for a taylor polynomial
f(x) = Sum a_n x^(n)
= Pn + epsilon
epsilon = Sum a_n+1 (x-x_0)^(n+1)
Lagrange remainder
a_n = f^(n) (x_0) / n!
|epsilon| < ∫ [x_0 to x] f^(n+1)(t) / (t - x_0)^n / n! dt
< max (|f^(n+1)|) (x-x_0)^n+1 / n+1!
In the Taylor expansion of sin x
sin x = x - x^3/3! ...
find n such that epsilon < 0.0001
epsilon < max (|f^(n+1)|) (x-x_0)^n+1 / n+1! < 0.0001
for sin x, max (|f^(n+1)|) < 1
(6/7) < 1 so (6/7)^n <1
1/(n+1)! < 0.0001
n+1! > 0.0001
(6/7)^7/(n+1)! < 0.0001
the coefficient of x^6 = 0 in the expansion of sin x
sin x = x - x^3 + x^5 +- 0.0001
