Power Series Representation Of F(x) = 18/(x+1)^3
Let's dive into finding the power series representation for the function f(x) = 18/(x+1)^3, centered at 0. This involves a bit of calculus and some series manipulation, but don't worry, we'll break it down step by step. We'll also figure out the interval of convergence, which tells us where our power series actually agrees with the original function. Think of power series as a way to express complicated functions as an infinite sum of simpler terms – polynomials, to be precise. They're super useful in various areas of math, physics, and engineering, especially when dealing with functions that are hard to work with directly. The key idea here is to relate our given function to a known power series, like the geometric series, and then use differentiation or integration to get the series we want. So, let's get started, guys!
Finding the Power Series
To find the power series for f(x) = 18/(x+1)^3, we'll use the hint provided and relate it to the derivative of a simpler function. Notice that:
f(x) = 18/(x+1)^3 = d2/dx2 [9/(x+1)]
This suggests that if we can find a power series for 9/(x+1), we can differentiate it twice to get the power series for f(x). The function 1/(1+x) looks a lot like the sum of a geometric series, which we know how to express as a power series. Let’s start there.
1. Geometric Series
Recall the geometric series formula:
1/(1-r) = Σ(n=0 to ∞) r^n, for |r| < 1
We can rewrite 9/(x+1) as 9/(1-(-x)). Now, it perfectly fits the form of a geometric series with r = -x. So, we can write:
9/(x+1) = 9 * Σ(n=0 to ∞) (-x)^n = 9 * Σ(n=0 to ∞) (-1)^n * x^n, for |-x| < 1 which implies |x| < 1
This gives us the power series representation for 9/(x+1). Now we need to differentiate it twice to get the power series for our original function, f(x). Remember, differentiating a power series term-by-term is a valid operation within its interval of convergence. This is a crucial property of power series that allows us to manipulate them like polynomials, making them incredibly useful for solving differential equations and approximating functions.
2. First Differentiation
Let's differentiate the series once with respect to x:
d/dx [9/(x+1)] = d/dx [9 * Σ(n=0 to ∞) (-1)^n * x^n] = 9 * Σ(n=1 to ∞) (-1)^n * n * x^(n-1)
Notice that the sum now starts from n=1 because the derivative of the constant term (when n=0) is zero. This is a common occurrence when differentiating power series – the index of summation often shifts. The key here is to understand that the power series representation is valid within a certain radius of convergence, and differentiation can affect this interval. However, in many cases, like this one, the interval of convergence remains the same. This makes power series a powerful tool for solving differential equations, as we can differentiate and integrate them term by term within their interval of convergence.
3. Second Differentiation
Now, let's differentiate again:
f(x) = d2/dx2 [9/(x+1)] = d/dx [9 * Σ(n=1 to ∞) (-1)^n * n * x^(n-1)] = 9 * Σ(n=2 to ∞) (-1)^n * n * (n-1) * x^(n-2)
Again, the sum now starts from n=2 because the derivative of the n=1 term is zero. We're getting closer to our power series representation for f(x)! The next step is to re-index the series to make it look cleaner and more standard. This involves a simple change of variable, which will help us express the power series in a more compact form. This is a common trick when working with power series, as it allows us to compare different series and perform algebraic manipulations more easily.
4. Re-indexing the Series
To re-index, let k = n - 2. Then n = k + 2, and the sum becomes:
f(x) = 9 * Σ(k=0 to ∞) (-1)^(k+2) * (k+2) * (k+1) * x^k
Since (-1)^(k+2) = (-1)^k * (-1)^2 = (-1)^k, we can simplify further:
f(x) = 9 * Σ(k=0 to ∞) (-1)^k * (k+2) * (k+1) * x^k
This is the power series representation for f(x) = 18/(x+1)^3 centered at 0. We can rewrite this, replacing k with n for a more standard notation:
f(x) = Σ(n=0 to ∞) 9 * (-1)^n * (n+2) * (n+1) * x^n
However, the function we want is 18/(x+1)^3, which is twice the result we have so far. This is a simple fix, we just need to multiply the entire series by 2.
5. Final Power Series
Therefore, the power series for f(x) = 18/(x+1)^3 is:
f(x) = Σ(n=0 to ∞) 18 * (-1)^n * (n+2) * (n+1) * x^n
So, the coefficient of x^n in the power series is 18 * (-1)^n * (n+2) * (n+1). This completes the first part of our task – we've successfully found the power series representation for the given function. The next crucial step is to determine the interval of convergence, which will tell us for which values of x this power series actually converges to the original function. This is essential because power series are infinite sums, and they don't always converge for every value of x. Understanding the interval of convergence is key to using power series effectively.
Determining the Interval of Convergence
Now, let's determine the interval of convergence for the power series we found:
f(x) = Σ(n=0 to ∞) 18 * (-1)^n * (n+2) * (n+1) * x^n
We'll use the Ratio Test to find the interval of convergence. The Ratio Test is a powerful tool for determining the convergence of series, especially power series. It involves taking the limit of the ratio of consecutive terms and comparing it to 1. If the limit is less than 1, the series converges; if it's greater than 1, the series diverges; and if it's equal to 1, the test is inconclusive. In the context of power series, the Ratio Test often leads us to an interval of convergence centered around the point where the series is expanded (in this case, 0).
1. Apply the Ratio Test
The Ratio Test states that if lim (n→∞) |a_(n+1) / a_n| < 1, then the series converges. Let's apply this to our power series. Let:
a_n = 18 * (-1)^n * (n+2) * (n+1) * x^n
Then:
a_(n+1) = 18 * (-1)^(n+1) * (n+3) * (n+2) * x^(n+1)
Now, let's find the ratio |a_(n+1) / a_n|:
|a_(n+1) / a_n| = |[18 * (-1)^(n+1) * (n+3) * (n+2) * x^(n+1)] / [18 * (-1)^n * (n+2) * (n+1) * x^n]| = |(-1) * (n+3) * x / (n+1)| = |(n+3) / (n+1)| * |x|
2. Calculate the Limit
Now, we need to find the limit as n approaches infinity:
lim (n→∞) |(n+3) / (n+1)| * |x| = |x| * lim (n→∞) (n+3) / (n+1) = |x| * 1 = |x|
3. Determine the Interval of Convergence
For the series to converge, we need this limit to be less than 1:
|x| < 1
This inequality tells us that the series converges for -1 < x < 1. This is our interval of convergence so far, but we need to check the endpoints to see if the series converges at x = -1 and x = 1. Checking the endpoints is a crucial step because the Ratio Test is inconclusive when the limit is equal to 1. The behavior of the series at the endpoints can be different from the behavior within the interval, so we need to investigate them separately.
4. Check Endpoints
Let's check the endpoints:
-
x = 1:
The series becomes:
Σ(n=0 to ∞) 18 * (-1)^n * (n+2) * (n+1)
This series diverges because the terms do not approach zero as n approaches infinity. The alternating sign might tempt us to think it could converge, but the polynomial term (n+2)(n+1) grows too quickly, preventing the terms from shrinking sufficiently.
-
x = -1:
The series becomes:
Σ(n=0 to ∞) 18 * (-1)^n * (n+2) * (n+1) * (-1)^n = Σ(n=0 to ∞) 18 * (n+2) * (n+1)
This series also diverges because the terms grow without bound as n approaches infinity. There's no alternating sign to help with convergence, and the polynomial term ensures that the terms keep increasing.
5. Final Interval of Convergence
Since the series diverges at both endpoints, the interval of convergence is:
-1 < x < 1
In interval notation, this is written as (-1, 1). So, the power series representation we found for f(x) = 18/(x+1)^3 is valid only for x values within this interval. Outside this interval, the series diverges and does not represent the function. This completes our analysis of the power series representation and its interval of convergence. We've successfully expressed the given function as an infinite sum of power terms and determined the range of x values for which this representation is valid. Guys, that’s a wrap!