método de la secante
TRANSCRIPT
![Page 1: Método de La Secante](https://reader036.vdocuments.pub/reader036/viewer/2022083018/577c7f8b1a28abe054a5095e/html5/thumbnails/1.jpg)
MÉTODO DE LA SECANTE
Usar el método de la secante para aproximar la raíz de:
Comenzando con x0=0 y x1=1, y hasta que |Ea|<1%
Usamos la fórmula:
Solución:
Primera Iteración
X0=0 F(x0) =1 X1=1 F(x1) = -0.632120559
X2 = X1 -F (x 1)(X1−X 0)F ( x 1 )−F (x0)
=1-−0.632120559(1−0)−0.632120559−1
=0.612699837
F(x2) =0.074313832
E(a2)=[0.612699837−10.612699837
X 100% ]
E(a2)= 63.21205588
Segunda Iteración
X1=1 F(x1) =-0.632120559 X2=0.612699837 F(x2) = 0.074313832
X3 = X2 - F (x 2)(X2−X 1)F (x 2 )−F (x1)
![Page 2: Método de La Secante](https://reader036.vdocuments.pub/reader036/viewer/2022083018/577c7f8b1a28abe054a5095e/html5/thumbnails/2.jpg)
=0.612699837 - 0.074313832(0.612699837−1)0.074313832−(−0.632120559)
=0.653442133%
F(x3) = -0.000969852
E(a3)=[0.653442133−0.612699837
0.653442133X100% ]
E(a3)= 6.235027469%
Tercera Iteración
X2=0.612699837 F(x2) =0.074313832 X3=0.653442133 F(x3) = -0.000969852
X4 = X3 - F (x 3)(X 3−X2)F (x 3 )−F (x 2)
=0.653442133- −0.000969852(0.653442133−0.612699837)
−0.000969852−0.074313832
=0.652917265
F(x4) =
0.000002548
E(a4)=[0.652917265−0.653442133
0.652917265X 100% ]
E(a4)= 0.080388139%
n Xn Fxn ea0 0 1 01 1 -0.632120559 1002 0.612699837 0.074313832 63.212055883 0.653442133 -0.000969852 6.2350274694 0.652917265 0.000002548 0.080388139
![Page 3: Método de La Secante](https://reader036.vdocuments.pub/reader036/viewer/2022083018/577c7f8b1a28abe054a5095e/html5/thumbnails/3.jpg)
Por tanto, concluimos que la aproximación de la raíz es x4=0.652917265 con un |Ea|=0.080388139%