Region Convergence 6Da330

1. **State the problem:** We want to find the region of convergence for the MacLaurin series of the function $$F(x) = \frac{x}{1 - x - x^2}$$. 2. **Recall the function's denominator roots:** The denominator is $$1 - x - x^2$$, which factors as $$-(x - r_1)(x - r_2)$$ where $$r_1 = \frac{1 + \sqrt{5}}{2}$$ and $$r_2 = \frac{1 - \sqrt{5}}{2}$$. 3. **Understand singularities and convergence:** The function $$F(x)$$ has singularities (points where it is not analytic) at $$x = r_1$$ and $$x = r_2$$ because the denominator is zero there. 4. **MacLaurin series convergence radius:** The MacLaurin series (Taylor series at $$x=0$$) converges within the radius equal to the distance from $$0$$ to the nearest singularity in the complex plane. 5. **Calculate distances:** Compute $$|r_1|$$ and $$|r_2|$$: $$r_1 = \frac{1 + \sqrt{5}}{2} \approx 1.618$$ $$r_2 = \frac{1 - \sqrt{5}}{2} \approx -0.618$$ So, $$|r_1| \approx 1.618$$ $$|r_2| \approx 0.618$$ 6. **Determine the radius of convergence:** The radius is the smaller of these distances, so $$R = \min(|r_1|, |r_2|) = 0.618$$ 7. **Interpretation:** The MacLaurin series converges for all $$x$$ such that $$|x| < R = 0.618$$. **Final answer:** The region of convergence for the MacLaurin series of $$F(x)$$ is the open interval $$|x| < \left| \frac{1 - \sqrt{5}}{2} \right|$$, which is approximately $$|x| < 0.618$$.