Constructing Power Series: Algebraic & Trigonometric Expansions

If the series converges at x = 4, and the center is c = 1, then the distance from the center is |4-1| = 3. By the properties of power series, if it converges at a point distance d from the center, then it converges absolutely for all points within a distance less than d. Therefore, the radius of convergence R is at least 3. We cannot conclude anything specific about other points like x=-2 or x=0 without knowing R exactly; they could be inside or outside the interval.

For the sum of two series to converge, both individual series must converge at that point. The domain is the intersection of the two intervals. The intersection of (-2, 2) and (-5, 5) is (-2, 2).

The condition for convergence is L < 1, so 2|x+3| < 1. Divide both sides by 2: |x+3| < 1/2. This inequality states that the distance from the center (x = -3) is less than 1/2. Therefore, the radius of convergence is R = 1/2.

Since tan(x) = sin(x)/cos(x), we can perform polynomial long division on their power series. Write sin(x) = x - x³/6 + ... and cos(x) = 1 - x²/2 + ... Then divide the series for sin(x) by the series for cos(x) to find the quotient series for tan(x), which starts as x + x³/3 + ...

The interval of convergence is symmetric about the center c. The midpoint of (-2, 4] is calculated as ( (-2) + 4 ) / 2 = 2/2 = 1. Even though the right endpoint is included and the left is not, the center is still the midpoint of the interval. Therefore, c = 1.

Integrate both sides of 1/(1+x) = sum of (-1)ⁿ xⁿ with respect to x. The integral of the left side is ln(1+x). Integrate the right side term-by-term: integral of (-1)ⁿ xⁿ dx = (-1)ⁿ * xⁿ⁺¹/(n+1). So, ln(1+x) = Σ  from n=0 to infinity of (-1)ⁿ xⁿ⁺¹/(n+1) + C. To find C, set x=0: ln(1)=0, so C=0. Therefore, the series is sum of (-1)ⁿ xⁿ⁺¹/(n+1). This can be re-indexed to match the common form.

Substituting 4x for x in the original series gives f(4x) = Σ aₙ (4x)ⁿ = Σ aₙ 4ⁿ xⁿ. The radius of convergence will change: if the original converges for |x| < R, then the new series converges for |4x| < R, or |x| < R/4.

Substitute u = -x² into the series for e^u. e^u = sum of uⁿ/n!. So e^(-x²) = sum of (-x²)ⁿ / n! = sum of (-1)ⁿ x^(2n) / n!. Both A and D are algebraically equivalent, but D is written in the directly substituted form. Option A is also correct, but D is the result of the substitution without simplification. Since the question asks for the series, either representation is fine, but D shows the substitution step explicitly.

Convergence at x=-3 means the distance from 0 is 3. Divergence at x=5 means the distance is 5. The radius R must be at least 3 (to include the convergent point within the interval) and at most 5 (since it diverges at a distance of 5). Therefore, the smallest R could be is 3.

Term-by-term integration does not change the radius of convergence. The integrated series will have the same radius, R=1. The interval of convergence may differ at the endpoints, but the distance from the center within which convergence is guaranteed remains 1.

At the endpoint x = 5, we substitute 5 into the expression: (5-4)ⁿ / √n = 1ⁿ / √n = 1 / n^(½). This is a p-series with p = 1/2. The p-series test states that Σ 1/nᵖ converges if p > 1 and diverges if p ≤ 1. Since 1/2 ≤ 1, the series diverges.

The interval (-3, 7) has midpoint = (-3+7)/2 = 4/2 = 2. So the center c = 2. The radius is the distance from the center to either endpoint: 7 - 2 = 5, or 2 - (-3) = 5. So radius R = 5.

At the boundary points where |x-c| = R, the behavior of a power series can vary. Different series will behave differently at these endpoints. Some may converge absolutely (if the terms decay rapidly enough), some may converge conditionally (like the alternating harmonic series), and some may diverge. Each endpoint must be tested individually using convergence tests for numerical series.

The Maclaurin series for e^u is n=0 uⁿ / n!. To find the series for e^(x³), we perform a substitution: let u = x³. We substitute x³ in place of u (or in place of x in the standard series) to get e^(x³) = Σ  from n=0 to infinity of (x³)ⁿ / n! = sum of x^(3n) / n!. This is a straightforward substitution method.

The condition from the Ratio Test is |x-7|/9 < 1. To find the radius of convergence R, we solve this inequality for |x-7|. Multiplying both sides by 9 gives |x-7| < 9. This inequality tells us that the series converges when the distance between x and the center (c=7) is less than 9. Therefore, the radius of convergence is R = 9.