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    For two functions, Leibniz's theorem says \displaystyle \frac{d^n}{dx^n}\left(u\cdot v\right) = \sum_{k=0}^{n}\binom{n}{k}u_{k}v  _{n-k}. What about when we have more than two functions - say three, four, or even more?
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    (Original post by PerigeeApogee)
    Any number of products of functions can be split into groups of two, then split again, and again, as necessary.
    But that gets tedious, as the number of functions gets large. I was sort of looking for a 'multinomial' version, but that's probably misguided.
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    Use what PerigeeApogee said apply Leibniz to what's inside and do some combinatorics.
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    (Original post by Liouville)
    But that gets tedious, as the number of functions gets large. I was sort of looking for a 'multinomial' version, but that's probably misguided.
    There is a multinomial Leibniz rule, comparable to the multinomial theorem.

    It just says (for example in the trinomial case)

     (uvw)^{(n)} = \sum  C^{n}_{k,l,m} u^{(k)} v^{(l)} w^{(m)}

    where the sum is taken over all nonnegative k,l,m such that k+l+m=n and

     C^{n}_{k,l,m} = \frac{n!}{k! l! m!}
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    \displaystyle (u_1\cdot u_{2}\cdots u_{k})^{(n)} = \sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m} u_1^{k_1} u_2^{k_2} \cdots u_m^{k_m}\,

    Is that correct? Never have done anything like that in my entire life. :eek:

    What I needed it for was to find the nth derivative of the product \displaystyle \prod_{k=1}^{n}(k-x)^{m}
    But now I realise it's sort of a massive overkill and even somewhat ineffective.
    Sure there must be better ways to find the nth derivative of the product, no?
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    (Original post by Liouville)
    ..
    I don't see an easy way of finding that product in the general case. (A quick play with Wolfram Alpha shows no obvious patterns).

    If this is a result you've found yourself wanting in the middle of solving some other problem, I suspect you'd be better off looking for a different approach.
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    (Original post by DFranklin)
    If this is a result you've found yourself wanting in the middle of solving some other problem, I suspect you'd be better off looking for a different approach.
    Thanks, that's what it was. I will leave it at that for now.
 
 
 
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