \(R_3\):\[\frac{1}{x+2} = T_2(x) + R_3 = 1 - (x+1) + (x+1)^2 - \frac{(x+1)^3}{(2+z)^4}\]This confirms that the initial statement is correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Remainder Term
The remainder term, often denoted as \( R_n \), plays a crucial role in approximating functions using Taylor series. When you approximate a function up to a certain degree \( n \), how close the approximation is to the actual function depends on this remainder. For smaller values of \( n \), the remainder may be significant, while for larger \( n \), it becomes negligible.In our exercise, the remainder term \( R_3 \) is given as: \[ R_3 = -\frac{(x+1)^3}{(2+z)^4} \] where \( z \) is a number between the point \( x \) and the center of the Taylor expansion, \( c \). This part of the formula fills in the gap between the real value of the function \( f(x) \) and its polynomial approximation. Understanding and calculating \( R_n \) allows us to estimate the error or difference between the actual function and its Taylor series representation.Key Takeaways:
- The remainder term is essential for understanding the accuracy of a Taylor approximation.
- It indicates the potential error margin in the approximation process.
- In practice, the smaller the remainder, the more accurate the approximation.
Polynomial Approximation
Polynomial approximation is the process by which we express a complex function in terms of polynomials. This is especially useful because polynomials are easier to work with, especially for calculations and analysis.In the given problem, the Taylor polynomial of degree 2 approximates the function \( f(x) = \frac{1}{x+2} \) around \( c = -1 \). The polynomial approximation we derived is:\[ T_2(x) = 1 - (x+1) + (x+1)^2 \]This is obtained by taking the function's value and its first few derivatives at the center \( c = -1 \), helping us to form a polynomial that mimics the function's behavior near this point.By using Taylor polynomials:
- We simplify complex functions into simpler polynomials.
- This enables more straightforward computations and predictions.
- It provides insights into the behavior of functions within specific ranges.
Derivative Calculation
Derivative calculation is foundational for developing a Taylor series or polynomial. It involves finding the derivatives of the function at the expansion center to get the coefficients of the polynomial terms.For the function \( f(x) = \frac{1}{x+2} \), we calculated the necessary derivatives up to the second order at \( c = -1 \):
- \( f(-1) = 1 \)
- First derivative: \( f'(x) = -\frac{1}{(x+2)^2} \) and \( f'(-1) = -1 \)
- Second derivative: \( f''(x) = \frac{2}{(x+2)^3} \) and \( f''(-1) = 2 \)
- They determine the polynomial's coefficients.
- Higher-order derivatives help refine the accuracy of the approximation.
- They offer insights into the function's rate of change near the expansion point.