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. $$ f(x)=\frac{x(1+x)}{(1-x)^{2}} $$

Short 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-by-step solution

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, \(\frac{d}{dx}x^n=nx^{n-1}\)

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 \(\frac{d}{dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty n x^{n-1}\).

Substituting the Values

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

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: \(f(x) = \sum_{n=0}^\infty (n+1) x^n + \sum_{n=0}^\infty (n+1) x^{n+1}\).

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: \(f(x) = \sum_{n=0}^\infty (2n+1)x^n = 1 +3x + 5x^2 + 7x^3 +... \).

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 \(|x|<1\).