Leonard of Pise - Fibonacci
(1180 - 1250)
Rather surprising results concerning Fibonacci's suite
*U*_{0} and *U*_{1} being strictly positive, (if equal to 1, it is the Fibonacci's suite !),
With the golden number,
Slice of life:
Fibonacci was born in Pise in *1180*. Son of a toscan
storekeeper (Bonaccio, from whichthe nickname), it is brought to travel much, notably
in the middle East. Fascinated by the Arabic numbering that he discovers, he introduces
it into the Westerner world by drawing up a work of explanation on his return *(Liber
abaci)*. This allows him to study more easily the equations of order *1* and *2* and, as it
is seen in the historic of records, to calculate some decimal of *Pi*!
Fibonacci remained very famous thanks to its series and to the fact that he is almost
the only Westerner mathematician of talent at this time.
The series was to resolve supposed problem according to:
*One considers a couple of rabbits which reproduces. How much we shall obtain from
couples of rabbits after a number given of month knowing that every couple produces
every month a new couple, which becomes productive only after two months (on *1202*, according
to *mathematicians of A in Z*, see Biblio)
About
Fibonacci never obviously wrote the previous result,
but his series and its derivatives are in the heart of this formula, and he began
a calculation of the decimal of Pi ( *3,1418* ); the interest for this constant deserved anyway a place
on this site!
Proof
it is essentially geometrical...
Given a recurring series of the type: *U*_{n}=U_{n-1}+U_{n-2}
*4* consecutive terms of this series are considered : *U*_{n},
U_{n+1}, U_{n+2}, U_{n+3}
Given *6* acute angles *ß, ß', ß",
µ, µ', µ"* contents
in the following figure, the rectangle of sides *U*_{n+3}=U_{n+2}+U_{n+1} and *U*_{n+2}=U_{n+1}+U_{n}
One has according to the plan* OM=ON* because *[OM] *and* [ON]* is the hypotenuses right-angled triangle of base *U*_{n+2} and height *U*_{n+1}.
The hatched triangles are the same by rotation of angle */2* so *µ"=µ'* now *ß"+µ'=/2* so *ß"+µ"=/2* and the triangle is isosceles rectangle in *O* (check, all
right, it is not flagrant on the picture!)
The other internal angles of the triangle have so for measure */4* so *µ-ß=/4* and *ß'+µ'=/4*
Now *tan(ß)=U*_{n }/ U_{n+3}, *tan(µ)=U*_{n+2}/U_{n+1}
and *tan(ß')=U*_{n} / U_{n+3}, *tan(µ')=U*_{n+1 }/ U_{n+2}
Where from */4=Arctan(U*_{n+2 }/ U_{n+1})-Arctan(U_{n }/ U_{n+3})
and */4=Arctan(U*_{n+1}/U_{n+2})+Arctan(U_{n}/U_{n+3})
Fibonacci's series being increasing, second equation is a sum of arctan numbers *1*. By applying development
limited of arctan in *0* (who is valid until *1*) in the second of the previous two equalities, we obtain
:
Even stronger!
Each knows that quotient *U*_{n+1}/U_{n} tends towards the golden section =. So it is easy to see that *U*_{n}/U_{n+3} _{ }tends towards *1/*^{3}. In passing in the limit for n in the last sum, one has
so :
After *e* and , here
is that and are
fervently linked !
Trials
We kwon very well the linear convergence of series
of Arctan, let us give rather some of the equalities that one can obtain thanks to
this formula :
*/4=Arctan(U*_{n+1}/U_{n+2})+Arctan(U_{n}/U_{n+3}) below noted *U*_{n+1}/U_{n+2}+U_{n}/U_{n+3} by simplicity !
*n=0* |
*1/2+1/3 *(Formule Euler's formula) |
*n=1* |
*2/3+1/5* |
*n=2* |
*3/5+1/4* |
*n=3* |
*5/8+3/13* |
*n=4* |
*8/13+5/21* |
*n=5* |
*13/21+4/17* |
*n=6* |
*21/34+13/55* |
*n=10* |
*144/233+89/377* |
One obtains a perfectly countable set of series converging towards ...
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