G i v e n e q u a t i o n i s x 2 − x − 3 = 0...... ( 1 ) T o i n i t i a t e t h e i t e r a t i v e m e t h o d , w e f i n d t h e i n t e r v a l i n w h i c h ( 1 ) h a s a r o o t b y u s i n g I n t e r m e t i a t e v a l u e t h e o r e m . C o n s i d e r v a l u e o f ( 1 ) a t x = 1 w h i c h i s ( 1 ) 2 − 1 − 3 = 1 − 4 = − 3 < 0 A t x = 2 , 2 2 − 2 − 3 = 4 − 5 = − 1 < 0 A t x = 3 , 3 2 − 3 − 3 = 3 > 0 B y I n t e r m e d i a t e v a l u e t h e o r e m t h e r e w h i c h s t a t e s t h a t , i f f ( x ) i s c o n t i n u o u s i n a n i n t e r v a l [ a , b ] a n d f ( a ) f ( b ) < 0 , t h e n t h e r e e x i s t s c i n ( a , b ) s u c h t h a t f ( c ) = 0. H e n c e , e q u a t i o n ( 1 ) h a s a r o o t i n ( 2 , 3 ) . L e t u s s e t u p t h e i t e r a t i v e e q u a t i o n f o r x F r o m ( 1 ) , w e h a v e x 2 = x + 3 ⇒ x = x + 3 ⇒ T h e i t e r a t i v e e q u a t i o n f o r t h e r o o t i s x n = x n − 1 + 3 , n = 1 , 2 , 3 , . . . . L e t x 0 = M i d v a l u e o f ( 2 , 3 ) = 2 + 3 2 = 1.5 ⇒ x 0 = 1.5 x 1 = 1.5 + 3 = 4.5 = 2.1213203436 ⇒ x 1 = 2.1213203436 x 2 = x 1 + 3 = 2.1213203436 + 3 = 5.1213203436 = 2.2630334385 ⇒ x 2 = 2.2630334385 x 3 = x 2 + 3 = 2.2630334385 + 3 = 5.2630334385 = 2.2941302139 ⇒ x 3 = 2.2941302139 x 4 = x 3 + 3 = 2.2941302139 + 3 = 5.2941302139 = 2.3008976974 ⇒ x 4 = 2.3008976974 x 5 = x 4 + 3 = 2.3008976974 + 3 = 5.3008976974 = 2.3023678458 ⇒ x 5 = 2.3023678458 x 6 = x 5 + 3 = 2.3023678458 + 3 = 5.3023678458 = 2.3026870925 ⇒ x 6 = 2.3026870925 x 7 = x 6 + 3 = 2.3026870925 + 3 = 5.3026870925 = 2.3027564119 ⇒ x 7 = 2.3027564119 x 8 = x 7 + 3 = 2.3027564119 + 3 = 5.3027564119 = 2.3027714632 ⇒ x 8 = 2.3027714632 x 9 = x 8 + 3 = 2.3027714632 + 3 = 5.3027714632 = 2.3027747313 ⇒ x 9 = 2.3027747313 S i n c e , x 8 , a n d x 9 a r e e q u a l u p t o 5 t h d e c i m a l t h e a p p r o x i m a t e r o o t o f ( 1 ) i s x = 2.30277 ( u p t o 5 d e c i m a l s ) 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) G i v e n e q u a t i o n i s x 2 − x − 3 = 0...... ( 1 ) T o ini t ia t e t h e i t er a t i v e m e t h o d , w e f in d t h e in t er v a l in w hi c h ( 1 ) ha s a roo t b y u s in g I n t er m e t ia t e v a l u e t h eore m . C o n s i d er v a l u e o f ( 1 ) a t x = 1 w hi c h i s ( 1 ) 2 − 1 − 3 = 1 − 4 = − 3 < 0 A t x = 2 , 2 2 − 2 − 3 = 4 − 5 = − 1 < 0 A t x = 3 , 3 2 − 3 − 3 = 3 > 0 B y I n t er m e d ia t e v a l u e t h eore m t h ere w hi c h s t a t es t ha t , i f f ( x ) i s co n t in u o u s in an in t er v a l [ a , b ] an df ( a ) f ( b ) < 0 , t h e n t h ere e x i s t s c in ( a , b ) s u c h t ha t f ( c ) = 0. He n ce , e q u a t i o n ( 1 ) ha s a roo t in ( 2 , 3 ) . L e t u s se t u p t h e i t er a t i v e e q u a t i o n f or x F ro m ( 1 ) , w e ha v e x 2 = x + 3 ⇒ x = x + 3 ⇒ T h e i t er a t i v e e q u a t i o n f or t h e roo t i s x n = x n − 1 + 3 , n = 1 , 2 , 3 , .... L e t x 0 = M i d v a l u e o f ( 2 , 3 ) = 2 2 + 3 = 1.5 ⇒ x 0 = 1.5 x 1 = 1.5 + 3 = 4.5 = 2.1213203436 ⇒ x 1 = 2.1213203436 x 2 = x 1 + 3 = 2.1213203436 + 3 = 5.1213203436 = 2.2630334385 ⇒ x 2 = 2.2630334385 x 3 = x 2 + 3 = 2.2630334385 + 3 = 5.2630334385 = 2.2941302139 ⇒ x 3 = 2.2941302139 x 4 = x 3 + 3 = 2.2941302139 + 3 = 5.2941302139 = 2.3008976974 ⇒ x 4 = 2.3008976974 x 5 = x 4 + 3 = 2.3008976974 + 3 = 5.3008976974 = 2.3023678458 ⇒ x 5 = 2.3023678458 x 6 = x 5 + 3 = 2.3023678458 + 3 = 5.3023678458 = 2.3026870925 ⇒ x 6 = 2.3026870925 x 7 = x 6 + 3 = 2.3026870925 + 3 = 5.3026870925 = 2.3027564119 ⇒ x 7 = 2.3027564119 x 8 = x 7 + 3 = 2.3027564119 + 3 = 5.3027564119 = 2.3027714632 ⇒ x 8 = 2.3027714632 x 9 = x 8 + 3 = 2.3027714632 + 3 = 5.3027714632 = 2.3027747313 ⇒ x 9 = 2.3027747313 S in ce , x 8 , an d x 9 a re e q u a l u pt o 5 t h d ec ima l t h e a pp ro x ima t e roo t o f ( 1 ) i s x = 2.30277 ( u pt o 5 d ec ima l s )
#339914 on Dec 2023
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