# Help finding a closed form

I have the following function: $$\frac{2e^x}{e^{2x}+1+2x}=\sum_{n=0}^\infty \varepsilon_n\frac{x^n}{n!}$$

I would like to find a closed form for the $\varepsilon_k$. One thing that I do know is that the $\varepsilon_n$ satisfy the following recurrence relation:

$$\varepsilon_n=1-2n\varepsilon_{n-1}-\sum_{k=0}^{n-2}\binom{n}{k}2^{n-k-1}\varepsilon_k$$

The first 11 numbers are below:

$$1,-1,3,-15,93,-725,6815, -74627, 933849, -13148361, 205690779$$

I've tried slowly plugging in small values of $n$, expanding the recurrence relation and then condensing it back to something smaller. I try to find patterns then amongst the summands and products but this method is very long and tedious, and the size of the summands expands quickly. What are some of the methods the have been used for closed forms similar to this and how can I better approach the problem?

• This generating function is rather close to that of the Bernoulli numbers, which don't have a better closed-form expression than a recursion like the one you have. Therefore it seems highly unlikely that you will have a closed form. Apr 10, 2015 at 23:07
• They are actually very similar to Euler Numbers and they do have a closed form, so I thought there might be a connection. Apr 10, 2015 at 23:12
• The coefficients can be extracted from $$f(x)=\frac{2e^x}{e^{2x}+1+2x}=2\sum_{n\geq 0}(-1)^n (1+2x)^n e^{-(2n+1)x}\tag{1}$$ but that leads to a rather convolved multiple sum. Apr 11, 2015 at 7:47
• $\varepsilon_n=n!\cdot\sum_{r=0}^{n}\sum_{m=0}^{n}\sum_{k=0}^{r}\frac{(r+m)!(2k+1)^{n-m}{\cdot}(-1)^{m+k}}{m!{\cdot}(r-k)!{\cdot}k!{\cdot}(n-m)!{\cdot}2^{r}}$ May 5, 2015 at 0:44
• @nczksv, how did you approach the problem and how did you solve it?? May 7, 2015 at 1:08

$$\varepsilon_n=n![x^n]\left(\frac{2e^{x}}{e^{2x}+1+2x}\right)$$

$$=n![x^n]\left(\frac{2e^{x}}{e^{2x}-1+2+2x}\right)$$ $$=n![x^n]\left(\frac{\frac{2e^{x}}{2+2x}}{1-\frac{1-e^{2x}}{2+2x}}\right)$$

$$=n![x^n]\left(\left(\frac{2e^{x}}{2(1+x)}\right)\sum_{r=0}^{\infty}\left(\frac{1-e^{2x}}{2(1+x)}\right)^{r}\right)$$

$$=n![x^n]\left(2e^{x}\sum_{r=0}^{\infty}\frac{(1-e^{2x})^{r}}{\left(2(1+x)\right)^{r+1}}\right)$$ $$=n![x^n]\left(2e^{x}\sum_{r=0}^{\color{red}{n}}\frac{(1-e^{2x})^{r}}{\left(2(1+x)\right)^{r+1}}\right)$$

$$=n![x^n]\left(e^x\sum_{r=0}^{n}\sum_{k=0}^{r}\binom{r}{k}(-e^{2x})^{k}\left(\frac{1}{2^r}\right)\sum_{m=0}^{\infty}\binom{r+m}{m}(-x)^m\right)$$

$$=n![x^n]\left(\sum_{r=0}^{n}\sum_{k=0}^{r}\binom{r}{k}(-1)^{k}\left(e^{(2k+1)x}\right)\left(\frac{1}{2^r}\right)\sum_{m=0}^{\infty}\binom{r+m}{m}(-x)^m\right)$$

$$=n![x^n]\left(\sum_{r=0}^{n}\sum_{k=0}^{r}\binom{r}{k}(-1)^{k}\sum_{h=0}^{\infty}\frac{(2k+1)^{h}x^{h}}{h!}\left(\frac{1}{2^r}\right)\sum_{m=0}^{\infty}\binom{r+m}{m}(-x)^m\right)$$

$$=n![x^n]\left(\sum_{r=0}^{n}\sum_{k=0}^{r}\sum_{h=0}^{\infty}\sum_{m=0}^{\infty}\binom{r}{k}(-1)^{k+m}\left(\frac{1}{2^r}\right)\frac{(2k+1)^{h}}{h!}\binom{r+m}{m}x^{h+m}\right)$$

$$=n!\sum_{r=0}^{n}\sum_{k=0}^{r}\sum_{m=0}^{n}\binom{r}{k}\binom{r+m}{m}(-1)^{k+m}\left(\frac{1}{2^r}\right)\frac{(2k+1)^{n-m}}{(n-m)!}$$

$$=n!\sum_{r=0}^{n}\sum_{k=0}^{r}\sum_{m=0}^{n}\frac{(r+m)!{\cdot}(2k+1)^{n-m}{\cdot}(-1)^{m+k}}{m!{\cdot}(r-k)!{\cdot}k!{\cdot}(n-m)!{\cdot}2^{r}}$$.

• Wow, that's great. I understand just about all of it except one part. The red n. How did you change the bound from $\infty$ to $n$? May 7, 2015 at 12:47
• @Iceman,$(1-e^{2x})^r=(-\frac{2x}{1!}-\frac{(2x)^2}{2!}-\frac{(2x)^3}{3!}-...)^r.$ $\quad$ Then, for $r{\gt}n$, $[x^n](1-e^{2x})^r=0.$ May 7, 2015 at 13:12
• Next question. You have $\frac{1}{(1+x)^{r+1}}=\sum_{m=0}^\infty\binom{r+m}{m}(-x)^m$. Wouldn't you get from the binomal theorem that it is equal to $\sum_{m=0}^{-r-1}\binom{-r-1}{m}x^m$ which seems nonsensical if $r>0$. May 7, 2015 at 13:27
• Newton's Binomial Theorem:$\quad$ whitman.edu/mathematics/cgt_online/section03.01.html May 7, 2015 at 14:13
• Awesome! Now it make sense. That's all i needed. May 7, 2015 at 17:04