f ( x ) = e x + x − 4 f(x)=e^x+x-4 f ( x ) = e x + x − 4
f ′ ( x ) = e x + 1 f'(x)=e^x+1 f ′ ( x ) = e x + 1
x n + 1 = x n − f ( x n ) f ′ ( x n ) x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)} x n + 1 = x n − f ′ ( x n ) f ( x n ) Initial solution x 0 = 1 x_0 =1 x 0 = 1
n x n f ( x n ) x n + 1 0 1 − 0.281718 1.075766 1 1.075766 0.008003 1.073729 2 1.073729 0.000006 1.073729 \begin{matrix}
n & x_n & f(x_n) & x_{n+1} \\
0 & 1 & -0.281718 & 1.075766\\
1 & 1.075766 & 0.008003 & 1.073729 \\
2 & 1.073729 & 0.000006 & 1.073729 \\
\end{matrix} n 0 1 2 x n 1 1.075766 1.073729 f ( x n ) − 0.281718 0.008003 0.000006 x n + 1 1.075766 1.073729 1.073729
ε = ∣ x n + 1 − x n x n ∣ ⋅ 100 % \varepsilon =\big|\dfrac{x_{n+1}-x_n}{x_n}\big|\cdot 100\% ε = ∣ ∣ x n x n + 1 − x n ∣ ∣ ⋅ 100%
ε = ∣ 1.075766 − 1 1 ∣ ⋅ 100 % = 7.5766 % \varepsilon =\big|\dfrac{1.075766-1}{1}\big|\cdot 100\%=7.5766\% ε = ∣ ∣ 1 1.075766 − 1 ∣ ∣ ⋅ 100% = 7.5766%
ε = ∣ 1.073729 − 1.075766 1.075766 ∣ ⋅ 100 % = 0.1894 % \varepsilon =\big|\dfrac{1.073729-1.075766}{1.075766}\big|\cdot 100\%=0.1894\% ε = ∣ ∣ 1.075766 1.073729 − 1.075766 ∣ ∣ ⋅ 100% = 0.1894%
ε = ∣ 1.073729 − 1.073729 1.073729 ∣ ⋅ 100 % = 0.000000 % \varepsilon =\big|\dfrac{1.073729-1.073729}{1.073729}\big|\cdot 100\%=0.000000\% ε = ∣ ∣ 1.073729 1.073729 − 1.073729 ∣ ∣ ⋅ 100% = 0.000000%
x = 1.073729 x=1.073729 x = 1.073729
The root is 1.073729 1.073729 1.073729
Comments