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

```

• Prev by Date: About Parallel Computing
• Next by Date: DirectKinescopage2