Problem 47
Question
Find the third-order Maclaurin polynomial for $$ (1+x)^{-1 / 2} $$ and bound the error \(R_{3}(x)\) if \(-0.05 \leq x \leq 0.05\).
Step-by-Step Solution
Verified Answer
The third-order polynomial is \( P_3(x) = 1 - \frac{1}{2}x + \frac{3}{8}x^2 - \frac{5}{16}x^3 \), with error bound \(|R_3(x)| \leq 3.2268 \times 10^{-7}\).
1Step 1: Review the Maclaurin Series
The Maclaurin series is a specific type of Taylor series expansion of a function \( f(x) \) about 0. It is given by: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots\] Our task is to find the third-order polynomial, which includes terms up to \( x^3 \).
2Step 2: Calculate Derivatives of the Function
The function under consideration is \( g(x) = (1+x)^{-1/2} \). We first find the derivatives:1. \( g(x) = (1+x)^{-1/2} \)2. \( g'(x) = -\frac{1}{2}(1+x)^{-3/2} \)3. \( g''(x) = \frac{3}{4}(1+x)^{-5/2} \)4. \( g'''(x) = -\frac{15}{8}(1+x)^{-7/2} \)
3Step 3: Evaluate Derivatives at x = 0
Evaluate each derivative at \( x = 0 \): - \( g(0) = 1 \) - \( g'(0) = -\frac{1}{2} \) - \( g''(0) = \frac{3}{4} \) - \( g'''(0) = -\frac{15}{8} \)
4Step 4: Construct the Third-Order Polynomial
Substitute the evaluated derivatives into the Maclaurin series formula up to the third order:\[ P_3(x) = 1 - \frac{1}{2}x + \frac{3}{4}\frac{x^2}{2!} - \frac{15}{8}\frac{x^3}{3!}\] Simplify this expression:\[ P_3(x) = 1 - \frac{1}{2}x + \frac{3}{8}x^2 - \frac{5}{16}x^3\]
5Step 5: Error Bound for the Maclaurin Polynomial
The error for the Maclaurin polynomial of order \( n \) is given by the remainder term \( R_n(x) \) which can be expressed as \[ R_3(x) = \frac{g^{(4)}(c)}{4!}x^4\]where \( c \) is some value between 0 and \( x \). Find \( g^{(4)}(x) = \frac{105}{16}(1+x)^{-9/2} \). Evaluate it for bounds \( -0.05 \leq x \leq 0.05 \). The maximum occurs at \( x = -0.05 \).
6Step 6: Calculate Maximum Error
The absolute maximum of \( g^{(4)}(x) \) over \( -0.05 \leq x \leq 0.05 \) is \( \frac{105}{16}(1+0.05)^{-9/2} \approx 6.2129 \). Thus, \[ |R_3(x)| \leq \frac{6.2129}{24} (0.05)^4 \approx 3.2268 \times 10^{-7} \]. This represents the maximum error when \(-0.05 \leq x \leq 0.05\).
Key Concepts
Taylor Series ExpansionThird-order PolynomialError BoundDerivatives Evaluation
Taylor Series Expansion
The Taylor series is a powerful tool used in mathematics to approximate functions by polynomials. This expansion expresses a function as an infinite sum of terms calculated from the values of its derivatives at a single point. The Maclaurin series is a special case where this point is zero. For a given function, the Maclaurin series can be written as:
- \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \)
Third-order Polynomial
The third-order polynomial refers to truncating the Maclaurin series expansion at the third derivative term. This means the polynomial includes all terms from the series up to and including the one containing \( x^3 \). For the function \( g(x) = (1+x)^{-1/2} \), the third-order polynomial approximation is achieved by evaluating the first, second, and third derivatives at the point \( x = 0 \).
- First, evaluate \( g(0) = 1 \).
- Next, find \( g'(0) = -\frac{1}{2} \).
- Then, \( g''(0) = \frac{3}{4} \).
- Finally, evaluate \( g'''(0) = -\frac{15}{8} \).
- \( P_3(x) = 1 - \frac{1}{2}x + \frac{3}{8}x^2 - \frac{5}{16}x^3 \)
Error Bound
When approximating functions using polynomial series like the Maclaurin series, understanding the error bound is crucial. The error bound is an estimate of how far the polynomial approximation deviates from the actual function. For the third-order Maclaurin polynomial, this error is denoted by the remainder term \( R_3(x) \). The formula for computing the error bound is:
- \( R_3(x) = \frac{g^{(4)}(c)}{4!}x^4 \)
Derivatives Evaluation
Evaluating derivatives is a key step in constructing a Taylor or Maclaurin series. Let's take the function \( g(x) = (1+x)^{-1/2} \) as an example. To find its third-order Maclaurin polynomial, we need to determine the first few derivatives at \( x = 0 \).
- The first derivative is \( g'(x) = -\frac{1}{2}(1+x)^{-3/2} \).
- The second derivative is \( g''(x) = \frac{3}{4}(1+x)^{-5/2} \).
- The third derivative is \( g'''(x) = -\frac{15}{8}(1+x)^{-7/2} \).
- \( g(0) = 1 \)
- \( g'(0) = -\frac{1}{2} \)
- \( g''(0) = \frac{3}{4} \)
- \( g'''(0) = -\frac{15}{8} \)
Other exercises in this chapter
Problem 46
Show that it is possible for \(\Sigma a_{n}\) and \(\Sigma b_{n}\) both to diverge and yet for \(\Sigma\left(a_{n}+b_{n}\right)\) to converge.
View solution Problem 47
Use a CAS to find the first four nonzero terms in the Maclaurin series for each of the following. Check Problems \(43-48\) to see that you get the same answers
View solution Problem 47
Let \(S=\left\\{x: x\right.\) is rational and \(\left.x^{2}
View solution Problem 48
Use a CAS to find the first four nonzero terms in the Maclaurin series for each of the following. Check Problems \(43-48\) to see that you get the same answers
View solution