Alright, testing the left hand right hand limit I get different values....
I'll refer to \( \vartheta(a, b, \sigma) = a\,\,\bigtriangle_\sigma\,\,b \) \( \{a,b, \sigma| \R(a), \R(b) > e\,;\,a,b,\sigma \in C\} \) from now on.
So therefore:
\( \lim_{h\to 0^+} \frac{\large \vartheta(3, 4, 1 + h) - \vartheta(3, 4, 1)}{h} = 10.7633\\
\lim_{h\to 0^-} \frac{\large \vartheta(3, 4, 1 + h) - \vartheta(3, 4, 1)}{h} = 9.9206 \)
and
\( \lim_{h\to 0^+} \frac{\large \vartheta(3, 3, 1 + h) - \vartheta(3, 3, 1)}{h} = 5.4105\\
\lim_{h\to 0^-} \frac{\large \vartheta(3, 3, 1 + h) - \vartheta(3, 3, 1)}{h} = 5.3674 \)
And finally, I'd like to post a question for anyone more familiar with iteration dynamics than me. This is the open problem to extend \( \R(\sigma) \in [2, 3] \)
if \( f(z) = a \,\,\bigtriangleup_{1+q}\,\, z = \vartheta(a, z, 1+q) \)
then:
\( \vartheta(a, t, 2+q) = f^{\circ t}(1) \)
and so solving for [2, 3] is solving for the iterate of f(1).
f(z) is an analytic function since it's composed of analytic functions and q is restricted to [0,1].
hmmm, so the values come close to each other but don't quite make it. I wonder, does this disqualify it considering it's not analytic at t=1 and the derivative is undefined?
The only case where it is analytic and it is defined is \( \vartheta(e, e, \sigma) = \text{cheta}(\sigma)\,\,\,\, \R(\sigma) \le 2 \)
Which of course I tested using code over the exponential period [1, 2].
i.e.: \( 0 \le q \le 1 \)
\( \lim_{h\to 0}\, \vartheta(e+h, e+h, 1 + q)\, =\, \text{cheta}(1+q) \)
But I think that makes it even more beautiful, the fact that it's only analytic at 1 when a, b = e
And finally, I'd like to post a question for anyone more familiar with iteration dynamics than me. This is the open problem to extend \( \R(\sigma) \in [2, 3] \)
if \( f(z) = a \,\,\bigtriangleup_{1+q}\,\, z = \vartheta(a, z, 1+q) \)
then:
\( \vartheta(a, t, 2+q) = f^{\circ t}(1) \)
and so solving for [2, 3] is solving for the iterate of f(1).
f(z) is an analytic function since it's composed of analytic functions.
I'll refer to \( \vartheta(a, b, \sigma) = a\,\,\bigtriangle_\sigma\,\,b \) \( \{a,b, \sigma| \R(a), \R(b) > e\,;\,a,b,\sigma \in C\} \) from now on.
So therefore:
\( \lim_{h\to 0^+} \frac{\large \vartheta(3, 4, 1 + h) - \vartheta(3, 4, 1)}{h} = 10.7633\\
\lim_{h\to 0^-} \frac{\large \vartheta(3, 4, 1 + h) - \vartheta(3, 4, 1)}{h} = 9.9206 \)
and
\( \lim_{h\to 0^+} \frac{\large \vartheta(3, 3, 1 + h) - \vartheta(3, 3, 1)}{h} = 5.4105\\
\lim_{h\to 0^-} \frac{\large \vartheta(3, 3, 1 + h) - \vartheta(3, 3, 1)}{h} = 5.3674 \)
And finally, I'd like to post a question for anyone more familiar with iteration dynamics than me. This is the open problem to extend \( \R(\sigma) \in [2, 3] \)
if \( f(z) = a \,\,\bigtriangleup_{1+q}\,\, z = \vartheta(a, z, 1+q) \)
then:
\( \vartheta(a, t, 2+q) = f^{\circ t}(1) \)
and so solving for [2, 3] is solving for the iterate of f(1).
f(z) is an analytic function since it's composed of analytic functions and q is restricted to [0,1].
hmmm, so the values come close to each other but don't quite make it. I wonder, does this disqualify it considering it's not analytic at t=1 and the derivative is undefined?
The only case where it is analytic and it is defined is \( \vartheta(e, e, \sigma) = \text{cheta}(\sigma)\,\,\,\, \R(\sigma) \le 2 \)
Which of course I tested using code over the exponential period [1, 2].
i.e.: \( 0 \le q \le 1 \)
\( \lim_{h\to 0}\, \vartheta(e+h, e+h, 1 + q)\, =\, \text{cheta}(1+q) \)
But I think that makes it even more beautiful, the fact that it's only analytic at 1 when a, b = e
And finally, I'd like to post a question for anyone more familiar with iteration dynamics than me. This is the open problem to extend \( \R(\sigma) \in [2, 3] \)
if \( f(z) = a \,\,\bigtriangleup_{1+q}\,\, z = \vartheta(a, z, 1+q) \)
then:
\( \vartheta(a, t, 2+q) = f^{\circ t}(1) \)
and so solving for [2, 3] is solving for the iterate of f(1).
f(z) is an analytic function since it's composed of analytic functions.