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I want to see if it is possible to write $$ \left(\frac{x}{e^x-1}\right) \left(\frac{x^2/2! }{e^x-1-x}\right) \left(\frac{x^3/3!}{e^x-1-x-x^2/2}\right)$$ as a linear combination of the factors $$p(x)\frac{x}{e^x-1}+q(x)\frac{x^2/2}{e^x-1-x}+r(x)\frac{x^3/3!}{e^x-1-x-x^2/2}$$

I'm specifically looking for the functions $p(x), q(x)$ and $r(x)$ so I thought I would try a partial fraction decomposition method and this would imply that $$\frac{A}{e^x-1}+\frac{B}{e^x-1-x}+\frac{C}{e^x-1-x-x^2/2}$$ where $A=xp(x), B=\frac{1}{2}x^2q(x), C=\frac{1}{6}x^3r(x)$. So this means that $$A(e^x-1-x)(e^x-1-x-x^2/2)+B(e^x-1)(e^x-1-x-x^2/2)+C(e^x-1)(e^x-1-x)=\frac{x^6}{12}$$ If we multiply through $$A\left(e^{2x}-2e^x-2xe^x-\frac{x^2}{2}e^x+1+2x+\frac{3}{2}x^2+\frac{x^3}{2}\right)$$ $$+B\left(e^{2x}-2e^x-xe^x-\frac{x^2}{2}e^x+1+x+\frac{x^2}{2}\right)$$ $$+C\left(e^{2x}-2e^x-xe^x+1+x\right)$$ Thus we get $$(A+B+C)(e^{2x}-2e^x)-(2A+B+C)xe^x-(A+B)\frac{x^2}{2}e^x+(A+B+C)+(2A+B+C)x+\left(\frac{3}{2}A+\frac{1}{2}B\right)x^2+\frac{1}{2}Ax^3=\frac{x^6}{12}$$

To me, now it seems that there is no solution since both $A+B+C=2A+B+C=0$ which means $A=0$ which then implies $B=C=0$. Is this method incorrect for such a problem and if so is there another way to approach this problem? Or did I make an error?

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I think that your idea to decompose in simple fractions is good. You can put $K=\mathbb{Q}(x)$, replace $\exp(x)$ by $y$, and decompose the following fraction in $K(y)$: $$\left(\frac{x}{y-1}\right) \left(\frac{x^2/2! }{y-1-x}\right) \left(\frac{x^3/3!}{y-1-x-x^2/2}\right)=\frac{A}{y-1}+\frac{B}{y-1-x}+\frac{C}{y-1-x-x^2/2}$$ To get $A$, you multiply by $y-1$ both side, simplify and put $y=1$, to get $B$, you multiply by $y-1-x$ both side, simplify and put $y=1+x$, etc .

When you are done, simply replace $y$ by $\exp(x)$.

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  • $\begingroup$ That is what I was looking for. I haven't done it yet, but thank you so much $\endgroup$ – Eleven-Eleven Dec 18 '14 at 18:44

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