I'm struggling to figure out how to find a bound on my error for this problem:

Let T_{6}(x) be the Taylor polynomial of degree 6 based at a = 0 for the function f(x)=\cos(x). Suppose you approximate f(x) by T_{6}(x). If |x|\leq 1, find a bound on the error in your approximation by using the alternating series estimate.

So far, I have

f(x) = cos(x), f(0) = 1

f^(1)(x) = -sin(x), f^(1)(0) = 0

f^(2)(x) = -cos(x), f^(2)(0) = -1

f^(3)(x) = sin(x), f^(3)(0) = 0

f^(4)(x) = cos(x), f^(4)(0) = 1


T6(x) = 1 - x^2/4! + x^4/8! -x^6/12!

and that [b_n] = x^8/16!

I'll be honest, this section is way over my head and I'm struggling to make any sense out of it, so I'm not sure that I'm taking the correct approach here in general.

Any help is appreciated!


2 Answers 2


the mcclaurin series $\cos x$ is $$\cos x = 1 - \frac1{2!} x^2 + \frac 1{4!}x^4 - \frac 1{6!}x^6 + \frac 1{8!} x^8 + \cdots $$ now, if you truncate this alternating series by $T_6,$ then the error committed is of the same sign as the next term and smaller in absolute value. therefore $$0 \le \cos x - T_6(x)\le \frac1{8!} x^8 \le \frac 18 \text{ for } |x| \le 1.$$

  • $\begingroup$ Much appreciated! I should also point out that in order for my HW site to accept my answer, I had to plug in 1 for x (since |x| <= 1) to get the value of the bound. $\endgroup$
    – WTFranklin
    Commented Nov 9, 2015 at 2:09
  • $\begingroup$ @FrankA., you are right. will edit my answer. $\endgroup$
    – abel
    Commented Nov 9, 2015 at 2:12

Your denominators are incorrect. The factorials should be the same as the exponents of $x$. Also the alternating series error bound says that your $b_n$ term is an upper bound for the error of the 6th degree polynomial.


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