Use Newton Raphson Method to solve the equation cos x − x e
x = 0
correct to four decimal places.
Answers
Answer:
ok
Step-by-step explanation:
MethodExample 1:Find a real root of the equationx3−6x−4=0 by bisection method.[JNTU 2006, Set No. 1]Solution:x:0123f(x)=x3−6x−4:−4−9−85...one root lies between 2 and 3. Takex0=2,x1=3. By bisection method, the next approxima-tion isx2=x0+x12=2+32=2.5. Now atx2=2.5,f(2.5)= −3.375<0. Thus, a root lies between 2.5and 3. So the next approximation isx3=2.5+32=5.52=2.75. Atx3=2.75,f(2.75)=0.2968>0.Thus a root lies between 2.5 and 2.75. Next approx-imation isx4=2.5+2.752=2.625.Nowf(2.625)= −1.6621<0. The root lies be-tween2.625and2.75.Thenx5=2.625+2.752=2.6875 andf(2.6875)= −0.7141<0. So root liesbetween 2.6875 and 2.75. The approximate value ofthe real root of the given equation is2.6875+2.752=2.71875.Example 2:Find a real root ofx3−5x+3=0using bisection method.[JNTU 2006, Set No. 2]Solution:x:012f(x)=x3−5x+3:3−11Thus, a root lies between 0 and 1 and another rootlies between 1 and 2.Choosex0=1 andx1=2. Thenx2=x0+x12=1+22=1.5,f(1.5)= −1.125<0.So root lies in between 1.5 and 2. Sox3=x2+x12=1.5+22=1.75,f(1.75)= −0.3906<0.So root lies between 1.75 and 2. Thenx4=1.75+22=1.875,f(1.875)=0.2167>0.So root lies between 1.75 and 1.875. So the approx-imate root of the given equation isx5=1.75+1.8752=1.8125.Example 3:Find a real root of the Equationx3−x−11=0 by bisection method.[JNTU 2007, Set No. 2]Solution:x:0123f(x)=x3−x−11:−11−11−513A root lies between 2 and 3. Then the first approxi-mation isx2=x0+x12=2+32=2.5. Nowf(2.5)=2.125>0. So root lies between 2 and 2.5. Sothe next approximation isx3=2+2.52=2.25. Nowf(2.25)= −1.859<0. So next approximation isx4=2.25+2.52=2.375. Nowf(2.375)=0.0215>0 andf(2.25)<0....Root lies between 2.25 and 2.375. Sox5=2.25+2.3752= 2.3125. Nowf(2.3125)= −0.946<0andf(2.375)>0.17.1