Answer to Question #180294 in Quantitative Methods for sarath lal

Question

x2-2x-3=0 for the positive root by iteration method.

Answer 1

$Given\ equation \ is \ x^2-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<0\\ At \ x= 2\ , 2^2 -2-3=4-5=-1<0\\ At \ x=3, \ 3^2 -3-3=3>0\\ By \ Intermediate \ value \ theorem \ there \ which\ states\ that, \ if \ f(x) \ is \\continuous \ in \ an \ interval [a,b] and f(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 \ x\\ From \ (1),\ we \ have \ x^2 =x+3\Rightarrow x=\sqrt{x+3}\\ \Rightarrow\ The \ iterative \ equation \ for \ the \ root \ is \ x_n=\sqrt{x_{n-1}+3},\ n=1, 2,3 , ....\\ Let \ x_{0}=Midvalue \ of\ (2,3)=\frac{2+3}{2}=1.5\Rightarrow x_{0}=1.5\\ x_{1}=\sqrt{1.5+3}\\ =\sqrt{4.5}\\ =2.1213203436 \Rightarrow x_{1}=2.1213203436\\ x_2=\sqrt{x_{1}+3}\\ =\sqrt{2.1213203436+3}\\ =\sqrt{5.1213203436}\\ =2.2630334385 \Rightarrow \ x_{2}= 2.2630334385\\ x_3=\sqrt{x_{2}+3}\\ =\sqrt{2.2630334385+3}\\ =\sqrt{5.2630334385}\\ =2.2941302139 \Rightarrow \ x_{3}= 2.2941302139\\ x_4=\sqrt{x_{3}+3}\\ =\sqrt{2.2941302139+3}\\ =\sqrt{5.2941302139}\\ =2.3008976974 \Rightarrow \ x_{4}= 2.3008976974 \\ x_5=\sqrt{x_{4}+3}\\ =\sqrt{2.3008976974+3}\\ =\sqrt{5.3008976974}\\ =2.3023678458 \Rightarrow \ x_{5}= 2.3023678458 \\ x_6=\sqrt{x_{5}+3}\\ =\sqrt{2.3023678458+3}\\ =\sqrt{5.3023678458}\\ =2.3026870925 \Rightarrow \ x_{6}= 2.3026870925\\ x_7=\sqrt{x_{6}+3}\\ =\sqrt{ 2.3026870925+3}\\ =\sqrt{ 5.3026870925}\\ =2.3027564119 \Rightarrow \ x_{7}= 2.3027564119\\ x_8=\sqrt{x_{7}+3}\\ =\sqrt{2.3027564119+3}\\ =\sqrt{5.3027564119}\\ =2.3027714632 \Rightarrow \ x_{8}= 2.3027714632\\ x_9=\sqrt{x_{8}+3}\\ =\sqrt{2.3027714632+3}\\ =\sqrt{5.3027714632}\\ =2.3027747313\Rightarrow \ x_{9}=2.3027747313\\ Since, \ x_8, and \ \ x_9 \ are\ equal \ upto \ 5th \ decimal\\ the\ approximate \ root\ of \ (1) \ is \ x= 2.30277(upto\ 5 \ decimals)$