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 $F_{5}(x)$, we need to use the Taylor series expansion of $f(x) = \sin(x)$ centered at $x_{0} = 0$. The Taylor series expansion is given by:<br /><br />$$F_{n}(x) = f(x_{0}) + f'(x_{0})(x - x_{0}) + \frac{f''(x_{0})}{2!}(x - x_{0})^2 + \frac{f'''(x_{0})}{3!}(x - x_{0})^3 + \cdots + \frac{f^{(n)}(x_{0})}{n!}(x - x_{0})^n$$<br /><br />For $n = 5$, we have:<br /><br />$$F_{5}(x) = \sin(0) + \cos(0)(x - 0) + \frac{-\sin(0)}{2!}(x - 0)^2 + \frac{-\cos(0)}{3!}(x - 0)^3 + \frac{\sin(0)}{4!}(x - 0)^4 + \frac{-\cos(0)}{5!}(x - 0)^5$$<br /><br />Simplifying, we get:<br /><br />$$F_{5}(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}$$<br /><br />To find $P_{7}(x)$ and $P_{9}(x)$, 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:<br /><br />$$P_{n}(x) = \sum_{k=0}^n \frac{f^{(k)}(x_{0})}{k!}(x - x_{0})^k$$<br /><br />For $n = 7$, we have:<br /><br />$$P_{7}(x) = \sin(0) + \cos(0)(x - 0) + \frac{-\sin(0)}{2!}(x - 0)^2 + \frac{-\cos(0)}{3!}(x - 0)^3 + \frac{\sin(0)}{4!}(x - 0)^4 + \frac{-\cos(0)}{5!}(x - 0)^5 + \frac{\sin(0)}{6!}(x - 0)^6 + \frac{-\cos(0)}{7!}(x - 0)^7$$<br /><br />Simplifying, we get:<br /><br />$$P_{7}(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}$$<br /><br />For $n = 9$, we have:<br /><br />$$P_{9}(x) = \sin(0) + \cos(0)(x - 0) + \frac{-\sin(0)}{2!}(x - 0)^2 + \frac{-\cos(0)}{3!}(x - 0)^3 + \frac{\sin(0)}{4!}(x - 0)^4 + \frac{-\cos(0)}{5!}(x - 0)^5 + \frac{\sin(0)}{6!}(x - 0)^6 + \frac{-\cos(0)}{7!}(x - 0)^7 + \frac{\sin(0)}{8!}(x - 0)^8 + \frac{-\cos(0)}{9!}(x - 0)^9$$<br /><br />Simplifying, we get:<br /><br />$$P_{9}(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!}$$<br /><br />(b) To show that the approximation $sin(x) \approx x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!}$ has the error bound $\vert E_{9}(x)\vert \lt \frac{1}{10!}$, we need to use the error bound formula for Taylor series:<br /><br />$$\vert E_{n}(x)\vert \leq \frac{M}{(n+1)!}\vert x - x_{0}\vert^{n+1}$$<br /><br />where $M$ is the maximum value of $\vert f^{(n+1)}(x)\vert$ on the interval $[x_{0} - h, x_{0} + h]$, and $h$ is the step size.<br /><br />In this case, $f(x)