Using power series remainder term

morenogabr
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Homework Statement


(For power series about x=1) Using the error formula, show that \left|ln(1.5)-p_{3}(1.5)\right|\leq\frac{(0.5)^{4}}{4}

Homework Equations


p_{3}(x) = x-1 - \frac{(x-1)^{2}}{2} + \frac{(x-1)^{3}}{3}
\\\epsilon_{n}(x)=\frac{f^{n+1}(\xi)}{(n+1)!}(x-x_{o})^{n+1}\\where \xi lies between x_{o} and x

The Attempt at a Solution


Im sure this is an easy one, but I can't think of any useful relationship between the difference |f(x)-p_3(x)| and that piece of the remainder function... any hints?
(excuse my latex crappiness)
 
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Isn't there supposed to be a sigma notation (for sum) on the left of the first term of the second equation (power series)?
 
No, its a 3rd order polynomial that approximates the power series, and its written out term by term so sigma is uneeded. Oh, if you mean for the epsilon, its the remainder term used to measure accuracy of the approximation. Its a single term, so no sigma needed. I guess I should say something like n=3 (bc of p_3(x)) so epsi_4(x)...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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