bo198214 Wrote:If \( -1<\log(h)<0 \) then we have the negative Eigenvalues \( \log(h)^{2n+1} \) in the power derivation matrix A of \( f(x)=b^x \). Now we compute \( A^t \). It has the Eigenvalues \( (\log(h)^n)^t \). Take for example \( t=\frac{1}{2} \) then we see that \( A^t \) has also non-real Eigenvalues and hence has also non-real entries. Supposed \( f^{\circ \frac{1}{2}} \) had only real coefficients then \( A^{\frac{1}{2} \) would have only real coefficients. So it is clear that \( f^{\circ \frac{1}{2} \) must have some non-real coefficients and is a non-real function.
However perhaps it could be that \( {}^{\frac{1}{2}}b=f^{\circ \frac{1}{2}}(1) \) is real or generally that \( {}^tb=f^{\circ t}(1) \) is real, which I dont believe. Can someone just compute it?
I computed for hl = log(h) = -1/2 , h = 0.60653066, b= 0.43851527
such that h^(1/h) = b
So b is our usual base-parameter, and it is made sure by the above computation, that it is in the admissible range.
Using my analytic solution:
Since hl is negative, we have for the half-iterate complex values for the eigenvalues of the constructed matrix-operator.
The half-iterate is (using 16 terms for the result)
\( \hspace{24} f^{\circ \frac{1}{2}}(b,1) \) = b^^(1/2) = 0.58983992+0.24626917*I
Using that result as input, the next half-iterate is
\( \hspace{24} f^{\circ \frac{2}{2}}(b,1) \) = b^^(2/2) = 0.43851527-2.1267671E-11*I
using 16 terms again.
Indeed, the result approximates b^^1 = b very good.
Gottfried
edit: I adapted the s^^ (1/2) expression to b^^(1/2) for consistency
Gottfried Helms, Kassel