6.3 Taylor and maclaurin series  (Page 5/13)

1. The fifth Maclaurin polynomial is
   
   $p_5\left(x\right)=x-\frac{x^3}{3\text{!}}+\frac{x^5}{5\text{!}}.$
   
   Using this polynomial, we can estimate as follows:
   
   $\begin{array}{cc}\hfill \text{sin}\left(\frac{\pi }{18}\right)& \approx p_5\left(\frac{\pi }{18}\right)\hfill \\ & =\frac{\pi }{18}-\frac{1}{3\text{!}}{\left(\frac{\pi }{18}\right)}^3+\frac{1}{5\text{!}}{\left(\frac{\pi }{18}\right)}^5\hfill \\ & \approx \mathrm{0.173648.}\hfill \end{array}$
   
   To estimate the error, use the fact that the sixth Maclaurin polynomial is $p_6\left(x\right)=p_5\left(x\right)$ and calculate a bound on $R_6\left(\frac{\pi }{18}\right).$ By [link] , the remainder is
   
   $R_6\left(\frac{\pi }{18}\right)=\frac{f^{\left(7\right)}\left(c\right)}{7\text{!}}{\left(\frac{\pi }{18}\right)}^7$
   
   for some c between 0 and $\frac{\pi }{18}.$ Using the fact that $\left|f^{\left(7\right)}\left(x\right)\right|\le 1$ for all x , we find that the magnitude of the error is at most
   
   $\frac{1}{7\text{!}}·{\left(\frac{\pi }{18}\right)}^7\le 9.8\times {10}^{\mathrm{-10}}.$
2. We need to find the values of x such that
   
   $\frac{1}{7\text{!}}{\left|x\right|}^7\le 0.0001.$
   
   Solving this inequality for x , we have that the fifth Maclaurin polynomial gives an estimate to within 0.0001 as long as $\left|x\right|<0.907.$

For $f\left(x\right)=\frac{1}{x},$ the values of the function and its first four derivatives at $x=1$ are

That is, we have $f^{\left(n\right)}\left(1\right)={\left(\mathrm{-1}\right)}^{n}n\text{!}$ for all $n\ge 0.$ Therefore, the Taylor series for $f$ at $x=1$ is given by

To find the interval of convergence, we use the ratio test. We find that

Thus, the series converges if $\left|x-1\right|<1.$ That is, the series converges for $0<x<2.$ Next, we need to check the endpoints. At $x=2,$ we see that

diverges by the divergence test. Similarly, at $x=0,$

diverges. Therefore, the interval of convergence is $\left(0,2\right).$