Question #139989
find the roots using newton raphson. Stop until at 15th iteration or when approximate percent relative error is below 0.05%.

e^x+x=4
1
Expert's answer
2020-10-25T18:51:42-0400
f(x)=ex+x4f(x)=e^x+x-4

f(x)=ex+1f'(x)=e^x+1

xn+1=xnf(xn)f(xn)x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}

Initial solution x0=1x_0 =1

nxnf(xn)xn+1010.2817181.07576611.0757660.0080031.07372921.0737290.0000061.073729\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}


ε=xn+1xnxn100%\varepsilon =\big|\dfrac{x_{n+1}-x_n}{x_n}\big|\cdot 100\%

ε=1.07576611100%=7.5766%\varepsilon =\big|\dfrac{1.075766-1}{1}\big|\cdot 100\%=7.5766\%

ε=1.0737291.0757661.075766100%=0.1894%\varepsilon =\big|\dfrac{1.073729-1.075766}{1.075766}\big|\cdot 100\%=0.1894\%

ε=1.0737291.0737291.073729100%=0.000000%\varepsilon =\big|\dfrac{1.073729-1.073729}{1.073729}\big|\cdot 100\%=0.000000\%

x=1.073729x=1.073729


The root is 1.0737291.073729


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