# 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. – Chappers Apr 10 '15 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. – Iceman Apr 10 '15 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. – Jack D'Aurizio Apr 11 '15 at 7:47
• $\varepsilon_n=n!\cdot\sum_{r=0}^{n}\sum_{m=0}^{n}\sum_{k=0}^{r}\frac{(r+m)!(2k+1)^{n-m}{\cdot}(-1)^{m+k}}{m!{\cdot}(r-k)!{\cdot}k!{\cdot}(n-m)!{\cdot}2^{r}}$ – nczksv May 5 '15 at 0:44
• @nczksv, how did you approach the problem and how did you solve it?? – Iceman May 7 '15 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 $n$? – Iceman May 7 '15 at 12:47
• @Iceman,$(1-e^{2x})^r=(-\frac{2x}{1!}-\frac{(2x)^2}{2!}-\frac{(2x)^3}{3!}-...)^r.$ $\quad$ Then, for $r{\gt}n$, $[x^n](1-e^{2x})^r=0.$ – nczksv May 7 '15 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 $r>0$. – Iceman May 7 '15 at 13:27
• Newton's Binomial Theorem:$\quad$ whitman.edu/mathematics/cgt_online/section03.01.html – nczksv May 7 '15 at 14:13
• Awesome! Now it make sense. That's all i needed. – Iceman May 7 '15 at 17:04