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I have the following recursive equation $$l_{n}=b_{n-1}+\sum_{k=0}^n l_k a_k b_{n-k},\ n\geq 1$$ where $b_n=1/n!,\ a_n=S_n(-1)$ where $S_n(x)=\sum_{k=0}^n \frac{x^k}{k!}\ \forall x\in \mathbb{R}$ and $l_0=0$. How to solve this?

I proceeded a little bit to solve it and here is what I did \begin{equation*} \begin{split} f(z)=&\sum_{n=0}^\infty l_n a_n z^n \\ =& \sum_{n=1}^\infty \left(b_{n-1}a_n z^n+\sum_{k=0}^n l_k a_k a_n z^n \right)\\ =& \phi(z,1)+\sum_{m=1}^\infty l_m a_m \phi(z,m) \end{split} \end{equation*}
Where $$\phi(z,m)=\sum_{n=m}^\infty b_{n-m}a_n z^n $$ Which I have shown to respect the identity $$ \phi(z,m) = z^m \left(e^{z-1}-\sum_{n=m+1}^\infty J_n(-2\sqrt{z})z^{-n/2}\right)$$ Where $J_n(x)$ is the $n$ th order Bessel function. But after that, I don't know how to proceed. Any idea?

Samrat Mukhopadhyay
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  • I don't know how to solve it from just looking at it. I would try both: 1) Using the recurrence to compute a generating function for $\ell_n$. or 2) Apply Z-transform (http://en.wikipedia.org/wiki/Z-transform). – OR. Jun 26 '13 at 14:31
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    Notice that the sum on the right is a discrete convolution. That is how the coefficients of the product of two series look like. So, with generating series you will get a product of function, hopefully. With the Z-transform you will also get a multiplication. The two methods are kind of equivalent. – OR. Jun 26 '13 at 14:33
  • I just noticed that probably there is a little problem with the subindexes on the right, inside the sum. Isn't it? I said it was a convolution thinking it was $\ell_k a_k b_{n-k}$. – OR. Jun 26 '13 at 14:33
  • @Franklin, I used the Z-transform technique, but as you correctly noticed, the problem is with the $l_k a_k$ term. I think since $a_k$ has this particular structure, we can use that somehow to simplify the expression, but I'm still unable to do that. – Samrat Mukhopadhyay Jun 26 '13 at 14:38
  • But if the 'recurrence' is just like it is written then you can simply solve for $l_n$ in the equation. Let me write it in an answer to have more space. – OR. Jun 26 '13 at 14:42
  • Sorry for posting it incorrectly before. Now I've edited the recurrence relation. The previous one made it very simple to solve. – Samrat Mukhopadhyay Jun 26 '13 at 14:54
  • Ok, that's how I imagined it was. Let me try. – OR. Jun 26 '13 at 14:55

1 Answers1

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Let us aim to find first $a_nl_n$ and then to get $l_n$ we just divide by $a_n$, which is given explicitly.

Let $f(z):=\sum_{k=0}^{\infty}l_ka_kz^k$.

Multiply the recurrence by $a_nz^n$ and add from $n=0$ to $\infty$

We get

$$f(z)=\sum_{k=0}^{\infty}\frac{a_nz^n}{(n-1)!}+f(z)e^{z}.$$

We can solve for $f$ here to get

$$f(z)=\frac{\sum_{n=0}^{\infty}\frac{a_nz^n}{(n-1)!}}{1-e^z}$$

Now, it would be useful to sum the numerator, if possible.

$$\sum_{n=0}^{\infty}\frac{a_nz^n}{(n-1)!}=\sum_{n=0}^{\infty}\sum_{i=0}^{n}\frac{(-1)^iz^n}{i!(n-1)!}$$

Uhm ... I don't see if it has a closed form. Maybe it doesn't. If it doesn't then the formula for $f$ above is the best I can give you.

OR.
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  • Are you trying to find the generating function? – Mhenni Benghorbal Jun 26 '13 at 15:02
  • Yes, but I am still writing the answer. Thinking on the computer directly. – OR. Jun 26 '13 at 15:03
  • Ok. Now I wrote all I could think for the moment. – OR. Jun 26 '13 at 15:24
  • @Franklin.vp you know I tried to find out the generating function for $l_n$ and I saw that you can get a closed form expression this if you can solve the following more general question: If $L(z)=\sum_{k=0}^\infty l_k z^k$ and $A(z)=\sum_{k=0}^\infty a_k z^k$ be the formal power series for ${l_n}$ and ${a_n}$ resp. then what is the formal power series for ${l_n a_n}$? – Samrat Mukhopadhyay Jun 26 '13 at 16:01
  • Look at the 'multiplication' entry in the table of properties of the Z-transform (http://en.wikipedia.org/wiki/Z-transform#Properties) It is given by a contour integral. It will turn the problem into solving an integral equation in $L$. For most sequences there is no closed form. The most useful way of presenting the solution depends on what you want to do with it afterwards. – OR. Jun 26 '13 at 16:28
  • Yes, I saw that, but I thought we could exploit the structure of $a_n$ to get some closed form expression, at least some expression for $l_n$ in series form. – Samrat Mukhopadhyay Jun 26 '13 at 16:49
  • Maybe. Now, $a_n$, by itself, doesn't have a nice closed form. – OR. Jun 26 '13 at 17:24