Problem 44 - Calculus Early Transcendentals

Question

Differentiating and integrating power series Find the power series representation for g centered at 0 by differentiating or integrating the power series for \(f\) (perhaps more than once). Give the interval of convergence for the resulting series. $$g(x)=\frac{x}{\left(1+x^{2}\right)^{2}} \text { using } f(x)=\frac{1}{1+x^{2}}$$

Step-by-Step Solution

Answer

Answer: The power series representation for \(g(x)\) is \(g(x)=x\sum_{n=1}^{\infty}(-1)^n (2n)x^{2(n-1)}\) and its interval of convergence is \(-1 < x < 1\).

Step 1: Find the power series representation for f(x)

We have the function f(x) as $$f(x) = \frac{1}{1+x^2}$$ Using geometric series, we can rewrite this as $$f(x) = \sum_{n=0}^{\infty}(-1)^n (x^2)^n = \sum_{n=0}^{\infty}(-1)^n x^{2n}$$ with the interval of convergence \(-1 < x < 1\).

Step 2: Differentiate the power series representation of f(x)

Now, differentiate f(x) with respect to x: $$f'(x) = \frac{d}{dx}\left( \sum_{n=0}^{\infty}(-1)^n x^{2n} \right) = \sum_{n=1}^{\infty}(-1)^n (2n)x^{2n-1}$$ The interval of convergence remains the same for the derivative, \(-1 < x < 1\).

Step 3: Compare \(f'(x)\) with \(g(x)\)

We have g(x) as $$g(x) = \frac{x}{\left(1+x^{2}\right)^{2}}$$ To check if \(f'(x)\) can be turned into g(x), we will factor out x from \(f'(x)\): $$f'(x) = x\sum_{n=1}^{\infty}(-1)^n (2n)x^{2n-2} = x\sum_{n=1}^{\infty}(-1)^n (2n)x^{2(n-1)}$$ Which is now identical to g(x) since \(g(x) = x\sum_{n=1}^{\infty}(-1)^n (2n)x^{2(n-1)}\).

Step 4: Determine the interval of convergence

The interval of convergence for the power series of \(f'(x)\) and \(g(x)\) is the same as the interval of convergence for the power series of \(f(x)\) which is \(-1 < x < 1\). Hence, the power series representation for \(g(x)\) is $$g(x)=x\sum_{n=1}^{\infty}(-1)^n (2n)x^{2(n-1)}$$ with an interval of convergence of \(-1 < x < 1\).