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 Derivative of E tetra x mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/12/2011, 09:56 PM (11/12/2011, 04:24 PM)Forehead Wrote: As a result, E^P == TE[x + 1]*P Now, we rearrange the equation a bit. Log[E^P] == Log[TE[x + 1]]*Log[P] I think this is the error. In going from the first to the second equation, it looks like you took, on the right hand side, $\log(ab) = \log(a) \log(b)$. But that is wrong. Instead, $\log(ab) = \log(a) + \log(b)$ and so your second equation should be Log[E^P] == Log[TE[x + 1]] + Log[P] If we continue your steps with this corrected equation, we get P == TE[x] + Log[P] E^P == E^(TE[x] + Log[P]) E^P == E^TE[x] E^Log[P] E^P == TE[x+1] P TE[x+1] P == TE[x+1] P a tautological equation. Though perhaps you could solve for P in the first equation via the Lambert function? « Next Oldest | Next Newest »

 Messages In This Thread Derivative of E tetra x - by Forehead - 11/12/2011, 07:08 AM RE: Derivative of E tetra x - by Gottfried - 11/12/2011, 09:27 AM RE: Derivative of E tetra x - by Forehead - 11/12/2011, 04:24 PM RE: Derivative of E tetra x - by Gottfried - 11/12/2011, 05:07 PM RE: Derivative of E tetra x - by mike3 - 11/12/2011, 09:56 PM RE: Derivative of E tetra x - by mike3 - 11/12/2011, 10:02 PM RE: Derivative of E tetra x - by Forehead - 11/13/2011, 04:21 PM RE: Derivative of E tetra x - by andydude - 12/25/2015, 03:59 AM

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