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Derivative of E tetra x
#5
(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, . But that is wrong. Instead, 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?
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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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