Problem 34 - Essential Calculus

Question

Use the power series $$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n},|x|<1.$$ Find the series representation of the function and determine its interval of convergence. $$ f(x)=\frac{x(1+x)}{(1-x)^{2}} $$

Step-by-Step Solution

Answer

The series representation of \(f(x)\) is \(1 +3x + 5x^2 + 7x^3 +...\), and the convergence interval of the function is \(|x|<1\).

Step 1: Expanding the Function

First rewrite the given function as \(f(x)=(1+x)\cdot\frac{x}{(1-x)^{2}}\). You can see that the second part is a derivative of \(\frac{1}{1-x}\). So we can apply the power rule on it, considering every term succeeding to represent \(x^n\). So, \(\frac{d}{dx}x^n=nx^{n-1}\)

Step 2: Applying Differentiation

Having recognized that \(\frac{x}{(1-x)^{2}}\) is a derivative of \(\frac{1}{1-x}\), differentiate \(\frac{1}{1-x}\) to get the representation for the second part of the function. By differentiating the power series representation of \(\frac{1}{1-x}\), we get \(\frac{d}{dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty n x^{n-1}\).

Step 3: Substituting the Values

Now substituting this into \(f(x)\), we have: \(f(x)= (1+x)\sum_{n=1}^\infty n x^{n-1} = \sum_{n=1}^\infty nx^{n-1} + \sum_{n=1}^\infty nx^{n}\).

Step 4: Adjusting the Summation Index

To simplify, adjust the summation index to go from 0 to infinity. We can rewrite the first series by replacing \(n\) and \(n-1\) in the exponents with \(n+1\), and adjust the second series by replacing \(n\) with \(n+1\). So, we obtain: \(f(x) = \sum_{n=0}^\infty (n+1) x^n + \sum_{n=0}^\infty (n+1) x^{n+1}\).

Step 5: Combining the Series

We can re-index and combine the two series into one. Note that the \(n=0\) term in the second series vanishes, so we can start both series at \(n=0\). In this way, we find: \(f(x) = \sum_{n=0}^\infty (2n+1)x^n = 1 +3x + 5x^2 + 7x^3 +... \).

Step 6: Determining the Convergence Interval

Finally, for the convergence interval, as the series derived from \(\frac{1}{1-x}\), its radius of convergence is the same as that of the power series for \(\frac{1}{1-x}\), which is given to be \(|x|<1\). Therefore, the sum converges as a power series for \(|x|<1\).

Key Concepts

Series Representation

A power series representation expresses a function as an infinite sum of terms, which can significantly simplify complex functions. By converting a function into its power series form, we break it down into simpler polynomial components. For the function \( f(x) = \frac{x(1+x)}{(1-x)^2} \), we need to represent it as a power series.
The key here is to recognize parts of the function that have known power series. The expression \((1-x)^{-2}\) is related to the power series \( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^{n} \). Differentiating this helps in obtaining the power series for \( (1-x)^{-2} \). \[(1-x)^{-2} = \frac{1}{(1-x)^2} = \sum_{n=1}^\infty n x^{n-1}\] Next, multiply this by \(x(1+x)\) as given in the function. Each term of the power series for \( \frac{x}{(1-x)^2} \) is then multiplied by \((1+x)\). By grouping terms and adjusting the indices, the final series representation becomes a single combined series. The result for \( f(x) \) is:\[f(x) = \sum_{n=0}^{\infty} (2n+1) x^n\] This approach allows for complex functions to be expressed and manipulated with ease.

Interval of Convergence

The interval of convergence of a power series tells us for which values of \(x\) the series converges to a finite number. Understanding this concept is crucial because it informs us about the range of inputs for which our series representation is valid.For the initial series \( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \), the interval of convergence is known to be \(|x|<1\). This interval is determined by the convergence tests applicable to power series (such as the ratio test). When constructing a related series by differentiating or manipulating terms of this power series, as in our function \( f(x) = \frac{x(1+x)}{(1-x)^2} \), the interval of convergence remains the same. Therefore:

• The power series \( f(x) = \sum_{n=0}^{\infty} (2n+1) x^n \) converges for \(|x|<1\).
• This means that \( f(x) \) is valid and meaningful for all \(x\) values between -1 and 1 (but not including -1 and 1).

Understanding convergence allows us to confidently use the series for approximations and calculations within its interval.

Differentiation

Differentiation on power series is a powerful tool that assists in forming derivatives of complex functions in an approachable format. Through differentiation, we can explore and manipulate series functions with accuracy and ease. In the exercise, differentiating the basic series \( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \) helps generate the series for a derivative or related expression:\[ \frac{d}{dx} \left( \frac{1}{1-x} \right) = \frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^{n-1}\]This derivative is needed when finding the series of the function \( f(x) = \frac{x(1+x)}{(1-x)^2} \). Differentiation breaks down the process into simpler steps:

• Apply the differentiation of powers: \( \frac{d}{dx}x^n = nx^{n-1} \).
• Multiply the result by the remaining part of the function \((1+x)\).
• Adjust indices to align with known series terms.

By using differentiation effectively, we simplify the task of handling compound functions and improve our comprehension of how different parts of the expression contribute to the whole.