Important, interesting, intermediate level exercise.
- To: mathgroup at smc.vnet.net
- Subject: [mg25878] Important, interesting, intermediate level exercise.
- From: Jack Goldberg <jackgold at math.lsa.umich.edu>
- Date: Sat, 4 Nov 2000 02:04:20 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Hi Group,
On pp 738-741 of Vol 107, Number 8, the October issue of the AMM, Tom
M. Apostle wrote an article entitled "Calculating Higher Derivatives of
Inverses". I think that his method involving a recursion is an
interesting and important exercise in Mathematica programming. For those who do
not get the AMM here is a very brief summary of the problem and its
mathematical solution. Let y = f[x] have a Taylor series expansion in
powers of x convergent in a neighborhood of x=0 with f'[0] not 0. Then
there is an inverse function x = g[y] expressible in a power series in y
about y=0. Let f[x,k] = the kth derivative of f at x, so that f'[x] =
f[x,1], f''[x]=f[x,2], etc. Then Apostol proves:
Theorem: Assume existence of all derivatives involved. Then for every
n>=1 we have
f[x,1]^(2n-1)*D[x,{y,n}] = Pn
where Pn is a polynomial in f[x,1], f[x,2],...,f[x,n] with integer
coefficients. These polynomials can be determined successively by the
recursion formula
P(n+1) = f[x,1]*D[Pn,x]-(2n-1)*f[x,2]*Pn,
where P1 = 1.
Note: P(n+1) is P subscripted by n+1. It is NOT P of n+1.
If you care to try programming this, here are some of the first few
formulas which you can use to check your work.
f[x,1]*D[x,y] = 1.
f[x,1]^3*D[x,{y,2}] = -f[x,2].
f[x,1]^5*D[x,{y,3}] = 3f[x,3]-f[x,1]*f[x,3].
f[x,1]^7*D[x,{y,4}] = -15f[x,2]^3+10*f[x,1]*f[x,2]*f[x,3]-f[x,1]^2*f[x,4]
I offer this as a substitute for the worn out example of the Fibonacci
numbers!
Jack