Asymptotic Rate of Growth

Question

How can I calculate the Asymptotic Rate of growth of a function, for instance like:

$X^3 - X^2 - X -1$

EDIT:

For instance, as you can see in this graph, after the 1200 the function approximates to the limit. I want to if there is a easy way to calculate the rate of grow, after 1200 for instance.

enter image description here

Edit

I'm trying to find a generalized way in order to graphically find it for Fibonacci, Tribonacci, Tetranacci sequences

Are you asking for a generalized way of determining which term, e.g. x^3, x^2, x, or 1 grows fastest as x increases? — Eric Brown, Oct 08 '12 at 01:14
For polynomials, this answer or this answer might be what you're looking for. Possible duplicate — rm -rf, Oct 08 '12 at 01:17
I'm trying to find a generalized way in order to graphically find it for fibonacci, tribonacci, tetranacci sequences — whynot, Oct 08 '12 at 01:30
It doesn't happen all of a sudden at 1200... it's just that the plot range is such that it seems it is that way. As a hint, look at LogPlot[Fibonacci[n], {n, 0, 1500}] — rm -rf, Oct 08 '12 at 01:48
Related http://mathoverflow.net/questions/47140/characteristic-polynomials-for-k-bonacci-numbers-whats-their-name — Dr. belisarius, Oct 08 '12 at 02:46
All the $n$-nacci sequences have exponential behavior; the base of their dominant exponential term is (expressed in Mathematica notation) Root[x^n - Sum[x^k, {k, 0, n - 1}], Mod[n, 2, 1]]. — J. M.'s missing motivation, Oct 08 '12 at 03:23

Sjoerd C. de Vries · Answer 1 · 2014-01-10T22:03:42.140

3

As J.M. mentioned in the comments:

All the n-nacci sequences have exponential behavior; the base of their dominant exponential term is (expressed in Mathematica notation)

Root[x^n - Sum[x^k, {k, 0, n - 1}], Mod[n, 2, 1]]

Demonstration:

With[{n = 2},
 base = Root[x^n - Sum[x^k, {k, 0, n - 1}], Mod[n, 2, 1]] // N;
 DiscretePlot[Fibonacci[x]/base^x, {x, 1, 40}, PlotRange -> {0, 0.7}]
 ]

Mathematica graphics

Daniel Lichtblau adds that Mathematica is able to calculate the Fibonacci series development at infinity:

Simplify[Normal[Series[FunctionExpand[Fibonacci[x]], {x,Infinity,2}]], 
         Assumptions->Element[x,Integers]]

Mathematica graphics

edited Jan 10 '14 at 22:03

answered Jan 09 '14 at 22:31

Sjoerd C. de Vries

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Could take the series at infinity for cases like Fibonacci, where we have a closed form. This gives some exponential stuff: Simplify[Normal[Series[FunctionExpand[Fibonacci[x]], {x,Infinity,2}]], Assumptions->Element[x,Integers]] – Daniel Lichtblau Jan 09 '14 at 23:40
PS Thanks for taking so much time to respond to several unanswered posts. I noticed you've handled quite a few in the past day or two. (Maybe others before that, but I wasn't noticing then.) – Daniel Lichtblau Jan 09 '14 at 23:42
@DanielLichtblau Thanks! I had some spare time which coincided nicely with rm-rf's clean-up call. – Sjoerd C. de Vries Jan 10 '14 at 22:05

Asymptotic Rate of Growth

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