Given equation is x2−x−3=0......(1)To initiate the iterative method ,we find the interval in which (1) has a root by using Intermetiate value theorem.Consider value of (1) at x=1 which is (1)2−1−3=1−4=−3<0At x=2 ,22−2−3=4−5=−1<0At x=3, 32−3−3=3>0By Intermediate value theorem there which states that, if f(x) iscontinuous in an interval[a,b]andf(a)f(b)<0,then there exists c in (a,b) such that f(c)=0.Hence,equation(1) has a root in (2,3).Let us set up the iterative equation for xFrom (1), we have x2=x+3⇒x=x+3⇒ The iterative equation for the root is xn=xn−1+3, n=1,2,3,....Let x0=Midvalue of (2,3)=22+3=1.5⇒x0=1.5x1=1.5+3=4.5=2.1213203436⇒x1=2.1213203436x2=x1+3=2.1213203436+3=5.1213203436=2.2630334385⇒ x2=2.2630334385x3=x2+3=2.2630334385+3=5.2630334385=2.2941302139⇒ x3=2.2941302139x4=x3+3=2.2941302139+3=5.2941302139=2.3008976974⇒ x4=2.3008976974x5=x4+3=2.3008976974+3=5.3008976974=2.3023678458⇒ x5=2.3023678458x6=x5+3=2.3023678458+3=5.3023678458=2.3026870925⇒ x6=2.3026870925x7=x6+3=2.3026870925+3=5.3026870925=2.3027564119⇒ x7=2.3027564119x8=x7+3=2.3027564119+3=5.3027564119=2.3027714632⇒ x8=2.3027714632x9=x8+3=2.3027714632+3=5.3027714632=2.3027747313⇒ x9=2.3027747313Since, x8,and x9 are equal upto 5th decimalthe approximate root of (1) is x=2.30277(upto 5 decimals)
Comments