Exercises for Taylor Series and Calculation of Functions 1. Let f(x)=sin(x) and apply Theorem 4.1 (a) Use x_(0)=0 and find F_(5)(x),P_(7)(x) and P_(9)(x) (b) Show that if vert xvert leqslant 1 then the approximation sin(x)approx x-(x^3)/(3!)+(x^5)/(5!)-(x^7)/(7!)+(x^9)/(9!) has the error bound vert E_(9)(x)vert lt 1/10!leqslant 2.75574times 10^-7 (c) Use x_(0)=pi /4 and find P_(5)(x) , which involves powers of (x-pi /4)

(a) To find , we need to use the Taylor series expansion of centered at . The Taylor series expansion is given by: F_{n}(x) = f(x_{0}) + f'(x_{0})(x - x_{0}) + \frac{f''(x_{0})}{2!}(x - x_{0})^2 + \frac{f'''(x_{0})}{3!}(x - x_{0})^3 + \cdots + \frac{f^{(n)}(x_{0})}{n!}(x - x_{0})^n For , we have: F_{5}(x) = \sin(0) + \cos(0)(x - 0) + \frac{-\sin(0)}{2!}(x - 0)^2 + \frac{-\cos(0)}{3!}(x - 0)^3 + \frac{\sin(0)}{4!}(x - 0)^4 + \frac{-\cos(0)}{5!}(x - 0)^5 Simplifying, we get: F_{5}(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} To find and , we need to use the Taylor series expansion up to the 7th and 9th terms, respectively. The general form of the Taylor series expansion is: P_{n}(x) = \sum_{k=0}^n \frac{f^{(k)}(x_{0})}{k!}(x - x_{0})^k For , we have: P_{7}(x) = \sin(0) + \cos(0)(x - 0) + \frac{-\sin(0)}{2!}(x - 0)^2 + \frac{-\cos(0)}{3!}(x - 0)^3 + \frac{\sin(0)}{4!}(x - 0)^4 + \frac{-\cos(0)}{5!}(x - 0)^5 + \frac{\sin(0)}{6!}(x - 0)^6 + \frac{-\cos(0)}{7!}(x - 0)^7 Simplifying, we get: P_{7}(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} For , we have: P_{9}(x) = \sin(0) + \cos(0)(x - 0) + \frac{-\sin(0)}{2!}(x - 0)^2 + \frac{-\cos(0)}{3!}(x - 0)^3 + \frac{\sin(0)}{4!}(x - 0)^4 + \frac{-\cos(0)}{5!}(x - 0)^5 + \frac{\sin(0)}{6!}(x - 0)^6 + \frac{-\cos(0)}{7!}(x - 0)^7 + \frac{\sin(0)}{8!}(x - 0)^8 + \frac{-\cos(0)}{9!}(x - 0)^9 Simplifying, we get: P_{9}(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} (b) To show that the approximation has the error bound , we need to use the error bound formula for Taylor series: \vert E_{n}(x)\vert \leq \frac{M}{(n+1)!}\vert x - x_{0}\vert^{n+1} where is the maximum value of on the interval , and is the step size.In this case, $f(x)