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?


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 $\exp(x)$.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.