Question #139992
find the roots using simple fixed point. Stop until at 15th iteration or when approximate percent relative error is below 0.05%.

2x^5+x^4-2x-1=0
1
Expert's answer
2020-10-26T18:55:50-0400
2x5+x42x1=x4(2x+1)(2x+1)=2x^5+x^4-2x-1=x^4(2x+1)-(2x+1)=

=(2x+1)(x41)=(2x+1)(x21)(x2+1)==(2x+1)(x^4-1)=(2x+1)(x^2-1)(x^2+1)=

=(2x+1)(x+1)(x1)(x2+1)=(2x+1)(x+1)(x-1)(x^2+1)

x1=1,x2=12,x3=1x1=-1, x2=-\dfrac{1}{2}, x3=1

We know that there is a solution for the equation 2x5+x42x1=02x^5+x^4-2x-1=0 in [0.5,1.5].[0.5,1.5].


x=(x4+2x+12)1/5x=\big(\dfrac{-x^4+2x+1}{2}\big)^{1/5}

xi+1=(xi4+2xi+12)1/5x_{i+1}=\big(\dfrac{-x_i^4+2x_i+1}{2}\big)^{1/5}

ixi00.510.99367021.00123930.99975141.00005050.99999061.00000271.00000081.000000\begin{matrix} i & x_i \\ 0 & 0.5\\ 1 & 0.993670 \\ 2 & 1.001239\\ 3 & 0.999751\\ 4 & 1.000050\\ 5 & 0.999990\\ 6 & 1.000002\\ 7 &1.000000 \\ 8 & 1.000000\\ \end{matrix}


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

ε=0.9936700.50.5100%=98.734%\varepsilon =\big|\dfrac{0.993670-0.5}{0.5}\big|\cdot 100\%=98.734\%

ε=1.0012390.9936700.993670100%=0.7617%\varepsilon =\big|\dfrac{1.001239-0.993670}{0.993670}\big|\cdot 100\%=0.7617\%

ε=0.9997511.0012391.001239100%=0.1486%\varepsilon =\big|\dfrac{0.999751-1.001239}{1.001239}\big|\cdot 100\%=0.1486\%

ε=1.0000500.9997510.999751100%=0.0299%\varepsilon =\big|\dfrac{1.000050-0.999751}{0.999751}\big|\cdot 100\%=0.0299\%

ε=0.9999901.0000501.000050100%=0.0060%\varepsilon =\big|\dfrac{0.999990-1.000050}{1.000050}\big|\cdot 100\%=0.0060\%

ε=0.9999980.9999900.999990100%=0.0016%\varepsilon =\big|\dfrac{0.999998-0.999990}{0.999990}\big|\cdot 100\%=0.0016\%

ε=1.0000020.9999980.999998100%=0.0012%\varepsilon =\big|\dfrac{1.000002-0.999998}{0.999998}\big|\cdot 100\%=0.0012\%

ε=1.0000001.0000021.000002100%=0.0002%\varepsilon =\big|\dfrac{1.000000-1.000002}{1.000002}\big|\cdot 100\%=0.0002\%

If ε0.05%,\varepsilon\leq0.05\%, then x=0.999998x=0.999998


The root is 1.000001.00000


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