1. Let f(x)=sin(x) and apply Theorem 4.1. (a) Use x_(0)=0 and find P_(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)^ast =pi /4 and find P_(5)(x) , which involves powers of (x-pi /4)

(a) To find $P_{5}(x)$, $P_{7}(x)$, and $P_{9}(x)$, we can use the Taylor polynomial approximation formula:<br /><br />$P_{n}(x) = f(x_{0}) + f'(x_{0})(x - x_{0}) + \frac{f''(x_{0})}{2!}(x - x_{0})^2 + \cdots + \frac{f^{(n)}(x_{0})}{n!}(x - x_{0})^n$<br /><br />For $x_{0} = 0$, we have $f(x) = \sin(x)$, $f'(x) = \cos(x)$, $f''(x) = -\sin(x)$, and so on.<br /><br />$P_{5}(x) = \sin(0) + \cos(0)(x - 0) + \frac{-\sin(0)}{2!}(x - 0)^2 + \frac{\cos(0)}{3!}(x - 0)^3 + \frac{-\sin(0)}{4!}(x - 0)^4 + \frac{\cos(0)}{5!}(x - 0)^5$<br /><br />$P_{7}(x) = \sin(0) + \cos(0)(x - 0) + \frac{-\sin(0)}{2!}(x - 0)^2 + \frac{\cos(0)}{3!}(x - 0)^3 + \frac{-\sin(0)}{4!}(x - 0)^4 + \frac{\cos(0)}{5!}(x - 0)^5 + \frac{-\sin(0)}{6!}(x - 0)^6 + \frac{\cos(0)}{7!}(x - 0)^7$<br /><br />$P_{9}(x) = \sin(0) + \cos(0)(x - 0) + \frac{-\sin(0)}{2!}(x - 0)^2 + \frac{\cos(0)}{3!}(x - 0)^3 + \frac{-\sin(0)}{4!}(x - 0)^4 + \frac{\cos(0)}{5!}(x - 0)^5 + \frac{-\sin(0)}{6!}(x - 0)^6 + \frac{\cos(0)}{7!}(x - 0)^7 + \frac{-\sin(0)}{8!}(x - 0)^8 + \frac{\cos(0)}{9!}(x - 0)^9$<br /><br />(b) To show that if $|x| \leq 1$, then 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 $|E_{9}(x)| < \frac{1}{10!} \leq 2.75574 \times 10^{-7}$, we can use the error bound formula for Taylor polynomials:<br /><br />$|E_{n}(x)| \leq \frac{M}{(n+1)!} |x - x_{0}|^{n+1}$<br /><br />where $M$ is the maximum value of $|f^{(n+1)}(x)|$ on $[x_{0} - h, x_{0} + h]$.<br /><br />In this case, $f(x) = \sin(x)$, $x_{0} = 0$, and $h = 1$. The maximum value of $|f^{(n+1)}(x)|$ on $[-1, 1]$ is $1$, so we have:<br /><br />$|E_{9}(x)| \leq \frac{1}{10!} |x - 0|^{10} = \frac{1}{10!} |x|^{10}$<br /><br />Since $|x| \leq 1$, we have $|x|^{10} \leq 1$, so $|E_{9}(x)| \leq \frac{1}{10!} \leq 2.75574 \times 10^{-7}$.<br /><br />(c) To find $P_{5}(x)$ when $x_{0}^{\ast} = \pi/4$, we can use the same Taylor polynomial approximation formula as in part (a), but with $x_{0} = \pi/4$:<br /><br />$P_{5}(x) = \sin(\pi/4) + \cos(\pi/4)(x - \pi/4) + \frac