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

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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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