Question #139397
find the roots using simple fixed point iteration. Stop until at 15th iteration or when approximate percent relative error is below 0.05%.

e^x+x=4
1
Expert's answer
2020-10-22T17:51:58-0400
ex+x=4e^x+x=4

We know that there is a solution for the equation ex+x4=0e^x+x-4=0 in [1,2].[1,2].


x=ln(4x)x=\ln(4-x)

xi+1=ln(4xi)x_{i+1}=\ln(4-x_i)

ixi0111.09861221.06518931.07664341.07273351.07406961.07361371.07376981.07371591.073734101.073727111.073729121.073729\begin{matrix} i & x_i \\ 0 & 1\\ 1 & 1.098612 \\ 2 & 1.065189\\ 3 & 1.076643\\ 4 & 1.072733\\ 5 & 1.074069\\ 6 & 1.073613\\ 7 & 1.073769 \\ 8 & 1.073715\\ 9 & 1.073734\\ 10 & 1.073727\\ 11 & 1.073729 \\ 12 & 1.073729\\ \end{matrix}


ε=ni+1nini100%\varepsilon =\big|\dfrac{n_{i+1}-n_i}{n_i}\big|\cdot 100\%

ε=1.09861211100%=9.8612%\varepsilon =\big|\dfrac{1.098612-1}{1}\big|\cdot 100\%=9.8612\%

ε=1.0651891.0986121.098612100%=3.0423%\varepsilon =\big|\dfrac{1.065189-1.098612}{1.098612}\big|\cdot 100\%=3.0423\%

ε=1.0766431.0651891.065189100%=1.0753%\varepsilon =\big|\dfrac{1.076643-1.065189}{1.065189}\big|\cdot 100\%=1.0753\%

ε=1.0727331.0766431.076643100%=0.3632%\varepsilon =\big|\dfrac{1.072733-1.076643}{1.076643}\big|\cdot 100\%=0.3632\%

ε=1.0740691.0727331.072733100%=0.1245%\varepsilon =\big|\dfrac{1.074069-1.072733}{1.072733}\big|\cdot 100\%=0.1245\%

ε=1.0737691.0737131.073713100%=0.005029%\varepsilon =\big|\dfrac{1.073769-1.073713}{1.073713}\big|\cdot 100\%=0.005029\%

ε=1.0740691.0727331.072733100%=0.1245%\varepsilon =\big|\dfrac{1.074069-1.072733}{1.072733}\big|\cdot 100\%=0.1245\%

ε=1.0736131.0740691.074069100%=0.0425%\varepsilon =\big|\dfrac{1.073613-1.074069}{1.074069}\big|\cdot 100\%=0.0425\%

ε=1.0737691.0736131.073613100%=0.0145%\varepsilon =\big|\dfrac{1.073769-1.073613}{1.073613}\big|\cdot 100\%=0.0145\%

ε=1.0737151.0737691.073769100%=0.0050%\varepsilon =\big|\dfrac{1.073715-1.073769}{1.073769}\big|\cdot 100\%=0.0050\%

x=1.073613x=1.073613


The root is 1.0741.074



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