Power Series Representation Of F(x) = 3/(4-81x^2)

Let's dive into the fascinating world of power series and explore how to represent a given function using an infinite sum of terms. In this article, we'll tackle the function $f(x) = \frac{3}{4 - 81x^2}$ and break down the steps to express it as a power series. Power series are incredibly useful in mathematics and physics, allowing us to approximate functions, solve differential equations, and much more. So, grab your mathematical toolkit, and let's get started!

Understanding Power Series

Before we jump into the specifics of our function, let's take a moment to understand what a power series actually is. At its core, a power series is an infinite series of the form:

$\sum_{n=0}^{\infty} c_n(x - a)^n = c_0 + c_1(x - a) + c_2(x - a)^2 + c_3(x - a)^3 + ...$

Where:

• $x$ is a variable.
• $n$ is the index of summation.
• $c_n$ are the coefficients of the series.
• $a$ is the center of the series.

The key idea here is that we are representing a function as an infinite sum of powers of $(x - a)$. The coefficients $c_n$ determine the shape and behavior of the function within its interval of convergence. A common and very important power series is the geometric series, which serves as the foundation for many power series representations. The geometric series is given by:

$\sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + ar^3 + ...$

This series converges to $\frac{a}{1 - r}$ if $|r| < 1$. We'll be using this formula to find the power series representation of our function.

Preparing the Function

Our goal is to express $f(x) = \frac{3}{4 - 81x^2}$ in the form of a power series. To do this, we'll manipulate the function to resemble the form of a geometric series. Let's start by factoring out a 4 from the denominator:

$f(x) = \frac{3}{4(1 - \frac{81}{4}x^2)}$

Now, we can rewrite this as:

$f(x) = \frac{3}{4} \cdot \frac{1}{1 - \frac{81}{4}x^2}$

Notice that the fraction $\frac{1}{1 - \frac{81}{4}x^2}$ now looks very similar to the form $\frac{1}{1 - r}$, which is the sum of a geometric series. This is a crucial step, as it allows us to apply the geometric series formula directly. By recognizing the structural similarity to the geometric series, we set the stage for a straightforward transformation.

Applying the Geometric Series Formula

Here comes the exciting part! We can now apply the formula for the sum of a geometric series. Comparing $\frac{1}{1 - \frac{81}{4}x^2}$ with $\frac{1}{1 - r}$, we can identify $r$ as $\frac{81}{4}x^2$. Plugging this into the geometric series formula, we get:

$\frac{1}{1 - \frac{81}{4}x^2} = \sum_{n=0}^{\infty} (\frac{81}{4}x^2)^n$

This simplifies to:

$\sum_{n=0}^{\infty} \left(\frac{81}{4}\right)^n x^{2n}$

Don't forget the $\frac{3}{4}$ factor we pulled out earlier! We need to multiply the entire series by this constant:

$f(x) = \frac{3}{4} \sum_{n=0}^{\infty} \left(\frac{81}{4}\right)^n x^{2n}$

Now, let's bring the constant inside the summation to get the final power series representation:

$f(x) = \sum_{n=0}^{\infty} \frac{3}{4} \left(\frac{81}{4}\right)^n x^{2n}$

This is the power series representation of $f(x)$. It expresses the function as an infinite sum of terms, each involving a power of $x$. This form is incredibly useful for various mathematical operations and approximations. We've successfully transformed a rational function into its equivalent power series representation, highlighting the power and versatility of power series.

Simplifying the Coefficient

To make our power series representation even cleaner and more readable, we can simplify the coefficient inside the summation. The coefficient is currently expressed as $\frac{3}{4} \left(\frac{81}{4}\right)^n$. Let's rewrite 81 as $3^4$, so we have:

$\frac{3}{4} \left(\frac{3^4}{4}\right)^n = \frac{3}{4} \cdot \frac{(3^4)^n}{4^n} = \frac{3}{4} \cdot \frac{3^{4n}}{4^n}$

Now, we can combine the powers of 4 in the denominator:

$\frac{3 \cdot 3^{4n}}{4^{n+1}}$

So, our simplified power series representation becomes:

$f(x) = \sum_{n=0}^{\infty} \frac{3^{4n+1}}{4^{n+1}} x^{2n}$

This is a more concise and elegant form of the power series. Simplifying coefficients not only improves readability but also makes it easier to perform further calculations or manipulations with the series. The clarity and conciseness of this representation are key advantages in more complex mathematical contexts.

Expressing in Summation Notation

We've arrived at the heart of the problem: expressing the power series in summation notation. Our power series representation is:

$f(x) = \sum_{n=0}^{\infty} \frac{3^{4n+1}}{4^{n+1}} x^{2n}$

This notation tells us a lot about the series. Let's break it down:

• $\sum_{n=0}^{\infty}$: This is the summation symbol, indicating that we are summing an infinite number of terms.
• $n = 0$: This tells us that the index variable $n$ starts at 0.
• $\infty$: This indicates that the sum goes on indefinitely.
• $\frac{3^{4n+1}}{4^{n+1}} x^{2n}$: This is the general term of the series. For each value of $n$, we plug it into this expression to get the corresponding term in the series.

To write out the first few terms of the series, we can substitute $n = 0, 1, 2, 3, ...$ into the general term:

• For $n = 0$: $\frac{3^{4(0)+1}}{4^{0+1}} x^{2(0)} = \frac{3}{4}$
• For $n = 1$: $\frac{3^{4(1)+1}}{4^{1+1}} x^{2(1)} = \frac{3^5}{4^2} x^2 = \frac{243}{16} x^2$
• For $n = 2$: $\frac{3^{4(2)+1}}{4^{2+1}} x^{2(2)} = \frac{3^9}{4^3} x^4 = \frac{19683}{64} x^4$
• For $n = 3$: $\frac{3^{4(3)+1}}{4^{3+1}} x^{2(3)} = \frac{3^{13}}{4^4} x^6 = \frac{1594323}{256} x^6$

So, the power series can be written as:

$f(x) = \frac{3}{4} + \frac{243}{16} x^2 + \frac{19683}{64} x^4 + \frac{1594323}{256} x^6 + ...$

This gives us a clear picture of how the power series expands and the contribution of each term. Understanding the structure and notation of power series is essential for advanced mathematical analysis.

Interval of Convergence

While we've found the power series representation, it's crucial to determine where this series actually converges. The interval of convergence is the set of $x$ values for which the series converges to a finite value. For a geometric series, the series converges if $|r| < 1$. In our case, $r = \frac{81}{4}x^2$, so we need to find the values of $x$ for which:

$\left| \frac{81}{4}x^2 \right| < 1$

This inequality can be rewritten as:

$\frac{81}{4}x^2 < 1$

$x^2 < \frac{4}{81}$

Taking the square root of both sides:

$|x| < \frac{2}{9}$

This means the interval of convergence is $-\frac{2}{9} < x < \frac{2}{9}$. The power series representation is valid only within this interval. Determining the interval of convergence is a critical step in working with power series, as it defines the region where the series provides a meaningful approximation of the function.

Conclusion

In this article, we successfully found the power series representation for the function $f(x) = \frac{3}{4 - 81x^2}$. We manipulated the function to resemble the form of a geometric series, applied the geometric series formula, simplified the coefficients, and expressed the result in summation notation. We also determined the interval of convergence for the power series. This process highlights the power and elegance of using power series to represent functions. By understanding and applying these techniques, you can tackle a wide range of problems in calculus, differential equations, and beyond.

To deepen your understanding of power series and related concepts, consider exploring resources like Khan Academy's Calculus series, which offers comprehensive lessons and practice exercises. Keep exploring, keep learning, and you'll continue to unlock the fascinating world of mathematics!