Can anyone explain if the Fibonnacci sequence can be placed into a parabolic or hyperbolic function?
If you're asking if there's a function such that the graph of is a parabola or hyperbola,
and that satistfies, for all positive integers , the Fibonnacci number, then the answer is no.
Parabolas and hyperbolas only have a few "free parameters", and the requirement would quickly exhaust them.
It would be possible that it algebraically worked out, but it doesn't. I'll do the parabola case, and leave to you the hyperbola ( ).
If a parabola is the graph of a function that satisfies , then let for some .
Then , so .
Thus , so
, so
The only way that could be true for all n is if each of those n-coefficients was 0. Thus .
That system of equations (3 equations in the 3 unknowns , and ) has only one solution: .
That means the only quadratic that has a prayer of working is the 0 quadratic .
But that obviously doesn't satisfy , and it was the only one that had a chance.
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I suppose I should note that even if you consider rotated parabolas/hyperbolas, you're still going to quickly run out of free parameters.
(The standard equation for the graph of a generic conic section in the plane has 6 free parameters. That's more than the 3 a quadratic
polynomial has, but it's still less than the infinity required.).