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. $$g(x)=\ln (1-3 x) \text { using } f(x)=\frac{1}{1-3 x}$$

Short answer

In summary, we found the power series representation for \(g(x) = \ln(1-3x)\) to be: $$g(x) = \sum_{n=0}^{\infty} \frac{(3x)^{n+1}}{n+1}$$ with an interval of convergence of \(|x|<\frac{1}{3}\).

Step-by-step solution

Find the power series representation of f(x)

First, let's find the power series representation for \(f(x) = \frac{1}{1-3x}\) using the geometric series formula: $$\frac{1}{1-r} = 1 + r + r^2 + r^3 + \cdots$$ Substitute \(r = 3x\) into the formula: $$f(x) = 1 + (3x) + (3x)^2 + (3x)^3 + \cdots$$ So, the power series representation of \(f(x)\) is: $$f(x) = \sum_{n=0}^{\infty} (3x)^n$$

Determine g(x) using differentiation or integration of the power series for f(x)

To find \(g(x)\), first notice that: $$g(x) = \ln (1 - 3x)$$ Taking the derivative of both sides with respect to x, we get: $$g'(x) = -\frac{3}{1-3x} \implies -\frac{1}{3}g'(x) = \frac{1}{1-3x}$$ Comparing the expressions obtained for \(f(x)\) and \(-\frac{1}{3}g'(x)\) we find that they are equal. $$-\frac{1}{3}g'(x) = f(x)$$ Using the power series of \(f(x)\): $$-\frac{1}{3}g'(x) = \sum_{n=0}^{\infty} (3x)^n$$ Now, to find the power series representation of \(g(x)\), we need to multiply both sides of the equation by \(-3\) and integrate both sides with respect to x: $$g(x) = -3 \int \sum_{n=0}^{\infty} (3x)^n dx$$ The power series can be integrated term by term: $$g(x) = \sum_{n=0}^{\infty} -3 \int (3x)^n dx$$ After integrating and replacing \(n\) by \((n+1)\): $$g(x) = \sum_{n=0}^{\infty} \frac{(3x)^{n+1}}{n+1} + C$$ Since \(g(0) = \ln(1) = 0\), \(C = 0\).

Find the interval of convergence for the power series

We are now going to find the interval of convergence for the power series. The original power series \(f(x)\) converges for \(|3x| < 1\), which gives \(|x|<\frac{1}{3}\). We will now check if the radius and the interval of convergence remain the same after integrating. The sum converges absolutely if: $$\sum_{n=0}^{\infty} \frac{|(3x)^{n+1}|}{n+1} \to 0$$ Applying the Ratio Test as follows $$\lim_{n\to\infty} \left\lvert\frac{(3x)^{n+2}/(n+2)}{(3x)^{n+1}/(n+1)}\right\rvert$$ Simplifying: $$\lim_{n\to\infty} \frac{n+1}{n+2}|3x| = |3x|$$ The interval of convergence is same as \(f(x)\), which is given by $$|3x| < 1 \implies |x|<\frac{1}{3}$$ Finally, the power series representation for \(g(x)\) is $$g(x) = \sum_{n=0}^{\infty} \frac{(3x)^{n+1}}{n+1}$$ and the interval of convergence is \(|x|<\frac{1}{3}\).