Tarea n° 1 metodos numericos
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The document describes calculating the root of the function F(x) = e^-x - x using the Newton-Raphson method. It runs the calculation iteratively until the relative error is less than 10^-8. After 3 iterations, a root of 0.5671 is found with a relative error of 0.000, satisfying the error criterion.
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![TAREA N° 1: METODO NEWTON - RAPSON
Sea la función F(x) = ℮−𝑥
− 𝑋, con valor inicial 𝑋0 = 0; calcular error relativo: Er<10−8
.
Para n=0
𝑋1 = 0 −
𝑒−0−0
−𝑒−0−1
𝑋1 = 0.5000
Er = [
0−0.5000
0.5000
] 𝑥 100 = 100%
Para n=1
𝑋2 = 0.5 −
𝑒−0.5−0.5
−𝑒−0.5−1
𝑋2 = 0.5663
Er = [
0.5−0.5663
0.5663
] 𝑥 100 = 11,70%
Para n=2
𝑋2 = 0.5663 −
𝑒−0.5663−0.5663
−𝑒−0.5663−1
𝑋3 = 0.5671
Er = [
0.5663−0.5671
0.5671
] 𝑥 100 = 0,14%
Para n=3
𝑋2 = 0.5671 −
𝑒−0.5671−0.5671
−𝑒−0.5671−1
𝑋4 = 0.5671
Er = [
0.5671−0.5671
0.5671
] 𝑥 100 = 0.000](https://image.slidesharecdn.com/tarean1metodosnumericos-170516204740/85/Tarea-n-1-metodos-numericos-1-320.jpg)
Tarea n° 1 metodos numericos
- 1. TAREA N° 1: METODO NEWTON - RAPSON Sea la función F(x) = ℮−𝑥 − 𝑋, con valor inicial 𝑋0 = 0; calcular error relativo: Er<10−8 . Para n=0 𝑋1 = 0 − 𝑒−0−0 −𝑒−0−1 𝑋1 = 0.5000 Er = [ 0−0.5000 0.5000 ] 𝑥 100 = 100% Para n=1 𝑋2 = 0.5 − 𝑒−0.5−0.5 −𝑒−0.5−1 𝑋2 = 0.5663 Er = [ 0.5−0.5663 0.5663 ] 𝑥 100 = 11,70% Para n=2 𝑋2 = 0.5663 − 𝑒−0.5663−0.5663 −𝑒−0.5663−1 𝑋3 = 0.5671 Er = [ 0.5663−0.5671 0.5671 ] 𝑥 100 = 0,14% Para n=3 𝑋2 = 0.5671 − 𝑒−0.5671−0.5671 −𝑒−0.5671−1 𝑋4 = 0.5671 Er = [ 0.5671−0.5671 0.5671 ] 𝑥 100 = 0.000