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Quotient rule for higher order derivatives

  1. Feb 2, 2009 #1
    what is quotient rule for higher order derivatives ? i mean the one analogous to http://en.wikipedia.org/wiki/Leibniz_rule_%28generalized_product_rule%29" [Broken] .
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 2, 2009 #2
    Why not just use the general form of the product rule? Division is multiplication.
  4. Feb 2, 2009 #3
    but then you gotta use the reciprocal rule for the nth derivative , with isn't easier from finding a generalization for the quotient rule !!
  5. Feb 3, 2009 #4


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    You have it backwards, womfalcs3 is correct. Using the product rule is so much easier. Test it for yourself
  6. Feb 3, 2009 #5


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    (u/v)"= (uv-1)"= (u'v-1- uv-2')'= u"v- v-2u'v'+ 2uv-3v".

    I presume you know that the product law can be generalized using the binomial theorem.
  7. Mar 10, 2009 #6
    Ok then ,

    [tex](\frac{f}{g})^{(n)}=\sum^{n}_{k=0} \binom{n}{k} f^{(n-k)} (g^{-1})^{(k)}[/tex]

    and by http://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula" [Broken]

    [tex] (g^{-1})^{(k)}=\sum^{k}_{r=0} (-1)^{r} \frac{r!}{g^{r+1}}B_{k,r}(g^{(1)},g^{(2)} , .... , g^{(k-r+1)}) [/tex]

    where [tex]B_{k,r}(g^{(1)},g^{(2)} , .... , g^{(k-r+1)}) [/tex] are the http://en.wikipedia.org/wiki/Bell_polynomial" [Broken]

    now this is tedious to evaluate , so isn't there any easier formula ??
    Last edited by a moderator: May 4, 2017
  8. Mar 10, 2009 #7
    i have come to find that

    [tex](\frac{f}{g})^{(n)}=\frac{1}{g} \sum^{n}_{k=0} (-1)^k \binom{n+1}{k+1} \frac{(fg^{k})^{(n)}}{g^{k}}[/tex]

    but again , this isn't easier to evaluate since we need to expand the term [tex] (fg^{k})^{(n)} [/tex] using both leibniz rule and Faà di Bruno's formula !!
  9. Mar 10, 2009 #8
    That's pretty :)
  10. Jun 19, 2011 #9
    It seems true at least up to n=10 checking with Mathematica, but it's not obvious how to derive it. What's the trick?
  11. Jun 19, 2011 #10
    a poor proof: let u = f/g then use leibniz rule tu expand (ug^(k+1))^n,noticing we only need to proove the factor of u^(m)=0 ,if m<n ; the factor of u^(m)=1 ,if m=n. By Faà di Bruno's formula we can get it...
  12. Jun 19, 2011 #11
    Could you show the details? Although, verifying the formula and deriving it are two different things.
  13. Jun 19, 2011 #12
    oh it's true ... trust me .... the derivation is tedious though ... i'll look in my papers , and post the derivation here .
  14. Jun 19, 2011 #13


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    You could always extract the n-th derivative of (f/g)(x) from the Taylor series for (f/g)(y) around x. (Which, in turn, can be computed by dividing the Taylor series for f(y) and g(y) around x)
  15. Jun 19, 2011 #14
    in the proof i mentioned before ,the factor of u^(m)=1 ,if m=n,is obvious.about the factor of u^(m),m<n,we can simplify it to the expansion of (1-1)^(m+1), By Faà di Bruno's formula .
    wait for bpet's derivation ~ and i've not catched the way of
    Hurkyl ...
    Last edited: Jun 19, 2011
  16. Jun 24, 2011 #15
    Ok here's a proof of the formula though it only verifies the formula (doesn't show how it was derived).

    Moving all terms to the LHS and substituting u=f/g, we need to show that

    [tex]\sum_{k=0}^{n+1}(-1)^k (^{n+1}C_k) g^{n+1-k}(g^ku)^{(n)} = 0[/tex]

    for all g and u. Expanding the product derivative, we need to show that

    [tex]\sum_{k=0}^{n+1}(-1)^k (^{n+1}C_k) g^{n+1-k}(g^k)^{(j)} = 0[/tex]

    for all g and all j<=n. This can be proved in two ways.

    First, using Faa di Bruno's formula, the LHS is

    [tex]\sum_{k=0}^{n+1}\sum_{i=0}^{\min(j,k)} (-1)^k (^{n+1}C_k) g^{n+1-k} \frac{k!}{(k-i)!}g^{k-i}B_{j,i}(g',...)[/tex]

    [tex]=\sum_{i=0}^{j}\sum_{k=i}^{n+1} (-1)^k (^{n+1}C_k)\frac{k!}{(k-i)!}g^{n+1-i}B_{j,i}(g',...)[/tex]

    The sum over k is

    [tex]\sum_{k=i}^{n+1}(-1)^k (^{n+1}C_k)\frac{k!}{(k-i)!} = (^{n+1}C_i)(-1)^i i!\sum_{k=0}^{n+1-i}(-1)^k(^{n+1-i}C_k) = (1-1)^{n+1-i} = 0[/tex]

    as required. For an alternative proof using induction, we use

    [tex]g^m(g^k)^{(j)} = (g^{k+m})^{(j)} - \sum_{h=1}^m\sum_{i=1}^j g^{m-h}(^jC_i)g^{(i)}(g^{k+h-1})^{(j-i)}[/tex]

    so the LHS is

    [tex]\sum_{k=0}^{n+1}(-1)^k(^{n+1}C_k)\left((g^{n+1})^{(j)}-\sum_{h=1}^{n+1-k}\sum_{i=1}^j (^jC_i)g^{n+1-k-h}g^{(i)}(g^{k+h-1})^{(j-i)}\right)[/tex]

    The first term is zero (binomial expansion of [itex](1-1)^{n+1}[/itex]) and the second term is zero if

    [tex]\sum_{k=0}^n \sum_{h=1}^{n+1-k}(-1)^k(^{n+1}C_k)g^{n+1-k-h}(g^{k+h-1})^{(j-i)}=0[/tex]

    whenever (j-i)<n (the sum has no terms when k=n+1). Substituting m=k+h-1, the sum is

    [tex]\sum_{m=0}^n \sum_{k=0}^m (-1)^k (^{n+1}C_k)g^{n-m}(g^m)^{(j-i)} = \sum_{m=0}^n (-1)^m (^nC_m) g^{n-m}(g^m)^{(j-i)}[/tex]

    which is true by induction on n (the case n=1 is easily verified).

    That proof was horribly complicated so it would be good to see an intuitive derivation.
  17. Jun 24, 2011 #16
    Yet another proof: the last term is the coefficient of [itex]h^j/j![/itex] of [itex](g(x)-g(x+h))^{n+1} = O(h^{n+1})[/itex], which is zero for j<=n.

    So the formula could be derived by working backwards from [itex](g(x)-g(x+h))^{n+1}f(x+h)/g(x+h)[/itex].
  18. Jan 11, 2012 #17
    [tex]\left( \frac{f}{g} \right)^{(n)}=\frac{1}{g} \sum^{n}_{k=0} \binom{n+1}{k+1} \frac{1}{(-g)^{k}} \sum_{l=0}^{n}\binom{n}{l}f^{(n-l)}\sum_{r=1}^{l}\frac{k!}{(k-r)!}g^{k-r}B_{l,r}(g^{(1)},g^{(2)},...,g^{(l-r+1)})[/tex]

    that's ugly !!
  19. Jan 12, 2012 #18
    here is what i was trying to do :

    assume [itex]x(t)[/itex] is a smooth function around 0 , then :

    [tex] x(t) = \sum_{n=0}^{\infty }\frac{x^{(n)}(0)}{n!}t^n [/tex]

    taking the laplace transform of both sides :

    [tex]X(s) =\frac{1}{s}\sum_{n=0}^{\infty }\frac{x^{(n)}(0)}{s^{n}}[/tex]
    [tex]=>\Gamma (s):=\frac{1}{s}X\left( \frac{1}{s}\right ) =\sum_{n=0}^{\infty }x^{(n)}s^{n}[/tex]
    [tex]=>\lim_{s→0}\frac{\Gamma^{(n)} (s)}{n!}=x^{(n)}(0) [/tex]
    [tex]=> x(t)=\lim_{s→0}\sum_{n=0}^{\infty }\frac{\Gamma^{(n)}(s)}{(n!)^2}t^n[/tex]

    now , assume [itex]X(s)[/itex] is a rational function of the form:
    [itex] X(s)=\frac{P(s)}{Q(s)} , P(s),Q(s)[/itex] are polynomials where [itex]deg[P]\leq deg[Q][/itex]

    [tex] \Gamma (s) = \frac{P(\frac{1}{s})}{sQ(\frac{1}{s})}:=\frac{p(s)}{q(s)}[/tex]
    [tex]\lim_{s→0}\Gamma^{(n)}(s)= \lim_{s→0} \left(\frac{p(s)}{q(s)}\right)^{(n)}= \lim_{s→0} \frac{1}{q(s)} \sum^{n}_{k=0} (-1)^k \binom{n+1}{k+1} \frac{(p(s)q(s)^{k})^{(n)}}{q(s)^{k}}[/tex]

    [tex] =>x(t)=\lim_{s→0}\frac{1}{q(s)} \sum_{n=0}^{\infty }\frac{t^n}{(n!)^2}\left(\sum^{n}_{k=0} (-1)^k \binom{n+1}{k+1} \frac{(p(s)q(s)^{k})^{(n)}}{q(s)^{k}} \right ) [/tex]

    i was hoping for a simpler /prettier algorithm to analytically calculate the impulse response - i.e [itex]x(t) [/itex] - of a system with a transfer function [itex]X(s)[/itex] without resorting to partial fraction decomposition , for which we need to know the roots of [itex]Q(s)[/itex] , which is not always analytically possible .
    Last edited: Jan 12, 2012
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