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 Actual formulas for tetration and its derivative andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 08/31/2007, 07:59 PM Oh, right. Thanks. All the magic is gone now... As for my information, I use logic and intuition; apparently both are on vacation today. Sorry about that. Andrew Robbins bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 08/31/2007, 08:05 PM andydude Wrote:Oh, right. Thanks. All the magic is gone now... That was not my intention, perhaps it helps you that the first formula still has the same magic to me, as the second had to you GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 08/31/2007, 10:21 PM Well, unfortunately, no magic at all. Formulas 1 and 2 can be easily obtained, taking into account the hyperoperational properties of tetration, e.g..: e#(x-1) = ln(e#x) as well as the definition of the product logarithm: if x . e^x = z then: x = ProductLog[z]. However, it is interesting to remind that the product logarithm is a complex function, with two real branches. I shall come back to it. Please see the attached short pdf comment. I am too lazy to insert it in this text. GFR Attached Files   No magic.pdf (Size: 5.96 KB / Downloads: 649) bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 08/31/2007, 10:26 PM Hm, Gianfranco seems still to be a bit uncomfortable with the tex formulas. So it shall be forgiven that it came way to late, when everything was already solved jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/31/2007, 11:29 PM (This post was last modified: 08/31/2007, 11:31 PM by jaydfox.) jaydfox Wrote:I've come up with these forumulae as well, though in a different form. They follow from basic principles and are "ignorant" of the underlying solution (i.e., the function for the critical interval doesn't matter). To start with, let's bear in mind the following identity: $T(b, x) = \exp_b(T(b, x-1))$ From this starting point, let's compute the first derivative using the chain rule: $ \begin{eqnarray} T'(b, x) & = & \text{D}_x \left(T(b, x)\right) \\ & = & \text{D}_x \left(\exp_b(T(b, x-1))\right) \\ & = & \exp_b(T(b, x-1))\text{D}_x\left(T(b, x-1)\right) \\ & = & T(b, x)\text{D}_x\left(T(b, x-1)\right) \\ & = & T(b, x)T'(b, x-1) \end{eqnarray}$ From here, we simply rearrange to get $T(b, x) = \frac{T'(b, x)}{T'(b, x-1)}$. As you can see, this is an identity, so there's no need to compute numerically to verify. If for some reason a particular solution doesn't satisfy this condition, then we can be sure it either doesn't satisfy the iterative exponential property, or it some other problem such as not being at least twice differentiable. I haven't tried with the W function, but I would assume that the process is similar.Oops. I wrote a subscript b, as though the base were arbitrary, but then I went ahead and assumed base e when I was solving. Amateur mistake. Let's try again: $ \begin{eqnarray} T'(b, x) & = & \text{D}_x \left(T(b, x)\right) \\ & = & \text{D}_x \left(\exp_b(T(b, x-1))\right) \\ & = & \ln(b)\exp_b(T(b, x-1))\text{D}_x\left(T(b, x-1)\right) \\ & = & \ln(b)T(b, x)\text{D}_x\left(T(b, x-1)\right) \\ & = & \ln(b)T(b, x)T'(b, x-1) \end{eqnarray}$ As before, we simply rearrange to get $T(b, x) = \frac{T'(b, x)}{T'(b, x-1)\ln(b)}$. Again, as before, we have a recurrence relation we can leverage: $ \begin{eqnarray} T'(b, x) & = & \ln(b)T(b, x)T'(b, x-1) \\ & = & \left(\ln(b)\right)^{2}T(b, x)T(b, x-1)T'(b, x-2) \\ & = & \left(\ln(b)\right)^{3}T(b, x)T(b, x-1)T(b, x-2)T'(b, x-3) \end{eqnarray}$ Et cetera. ~ Jay Daniel Fox « Next Oldest | Next Newest »

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