pts$.)$ The Taylor series for the function $arctan(1-x)$ centered at $1$ is:

${\displaystyle \underset{k=0}{\overset{}{}}}(-1{)}^{k+1}\frac{(x-1{)}^{2k+1}}{2k+1}$

Use the $3$rd Taylor polynomial for $arctan(1-x)$ centered at $1$ to estimate the value of $arctan(0\mathrm{.}8)$

$($a$)(0\mathrm{.}8)-\frac{(0\mathrm{.}8{)}^3}{3}$

$($b$)(0\mathrm{.}2)-\frac{(0\mathrm{.}2{)}^3}{3}+\frac{(0\mathrm{.}2{)}^5}{5}$

$($c$)(0\mathrm{.}8)-\frac{(0\mathrm{.}8{)}^3}{6}+\frac{(0\mathrm{.}8{)}^5}{120}$

$($d$)(0\mathrm{.}8)-\frac{(0\mathrm{.}8{)}^3}{3}+\frac{(0\mathrm{.}8{)}^5}{5}$

$($e$)(0\mathrm{.}2)-\frac{(0\mathrm{.}2{)}^3}{3}$

$6.(6$ pts$.)$ Using summation notation, write the ERROR for the estimate in the above problem as an infinite sum.

$($a$){\displaystyle \underset{k=3}{\overset{}{}}}(-1{)}^k\frac{(0\mathrm{.}8{)}^{2k+1}}{2k+1}$

$($b$){\displaystyle \underset{k=3}{\overset{}{}}}(-1{)}^k\frac{(0\mathrm{.}2{)}^{2k+1}}{2k+1}$

$($c$){\displaystyle \underset{k=4}{\overset{}{}}}(-1{)}^k\frac{(0\mathrm{.}2{)}^{2k+1}}{2k+1}$

$($d$){\displaystyle \underset{k=2}{\overset{}{}}}(-1{)}^k\frac{(0\mathrm{.}8{)}^{2k+1}}{2k+1}$

$($e$){\displaystyle \underset{k=2}{\overset{}{}}}(-1{)}^k\frac{(0\mathrm{.}2{)}^{2k+1}}{2k+1}$