computing the iterated exp(x)-1

[Gottfried]

\( f^n(z)=z+\frac{1}{2} n z^2 + \frac{1}{12} (3n^2-n) z^3 + \frac{1}{48} (6n^3-5n^2+n) z^4 + \cdots \) is the general solution

Which formula for the coefficients is this exactly?

The coefficients occur also by explicitely computing
the matrix-exponential via the exponential-series.

Let CL = log( C ) ,
where C is the iterable operator for exp(z)-1 as defined in my other post, then

Code:

.
     exp(n*CL)= (n Cl)^0 + (n CL)^1 + (n CL)^2/2! + (n CL)^3/3! ...

with as many terms as a given finite dimension requires
(CL is nilpotent to the order of its dimension)

Then we have (selecting dimension 5 for an example)

Code:

exponential-series
     tmpm = CL*n
     tmpe= matid(5)+ tmpm + tmpm^2/2! + tmpm^3/3! + tmpm^4/4!

and tmpe is

Code:

tmpe
  1                        .              .      .  .
  0                        1              .      .  .
  0                    1/2*n              1      .  .
  0           1/4*n^2-1/12*n              n      1  .
  0  1/8*n^3-5/48*n^2+1/48*n  3/4*n^2-1/6*n  3/2*n  1

The value for the half-iterate h(z) occurs then from the matrix-product

Code:

V(z)~ * tmpe [,1]

where as usual the second column of tmpe is the interesting one.

Code:

second column of tmpe:
    tmpe[,1]= [0, 1, 1/2*n, 1/4*n^2 - 1/12*n, 1/8*n^3 - 5/48*n^2 + 1/48*n]~

and with h(z) for the value of the half-iterate resp to parameter z in

Code:

.
    h(z) = V(z)~ * tmpe [,1]

we have

Code:

formula for half-iterate, using matrix dimension 5
h(z) = z^0 * 0
       +z^1 * 1
       +z^2 * 1/2 *(1*n)
       +z^3 * 1/12*(3*n^2 - 1*n)
       +z^4 * 1/48*(6*n^3 - 5*n^2 + 1*n]

which agrees with your formula.

Gottfried