09/08/2014, 10:37 AM

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 )

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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