Maclaurin Partial Fractions 849089

1. **State the problem:** We want to find the MacLaurin series coefficients of $$F(x) = \frac{x}{1 - x - x^2}$$ using partial fractions within its region of convergence. 2. **Identify the denominator roots:** Solve $$1 - x - x^2 = 0$$ for $$x$$. Rewrite as $$x^2 + x - 1 = 0$$. Use the quadratic formula: $$x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$$. So the roots are $$r_1 = \frac{-1 + \sqrt{5}}{2}$$ and $$r_2 = \frac{-1 - \sqrt{5}}{2}$$. 3. **Express the function as partial fractions:** We want to write $$\frac{x}{1 - x - x^2} = \frac{x}{-(x^2 + x - 1)} = \frac{x}{-(x - r_1)(x - r_2)} = \frac{A}{x - r_1} + \frac{B}{x - r_2}$$. Multiply both sides by $$-(x - r_1)(x - r_2)$$: $$x = -A(x - r_2) - B(x - r_1)$$. 4. **Find coefficients $$A$$ and $$B$$:** Set $$x = r_1$$: $$r_1 = -A(r_1 - r_2) - B(r_1 - r_1) = -A(r_1 - r_2)$$ $$\Rightarrow A = -\frac{r_1}{r_1 - r_2}$$. Set $$x = r_2$$: $$r_2 = -A(r_2 - r_2) - B(r_2 - r_1) = -B(r_2 - r_1)$$ $$\Rightarrow B = -\frac{r_2}{r_2 - r_1}$$. Note that $$r_1 - r_2 = \sqrt{5}$$ and $$r_2 - r_1 = -\sqrt{5}$$. So: $$A = -\frac{r_1}{\sqrt{5}}, \quad B = -\frac{r_2}{-\sqrt{5}} = \frac{r_2}{\sqrt{5}}$$. 5. **Rewrite partial fractions:** $$F(x) = \frac{A}{x - r_1} + \frac{B}{x - r_2} = \frac{A}{x - r_1} + \frac{B}{x - r_2}$$. Rewrite denominators as $$x - r_i = -(r_i - x)$$: $$F(x) = -\frac{A}{r_1 - x} - \frac{B}{r_2 - x}$$. 6. **Express as geometric series:** For $$|x| < |r_i|$$, expand: $$\frac{1}{r_i - x} = \frac{1}{r_i} \cdot \frac{1}{1 - \frac{x}{r_i}} = \frac{1}{r_i} \sum_{n=0}^\infty \left(\frac{x}{r_i}\right)^n = \sum_{n=0}^\infty \frac{x^n}{r_i^{n+1}}$$. So: $$F(x) = -A \sum_{n=0}^\infty \frac{x^n}{r_1^{n+1}} - B \sum_{n=0}^\infty \frac{x^n}{r_2^{n+1}} = \sum_{n=0}^\infty \left(-\frac{A}{r_1^{n+1}} - \frac{B}{r_2^{n+1}}\right) x^n$$. 7. **MacLaurin series coefficients:** The coefficient of $$x^n$$ is $$c_n = -\frac{A}{r_1^{n+1}} - \frac{B}{r_2^{n+1}}$$. Recall from step 4: $$A = -\frac{r_1}{\sqrt{5}}, \quad B = \frac{r_2}{\sqrt{5}}$$. Substitute: $$c_n = -\left(-\frac{r_1}{\sqrt{5}}\right) \frac{1}{r_1^{n+1}} - \frac{r_2}{\sqrt{5}} \frac{1}{r_2^{n+1}} = \frac{r_1}{\sqrt{5} r_1^{n+1}} - \frac{r_2}{\sqrt{5} r_2^{n+1}} = \frac{1}{\sqrt{5}} \left(r_1^{-n} - r_2^{-n}\right)$$. 8. **Simplify coefficients:** Since $$r_1$$ and $$r_2$$ are roots of $$x^2 + x - 1 = 0$$, their inverses satisfy the Fibonacci recurrence. Therefore, $$c_n = \frac{1}{\sqrt{5}} \left(r_1^{-n} - r_2^{-n}\right)$$ equals the $$n$$th Fibonacci number $$F_n$$. 9. **Final MacLaurin series:** $$F(x) = \sum_{n=0}^\infty F_n x^n$$ where $$F_0 = 0, F_1 = 1, F_2 = 1, F_3 = 2, \ldots$$ 10. **Region of convergence:** The series converges for $$|x| < \min(|r_1|, |r_2|) = \frac{\sqrt{5} - 1}{2}$$ (the smaller root in absolute value). **Summary:** The MacLaurin series coefficients of $$F(x) = \frac{x}{1 - x - x^2}$$ are the Fibonacci numbers $$F_n$$, and the series converges for $$|x| < \frac{\sqrt{5} - 1}{2}$$.