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I get stuck on problems about approximating values with Taylor's formula when it comes to determining the amount of taylor terms (n) required to satisfy a minumum error by just using Lagrange's remainder.

Example: Approximate $\sqrt[3]6$ (cube root) with an error that is less than $10^{-4}$.

I typically try to use the remainderterm and inequalities to determine n, but my success is infrequent so I'm not sure I do things right.

Any help is appreciated.

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Hint

Using your example : you can write $$\sqrt[3]6=\sqrt[3]{8-2}=2\sqrt[3]{1-\frac{1}{4}}$$ Now, consider the Taylor expansion of $(1-x)^{1/3}$ built at $x=0$ or use the binomial theorem to get $$2(1-x)^{1/3}=2-\frac{2 x}{3}-\frac{2 x^2}{9}-\frac{10 x^3}{81}-\frac{20 x^4}{243}-\frac{44 x^5}{729}-\frac{308 x^6}{6561}+O\left(x^7\right)$$ where $x=\frac{1}{4}$

I am sure that you can take from here.

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  • $\begingroup$ I am sorry but I don't understand how this ensures that the approximation is within the error limit. Can I use this information to acquire the amount of (n) taylor terms required with Lagrange's remainder? $\endgroup$ – Ergoprox Aug 21 '14 at 6:13
  • $\begingroup$ I know that I can just keep Taylor expanding until I hit the desirable maximum error. I just thought there would be a way to determine the amount (n) before Taylor expanding. If no one knows an answer by tommorow I'll just accept your answer. $\endgroup$ – Ergoprox Aug 21 '14 at 13:15

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