For the function $f(x) = cos(x)$ I want a taylor series approximation accurate to within 0.1 of the actual function on an interval $-5\le x \le 5$. How would I determine the number of terms necessary to do this?

  • $\begingroup$ Look up remainder of alternating series on google $\endgroup$ Apr 9 '17 at 21:25
  • $\begingroup$ @qbert I was under the impression that cos (x) was not an alternating series, despite having positive and negative values, because it does not alternate between them? $\endgroup$
    – JQQ
    Apr 9 '17 at 21:33
  • 2
    $\begingroup$ @JQQ It's Taylor power-series expansion is an alternating series. (Of course only on the even indices. On the odd it is 0) $\endgroup$
    – AsafHaas
    Apr 9 '17 at 21:35

The Lauren remainder in Lagrange form is $$R_n(x) = \frac {f^{(n+1)}(c)} {(n+1)!}x^{n+1}$$ for some point $c$. So seeking for an $n$ for which $|R_n(x)| \lt 0.1$ means:

$|\frac {{cos}^{(n+1)}(c)} {(n+1)!} x^{n+1}| \le | \frac {x^{n+1}} {(n+1)!}| \lt \frac {5^{n+1}} {(n+1)!} \lt 0.1$ first for $n=13$ (check that), so $n=13$ is guaranteed to work.


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