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taylor series error estimation?
Consider the nth order Taylor polynomial for cos x centered at 0 dented T(n) (x,0). How larger must we take n to guarantee that the error |cos x-T(n) (x,0) |is at most 10^-3 for x in [-pi/2,pi/2)
1 Answer
- kbLv 77 years agoFavorite Answer
Since the derivative of f(x) = cos x is either ±sin x or ±cos x,
we have |f^(n)(x)| ≤ 1 for all x.
Hence, the error is given by
|f^(n+1)(c) (x - 0)^(n+1)/(n+1)!| for some c between 0 and x
≤ 1 * |x|^(n+1)/(n+1)!, by the remarks above
≤ (π/2)^(n+1)/(n+1)!, since |x| ≤ π/2.
So, it suffices to solve (π/2)^(n+1)/(n+1)! < 10^(-3).
==> n = 7 (or higher), by trial and error.
I hope this helps!
