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Math Help - The Leibnitz rule for exponential function?

  1. #1
    Senior Member bkarpuz's Avatar
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    Exclamation [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?
    Last edited by bkarpuz; September 24th 2009 at 05:02 AM.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by bkarpuz View Post
    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 Fa 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.
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  3. #3
    Senior Member bkarpuz's Avatar
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    Red face

    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 Fa di Bruno's formula.
    I will be thinking on it.
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  4. #4
    Moo
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    Hi you
    Lady Moo, may be using the notation in your signature it can be shown simpler?
    Huuuu ?
    I don't see how it can be related ^^
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  5. #5
    Senior Member bkarpuz's Avatar
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    Cool

    Quote Originally Posted by bkarpuz View Post
    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?
    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):=<br />
\begin{cases}<br />
1,&k=0\\<br />
y_{k-1}^{\prime}(t)+y_{k-1}(t)y(t),&k\in\mathbb{N}<br />
\end{cases}\qquad\text{for}\ t\in[a,b]\ \text{and}\ k\in\mathbb{N}_{0},
    provided that y\in\mathrm{C}^{(n-1)}([a,b],\mathbb{R}).
    Last edited by bkarpuz; September 24th 2009 at 05:16 AM.
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