04/25/2010, 10:41 AM
Hey Kobi,
thanks for sharing your ideas.
The formula looks very interesting, however it seems only useful with the natural continuum product which you describe, which limits the base range to \( <e^{1/e} \). It seems indeed to yield regular iteration tetration, a nice finding.
For the case \( b>e^{1/e} \) we however lack an expression of \( f_k \) continuous in \( k \) (and if we had that then we would already have tetration). So we have no possibility to compute the continuum product in an alternate way, which however would be necessary for the case \( b>e^{1/e} \).
Actually I also have no experiences with partial differential equations. So we have to wait until someone comes with more knowledge about it and can make use of this equation.
I think this is easier derived via the standard methods of regular iteration at the fixed point.
Ansus, Taylor series are not the only allowed tools in computation of tetration! (Eg. would you object Dmitrii's Cauchy-integral computation only because it doesnt contain Taylor series?)
thanks for sharing your ideas.
(04/24/2010, 09:01 PM)kobi_78 Wrote: \( y' = f'_{\alpha}(x) = \left( \ln{a} \right)^\alpha \cdot \prod_{k=1}^{\alpha}{f_{k}(x)} = \left( \ln{a} \right)^\alpha \cdot b^{\alpha} \cdot \prod_{k=1}^{\infty}{\frac{f_{k}(x)}{f_{k+\alpha}(x)}} \)
The formula looks very interesting, however it seems only useful with the natural continuum product which you describe, which limits the base range to \( <e^{1/e} \). It seems indeed to yield regular iteration tetration, a nice finding.
For the case \( b>e^{1/e} \) we however lack an expression of \( f_k \) continuous in \( k \) (and if we had that then we would already have tetration). So we have no possibility to compute the continuum product in an alternate way, which however would be necessary for the case \( b>e^{1/e} \).
Quote:The second one is the PDE of \( f_{\alpha}(x) \) which you can see here. I will explain how I discovered it later.
Actually I also have no experiences with partial differential equations. So we have to wait until someone comes with more knowledge about it and can make use of this equation.
Quote:The third idea,
....
\( f'_{\alpha}(b) = \left( \ln{a} \right)^\alpha \cdot \prod_{k=1}^{\alpha}{f_{k}(b)} = \left( \ln{a} \right)^\alpha \cdot \prod_{k=1}^{\alpha}{b} = \left( \ln{a} \right)^\alpha \cdot b^{\alpha} \)
\( \frac{g'_{n}(x)}{g_{n}(x)} = \ln{a} \cdot \sum_{k = 1}^{n} { g_{k - 1}(x) } \)
\( f''_{\alpha}(b) = g'_{\alpha}(b) = g_{\alpha}(b) \cdot \ln{a} \cdot \sum_{k = 1}^{\alpha} { g_{k - 1}(b) } = \left( \ln{a} \right)^\alpha \cdot b^{\alpha} \ln{a} \cdot \sum_{k = 1}^{\alpha} { \left( \ln{a} \right)^{k - 1} \cdot b^{k - 1} } = \left( \ln{a} \right)^\alpha \cdot b^{\alpha} \cdot \ln{a} \cdot \frac{\left( \ln{a} \right)^{\alpha} \cdot b^{\alpha} - 1}{ \ln{a} \cdot b - 1} \)
I think we can continue in a similar fashion and find values in higher derivatives order, but I am not sure.
I think this is easier derived via the standard methods of regular iteration at the fixed point.
(04/25/2010, 09:34 AM)Ansus Wrote: Oh in that case if you obtain derivatives f'(x), f''(x) etc and build a Taylor series for \( f_\alpha(x) \), this would not help to find \( f'_x(1) \), \( f''_x(1) \) etc (derivatives by x) which are necessary to build Taylor series for tetration...
Ansus, Taylor series are not the only allowed tools in computation of tetration! (Eg. would you object Dmitrii's Cauchy-integral computation only because it doesnt contain Taylor series?)
