Problem 43 - 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. $$g(x)=\frac{1}{(1-x)^{4}} \text { using } f(x)=\frac{1}{1-x}$$

Step-by-Step Solution

Answer

The power series representation of g(x) is: $$g(x)=\frac{1}{6}\sum_{n=3}^{\infty}n(n-1)(n-2)x^{n-3}$$ And its interval of convergence is (-1,1).

Step 1: Refresh Properties of Power Series

Before working with power series, we should recall that the power series representation of \(\frac{1}{1-x}\) is \(\sum_{n=0}^{\infty}x^n\). This series converges for \(|x|<1\), which is the interval of convergence.

Step 2: Differentiate f(x) three times

Since g(x) is obtained by differentiating f(x) three times, we will now calculate the first three derivatives of f(x): 1. \(f'(x) = \frac{d}{dx}(\frac{1}{1-x}) = \frac{1}{(1-x)^2}\) 2. \(f''(x) = \frac{d^2}{dx^2}(\frac{1}{1-x}) = \frac{2}{(1-x)^3}\) 3. \(f'''(x) = \frac{d^3}{dx^3}(\frac{1}{1-x}) = \frac{6}{(1-x)^4}\)

Step 3: Scale the Third Derivative

We can notice that g(x) is a scaled version of the third derivative of f(x). To find the exact scaling factor, we need to equate their coefficients: \(g(x) = k \cdot f'''(x)\) \(\frac{1}{(1-x)^4} = k \cdot \frac{6}{(1-x)^4}\) Comparing the coefficients, we have \(k = \frac{1}{6}\). Thus, our desired function g(x) can be written as: \(g(x) = \frac{1}{6} \cdot f'''(x)\)

Step 4: Find the Power Series of g(x)

Now, using the power series representation of f(x), we can find the power series of its third derivative, and then scale it by the factor 1/6. We will differentiate the power series of f(x) three times: 1. Power series of f(x): \(\sum_{n=0}^{\infty}x^n\) 2. Power series of f'(x): \(\sum_{n=1}^{\infty}nx^{n-1}\) 3. Power series of f''(x): \(\sum_{n=2}^{\infty}n(n-1)x^{n-2}\) 4. Power series of f'''(x): \(\sum_{n=3}^{\infty}n(n-1)(n-2)x^{n-3}\) Now, we have the power series of f'(x), f''(x), and f'''(x). To get the power series representation of g(x), we need to multiply the power series of f'''(x) by the scaling factor \(\frac{1}{6}\): Power series of g(x): \(\frac{1}{6}\sum_{n=3}^{\infty}n(n-1)(n-2)x^{n-3}\)

Step 5: Find the Interval of Convergence

Since we now have the power series representation of g(x), we need to find its interval of convergence. As we know, the interval of convergence for the power series representation of f(x) is \(|x|<1\). Since we obtained g(x) by differentiating f(x) thrice and scaling it by a constant factor, the interval of convergence remains the same: |\(x\)|<1 which means the interval of convergence for g(x) is (-1,1). So, the power series representation of g(x) centered at 0 is: $$g(x)=\frac{1}{6}\sum_{n=3}^{\infty}n(n-1)(n-2)x^{n-3}$$ And its interval of convergence is (-1,1).