Respuesta :
Answer:
Iteration 1: [tex]x_{2}=1.446[/tex]
Iteration 2: [tex]x_{3}=1.337[/tex]
Step-by-step explanation:
Formula for Newton's method is,
[tex]x_{n+1}=x_n-\dfrac{f\left(x_n\right)}{f'\left(x_n\right)}[/tex]
Given the initial guess as [tex]x_{1}=1.6[/tex], therefore value of n = 1.
Also, [tex]f\left(x\right)=x^{9}-9[/tex].
Differentiating with respect to x,
[tex]\dfrac{d}{dx}\left(f\left(x\right)\right)=\frac{d}{dx}\left(x^9-9\right)[/tex]
Applying difference rule of derivative,
[tex]\dfrac{d}{dx}\left(f\left(x\right)\right)=\dfrac{d}{dx}\left(x^9\right)-\dfrac{d}{dx}\left(9\right)[/tex]
Applying power rule and constant rule of derivative,
[tex]\dfrac{d}{dx}\left(f\left(x\right)\right)=\left(9x^{9-1}\right)-0[/tex]
[tex]\dfrac{d}{dx}\left(f\left(x\right)\right)=9x^{8}[/tex]
Substituting the value,
[tex]x_{1+1}=x_1-\dfrac{f\left(x_1\right)}{f'\left(x_1\right)}[/tex]
[tex]x_{2}=1.6-\dfrac{f\left(1.6\right)}{f'\left(1.6\right)}[/tex]
Calculating the value of [tex]f\left(1.6\right)[/tex] and [tex]f'\left(1.6\right)[/tex]
Calculating [tex]f\left(1.6\right)[/tex]
[tex]f\left(1.6\right)=\left(1.6\right)^{9}-9[/tex]
[tex]f\left(1.6\right)=59.71947674[/tex]
Calculating [tex]f'\left(1.6\right)[/tex],
[tex]f'\left(1.6\right)=9\left(1.6\right)^{8}[/tex]
[tex]f'\left(1.6\right)=386.5470566[/tex]
Substituting the value,
[tex]x_{2}=1.6-\dfrac{59.71947674}{386.5470566}[/tex]
[tex]x_{2}=1.446[/tex]
Therefore value after second iteration is [tex]x_{2}=1.446[/tex]
Now use [tex]x_{2}=1.446[/tex] as the next value to calculate second iteration. Here n = 2
Therefore,
[tex]x_{2+1}=x_2-\dfrac{f\left(x_2\right)}{f'\left(x_2\right)}[/tex]
[tex]x_{3}=1.446-\dfrac{f\left(1.446\right)}{f'\left(1.446\right)}[/tex]
Calculating the value of [tex]f\left(1.446\right)[/tex] and [tex]f'\left(1.446\right)[/tex]
Calculating [tex]f\left(1.446\right)[/tex]
[tex]f\left(1.446\right)=\left(1.446\right)^{9}-9[/tex]
[tex]f\left(1.446\right)=18.63851065[/tex]
Calculating [tex]f'\left(1.446\right)[/tex],
[tex]f\left(1.446\right)=9\left(1.446\right)^{8}[/tex]
[tex]f\left(1.446\right)=172.0239252[/tex]
Substituting the value,
[tex]x_{3}=1.446-\dfrac{18.63851065}{172.0239252}[/tex]
[tex]x_{3}=1.337[/tex]
Therefore value after second iteration is [tex]x_{3}=1.337[/tex]
