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Kth derivative of the nth iterate

  1. Sep 12, 2004 #1
    I'm trying to find a Taylor series for g^n, the nth iterate of g, centered at a fixed point p of g.

    I know the first few terms:
    if g^n(x)=a_0 + a_1 (x-p) + a_2 (x-p)^2 + ..., then
    let s=g'(p)
    a_2=[s^(n-1) (s^n - 1) g''(p)] / (2(s-1)).

    It's safe to say that a_3 is as complex as a_2 is compared to a_1; it's not pretty.

    I have a feeling that FaĆ” di Bruno's formula (http://mathworld.wolfram.com/FaadiBrunosFormula.html) may be involved but I would prefer to avoid that.

    One of my goals is to find a formula in which I can let n be something other than an integer to at least approximate some fractional iterates of a function.

    Does anyone know a formula for the kth derivative of the nth iterate of g (assuming its sufficiently well behaved)? Thanks.
  2. jcsd
  3. Sep 13, 2004 #2
    So far, I have this...

    [tex]D\left( g^{m}\left( x\right) \right) =\prod_{j=1}^{m}g^{\prime }\left( g^{m-j}\left( x\right) \right) [/tex] and if p is a fixed point of g, then

    [tex]\left. D\left( g^{m}\left( x\right) \right) \right| _{x=p}=g^{\prime }\left( p\right) ^{m}[/tex].

    Then the first derivative of the nth iterate of g is this:
    [tex]D\left( g^{n}\left( x\right) \right) =\prod_{j=1}^{n}g^{\prime }\left( g^{n-j}\left( x\right) \right) [/tex]. Now we have the product of n functions and we have to take k-1 more derivatives to get the kth derivative. I'm guessing that a formula for the kth derivative of the product of n functions is this:
    [tex]D^{k}\left( \prod_{j=1}^{n}f_{j}\right) =\sum \frac{k!}{j_{1}!\cdot \cdots \cdot j_{n}!}f_{1}^{\left( j_{1}\right) }\cdot \cdots \cdot f_{n}^{\left( j_{n}\right) }[/tex], where the sum ranges over n-tuples [tex]\left( j_{1},\ldots ,j_{n}\right) [/tex] whose sum is k.

    Is that the correct formula for the kth derivative of the product of n functions?

    Well, I imagine that that or something like that is correct. Then we apply the formula so that [tex]f_{j}=g^{\prime }\circ g^{n-j}[/tex] but then we'd have to calculate the q-th derivative of the right hand side. ACK!
  4. Sep 16, 2004 #3
    I found a formula though it's not that pretty. See the attached pdf. Next I will try to work out the coefficients.

    Attached Files:

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