Hi again,

I have a problem as follows:

Let

$\displaystyle x(t):=\exp\bigg\{\int_{0}^{t}y(s)\mathrm{d}s\bigg\ }\quad\text{for}\ t\geq0.$

Is there a simple formula for $\displaystyle x^{(n+1)}$ in the terms of $\displaystyle x$ and $\displaystyle y^{(k)}$ only?

I did some calculation as follows:

$\displaystyle x^{\prime}=yx,$

and hence by the Leibnitz rule

$\displaystyle x^{(n+1)}=\sum_{i=0}^{n}\binom{n}{i}y^{(i)}x^{(n-i)}.$

Now, iterating the above formula, I get

$\displaystyle x^{(n+1)}=\sum_{i=0}^{n}\binom{n}{i}y^{(i)}\sum_{j =0}^{n-1-i}\binom{n-1-i}{j}y^{(j)}x^{(n-1-i-j)}.$

And repeating the recursion above as long as possible, I deduce

$\displaystyle x^{(n+1)}=x\sum_{j_{0}=0}^{n}\sum_{j_{1}=0}^{n-1-j_{0}}\cdots\sum_{j_{m}=0}^{n-1-\sum_{i=0}^{m-1}j_{i}=0}\prod_{k=0}^{m-1}\binom{n-\mathrm{sgn}(k)-\sum_{i=0}^{k}j_{i}}{j_{k}}y^{(j_{k})}.$

Particularly, we get

$\displaystyle x^{\prime\prime}=x\big(y^{\prime}+y^{2}\big)\quad\ text{and}\quad x^{(4)}=x\big(y^{(3)}+4y^{\prime\prime}y+3y^{\prim e}(y^{\prime}+2y^{2})+y^{4}\big).$

I really wonder to know if there is a simpler formula for this or not?

I feel that there must be a relation between this and the Stirling's numbers.

Lady

**Moo**, may be using the notation in your signature it can be shown simpler?