Introduction To Power Series: Convergence, Radius & Basic Manipulations

1) What is the general form of a power series centered at x = r?

Σ aₙ * (x - r)ⁿ from n=0 to infinity

Σ aₙ * xⁿ from n=0 to infinity

Σ aₙ * (x - n)^r from n=0 to infinity

Σ aₙ * (r - x)ⁿ from n=0 to infinity

A power series is an infinite series that represents a function. The general form is written as the sum from n equals 0 to infinity of a coefficient, aₙ, multiplied by (x - r) raised to the nth power. Here, 'r' is the center of the series. The expression (x - r)ⁿ shows that the series is built around the point x = r.

Explanation

A power series is an infinite series that represents a function. The general form is written as the sum from n equals 0 to infinity of a coefficient, aₙ, multiplied by (x - r) raised to the nth power. Here, 'r' is the center of the series. The expression (x - r)ⁿ shows that the series is built around the point x = r.

2) For what values of x does the geometric series Σn=0∞ xⁿ converge?

All real numbers

X = 0 only

X such that |x| < 1

X such that |x| > 1

The geometric series sum of xⁿ converges when the absolute value of the common ratio, which is x, is less than 1. This is the condition for an infinite geometric series to have a finite sum. If |x| is greater than or equal to 1, the terms do not approach zero and the series diverges.

Explanation

The geometric series sum of xⁿ converges when the absolute value of the common ratio, which is x, is less than 1. This is the condition for an infinite geometric series to have a finite sum. If |x| is greater than or equal to 1, the terms do not approach zero and the series diverges.

3) What is the radius of convergence for a power series that converges only when x = 5?

0

1

5

Infinity

The radius of convergence, R, defines the interval (r - R, r + R) within which the series converges absolutely. If a series converges only at its center (x = 5), then there is no positive distance from the center where it converges. Therefore, the radius of convergence is 0.

Explanation

The radius of convergence, R, defines the interval (r - R, r + R) within which the series converges absolutely. If a series converges only at its center (x = 5), then there is no positive distance from the center where it converges. Therefore, the radius of convergence is 0.

4) Which test is most commonly used to find the radius of convergence of a power series?

Alternating Series Test

Integral Test

Ratio Test

Comparison Test

The Ratio Test is the standard tool for finding the radius of convergence. For a power series Σ aₙ (x-r)ⁿ, we apply the Ratio Test to the absolute value of the terms. We compute the limit L = limn 🠒∞ |a_{n+1}(x-r)ⁿ⁺¹ / aₙ (x-r)ⁿ| = |x-r| * limn 🠒∞ |a_{n+1}/aₙ|. The series converges when L < 1, which leads to an inequality of the form |x-r| < R, where R is the radius.

Explanation

The Ratio Test is the standard tool for finding the radius of convergence. For a power series Σ aₙ (x-r)ⁿ, we apply the Ratio Test to the absolute value of the terms. We compute the limit L = limn 🠒∞ |a_{n+1}(x-r)ⁿ⁺¹ / aₙ (x-r)ⁿ| = |x-r| * limn 🠒∞ |a_{n+1}/aₙ|. The series converges when L < 1, which leads to an inequality of the form |x-r| < R, where R is the radius.

5) TRUE or FALSE: The radius of convergence of the power series Σ n!xⁿ is 1.

True

False

Using the Ratio Test, we find lim(n→∞)|(n+1)!xⁿ⁺¹/n!xⁿ| = lim(n→∞)|(n+1)x| = ∞ for any x ≠ 0. This means the series only converges at x = 0, giving a radius of convergence of 0, not 1.

Explanation

Using the Ratio Test, we find lim(n→∞)|(n+1)!xⁿ⁺¹/n!xⁿ| = lim(n→∞)|(n+1)x| = ∞ for any x ≠ 0. This means the series only converges at x = 0, giving a radius of convergence of 0, not 1.

6) Given the known series for eˣ = Σ xⁿ / n!, find the power series for e²ˣ.

(2x)ⁿ / n!

2ⁿ xⁿ / n!

Xⁿ / (2n)!

2xⁿ / n!

To find the series for e²ˣ, we take the known series for e^u and substitute u = 2x. This means everywhere we see 'x' in the original series, we replace it with '2x'. The original series is sum of xⁿ / n!. Substituting gives sum of (2x)ⁿ / n!. This can also be written as sum of 2ⁿ * xⁿ / n!, but option A presents it in the explicitly substituted form.

Explanation

To find the series for e²ˣ, we take the known series for e^u and substitute u = 2x. This means everywhere we see 'x' in the original series, we replace it with '2x'. The original series is sum of xⁿ / n!. Substituting gives sum of (2x)ⁿ / n!. This can also be written as sum of 2ⁿ * xⁿ / n!, but option A presents it in the explicitly substituted form.

7) The series for sin(x) is Σn=0∞ (-1)ⁿ x^(2n+1) / (2n+1)!. What is the power series for the derivative, d/dx [sin(x)]

X^(2n)/(2n)!

X^(2n+1)/(2n)!

(2n+1)x^(2n)/(2n+1)!

X^(2n)/(2n+1)!

To differentiate a power series term-by-term, we apply the power rule to each term: d/dx [x^(2n+1)] = (2n+1) x^(2n). So, the differentiated term becomes (-1)ⁿ * (2n+1) * x^(2n) / (2n+1)!. Notice (2n+1)/(2n+1)! = 1/(2n)!. Therefore, the derivative series simplifies to sum of (-1)ⁿ x^(2n) / (2n)!. This is indeed the known series for cos(x), which is the derivative of sin(x).

Explanation

To differentiate a power series term-by-term, we apply the power rule to each term: d/dx [x^(2n+1)] = (2n+1) x^(2n). So, the differentiated term becomes (-1)ⁿ * (2n+1) * x^(2n) / (2n+1)!. Notice (2n+1)/(2n+1)! = 1/(2n)!. Therefore, the derivative series simplifies to sum of (-1)ⁿ x^(2n) / (2n)!. This is indeed the known series for cos(x), which is the derivative of sin(x).

8) If a power series Σ aₙ (x-3)ⁿ has a radius of convergence R=4, what can we say about the series obtained by integrating it term-by-term?

It becomes 0

It becomes <4

It remains 4

It becomes >4

A key property of power series is that term-by-term differentiation and term-by-term integration do not change the radius of convergence. The new series (integrated or differentiated) will converge for the same values of x that are strictly within the original interval of convergence, meaning the distance from the center x=3 is less than 4. The behavior at the endpoints (x = -1 and x = 7) may change and must be tested separately, but the radius R=4 remains the same.

Explanation

A key property of power series is that term-by-term differentiation and term-by-term integration do not change the radius of convergence. The new series (integrated or differentiated) will converge for the same values of x that are strictly within the original interval of convergence, meaning the distance from the center x=3 is less than 4. The behavior at the endpoints (x = -1 and x = 7) may change and must be tested separately, but the radius R=4 remains the same.

9) The power series for f(x) = 1/(1 - x) is Σn=0∞ xⁿ, valid for |x|<1. Using this, find the series for 1/(1 - 5x).

5ⁿ xⁿ

Xⁿ/5ⁿ

(5x)ⁿ

X^(5n)

We use the known series for 1/(1 - u) = Σn=0∞ uⁿ, with |u| < 1. To get 1/(1 - 5x), we set u = 5x. Making this substitution into the series gives Σn=0∞ (5x)ⁿ. This is valid when |5x| < 1, or |x| < 1/5.

Explanation

We use the known series for 1/(1 - u) = Σn=0∞ uⁿ, with |u| < 1. To get 1/(1 - 5x), we set u = 5x. Making this substitution into the series gives Σn=0∞ (5x)ⁿ. This is valid when |5x| < 1, or |x| < 1/5.

10) Applying the Ratio Test to the power series Σn=1∞ (x-2)ⁿ / n leads to the limit L = |x-2|. What is the radius of convergence?

0

1

2

Inconclusive

The Ratio Test tells us the series converges when L < 1. Here, L = |x-2|. The condition for convergence is |x-2| < 1. This inequality describes an interval centered at x=2 with a radius of 1. Therefore, the radius of convergence, R, is 1.

Explanation

The Ratio Test tells us the series converges when L < 1. Here, L = |x-2|. The condition for convergence is |x-2| < 1. This inequality describes an interval centered at x=2 with a radius of 1. Therefore, the radius of convergence, R, is 1.

11) Consider the power series Σn=0∞ (x+1)ⁿ / 3ⁿ. After applying the Ratio Test and finding the condition |x+1|/3 < 1, what is the radius of convergence?

1

3

1/3

X = -1

The condition from the Ratio Test is |x+1|/3 < 1. Multiplying both sides by 3 gives |x+1| < 3. This inequality states that the distance between x and the center (x = -1) is less than 3. Therefore, the radius of convergence is R = 3.

Explanation

The condition from the Ratio Test is |x+1|/3 < 1. Multiplying both sides by 3 gives |x+1| < 3. This inequality states that the distance between x and the center (x = -1) is less than 3. Therefore, the radius of convergence is R = 3.

12) For the series Σn=0∞ (x+1)ⁿ / 3ⁿ, if you already know that the radius of convergence is 3, which describes the next step to find the full interval of convergence?

Interval = (-4,2)

Test x = -4 and x = 2

Interval = (-3,3)

No steps needed

The radius tells us the series converges for |x+1| < 3, which is the open interval (-4, 2). To determine the interval of convergence, we must test the endpoints where |x+1| = 3, namely x = -4 and x = 2. We substitute these values into the original series and use other convergence tests (like the nth term test, geometric series test, or p-series test) to see if the series converges or diverges at these points. The final interval could be [-4, 2), (-4, 2], or [-4, 2].

Explanation

The radius tells us the series converges for |x+1| < 3, which is the open interval (-4, 2). To determine the interval of convergence, we must test the endpoints where |x+1| = 3, namely x = -4 and x = 2. We substitute these values into the original series and use other convergence tests (like the nth term test, geometric series test, or p-series test) to see if the series converges or diverges at these points. The final interval could be [-4, 2), (-4, 2], or [-4, 2].

13) If a function f(x) is represented by a power series Σ aₙ (x-c)ⁿ with a positive radius of convergence, what is the relationship between this power series and the Taylor series of f(x) centered at c?

They are different

They are the same

One is integral of other

Taylor has different radius

A fundamental result in calculus is that if a function f(x) can be represented by a power series centered at c on some interval (c-R, c+R), then that power series must be the Taylor series of f(x) centered at c. The coefficients aₙ will be exactly f⁽ⁿ⁾(c)/n!. Therefore, within the interval of convergence, the power series representation and the Taylor series are identical.

Explanation

A fundamental result in calculus is that if a function f(x) can be represented by a power series centered at c on some interval (c-R, c+R), then that power series must be the Taylor series of f(x) centered at c. The coefficients aₙ will be exactly f⁽ⁿ⁾(c)/n!. Therefore, within the interval of convergence, the power series representation and the Taylor series are identical.

14) Using the known series for sin(x) = x - x³/3! + x⁵/5! - ..., and for cos(x) = 1 - x²/2! + x⁴/4! - ..., how would you find the first three non-zero terms of the series for sin(x) * cos(x)?

Differentiate sin(x)

Integrate cos(x)

Multiply series

Substitute x²

To find the series for a product of two functions, we can multiply their respective power series as if they were polynomials. We take the series for sin(x) and cos(x), align terms by powers of x, and multiply, collecting all terms that contribute to each power. For example, the constant term comes from 0*1=0. The x term comes from x*1 = x. The x² term comes from x*0 + 0*1 = 0, and so on. This process yields the series for (½)sin(2x).

Explanation

To find the series for a product of two functions, we can multiply their respective power series as if they were polynomials. We take the series for sin(x) and cos(x), align terms by powers of x, and multiply, collecting all terms that contribute to each power. For example, the constant term comes from 0*1=0. The x term comes from x*1 = x. The x² term comes from x*0 + 0*1 = 0, and so on. This process yields the series for (½)sin(2x).

15) What is the radius of convergence for the power series Σn=1∞ n! * xⁿ?

0

1

E

Infinity

We use the Ratio Test. Let aₙ = n!. Then |a_{n+1} / aₙ| = (n+1)!/n! = n+1. The limit L = limn 🠒∞ |a_{n+1}/aₙ| * |x| = limit of (n+1)|x|. This limit is infinity for any x ≠ 0. The condition for convergence (L < 1) will only be satisfied if the limit is 0, which only happens when x = 0. Therefore, the series converges only at its center (x=0), giving a radius of convergence R = 0.

Explanation

We use the Ratio Test. Let aₙ = n!. Then |a_{n+1} / aₙ| = (n+1)!/n! = n+1. The limit L = limn 🠒∞ |a_{n+1}/aₙ| * |x| = limit of (n+1)|x|. This limit is infinity for any x ≠ 0. The condition for convergence (L < 1) will only be satisfied if the limit is 0, which only happens when x = 0. Therefore, the series converges only at its center (x=0), giving a radius of convergence R = 0.

Introduction to Power Series: Convergence, Radius & Basic Manipulations