We know that $ \ \ \sin \theta \approx \theta \ \ $ if the angle be small. If error tolerance is $ 10^{-n} $ , find the maximum allowable angle $ \ \theta \ $ in degrees. $$ $$ To get this , we need to expand $ \ \sin \theta= \theta-\frac{\theta^{3}}{3!}+ ...... (-1)^{n-1} \frac{\theta^{n}}{n!}. $ . But then how to approach. I have thought another way of doing this- Let $ \ \sin x = x $ and let f(x)=sin x, then if R is the error we must have , $ R < 10^{-n} $
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1Hint: In an alternating series the error is always no greater than the absolute value of the next term. – Ethan Bolker Apr 07 '17 at 12:37
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sir can you suggest me the error term in $ sin x \approx x $ – MAS Apr 07 '17 at 12:40
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My comment does just that: the error is bounded by the first term after you stop. If you can't make that idea work you will have to wait for someone to answer the question for you. – Ethan Bolker Apr 07 '17 at 12:45
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ok , I now understood , many thanks – MAS Apr 07 '17 at 12:47
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When you work out the details you can post an answer here to your own question. – Ethan Bolker Apr 07 '17 at 12:50
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For small (and even not so small angles) you get from the convergence behavior of alternating series $$ θ\ge\sinθ\ge θ-\frac16θ^3 $$ For the relative error of the linear approximation you get thus $$ 0\ge\frac{\sinθ-θ}θ\ge -\frac16θ^2 $$ To get that smaller in absolute value than $10^{-n}$, you need $|θ|\le\sqrt{6·10^{-n}}$
Lutz Lehmann
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