0
$\begingroup$

As part of my homework I was asked to expand this function by powers of $x $:

$$f(x)=\frac{e^{\pi x}}{(e^{\pi x}-1)^2}$$

The answer is given as: $$\frac{1}{\pi^2 x^2}-\frac{1}{12}+O(x^2)$$

I've tried expanding the denominator and numerator around zero, separately and then dividing them but I only derived the first term. I've also tried writing the function in partial fractions and I reached this: $$\frac{1}{(e^{\pi x}-1)}+\frac{1}{(e^{\pi x}-1)^2}$$ I attempted to use the Taylor series for $(e^{\pi x}-1)$ and $(e^{\pi x}-1)^2$ but again I couldn't reach the answer. Wolfram Alpha gave the answer with even more terms than I needed but not the steps on how it got them. I would very much appreciate it if someone can enlighten me on how to reach the answer. Thank you!

| cite | improve this question | | | | |
$\endgroup$
1
$\begingroup$

Hint: Your function has a double pole in $0$, so multiply the function with $x^2$ and then Taylor expand.

| cite | improve this answer | | | | |
$\endgroup$
1
$\begingroup$

Considering $$f(x)=\frac{e^{\pi x}}{(e^{\pi x}-1)^2}$$ let $t=\pi x$ and use the normal expansions $$e^{t}=1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}+\frac{t^5}{120}+\frac{t^6}{720}+ O\left(t^7\right)$$ $$e^{t}-1=t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}+\frac{t^5}{120}+\frac{t^6}{720}+ O\left(t^7\right)$$ $$\left(e^t-1\right)^2=t^2+t^3+\frac{7 t^4}{12}+\frac{t^5}{4}+\frac{31 t^6}{360}+\frac{t^7}{40}+O\left(t^8\right)$$ $$\frac{e^t}{\left(e^t-1\right)^2}=\frac{1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}+\frac{t^5}{120}+\frac{t^6}{720}+ O\left(t^7\right)} {t^2+t^3+\frac{7 t^4}{12}+\frac{t^5}{4}+\frac{31 t^6}{360}+\frac{t^7}{40}+O\left(t^8\right) }$$ Now, long division $$\frac{e^t}{\left(e^t-1\right)^2}=\frac{1}{t^2}-\frac{1}{12}+\frac{t^2}{240}+O\left(t^4\right)$$

Just replace $t$ by $\pi x$.

| cite | improve this answer | | | | |
$\endgroup$
0
$\begingroup$

Note that $$ f(x)=\frac{e^{\pi x}}{(e^{\pi x}-1)^2}=\frac{e^{\pi x}}{\pi^2x^2(\frac{e^{\pi x}-1}{\pi x})^2}=\frac{1}{\pi^2x^2}g(\pi x)$$ where $$ g(x)= \frac{e^{x}}{(\frac{e^{x}-1}{x})^2}.$$ Clearly $$ g(0)=\lim_{x\to0}\frac{e^{x}}{(\frac{e^{x}-1}{x})^2}=1 $$ and $$ g'(0)=\lim_{x\to0}g'(x)=\lim_{x\to0}-\frac{xe^x(2+e^x(x-2)+x)}{(e^x-1)^3}=\lim_{x\to0}-\frac{e^xx^3}{(e^x-1)^3}\cdot\frac{2+e^x(x-2)+x}{x^2}=0.$$ You can check that $$ g''(0)=\lim_{x\to0}g''(x)=\cdots=-\frac1{12}, g'''(0)=0.$$ So $$ g(x)=g(0)+g'(0)x+\frac12g''(0)x^2+\frac1{3!}x^3+O(x^4)=1-\frac1{12}x^2++O(x^4)$$ and hence $$ f(x)=\frac{1}{\pi^2x^2}g(\pi x)=\frac{1}{\pi^2x^2}(1-\frac1{12}\pi^2x^2+O(x^4))=\frac{1}{\pi^2x^2}-\frac1{12}+O(x^2)). $$

| cite | improve this answer | | | | |
$\endgroup$

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.