Answer to Question #139999 in Quantitative Methods for xerin

Question

find the roots using bisector method Stop until at 15th iteration or when approximate percent relative error is below 0.05%.

2x^5+x^4-2x-1=0

Answer 1

$$f(x)=2x^5+x^4-2x-1$$

$a=0, b=1.5$

$f(a)=f(0)=2(0)^5+(0)^4-2(0)-1=-1$

$f(b)=f(1.5)=2(1.5)^5+(1.5)^4-2(1.5)-1=16.25$

$f(a)f(b)=f(0)f(2)=-1(16.25)<0$

$$x_n=\dfrac{a_n+b_n}{2}$$

$a_{n+1}=x_n, b_{n+1}=b_n, f(a_n)f(x_n)\geq0$

$a_{n+1}=a_n, b_{n+1}=x_n, f(b_n)f(x_n)\geq0$

$|f(x_n)|\leq \varepsilon=>answer =x_n$

$$\begin{matrix} n & x_n & f(x_n)\\ 0 & 0.75 & -1.708984375 \\ 1 & 1.125 & 1.955871582 \\ 2 & 0.9375 & -0.654130936 \\ 3 & 1.03125 & 0.401133597 \\ 4 & 0.984375 & -0.181243243 \\ 5 & 1.0078125 & 0.095348399 \\ 6 & 0.99609375 & -0.046479699 \\ 7 & 1.001953125 & 0.023536861 \\ 8 & 0.9990234375 & -0.011693977 \\ 9 & 1.00048828125 & 0.005865577 \\ 10 & 0.999755859375 & -0.002928138 \\ 11 & 1.0001220703125 & 0.001465231 \\ 12 & 0.99993896484375 & -0.000732325 \\ 13 & 1.000030517578125 & 0.000366235 \\ 14 & 0.9999847412109375 & -0.000183099 \\ 15 & 1.00000762939453125 & 0.000091554 \\ \end{matrix}$$

$$\big|\dfrac{0.999755859375-1.00048828125}{1.00048828125}\big|\cdot100\%\approx$$

$$\approx0.073\%>0.05\%$$

$$\big|\dfrac{1.0001220703125-0.999755859375}{0.999755859375}\big|\cdot100\%\approx$$

$$\approx0.037\%<0.05\%$$

$$\big|\dfrac{1.000030517578125-0.99993896484375}{0.99993896484375}\big|\cdot100\%\approx$$

$$\approx0.009\%<0.05\%$$

Root is $1.000122$