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 $?

  • $\begingroup$ Since you develop up to the $n^{th}$ term, the reminder is defined by the $(n+1)^{th}$ term. $\endgroup$ – Claude Leibovici Dec 17 '14 at 8:36
  • $\begingroup$ @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. $\endgroup$ – Sophie Clad Dec 17 '14 at 8:39
  • $\begingroup$ You just got the answer from gammatester ! You need to bound $R_3$. $\endgroup$ – Claude Leibovici Dec 17 '14 at 8:40
  • $\begingroup$ But i am not still clear about . Not the bound but what to bound $\endgroup$ – Sophie Clad Dec 17 '14 at 8:41
  • $\begingroup$ $R_3$ as gammatester answered. Have a look at en.wikipedia.org/wiki/Taylor's_theorem $\endgroup$ – 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 $(-0.6, +0.6).$

  • $\begingroup$ Thankyou for answer , my main concern is whether i bound R2 or R3 . $\endgroup$ – Sophie Clad Dec 17 '14 at 8:41
  • $\begingroup$ $R_3$ is OK, see my edit. $\endgroup$ – gammatester Dec 17 '14 at 9:44
  • $\begingroup$ Taylor series restricted to first two terms means i have to find first two non zero terms from taylor series which occur at second derivative as first derivative vanishes .In general form they have stoped at nth derivative and have written Rn. Herei stopped at second derivative so it should be R2 .please help i am getting confused $\endgroup$ – Sophie Clad Dec 18 '14 at 2:12
  • $\begingroup$ 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). $\endgroup$ – gammatester Dec 18 '14 at 8:11
  • $\begingroup$ @gammatester shouldn't we bound R2 here instead of R3 $\endgroup$ – godonichia Jan 7 '15 at 17:29

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