Use an appropriate Taylor polynomial about 0 and the Lagrange Remainder Formula to approximate sin(4/7) with an error less than
0.0001.
1 answer:
For a taylor polynomial
<span>f(x) = Sum a_n x^(n) </span>
<span>= Pn + epsilon </span>
<span>epsilon = Sum a_n+1 (x-x_0)^(n+1) </span>
<span>Lagrange remainder </span>
<span>a_n = f^(n) (x_0) / n! </span>
<span>|epsilon| < ∫ [x_0 to x] f^(n+1)(t) / (t - x_0)^n / n! dt </span>
<span>< max (|f^(n+1)|) (x-x_0)^n+1 / n+1! </span>
<span>In the Taylor expansion of sin x </span>
<span>sin x = x - x^3/3! ... </span>
<span>find n such that epsilon < 0.0001 </span>
<span>epsilon < max (|f^(n+1)|) (x-x_0)^n+1 / n+1! < 0.0001 </span>
<span>for sin x, max (|f^(n+1)|) < 1 </span>
<span>(6/7) < 1 so (6/7)^n <1 </span>
<span>1/(n+1)! < 0.0001 </span>
<span>n+1! > 0.0001 </span>
<span>(6/7)^7/(n+1)! < 0.0001 </span>
<span>the coefficient of x^6 = 0 in the expansion of sin x </span>
<span>sin x = x - x^3 + x^5 +- 0.0001</span>
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