I was thinking about (real) interpolation of the tribonacci sequence.
Of course there is the real part of the binet like analogue.
But I do not want that solution or at least I want to arrive at it in a different way.
Instead of considering asymptotics , positive derivatives , fake function theory and the recent alike stuff , I was more intrested in using more "classical" stuff.
Considering that a recursion is close to an iteration I got the idea to use superfunctions.
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I must note however that a continuum sum might also be usefull here !!
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But then I encountered a problem.
It can probably be fixed though.
This is the idea I had and the trouble I encountered ;
trib(x) = trib(x-1) + trib(x-2) + trib(x-3)
Define rat(x) = trib(x+1)/trib(x).
Now
rat(x+1) = ( trib(x+1) + trib(x) + trib(x-1) ) / trib(x+1)
= 1 + (trib(x) + trib(x-1))/trib(x+1)
= 1 + 1/rat(x-1) + rat(x-2)/rat(x-1)
This looks familiar ...
Somos , Fibonacci ... hmm.
( still thinking , might edit )
---
Notice the analogue for Fibonacci works : we get
ratfibo(x+1) = 1 + 1/ratfibo(x) which has the golden mean and 1 - golden mean as its fixpoints.
Im trying to get the tribonacci constant as a fixpoint here ...
---
regards
tommy1729
Of course there is the real part of the binet like analogue.
But I do not want that solution or at least I want to arrive at it in a different way.
Instead of considering asymptotics , positive derivatives , fake function theory and the recent alike stuff , I was more intrested in using more "classical" stuff.
Considering that a recursion is close to an iteration I got the idea to use superfunctions.
---
I must note however that a continuum sum might also be usefull here !!
---
But then I encountered a problem.
It can probably be fixed though.
This is the idea I had and the trouble I encountered ;
trib(x) = trib(x-1) + trib(x-2) + trib(x-3)
Define rat(x) = trib(x+1)/trib(x).
Now
rat(x+1) = ( trib(x+1) + trib(x) + trib(x-1) ) / trib(x+1)
= 1 + (trib(x) + trib(x-1))/trib(x+1)
= 1 + 1/rat(x-1) + rat(x-2)/rat(x-1)
This looks familiar ...
Somos , Fibonacci ... hmm.
( still thinking , might edit )
---
Notice the analogue for Fibonacci works : we get
ratfibo(x+1) = 1 + 1/ratfibo(x) which has the golden mean and 1 - golden mean as its fixpoints.
Im trying to get the tribonacci constant as a fixpoint here ...
---
regards
tommy1729