Approximate solutions are the estimate values of an equation
The approximate solution of the equation is x = 0.45
How to determine the approximate solution
The equation is given as:
[tex]x^3 + 2x- 1= 0[/tex]
The iteration is given as:
[tex]x_{n+1} = \frac{1}{x_n^2 + 2}[/tex]
To start with, we have:
[tex]x_1 = 1[/tex]
So, we have:
[tex]x_2 = \frac{1}{1^2 + 2} = \frac 13 =0.33333333[/tex]
The next iteration is:
[tex]x_3 = \frac{1}{0.33333333^2 + 2} = 0.47368421102[/tex]
The next iteration is:
[tex]x_4 = \frac{1}{0.47368421102^2 + 2} = 0.4495641344[/tex]
The next iteration is:
[tex]x_5 = \frac{1}{0.4495641344^2 + 2} = 0.45411035264[/tex]
The next iteration is:
[tex]x_6 = \frac{1}{0.45411035264^2 + 2} = 0.45326473189[/tex]
Notice that:
x5 and x6 have the same value to 2 decimal places.
i.e. [tex]x_5 \approx x_6 = 0.45[/tex]
Hence, the approximate solution of the equation is x = 0.45
Read more about approximate solutions at:
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