What is the Maclaurin Series for f(x)=x/(e^x-1) up to x^2?

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SUMMARY

The Maclaurin Series for the function f(x) = x/(e^x - 1) up to the term involving x^2 can be derived without computing derivatives. By utilizing the Taylor series expansion for e^x, specifically 1 - e^x = x + x^2/2! + x^3/3! + ..., the series can be simplified. The key steps involve dividing the numerator and denominator by x and applying the expansion for 1/(1+a) while discarding higher-order terms. The resulting series provides the necessary terms for analysis.

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  • Understanding of Taylor Series and Maclaurin Series
  • Familiarity with the exponential function e^x
  • Basic algebraic manipulation of series
  • Knowledge of piecewise functions
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  • Study the derivation of Taylor Series for various functions
  • Learn about the properties and applications of piecewise functions
  • Explore the convergence of series and their implications in calculus
  • Investigate the use of series expansions in solving differential equations
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Students and educators in calculus, mathematicians focusing on series expansions, and anyone interested in the applications of Taylor and Maclaurin series in mathematical analysis.

DjDriftX
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Find the first few terms of the Taylor Series around x=0...?
of the function

f(x)= {x/(e^x - 1) , x =/ 0}
{1 , x=0}

the function is piecewise.
up to and including the term involving x^2

It says to not compute derivatives of f but to use the formula for the Taylor series of e^x


x/(e^x-1) = x (1/(e^x-1)
so.
would that be x (1/ \sum(xn/n!) - 1)
or maybe x \sum (1 / (xn/n!) - 1))

I'm not really sure where to start
 
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Remember you only need terms up to and including x^2. 1-e^x=x+x^2/2!+x^3/3!+... Divide numerator and denominator by x. Now you've got 1/(1+x/2!+x^2/3!+...). You know an expansion for 1/(1+a), right? Use it. Throw away terms that are higher power than you are looking for.
 

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