Quotient rule for higher order derivatives

1. Feb 2, 2009

mmzaj

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: Apr 24, 2017 at 10:50 AM
2. Feb 2, 2009

womfalcs3

Why not just use the general form of the product rule? Division is multiplication.

3. Feb 2, 2009

mmzaj

but then you gotta use the reciprocal rule for the nth derivative , with isn't easier from finding a generalization for the quotient rule !!

4. Feb 3, 2009

djeitnstine

You have it backwards, womfalcs3 is correct. Using the product rule is so much easier. Test it for yourself

5. Feb 3, 2009

HallsofIvy

Staff Emeritus
(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.

6. Mar 10, 2009

mmzaj

Ok then ,

$$(\frac{f}{g})^{(n)}=\sum^{n}_{k=0} \binom{n}{k} f^{(n-k)} (g^{-1})^{(k)}$$

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

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

where $$B_{k,r}(g^{(1)},g^{(2)} , .... , g^{(k-r+1)})$$ 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: Apr 24, 2017 at 11:50 AM
7. Mar 10, 2009

mmzaj

i have come to find that

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

but again , this isn't easier to evaluate since we need to expand the term $$(fg^{k})^{(n)}$$ using both leibniz rule and Faà di Bruno's formula !!

8. Mar 10, 2009

John Creighto

That's pretty :)

9. Jun 19, 2011

bpet

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?

10. Jun 19, 2011

yanshu

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...

11. Jun 19, 2011

bpet

Could you show the details? Although, verifying the formula and deriving it are two different things.

12. Jun 19, 2011

mmzaj

oh it's true ... trust me .... the derivation is tedious though ... i'll look in my papers , and post the derivation here .

13. Jun 19, 2011

Hurkyl

Staff Emeritus
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)

14. Jun 19, 2011

yanshu

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
15. Jun 24, 2011

bpet

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

$$\sum_{k=0}^{n+1}(-1)^k (^{n+1}C_k) g^{n+1-k}(g^ku)^{(n)} = 0$$

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

$$\sum_{k=0}^{n+1}(-1)^k (^{n+1}C_k) g^{n+1-k}(g^k)^{(j)} = 0$$

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

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

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

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

The sum over k is

$$\sum_{k=i}^{n+1}(-1)^k (^{n+1}C_k)\frac{k!}{(k-i)!} = (^{n+1}C_i)(-1)^i i!\sum_{k=0}^{n+1-i}(-1)^k(^{n+1-i}C_k) = (1-1)^{n+1-i} = 0$$

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

$$g^m(g^k)^{(j)} = (g^{k+m})^{(j)} - \sum_{h=1}^m\sum_{i=1}^j g^{m-h}(^jC_i)g^{(i)}(g^{k+h-1})^{(j-i)}$$

so the LHS is

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

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

$$\sum_{k=0}^n \sum_{h=1}^{n+1-k}(-1)^k(^{n+1}C_k)g^{n+1-k-h}(g^{k+h-1})^{(j-i)}=0$$

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

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

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.

16. Jun 24, 2011

bpet

Yet another proof: the last term is the coefficient of $h^j/j!$ of $(g(x)-g(x+h))^{n+1} = O(h^{n+1})$, which is zero for j<=n.

So the formula could be derived by working backwards from $(g(x)-g(x+h))^{n+1}f(x+h)/g(x+h)$.

17. Jan 11, 2012

mmzaj

$$\left( \frac{f}{g} \right)^{(n)}=\frac{1}{g} \sum^{n}_{k=0} \binom{n+1}{k+1} \frac{1}{(-g)^{k}} \sum_{l=0}^{n}\binom{n}{l}f^{(n-l)}\sum_{r=1}^{l}\frac{k!}{(k-r)!}g^{k-r}B_{l,r}(g^{(1)},g^{(2)},...,g^{(l-r+1)})$$

that's ugly !!

18. Jan 12, 2012

mmzaj

here is what i was trying to do :

assume $x(t)$ is a smooth function around 0 , then :

$$x(t) = \sum_{n=0}^{\infty }\frac{x^{(n)}(0)}{n!}t^n$$

taking the laplace transform of both sides :

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

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

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

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

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

Last edited: Jan 12, 2012