Answer to Question #139998 in Quantitative Methods for xerin

Question #139998
find the roots using bisector. Stop until at 15th iteration or when approximate percent relative error is below 0.05%.

e^x+x=4
1
Expert's answer
2020-10-26T20:10:49-0400
f(x)=ex+x4f(x)=e^x+x-4

a=0,b=2a=0, b=2

f(a)=f(0)=e0+04=3f(a)=f(0)=e^0 +0-4=-3

f(b)=f(2)=e2+24=e22f(b)=f(2)=e^2 +2-4=e^2-2

f(a)f(b)=f(0)f(2)=3(e22)<0f(a)f(b)=f(0)f(2)=-3(e^2-2)<0


xn=an+bn2x_n=\dfrac{a_n+b_n}{2}

an+1=xn,bn+1=bn,f(an)f(xn)0a_{n+1}=x_n, b_{n+1}=b_n, f(a_n)f(x_n)\geq0

an+1=an,bn+1=xn,f(bn)f(xn)0a_{n+1}=a_n, b_{n+1}=x_n, f(b_n)f(x_n)\geq0

f(xn)ε=>answer=xn|f(x_n)|\leq \varepsilon=>answer =x_n


nxnf(xn)010.2817181711.51.986890721.250.7403429631.1250.2052168541.06250.0439040651.093750.0791985461.0781250.0172884571.07031250.0133968081.074218750.0019234991.0722656250.00574223101.07324218750.00191077111.073730468750.00000601\begin{matrix} n & x_n & f(x_n)\\ 0 & 1 & -0.28171817 \\ 1 & 1.5 & 1.9868907 \\ 2 & 1.25 & 0.74034296 \\ 3 & 1.125 & 0.20521685 \\ 4 & 1.0625 & -0.04390406 \\ 5 & 1.09375 & 0.07919854 \\ 6 & 1.078125 & 0.01728845 \\ 7 & 1.0703125 & -0.01339680 \\ 8 & 1.07421875 & 0.00192349 \\ 9 & 1.072265625 & -0.00574223 \\ 10 & 1.0732421875 & -0.00191077 \\ 11 & 1.07373046875 & 0.00000601 \\ \end{matrix}


1.073730468751.07324218751.0732421875100%\big|\dfrac{1.07373046875-1.0732421875}{1.0732421875}\big|\cdot100\%\approx

0.0455%<0.05%\approx0.0455\%<0.05\%

Root is 1.073730468751.07373046875



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