Gottfried Wrote:I just computed this with the function
U_t (x) = t^x - 1
T_b(x) = b^x
U_t°h(x) and T_b°h(x) as their iterates of general height h
and the shift
T_b°h(x) = (U_t°h(x/t-1) +1)*t
and the height-iteration by applying powers to the diagonal of the diagonalized operator for U_t(x) with 96 terms, using b=sqrt(2),t=2,x=1
But I still dont get how you calculated the iterates. Let me try to reconstruct:
You consider the function \( U(x)=a^x-1 \), where \( a \) is the fixed point of \( b^x \), so \( b=a^{1/a} \). (Its good to stay a bit with useful conventions: \( b \) is the letter for the base, \( a \) is mostly used for the fixed point, \( t \) is mostly used as the iteration variable.) If you now use the transformation \( \tau(x)=x/a-1 \), \( \tau^{-1}(x)=a(x+1) \) then \( \tau^{-1}(U(\tau(x)) = (a^{x/a-1}-1+1)t=a^{x/a}=b^x \).
And then you compute the regular iteration powerseries of \( U \) at its fixed point \( 0 \). And so get the regular iteration of \( b^x \) as \( \exp_b^{\circ t}=\tau^{-1}\circ U^{\circ t}\circ \tau \). It is regular as we have seen that additional (to translating the fixed point to 0) multiplicative constants in the transformation do not change the result.
My formula also computes the regular iteration at the lower fixed point, but in an iterative way, not by power series. So at a theoretical level both results must be equal! The differences are of purely numeric nature.
Euler summation should only be needed for \( a=e \), \( U(x)=e^x-1 \) otherwise it should converge/is analytic at the fixed point. (Which does not mean that you shouldnt use it for acceleration of convergence
)edit: Ah now I see, what puzzled me was U_t°h(x) and T_b°h(x). Gottfried, this is very prone to misinterpretation, my first reading was \( U_t\circ h (x) \), you know \( h(x) \) is used for the inversion of \( x^{1/x} \), but what you meant was \( U_t^{\circ h}(x) \)! If you dont use tex you need another presentation of iteration. The non-superscripted \( \circ \) has the completely different meaning of composition!
