# Taylor series error estimation question

Question is that Taylor series of cosx is restricted to only first two terms and permissible error is 0.54 × 10^(-2) then x can atmost be A) 0.6 B) 0.5 C) 0.4 D) 0.3

My atempt is as follows we need first two terms so expansion is as follows , cosx=1 - x^(2)\2! . I also know formula for remainder in taylor series but i am just confused that do i have to bound $R_2$ or$R_3$?

• Since you develop up to the $n^{th}$ term, the reminder is defined by the $(n+1)^{th}$ term. – Claude Leibovici Dec 17 '14 at 8:36
• @Claude Leibovici Yes here i stoped at second derivative .soi have tofind R3 .am i correct . I am little bit confused in this but i know how to bound error. – Sophie Clad Dec 17 '14 at 8:39
• You just got the answer from gammatester ! You need to bound $R_3$. – Claude Leibovici Dec 17 '14 at 8:40
• But i am not still clear about . Not the bound but what to bound – Sophie Clad Dec 17 '14 at 8:41
• $R_3$ as gammatester answered. Have a look at en.wikipedia.org/wiki/Taylor's_theorem – Claude Leibovici Dec 17 '14 at 8:43

Using the Wiki formula for the interval $(-r,r)$ $$|R_3(x)| \le M \frac{r^4}{4!}$$
and the conservative estimate $M\le 1$, you get for the remainders in the case A: 0.54E-2, B: 0.26e-2, so the correct choice is A.
Edit: In your case you use $R_3$ because the Taylor polynom $P_3(x)$ is the same as $P_2(x) = 1 -\tfrac{1}{2}x^2.\;$ Note that even the $R_3$ term slighty overestimates the actual error, which is about $0.53356\cdot 10^{-2}\;$ in the intervall $(-0.6, +0.6).$
• $R_3$ is OK, see my edit. – gammatester Dec 17 '14 at 9:44
• You may use $R_2$ but it is obviously suboptimal. Because the $x^3$ term in the Taylor expansion of $\cos(x)\;$ is zero you actually compute the $P_3$ polynomial and the $R_3$ error estimate applies, and it is fairly good (only 1.2% too large). – gammatester Dec 18 '14 at 8:11