The Maclaurin Series of an inverse polynomial function

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SUMMARY

The discussion focuses on deriving the Maclaurin series for the function f(x) = 1/(x^2 + x + 1). Participants analyze the series representation, arriving at f(x) = Σx^(3n) - Σx^(3n+1). The key coefficients identified are c_{3n} = 1, c_{3n+1} = -1, and c_{3n+2} = 0. The specific task is to compute the value of c_{36} - c_{37} + c_{38}, leading to a deeper understanding of polynomial series expansions.

PREREQUISITES
  • Understanding of Maclaurin series and their derivation
  • Familiarity with polynomial functions and their properties
  • Knowledge of infinite series and convergence
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the derivation of Maclaurin series for various functions
  • Explore the convergence criteria for infinite series
  • Learn about polynomial long division and its applications
  • Investigate the relationship between power series and Taylor series
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Mathematicians, students studying calculus, and anyone interested in series expansions and polynomial functions will benefit from this discussion.

kudoushinichi88
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Let

f(x)=\frac{1}{x^2+x+1}

Let f(x)=\sum_{n=0}^{\infty}c_nx^n be the Maclaurin series representation for f(x). Find the value of c_{36}-c_{37}+c_{38}.

After working out the fraction, I arrived at the following,

f(x)=\sum_{n=0}^{\infty}x^{3n}-\sum_{n=0}^{\infty}x^{3n+1}

But I dun get how to compare this to the the form given in the question to get the answer...
 
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hi kudoushinichi88! :wink:
kudoushinichi88 said:
After working out the fraction

you mean (1-x)/(1 - x3)? :smile:
I arrived at the following,

f(x)=\sum_{n=0}^{\infty}x^{3n}-\sum_{n=0}^{\infty}x^{3n+1}

But I dun get how to compare this to the the form given in the question to get the answer...

but isn't that just c3n = 1, c3n+1 = -1, c3n+2 = 0 ? :confused:
 

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