Taller Matemáticas y Plantas
Guanajuato, Julio 2016 - Christophe Golé (Smith College)

Calcular la sequencia de Fibonacci

Hay multiples maneras, mas o menos eficientes

F[1]=1;
F[2]=1;
F[n_]:= F[n-1]+F[n-2]

F[10]

Clear[a]
RecurrenceTable[{a[n+2]==a[n+1]+a[n], a[1]==1, a[2]== 1}, a, {n,1,10}]

Fibonacci[10]

Numero y ángulo de oro

GoldenRatio//N

(1/2) (1+(Sqrt[5]))//N

Simplify[Limit[Fibonacci[n+2]/Fibonacci[n], n->Infinity]*goldenangle]

goldenangle=Limit[Fibonacci[n-2]/Fibonacci[n], n->Infinity]

El ángulo de oro es = 1/(ϕ^2)

goldenangle==1/GoldenRatio^2

goldenangle*(GoldenRatio)^2//N

N[360*(1/2) (3-(Sqrt[5])), 20]

137.50776405003785`

ContinuedFraction[goldenangle]

Convergents[goldenangle, 15]

Convergents[{0,2,{1}}]

LucasL[8]
LucasL[6]
ContinuedFraction[LucasL[8]/LucasL[6],6]
Convergents[LucasL[8]/LucasL[6],6]

Length[%]

13*18-5*47

FromContinuedFraction[{2,1,1,1,1}]

Fibonacci[7]

360Limit[Fibonacci[n-1]/LucasL[n], n->Infinity]//N

Números complejos

(2+3I )(-1 + 5 I)

Solve[58-8 x+2 (x^2)==0, x]

Formula de Euler (e^(iθ))= cos(θ) + i sin(θ)

es justificada por las “series de Taylor” de las funciónes en ambos lados:

Series[Cos[x], {x,0, 10}]
Series[Sin[x], {x,0, 10}]
Series[Cos[x],  {x,0, 10}]+I Series[Sin[x], {x,0, 10}]
Series[Exp[I x], {x,0, 10}]

Números irracionales y sus approximaciónes racionales

N[Pi, 10]

ContinuedFraction[Pi, 10]

FromContinuedFraction[{3,7,15,1,292,1,1,1,2,1}]

Convergents[Pi, 10]

error = Convergents[Pi, 20]-Pi//N

hur[q_]:=1/(Sqrt[5]q^2)//N

h=Map[hur, Denominator[Convergents[Pi, 20]]];
errorpi= Abs[error/h]

Para GoldenRatio

ContinuedFraction[GoldenRatio, 20]

Convergents[GoldenRatio, 20]

error = Convergents[GoldenRatio, 20]-GoldenRatio//N

hur[q_]:=1/(Sqrt[5]q^2)//N

h=Map[hur, Denominator[Convergents[GoldenRatio, 20]]]

errorphi=Abs[error/h]

Show[ListLinePlot[errorpi, PlotRange->All], ListLinePlot[errorphi,  PlotRange->All, PlotStyle->Red]]