Quotient rule for higher order derivatives

[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" .

 

[womfalcs3]

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

 

[mmzaj]

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

 

[djeitnstine]

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

 

[HallsofIvy]

(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.

 

[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"

(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"

now this is tedious to evaluate , so isn't there any easier formula ??

 

[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 !

 

[John Creighto]

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 !

That's pretty :)

 

[bpet]

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 !

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?

 

[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...

 

[bpet]

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...

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

 

[mmzaj]

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?

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

 

[Hurkyl]

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)

 

[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 ...

 

[bpet]

i have come to find that

(\frac{f}{g})^{(n)}=\frac{1}{g} \sum^{n}_{k=0} (-1)^k (^{n+1}C_k) \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 !

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&#039;,...)

=\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&#039;,...)

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.

 

[bpet]

...
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.
...

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).

 

[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 !

 

[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}}
=&gt;\Gamma (s):=\frac{1}{s}X\left( \frac{1}{s}\right ) =\sum_{n=0}^{\infty }x^{(n)}s^{n}
=&gt;\lim_{s→0}\frac{\Gamma^{(n)} (s)}{n!}=x^{(n)}(0)
=&gt; 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}}

=&gt;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 .