The Correct Answer and Explanation is:
Based on the provided solution for part (a), there appears to be a typo in the problem statement for f(x). The function as written, f(x) = Σ [(-1)^(n+1)(x-2)^n] / n, has a radius of convergence of R=1, leading to an interval of (1, 4), which contradicts the given answer of (0, 4].
For the interval to be (0, 4], the center must be a=2 and the radius must be R=2. This occurs if the function is:
f(x) = Σ (from n=1 to ∞) [(-1)^(n+1)(x-2)^n] / (n * 2^n)
We will proceed using this corrected function.
(a) Interval of convergence for f(x)
The interval of convergence for this corrected function is indeed (0, 4]. At x=0, the series becomes -Σ 1/n, the divergent harmonic series. At x=4, it becomes Σ (-1)^(n+1)/n, the convergent alternating harmonic series.
(b) Interval of convergence for f'(x)
Term-by-term differentiation gives:
f'(x) = Σ (from n=1 to ∞) [(-1)^(n+1) * n(x-2)^(n-1)] / (n * 2^n) = Σ (from n=1 to ∞) [(-1)^(n+1)(x-2)^(n-1)] / 2^n
The radius of convergence remains R=2, so the open interval is (0, 4). We check the endpoints:
- At x=0: The series is Σ [(-1)^(n+1)(-2)^(n-1)] / 2^n = Σ [(-1)^(2n)] / 2 = Σ 1/2. This diverges by the n-th Term Test.
- At x=4: The series is Σ [(-1)^(n+1)(2)^(n-1)] / 2^n = Σ (-1)^(n+1) / 2. This diverges as its partial sums oscillate.
The interval of convergence for f'(x) is (0, 4).
(c) Interval of convergence for f”(x)
Differentiating f'(x) gives:
f”(x) = Σ (from n=2 to ∞) [(-1)^(n+1)(n-1)(x-2)^(n-2)] / 2^n
The radius is still R=2. We check the endpoints of (0, 4):
- At x=0: The series is Σ [(-1)^(n+1)(n-1)(-2)^(n-2)] / 2^n = Σ -(n-1)/4. This diverges.
- At x=4: The series is Σ [(-1)^(n+1)(n-1)(2)^(n-2)] / 2^n = Σ (-1)^(n+1)(n-1)/4. This diverges.
The interval of convergence for f”(x) is (0, 4).
(d) Interval of convergence for ∫f(x) dx
Term-by-term integration gives:
∫f(x) dx = C + Σ (from n=1 to ∞) [(-1)^(n+1)(x-2)^(n+1)] / (n(n+1) * 2^n)
The radius is still R=2. We check the endpoints of (0, 4):
- At x=0: The series is Σ [(-1)^(n+1)(-2)^(n+1)] / (n(n+1)2^n) = Σ 2/(n(n+1)). This series converges by comparison with the p-series Σ 1/n².
- At x=4: The series is Σ [(-1)^(n+1)(2)^(n+1)] / (n(n+1)2^n) = Σ 2(-1)^(n+1)/(n(n+1)). This series converges absolutely (as shown for x=0), so it converges.
The interval of convergence for ∫f(x) dx is [0, 4].
Final Answers:
(a) f(x): (0, 4]
(b) f'(x): (0, 4)
(c) f”(x): (0, 4)
(d) ∫ f(x) dx: [0, 4]
