Power series from the geometric series Use the geometric serie...

Question

Power series from the geometric series Use the geometric series a Σₖ ₌ ₀ ∞ (x)ᵏ = 1/(1 - x), for |x| < 1, to determine the Maclaurin series and the interval of convergence for the following functions.

ƒ(x) = 1/(1 + 5x)

ƒ(x) = 1/(1 + 5x)

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to find the McClaurin series. The power series about equal to 0, and the interval of convergence 4 F of X, which is equal to 1 divided by the quantity of 3 minus 2 X in quantity. Using the geometric series formula, the sum of K equal to 0 to infinity of T rates to the power K, which is equal to 1 divided by the quantity of 1 minus T in quantity, where the absolute value of T is less than 1. So for the first part of the problem, we need to find our McClaurin series for our function F of X. And here we're actually going to begin by rewriting our function F of X. So F of X is going to be equal to, and here we're going to factor out a 3 out of the denominator. So we'll have 1 divided by 3, multiplied by the quantity of 1 divided by the quantity of 1 minus 2 X divided by 3 in quantity. And if we look at this, we can now use our geometric series formula. And here, to use that formula, we're gonna set T equal to 2 X divided by 3. So that means that F of X is going to be equal to 1 divided by 3, multiplied by the sum of K equal to 0 to infinity of the quantity of 2 X divided by 3 in quantity raised to the power K. And we can simplify this expression, and so this is going to be equal to the sum of K equal to 0 to infinity of 2 to the power K divided by 3 rates of the quantity of K + 1. In quantity multiplied by X-rays to the power K. So this here is the McLaurin series that we're looking for for this problem. Next, we need to find the interval of convergence. Now, since the absolute value of T has to be less than 1, that means that the absolute value. Of 2 X divided by 3 has to be less than 1. Now, if you multiply both sides of this inequality by 3 and divide both sides by 2, we get that the absolute value of X has to be less than 3 divided by 2. Or another way to write this is that minus 3 divided by 2 has to be less than X, which is less than 3 divided by 2. Now, to determine our interval of convergence, we need to check the end points of this interval here to see if our series converges. And so, we're gonna look at first, the case when X is equal to -3 divided by 2. In this case, our series becomes the sum of K equal to 0 to infinity of 1 divided by 3, multiplied by the quantity of minus 1, in quantity raised to the power K. Now, this here is an alternating series, and the partial sums alternate between 1 divided by 3 and 0. And since the partial sums are oscillating between those two values, that means that this series does not converge. And since it does not converge, that means we're not going to include it in this end point in our interval of convergence. Next we will look at the case when X is equal to 3 divided by 2. And so in this case, our sum becomes the sum of K equal to 0 to infinity of 1 divided by 3, multiplied by 1 raised to the power K, which is just going to be equal to. The sum Of K equal to 0 to infinity of 1 divided by 3. And we know that this series is actually going to diverge, since we're just adding the same value, an infinite amount of times. So when, when we add 1/3 an infinite amount of times, the series here is going to diverge to infinity. And so since the series here diverges, that means we're not going to include this endpoint in our interval of convergence. So that means that our interval of convergence is going to be the open interval from -3 divided by 2 to 3 divided by 2. And so that is the answer that we're looking for for the second part of this problem. So I hope this explanation has been useful, and I'll see you in the next video.

Power series from the geometric series Use the geometric series a Σₖ ₌ ₀ ∞ (x)ᵏ = 1/(1 - x), for |x| < 1, to determine the Maclaurin series and the interval of convergence for the following functions.

ƒ(x) = 1/(1 + 5x)

Key Concepts

Geometric Series and Its Sum Formula

Maclaurin Series

Interval of Convergence

Watch next

Power series from the geometric series Use the geometric series a Σₖ ₌ ₀ ∞ (x)ᵏ = 1/(1 - x), for |x| < 1, to determine the Maclaurin series and the interval of convergence for the following functions.ƒ(x) = 1/(1 + 5x)

Key Concepts

Geometric Series and Its Sum Formula

Maclaurin Series

Interval of Convergence

Watch next

Power series from the geometric series Use the geometric series a Σₖ ₌ ₀ ∞ (x)ᵏ = 1/(1 - x), for |x| < 1, to determine the Maclaurin series and the interval of convergence for the following functions.

ƒ(x) = 1/(1 + 5x)