Using power series remainder term

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SUMMARY

The discussion focuses on using the error formula for power series to demonstrate that the absolute difference between the natural logarithm of 1.5 and its third-degree polynomial approximation, p₃(1.5), is bounded by \(\frac{(0.5)^{4}}{4}\). The polynomial p₃(x) is defined as \(p_{3}(x) = x-1 - \frac{(x-1)^{2}}{2} + \frac{(x-1)^{3}}{3}\). The remainder term, \(\epsilon_{n}(x)\), is utilized to measure the accuracy of the approximation, specifically indicating that for n=3, the remainder term is \(\epsilon_{4}(x)\).

PREREQUISITES
  • Understanding of power series expansion
  • Familiarity with Taylor series and polynomial approximations
  • Knowledge of error analysis in numerical methods
  • Basic proficiency in calculus, particularly derivatives and factorials
NEXT STEPS
  • Study the derivation of Taylor series and their applications
  • Learn about the concepts of convergence and error bounds in power series
  • Explore the use of remainder terms in polynomial approximations
  • Investigate the properties of logarithmic functions and their series expansions
USEFUL FOR

Students and educators in mathematics, particularly those studying calculus and numerical analysis, as well as anyone interested in understanding polynomial approximations and error analysis in power series.

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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)...
 

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