Constructing Power Series: Algebraic & Trigonometric Expansions

1) The power series Σ aₙ (x-1)ⁿ converges at x = 4. What can we conclude?

It diverges at x = -2.

It converges at x = 0.

It converges at x = -1.

The radius is at least 3.

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.

Explanation

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.

2) The power series for f(x) converges on (-2, 2) and the power series for g(x) converges on (-5, 5). What is the interval of convergence for the sum f(x) + g(x)?

(-5, 5)

(-2, 2)

[-2, 5]

[-2, 2]

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).

Explanation

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).

3) For the series Σ from n=0 to infinity of (-2)ⁿ (x+3)ⁿ, applying the Ratio Test gives L = 2|x+3|. What is the radius of convergence?

1/2

1

2

3

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.

Explanation

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.

4) Using the known series for sin(x) and cos(x), how would you find the series for tan(x) up to the x³ term?

Differentiate sin(x).

Integrate cos(x).

Divide the series for sin(x) by the series for cos(x) using polynomial long division.

Multiply the series for sin(x) and cos(x).

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 + ...

Explanation

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 + ...

5) The power series for f(x) has an interval of convergence (-2, 4]. What is the center of the series?

-2

1

2

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.

Explanation

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.

6) Integrate the series for 1/(1+x) = Σ from n=0 to infinity of (-1)ⁿ xⁿ term-by-term to find a series for ln(1+x). What is the result?

Σ from n=0 to infinity of (-1)ⁿ xⁿ⁺¹ / (n+1)

Sum from n=1 to infinity of (-1)ⁿ⁺¹ xⁿ / n

Σ from n=0 to infinity of (-1)ⁿ xⁿ / n

Sum from n=1 to infinity of (-1)ⁿ xⁿ⁻¹ / n

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.

Explanation

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.

7) A power series Σ aₙ xⁿ converges to f(x). What is the power series representation for f(4x)?

Σ aₙ 4ⁿ xⁿ

Σ aₙ x^(4n)

Sum of 4aₙ xⁿ

Σ aₙ xⁿ / 4ⁿ

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.

Explanation

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.

8) The power series for eˣ is sum of xⁿ/n!. Find the series for e^(-x²).

Σ from n=0 to infinity of (-1)ⁿ x^(2n) / n!

Σ from n=0 to infinity of (-1)ⁿ x^(2n) / (2n)!

Σ from n=0 to infinity of x^(2n) / n!

Σ from n=0 to infinity of (-x²)ⁿ / n!

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.

Explanation

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.

9) If a power series converges at x = -3 and diverges at x = 5, and its center is 0, what is the minimum possible radius of convergence?

3

4

5

8

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.

Explanation

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.

10) The power series n=0xⁿ / n² has radius of convergence 1. After integrating term-by-term, what is the radius of convergence of the new series?

0

1/2

1

2

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.

Explanation

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.

11) The series Σ (x-4)ⁿ / √n converges on the interval [3, 5). Why does it diverge at x = 5?

It becomes the harmonic series.

It becomes a p-series with p = 1/2.

It becomes the alternating harmonic series.

The terms do not approach zero.

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.

Explanation

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.

12) If the interval of convergence for a power series is (-3, 7), what is the center and radius?

Center 2, radius 5

Center 2, radius 10

Center 5, radius 2

Center 5, radius 8

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.

Explanation

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.

13) A power series converges absolutely for |x-c| < R. What can happen at |x-c| = R?

It always converges.

It always diverges.

It may converge absolutely, converge conditionally, or diverge.

It converges if R>0.

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.

Explanation

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.

14) Using the known Maclaurin series for eˣ, which operation would you use to find the series for the function f(x) = e^(x³)?

Differentiation

Integration

Substitution (replacing x with x³)

Multiplying by x³

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.

Explanation

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.

15) Consider the power series n=0 (x-7)ⁿ / (n * 9ⁿ). The Ratio Test gives the condition |x-7|/9 < 1 for convergence. What is the radius of convergence?

7

9

1/9

2

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.

Explanation

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.