# The Leibnitz rule for exponential function?

• Feb 22nd 2009, 04:17 AM
bkarpuz
[SOLVED] The Leibnitz rule for exponential function?
Hi again,

I have a problem as follows:
Let
$x(t):=\exp\bigg\{\int_{0}^{t}y(s)\mathrm{d}s\bigg\ }\quad\text{for}\ t\geq0.$
Is there a simple formula for $x^{(n+1)}$ in the terms of $x$ and $y^{(k)}$ only?

I did some calculation as follows:
$x^{\prime}=yx,$
and hence by the Leibnitz rule
$x^{(n+1)}=\sum_{i=0}^{n}\binom{n}{i}y^{(i)}x^{(n-i)}.$
Now, iterating the above formula, I get
$x^{(n+1)}=\sum_{i=0}^{n}\binom{n}{i}y^{(i)}\sum_{j =0}^{n-1-i}\binom{n-1-i}{j}y^{(j)}x^{(n-1-i-j)}.$
And repeating the recursion above as long as possible, I deduce
$x^{(n+1)}=x\sum_{j_{0}=0}^{n}\sum_{j_{1}=0}^{n-1-j_{0}}\cdots\sum_{j_{m}=0}^{n-1-\sum_{i=0}^{m-1}j_{i}=0}\prod_{k=0}^{m-1}\binom{n-\mathrm{sgn}(k)-\sum_{i=0}^{k}j_{i}}{j_{k}}y^{(j_{k})}.$
Particularly, we get
$x^{\prime\prime}=x\big(y^{\prime}+y^{2}\big)\quad\ text{and}\quad x^{(4)}=x\big(y^{(3)}+4y^{\prime\prime}y+3y^{\prim e}(y^{\prime}+2y^{2})+y^{4}\big).$

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? :p
• Feb 22nd 2009, 04:46 AM
Laurent
Quote:

Originally Posted by bkarpuz
Hi again,

I have a problem as follows:
Let
$x(t):=\exp\bigg\{\int_{0}^{t}y(s)\mathrm{d}s\bigg\ }\quad\text{for}\ t\geq0.$
Is there a simple formula for $x^{(n+1)}$ in the terms of $x$ and $y^{(k)}$ only?

There is a formula for iterates of Leibniz' rule, it is called Faà di Bruno's formula. It will probably help you find a shorter formula.

For instance, $(e^y)^{(n)}=e^y\sum_{\pi\in\Pi(n)}\prod_{B\in\pi}y ^{(|B|)}$ where $\Pi(n)$ is the set of partitions of $\{1,\ldots,n\}$, like $\pi=\{\{1,5\},\{3\},\{4,2,6\}\}$ for $n=6$. You can also group together in the sum the terms where the elements of the partition have the same size. This gives you what the coefficients are: in the expansion of $(e^y)^{(n)}e^{-y}$, the coefficient before $(y')^{m_1}\cdots (y^{(n)})^{m_n}$ is the number of partitions of $\{1,\ldots,n\}$ with $m_1$ blocks of size 1, $m_2$ blocks of size 2, etc.. (so that $1m_1+\cdots+n m_n=n$). In other words, the coefficient is $\frac{n!}{m_1!\,1!^{m_1}\,m_2!\,2!^{m_2}\,\cdots\, m_n!\,n!^{m_n}}$.

This is connected to Bell polynomials. The Wikipedia gives other formulations.
• Feb 22nd 2009, 04:55 AM
bkarpuz
Many thanks Laurent, this looks very interesting.
However, I think it will be more complicated with this formula, to take out the term $x$ out of the sum by iterating Faà di Bruno's formula. (Thinking)
I will be thinking on it.
• Feb 22nd 2009, 05:05 AM
Moo
Hi you :D
Quote:

Lady Moo, may be using the notation in your signature it can be shown simpler?
Huuuu ? :o
I don't see how it can be related ^^
• Sep 24th 2009, 06:01 AM
bkarpuz
Quote:

Originally Posted by bkarpuz
Hi again,

I have a problem as follows:
Let
$x(t):=\exp\bigg\{\int_{0}^{t}y(s)\mathrm{d}s\bigg\ }\quad\text{for}\ t\geq0.$
Is there a simple formula for $x^{(n+1)}$ in the terms of $x$ and $y^{(k)}$ only?

I did some calculation as follows:
$x^{\prime}=yx,$
and hence by the Leibnitz rule
$x^{(n+1)}=\sum_{i=0}^{n}\binom{n}{i}y^{(i)}x^{(n-i)}.$
Now, iterating the above formula, I get
$x^{(n+1)}=\sum_{i=0}^{n}\binom{n}{i}y^{(i)}\sum_{j =0}^{n-1-i}\binom{n-1-i}{j}y^{(j)}x^{(n-1-i-j)}.$
And repeating the recursion above as long as possible, I deduce
$x^{(n+1)}=x\sum_{j_{0}=0}^{n}\sum_{j_{1}=0}^{n-1-j_{0}}\cdots\sum_{j_{m}=0}^{n-1-\sum_{i=0}^{m-1}j_{i}=0}\prod_{k=0}^{m-1}\binom{n-\mathrm{sgn}(k)-\sum_{i=0}^{k}j_{i}}{j_{k}}y^{(j_{k})}.$
Particularly, we get
$x^{\prime\prime}=x\big(y^{\prime}+y^{2}\big)\quad\ text{and}\quad x^{(4)}=x\big(y^{(3)}+4y^{\prime\prime}y+3y^{\prim e}(y^{\prime}+2y^{2})+y^{4}\big).$

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? :p

The discussion due to Opalg located at here also delivers an answer to this problem.

For any fixed $n\in\mathbb{N}_{0}$, we have $x^{(n)}(t)=y_{n}(t)x(t)$ for all $t\in(a,b)$, where
$y_{k}(t):=
\begin{cases}
1,&k=0\\
y_{k-1}^{\prime}(t)+y_{k-1}(t)y(t),&k\in\mathbb{N}
provided that $y\in\mathrm{C}^{(n-1)}([a,b],\mathbb{R})$.