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Take the following recurrence relation into account:

$$ a_{n+2} = \frac{1}{(n+1)(n+2)} \sum_{k=0}^n (s_k - (k+1)a_{k+1})(n-k+1)a_{n-k+1} $$

I know that:

$$ s_{2m+1} = \frac{(-1)^m}{(2m+1)!} $$

and

$$ s_{2m} = 0 $$

My goal is to find out, In terms of $a_0$ and $a_1$:

$$ \sum_{n=0}^\infty a_n x^n $$

I have came out with this first few $a$ terms:

$$ a_2 = - \frac{a_1^2}{2} $$

$$ a_3 = \frac{a_1^3}{3} + \frac{a_1}{6} $$

$$ a_4 = -\frac{a_1^4}{4} $$

The $a$ terms seem to alternate signs, and it also appears there is a connection between the exponent of the $a_1$ terms and the denominator. This is all I could come up with thus far. My guess would be a natural log result of some kind due to these factors.

Edit (more on the side of how I am seeing the coefficient):

If you set $f(n) = \sum_{k=1}^{[\frac{n}{2}]-1} [s_{2k+1} a_{n-2k+1}]$ and $g(x) = \sum_{k=1}^{n-1}[k(n-k) a_k a_{n-k}]$, It seems as though $g(x)$ is just a way to convolute the terms that are already known, in which you would find a new $f(n)$ which would basically be alternating from even and odd $n$ values for corrisponding even or odd value $a_{n+2}$. This new $f(n)$ function would take the corrisponding (even or odd value that $a_{n+2}$ would take and have those terms desend, as the sequence desends, the terms would alternate signs and decrease due to the properties of $s_n$.

The purpose of doing this is to solve the following differential equation using series solutions:

$$ \sin(x) = \frac{y''}{y'} + y' $$

Setting

$$ f(x) = \sum_{n=0}^\infty s_n x^n $$

and

$$ y = \sum_{n=0}^\infty a_n x^n $$

By solving in terms of $a_1$ i can group the terms in a specific way, by doing so, with the first few terms (meaning about the first 25) I could group the $x$ terms with respect to its $a_1^n$ terms such that:

$$ y = \sum_{n=0}^\infty a_1^n f_n(x) $$ Perhaps by doing this, something interesting can come out of doing this, this intrigues me because I see that the solution to such a differential equation is:

$$ e^y = \int e^{\int \sin(x) dx} dx $$

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  • $\begingroup$ You already have $a_3$ wrong: The $s_1$ in the $k=0$ term is multiplied by $a_2$ giving a contribution to $a_3 of $\frac{1}{3}a_1^2$. I think this problem could be attack by convolutions of generator functions, but it would not be at all easy! $\endgroup$ – Mark Fischler Jan 30 '15 at 1:21
  • $\begingroup$ @MarkFischler Thanks for the heads up, I don't mind a challenging problem. Also my work for $a_3$ is: $$ a_3 = \frac{1}{6}[2(s_0 - a_1)a_2 + (s_1 - 2a_2)a_1] $$ $$ a_3 = \frac{1}{6}[a_1 - 4 a_2 a_1] $$ $$ 4a_1 a_2 = -2a_1^3 $$ $$ a_3 = \frac{1}{6} [ a_1 + 2 a_1^3] $$ $\endgroup$ – Eric Lawson Jan 30 '15 at 1:46
  • $\begingroup$ THe work you show in this comment is different than the expression appearing in the problem inn 2 ways: You have two appearences of $s_r$ fgor some index $r$ but the problem has only one, and the problem uses only a static $s_n$ that does not depend on $k$, yet the work you show uses two values that do depend on $k$, agin in a manner suggesting a convolution. If you have stated the problem erroneously, please correct the statement. $\endgroup$ – Mark Fischler Jan 31 '15 at 20:01
  • $\begingroup$ @MarkFischler That was a mistkae, Sorry for the confusion. $\endgroup$ – Eric Lawson Feb 1 '15 at 21:34
  • $\begingroup$ One approach could be to try to expand the solution $y(x) = a_0 + \log\left(1 + a_1\int_0^xe^{1-\cos(t)}dt\right)$ into a series using the $\log$-series. From this I get $a_k = k!\sum_{n=1}^\infty \frac{(-1)^na_1^n}{n}\left(\frac{d^kY^n(x)}{dx^k}\right)_{x=0}$ where $Y = \int_0^xe^{1-\cos(t)}dx$. But finding the derivative in the sum seems hard... $\endgroup$ – Winther Feb 4 '15 at 22:43

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