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exp^(1/2)[ln^(1/2)[x]+1] - x

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exp^(1/2)[ln^(1/2)[x]+1] - x
tommy1729 Offline
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#4
12/14/2012, 11:49 PM
It might be a mistake. further numerical , statistical and theoretical ideas are not supportive.

But maybe with other functions ...

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tommy1729
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Messages In This Thread
exp^(1/2)[ln^(1/2)[x]+1] - x - by tommy1729 - 12/04/2012, 12:07 AM
RE: exp^(1/2)[ln^(1/2)[x]+1] - x - by tommy1729 - 12/08/2012, 09:22 PM
RE: exp^(1/2)[ln^(1/2)[x]+1] - x - by tommy1729 - 12/08/2012, 09:32 PM
RE: exp^(1/2)[ln^(1/2)[x]+1] - x - by tommy1729 - 12/14/2012, 11:49 PM



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