My Math Forum power series approximation

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 November 27th, 2011, 08:09 PM #1 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 power series approximation Use the formula to compute ln(1.07) correct to five decimal places. $ln(1-x) - \sum _{n=1} ^\infty \dfrac{x^n}{n}$
 November 28th, 2011, 03:04 AM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: power series approximation [color=#000000]I think you mean $\ln(1-x)=-\sum_{k=1}^{\infty}\frac{x^k}{k}$ for $|x|<1$. Now for $x=-0.07$, $\hspace{350pt}\ln(1.07)=-\sum_{k=1}^{\infty}\frac{(-0.07)^{k}}{k}$ and so let $s_n=-\sum_{k=1}^{n}\frac{(-0.07)^k}{k}$, $\hspace{350pt}\ln(1.07)\approx s_{2}=\frac{(0.07)}{1}-\frac{(0.07)^{2}}{2}=0.06755$ .[/color]

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### using taylor's formula compute square root 1.07

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