Problem 14
Question
a. Find the first four nonzero terms of the Maclaurin series for the given function. b. Write the power series using summation notation. c. Determine the interval of convergence of the series. $$f(x)=(1+2 x)^{-1}$$
Step-by-Step Solution
Verified Answer
Answer: $$-\frac{1}{4}
1Step 1: Find the first four nonzero terms of the Maclaurin series
Recall that a Maclaurin series is a Taylor series expansion of a function about x=0. The Maclaurin series representation of a function f(x) is:
$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$$
Given the function $$f(x)=(1+2 x)^{-1}$$, we will find its first four derivatives and evaluate them at x=0. Then substitute the derivatives and values into the formula.
1. \(f(x) = (1+2x)^{-1}\)
2. \(f'(x) = -2(1+2x)^{-2}(2)\)
3. \(f''(x) = 12(1+2x)^{-3}(2^2)\)
4. \(f'''(x) = -120(1+2x)^{-4}(2^3)\)
5. \(f^{(4)}(x) = 1680(1+2x)^{-5}(2^4)\)
Now, evaluate the derivatives at x=0:
1. \(f(0) = (1+2(0))^{-1} = 1\)
2. \(f'(0) = -2(1+2(0))^{-2}(2) = -2\)
3. \(f''(0) = 12(1+2(0))^{-3}(2^2) = 12\)
4. \(f'''(0) = -120(1+2(0))^{-4}(2^3) = -120\)
5. \(f^{(4)}(0) = 1680(1+2(0))^{-5}(2^4) = 1680\)
We can now use the Maclaurin series formula to find the first four nonzero terms:
$$f(x) \approx f(0) + \frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3$$
$$f(x) \approx 1 - 2x + 6x^2 - 20x^3$$
2Step 2: Write the power series using summation notation
The Maclaurin series for the given function, in summation notation, is:
$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$$
As we have already evaluated the derivatives of the function and their corresponding values at x=0, we can rewrite the summation notation in terms of a general formula:
$$f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n (2n)!}{(1-2n)(n!)^2} x^n$$
3Step 3: Determine the interval of convergence of the series
To find the interval of convergence, we will apply the Ratio Test. Using the absolute value of the ratio of consecutive terms and the n-th term:
$$\lim_{n\rightarrow \infty} \frac{(a_{n+1})}{(a_n)}=\lim_{n\rightarrow \infty} \frac{|(-1)^{n+1} (2(n+1))!x^{n+1}|}{|(1-2(n+1))((n+1)!)^2(-1)^n(2n)!x^n|}$$
After some algebraic simplification:
$$\lim_{n\rightarrow \infty} \frac{(2(n+1))(2(n+1)-1)|x|}{|\big(1-2(n+1)\big)\big(n+1\big)^2|}$$
The series converges if the limit is less than 1. Solving for x:
$$\lim_{n\rightarrow \infty} \frac{(2(n+1))(2(n+1)-1)|x|}{|\big(1-2(n+1)\big)\big(n+1\big)^2|}<1$$
As n approaches infinity, the limit becomes:
$$\lim_{n\rightarrow \infty} |x|=\frac{1}{4}$$
Thus, the interval of convergence for the series is:
$$-\frac{1}{4}
Other exercises in this chapter
Problem 13
a. Find the first four nonzero terms of the Maclaurin series for the given function. b. Write the power series using summation notation. c. Determine the interv
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