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For small values of $x$, the approximation $\sin(x)\approx x$ is often used. Estimate the error using this formula with the aid of Taylor's Theorem. For what range of values of $x$ will this approximation give results correct to six decimal places?

I'm not sure how to do this. Any solutions or hints are greatly appreciated.

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  • $\begingroup$ With $ sin(x)=f(x)$ ,we have $f(x)=f(x)-f(0)= xf'(0) +x^2f"(0)+x^3f^{'''}(y)/6=x-y^3/6$ where $y$ is between $ 0$ and $ x$ when $ x \ne 0$, so $|y|<|x| $ for $ x \ne 0$. $\endgroup$ – DanielWainfleet Aug 30 '15 at 3:10
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Lagrange's estimate using a Taylor approximation centered at $c = 0$ gives $$\sin(x) = 0 + x + 0\cdot \frac{x^2}{2!} - \cos(\zeta_x)\cdot\frac{x^3}{3!}$$ for some $\zeta$ between $0$ and $x$. "Correct to six decimal places" means the error is less than $10^{-6}$. Rewriting the equation above, we want $$\left|\,\sin(x) - x\,\right| = \frac{1}{6} \left|\,\cos(\zeta_x)\cdot x^3\, \right| < 10^{-6},$$ or $$\left|\,\cos(\zeta_x)\cdot x^3\, \right| < 6\cdot 10^{-6}.$$ For any $\zeta_x$, $\left|\,\cos(\zeta_x)\,\right| \leq 1$, so we want $$|\,x^3\,| < 6\cdot 10^{-6}.$$ It should be clear what to do from here. Note that this is a bit conservative of an estimate since we approximated cosine as 1, but it's certainly a sufficient condition for the conclusion (and most likely what your instructor would like).

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