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