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 Just asking... bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 07/16/2008, 10:36 PM martin Wrote:Sorry, I thought you already considered this type of formula or made a quick check about derivatives and suchNo, absolutely not. New formula, new luck Quote: ... anyway, my "function" here is continuous and differentiable (as far as I understand what that means), but not infinitely often differentiable. The 3rd derivative has a slight peak near the integer points that I didn't manage to wipe away with a higher accuracy of my "parameter" 0.345627(2-x), or maybe Excel is just too inaccurate there. I've tried similar formulae before, but never got this close (e.g. had this peak already in the 2nd derivative). I really wonder how you actually came to *this* formula. If you are after finite differentiability, one rather would consider polynomials. With a polynomial of degree $d$ you can always get a $d$ times continuously differentiable function. One just have to determine numerically the coefficients. For example for $d=2$ let $p(x)=a+bx+cx^2$ the polynomial. We have to satisfy $p^{(k)}(1)=\left(u^p\right)^{(k)}(0)$, for $k=0,1,2$, where $u$ is the base of our tetration. So well for $k=0$: (*) $a+b+c = u^a$ then $k=1$ $(b+2cx)|_{x=1} = u^p \ln(u) (b+2cx) |_{x=0}$ (**) $b+2c = u^a \ln(u) b$ and now $k=2$ (***) $2c = u^p (\ln(u)(b+2cx))^2 + u^p \ln(u) 2c|_{x=0} = u^a (\ln(u)b)^2 + u^a \ln(u) 2c$ So these are 3 equations with 3 variables $a$, $b$, $c$ you can solve them numerically and voila you obtain a polynomial which gives a 2-times differentiable tetration. Quote:I had in mind that this might be a formula, even if it only gives approximated values, that is not too complicated to calculate with. (Okay, I still would need to find a way to get a general formula to substitute this 0.345627-thingy for other bases.) Or, if there is a way to get that peak out of the 3rd derivative and eventually make it infinitely differentiable, whether the results are congruent to already existing results or they aren't. Polynomials are not complicated to calculate with either, are they? « Next Oldest | Next Newest »