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
Related questions
Question
100%
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 için
Bisection 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, f
fonksiyonunun [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>⁹
olduğundan 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, f
fonksiyonunun [1.25,1.5] aralığında bir çözümü mevcuttur.
X4 = ((x2+x3)/2)=1.375,
Hata = |x4-X₂1=0.125>⁹
olduğundan 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, f
fonksiyonunun [1.375,1.5] aralığında bir çözümü mevcuttur.
X5 = ((X3+X4)/2) = (1.375+1.5)/2= 1.4375
Hata = |x4-X₂|=0.0625<=10¹
olduğundan iterasyon sonlandırılır. Aradığımız çözüm
X=X5= 1.4375](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2cd79f82-3730-49d1-8559-a01cc4e6e9ca%2Fd64a009d-fc01-40d4-a51c-723b7e8d0da9%2F0h97g9g_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Örnek 5. f(x)=x²-2=0 denkleminin [1,2] aralığındaki yaklaşık çözümünü &-10¹ hata toleransı için
Bisection 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, f
fonksiyonunun [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>⁹
olduğundan 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, f
fonksiyonunun [1.25,1.5] aralığında bir çözümü mevcuttur.
X4 = ((x2+x3)/2)=1.375,
Hata = |x4-X₂1=0.125>⁹
olduğundan 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, f
fonksiyonunun [1.375,1.5] aralığında bir çözümü mevcuttur.
X5 = ((X3+X4)/2) = (1.375+1.5)/2= 1.4375
Hata = |x4-X₂|=0.0625<=10¹
olduğundan iterasyon sonlandırılır. Aradığımız çözüm
X=X5= 1.4375
Expert Solution

This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 12 images

Recommended textbooks for you

Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated

Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education

Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY

Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated

Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education

Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY

Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,

