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Hi i am new here
I have a question concerning the number 2.7767750400967...
what is 2.7767750400967^^1/2 ?
Some background on the number in case you did not know:
This number satisfies the equation (x1)^^2 =x or more importantly that its super square root is 1.7767750400967...
Also 1.7767750400967... satisfies several interesting relations
1. (x^^2)^^2 = x^^4
2. x^^2  x^^1  x^^0 = 0
(note that the second equation is like the tetration equivalent of the equation for the golden ratio)
hopfully you can help me solve this
lastly I kinda cheated in finding the value 2.7767750400967 so I dont know if it some other equation that has been simplified.
Thanks
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(10/22/2009, 12:12 AM)dantheman163 Wrote: what is 2.7767750400967^^1/2 ?
Using intuitive/natural iteration, my estimate would be about 1.65955
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While I was doing this, I thought it might be nice to graph x^^1/2, so I did. The black line is x^^1/2, the blue line is the real part and the red line is the imaginary part. Also, the green line is 2superroot to show that it is a slightly different function.
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(10/22/2009, 07:31 AM)andydude Wrote: While I was doing this, I thought it might be nice to graph x^^1/2, so I did. The black line is x^^1/2, the blue line is the real part and the red line is the imaginary part. Also, the green line is 2superroot to show that it is a slightly different function.
Thank you very much for the graph and the value, they are very usefull.
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If you liked that, you might enjoy this graph too. This is x^^(1/2), same color scheme.
It appears that the limit which I was surprised to see. It really is true that a picture is worth 1000 words.
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I dont want to weaken your engagement.
But everybody seems already to have forgotten that we dont have "the" tetration.
So whenever you write x^^t for some noninteger t, you have to add which tetration you mean. (Really, guys, how often did I repeat that already?!)
Andrew, can you reformulate your conjecture accordingly? (Otherwise I just dont know what you mean. Do you mean "for all tetrations", or do you mean a particular tetration or do you mean a certain class of tetrations?)
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(10/24/2009, 09:09 AM)bo198214 Wrote: Andrew, can you reformulate your conjecture accordingly?
OK, done. I haven't forgotten that there are many tetrations, you don't need to remind me, in fact I have reminded others on several occasions.
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(10/25/2009, 07:47 AM)andydude Wrote: OK, done. I haven't forgotten that there are many tetrations, you don't need to remind me, in fact I have reminded others on several occasions.
Thank you. Yes, actually I saw that you described it in your recent posts. *takes his blame back*
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(10/23/2009, 01:18 AM)andydude Wrote: If you liked that, you might enjoy this graph too. This is x^^(1/2), same color scheme.
It appears that the limit which I was surprised to see. It really is true that a picture is worth 1000 words.
this is an old post , but has this been explained already ?
maybe its just that x^^y = 1 for all x = 1.
or is that too simple ?
i really feel that  after being some time on the forum  the most intresting stuff , closed form but unproven , are almost always limits.
limits are gaining territory in tetration.
one could almost split up the forum into
matrix
contour integral
sums and products
limits
are all solutions to tetration consistant with x^^y = 1 for all x = 1 ?
andrew's method is not ?
( my own method is designed for bases > eta )
regards
tommy1729
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(04/25/2010, 03:19 PM)tommy1729 Wrote: are all solutions to tetration consistant with x^^y = 1 for all x = 1 ?
There is one point that is different. (1^^1) is undefined/indeterminate.
This is because we have a rule that states (x^^1 == 0) for (almost) all x, and (1^^y == 1) for (almost) all y, and since both of them cannot be true, we leave it as undefined so that there is no conflict.
Also, to clarify what surprised me (I didn't say the right thing before), I wasn't so surprised about the limit to 1 as I was the limit to 0. Granted, the graph stops way before it gets to zero, but the indication is that it seems to have a limit of zero as well. This is interesting because (0^^y) is 0 for odd y and 1 for even y, and _undefined_ for all other y. This limittozero may provide an insight that will allow (0^^y) to be defined for noninteger y.
Andrew Robbins
