1. Problem a) I: Find the convergence and interval of convergence for the power series: $$\sum_{n=1}^\infty \frac{(-1)^n 10^n}{n!} (x-10)^n$$ 2. Use the Ratio Test to determine convergence: $$L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n\to\infty} \left| \frac{(-1)^{n+1} 10^{n+1} (x-10)^{n+1} / (n+1)!}{(-1)^n 10^n (x-10)^n / n!} \right| = \lim_{n\to\infty} \left| \frac{10 (x-10)}{n+1} \right| = 0$$ 3. Since $$L=0$$ for all $$x$$, the series converges for all real $$x$$, so the interval of convergence is $$(-\infty, \infty)$$. 4. Problem a) II: Find convergence interval for: $$\sum_{n=1}^\infty \frac{(-1)^n}{n 10^n} (x-2)^n$$ 5. Apply Ratio Test: $$L = \lim_{n\to \infty} \left| \frac{(-1)^{n+1} (x-2)^{n+1} / ((n+1)10^{n+1})}{(-1)^n (x-2)^n / (n 10^n)} \right| = \lim_{n\to\infty} \left| \frac{n}{n+1} \cdot \frac{|x-2|}{10} \right| = \frac{|x-2|}{10}$$ 6. Convergence when $$L<1 \Rightarrow \frac{|x-2|}{10} <1 \Rightarrow |x-2|<10$$. 7. So the interval of convergence is $$(-8,12)$$. Check endpoints: At $$x=-8$$: series becomes $$\sum \frac{(-1)^n (-10)^n}{n 10^n} = \sum \frac{1}{n} (-1)^n (-1)^n = \sum \frac{1}{n}$$ diverges. At $$x=12$$: series becomes $$\sum \frac{(-1)^n (10)^n}{n 10^n} = \sum \frac{(-1)^n}{n}$$ converges by alternating series test. 8. Thus, interval of convergence is $$(-8, 12]$$. 9. Problem b): Find Maclaurin series for $$f(x) = \frac{1}{1 + 2x^2}$$ and interval of convergence. 10. Note $$f(x) = \frac{1}{1 - (-2x^2)}$$, geometric series form with ratio $$r = -2x^2$$. 11. Maclaurin series is: $$\sum_{n=0}^\infty (-2x^2)^n = \sum_{n=0}^\infty (-1)^n 2^n x^{2n}$$. 12. Interval of convergence requires $$|r|<1 \Rightarrow | -2x^2 | < 1 \Rightarrow 2x^2 <1 \Rightarrow |x| < \frac{1}{\sqrt{2}}$$. 13. So interval of convergence is $$\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$$. 14. Problem c) I: Maclaurin expansion for $$f(x) = \cos x$$. 15. Known Maclaurin series: $$\cos x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$$ 16. Problem c) II: Deduce Maclaurin series for $$f(x) = \sin^2 x$$. 17. Using identity: $$\sin^2 x = \frac{1 - \cos(2x)}{2}$$. 18. Use Maclaurin of $$\cos(2x)$$: $$\cos(2x) = \sum_{n=0}^\infty (-1)^n \frac{(2x)^{2n}}{(2n)!} = 1 - 2^2 \frac{x^2}{2!} + 2^4 \frac{x^4}{4!} - 2^6 \frac{x^6}{6!} + \cdots$$ 19. So $$\sin^2 x = \frac{1}{2} - \frac{1}{2}\cos(2x) = \frac{1}{2} - \frac{1}{2} \left(1 - 2x^2 + \frac{2^4 x^4}{24} - \frac{2^6 x^6}{720} + \cdots \right)$$ 20. Simplify: $$= \frac{1}{2} - \frac{1}{2} + x^2 - \frac{2^3 x^4}{24} + \frac{2^5 x^6}{720} - \cdots = x^2 - \frac{4 x^4}{24} + \frac{32 x^6}{720} - \cdots$$ 21. Simplify coefficients: $$\sin^2 x = x^2 - \frac{x^4}{6} + \frac{4 x^6}{90} - \cdots = x^2 - \frac{x^4}{6} + \frac{2 x^6}{45} - \cdots$$ 22. Problem c) III: Find integral $$\int_0^{\frac{\pi}{6}} \sin^2 x \ dx$$ using Maclaurin approximation with 3 terms. 23. Use approximation: $$\sin^2 x \approx x^2 - \frac{x^4}{6} + \frac{2 x^6}{45}$$ 24. Integrate termwise: $$\int_0^{\pi/6} \left(x^2 - \frac{x^4}{6} + \frac{2 x^6}{45} \right) dx = \left[ \frac{x^3}{3} - \frac{x^5}{30} + \frac{2 x^7}{315} \right]_0^{\pi/6}$$ 25. Evaluate: $$= \frac{(\pi/6)^3}{3} - \frac{(\pi/6)^5}{30} + \frac{2 (\pi/6)^7}{315}$$ 26. Calculate values: $$\frac{(\pi/6)^3}{3} = \frac{\pi^3}{3 \cdot 6^3} = \frac{\pi^3}{648}$$ $$\frac{(\pi/6)^5}{30} = \frac{\pi^5}{30 \cdot 6^5} = \frac{\pi^5}{38880}$$ $$\frac{2 (\pi/6)^7}{315} = \frac{2 \pi^7}{315 \cdot 6^7} = \frac{2 \pi^7}{68024448}$$ 27. So approximate value of integral: $$\frac{\pi^3}{648} - \frac{\pi^5}{38880} + \frac{2 \pi^7}{68024448}$$ 28. Problem c) IIIa: Evaluate the integral exactly. 29. Use identity: $$\sin^2 x = \frac{1 - \cos(2x)}{2}$$ 30. Integral: $$\int_0^{\pi/6} \sin^2 x\ dx = \int_0^{\pi/6} \frac{1 - \cos(2x)}{2} dx = \frac{1}{2} \int_0^{\pi/6} 1\ dx - \frac{1}{2} \int_0^{\pi/6} \cos(2x) dx$$ 31. Integrate: $$= \frac{1}{2} \left[ x \right]_0^{\pi/6} - \frac{1}{2} \left[ \frac{\sin(2x)}{2} \right]_0^{\pi/6} = \frac{\pi}{12} - \frac{1}{4} \sin(\frac{\pi}{3}) = \frac{\pi}{12} - \frac{1}{4} \cdot \frac{\sqrt{3}}{2} = \frac{\pi}{12} - \frac{\sqrt{3}}{8}$$ 32. Problem c) IIIb: Confirm estimate consistency with alternating series estimation theorem. 33. The fourth term of Maclaurin expansion is given by: $$-\frac{x^8}{315}$$ 34. This term bounds the error of the approximation in absolute value. 35. Since the integral is the integral of the truncated series plus an error within the integral of this term, the error estimate for the integral is: $$\left| \int_0^{\pi/6} -\frac{x^8}{315} dx \right| = \frac{1}{315} \int_0^{\pi/6} x^8 dx = \frac{1}{315} \cdot \frac{(\pi/6)^9}{9} = \frac{\pi^9}{9 \cdot 315 \cdot 6^9}$$ 36. Since this error is very small, the alternating series estimation theorem confirms that the approximation is consistent and the error bound is valid. Final answers: - a) I: Interval = $$(-\infty, \infty)$$ - a) II: Interval = $$(-8, 12]$$ - b) Maclaurin series: $$\sum_{n=0}^\infty (-1)^n 2^n x^{2n}$$ with interval $$(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$$ - c) I: $$\cos x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}$$ - c) II: $$\sin^2 x = x^2 - \frac{x^4}{6} + \frac{2 x^6}{45} - \cdots$$ - c) III: Approximate integral $$= \frac{\pi^3}{648} - \frac{\pi^5}{38880} + \frac{2 \pi^7}{68024448}$$ - c) IIIa: Exact integral $$= \frac{\pi}{12} - \frac{\sqrt{3}}{8}$$ - c) IIIb: Error estimate consistent with alternating series test.