Numerical analysis In this example is using bisection method So isn’t x0=1 and x1=2 since it’s [1,2] then x2=1+2/2 which gives us 1.5 So x2=1.5 and x3=x2+x1 means x3=1.5+2/2 =1.75 but in the example it says 1.25 am I wrong if I am wrong which place Ia m wrong at ?? Please correct me if possible Örnek 5. f(x)=x²-2=0 denkleminin [1,2] aralığındaki yaklaşık çözümünü &-10¹ hata toleransı içinBisection yöntemini kullanarak bulunuz.Çözüm. f(xo)-f(1)-12-2-1-2--1<0,f(x₁)=f(2)=22-2-4-2-2>0. Yani f(1) f(2)<0 olduğundan, ffonksiyonunun [1,2] aralığında bir çözümü mevcuttur.1.ci iterasyon;X2 = ((Xo+x1)/2)-1.5,Hata | X₂-Xol=0.5>€olduğundan iterasyona devam edilir.2.ci iterasyon; f(x₂)=f(1.5)-(1.5)²-2-0.25>0 dir. f(1) f(1.5) <0 olduğundan, f fonksiyonunun[1,1.5] aralığında bir çözümü mevcuttur.X3 = ((x₁+x₂)/2)=1.25,Hata | X3-X₁|=0.25>⁹olduğundan iterasyona devam edilir.3.ci iterasyon; f(x3)=f(1.25)=(1.25)²-2= -0.4375<0 dir. f(1.25) *f(1.5) <0 olduğundan, ffonksiyonunun [1.25,1.5] aralığında bir çözümü mevcuttur.X4 = ((x2+x3)/2)=1.375,Hata = |x4-X₂1=0.125>⁹olduğundan iterasyona devam edilir.4.ci iterasyon; f(x3)-f(1.375)-(1.375)²-2= -0.10938<0 dir. f(1.375) *f(1.5) <0 olduğundan, ffonksiyonunun [1.375,1.5] aralığında bir çözümü mevcuttur.X5 = ((X3+X4)/2) = (1.375+1.5)/2= 1.4375Hata = |x4-X₂|=0.0625<=10¹olduğundan iterasyon sonlandırılır. Aradığımız çözümX=X5= 1.4375

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Numerical analysis In this example is using bisection method So isn’t x0=1 and x1=2 since it’s [1,2] then x2=1+2/2 which gives us 1.5 So x2=1.5 and x3=x2+x1 means x3=1.5+2/2 =1.75 but in the example it says 1.25 am I wrong if I am wrong which place Ia m wrong at ?? Please correct me if possible

Numerical analysis In this example is using bisection method So isn’t x0=1 and x1=2 since it’s [1,2] then x2=1+2/2 which gives us 1.5 So x2=1.5 and x3=x2+x1 means x3=1.5+2/2 =1.75 but in the example it says 1.25 am I wrong if I am wrong which place Ia m wrong at ?? Please correct me if possible

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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