Quotient rule for higher order derivatives

In summary, the quotient rule for higher order derivatives is a generalized form of the product rule, analogous to the Leibniz rule. There are multiple ways to derive this rule, including using the binomial theorem and Faà di Bruno's formula, or through induction. However, it is a tedious process and there may not be a simpler formula for evaluating it.
  • #1
mmzaj
107
0
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" .
 
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  • #2
Why not just use the general form of the product rule? Division is multiplication.
 
  • #3
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 !
 
  • #4
You have it backwards, womfalcs3 is correct. Using the product rule is so much easier. Test it for yourself
 
  • #5
(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
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"

[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"

now this is tedious to evaluate , so isn't there any easier formula ??
 
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  • #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 !
 
  • #8
mmzaj said:
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 !

That's pretty :)
 
  • #9
mmzaj said:
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 !

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
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
yanshu said:
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.
 
  • #12
bpet said:
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 .
 
  • #13
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
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 ...
 
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  • #15
mmzaj said:
i have come to find that

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

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.
 
  • #16
bpet said:
...
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.
...

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].
 
  • #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 !
 
  • #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 .
 
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Related to Quotient rule for higher order derivatives

1. What is the quotient rule for higher order derivatives?

The quotient rule for higher order derivatives is a mathematical rule used to find the derivative of a function that is expressed as the ratio of two other functions. It is an extension of the quotient rule for first derivatives, and it is used to find the derivatives of functions with multiple variables.

2. How do you apply the quotient rule for higher order derivatives?

To apply the quotient rule for higher order derivatives, you first need to find the first and second derivatives of the numerator and denominator. Then, plug those values into the formula: (f'g - g'f) / g^2. This will give you the derivative of the original function.

3. What is the purpose of the quotient rule for higher order derivatives?

The purpose of the quotient rule for higher order derivatives is to help find the rate of change of a function that is expressed as a ratio of two other functions. This is useful in many scientific fields, such as physics, economics, and engineering.

4. Can the quotient rule for higher order derivatives be applied to any function?

Yes, the quotient rule for higher order derivatives can be applied to any function that is expressed as a ratio of two other functions. However, it is important to note that the function must be differentiable for the rule to be applicable.

5. Are there any common mistakes when applying the quotient rule for higher order derivatives?

Yes, some common mistakes when applying the quotient rule for higher order derivatives include incorrectly finding the first and second derivatives of the numerator and denominator, failing to simplify the final answer, and forgetting to use the chain rule when necessary.

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