Answer to Question #229331 in Electrical Engineering for Alock kumar

Question

Determine a real root accurate to four decimal places of the following equation by using (i) Bisection Method (ii) Fixed Point Iteration Method (iii) Method of False Position and (iv) Newton-Raphson.

16-9(x - sin x) = 0; interval [2, 3]

Answer 1

1)     Using Bisection Method:

Here 16-9(x-sinx)=0

∴-9⋅(x-sin(x))+16=0

Let f(x)=-9⋅(x-sin(x))+16

1st iteration :

Here f(2)=6.1837>0 and f(3)=-9.7299<0

∴ Now, Root lies between 2 and 3

x0=2+32=2.5

f(x0)=f(2.5)=-9â‹…(2.5-sin(2.5))+16=-1.1138<0

2nd iteration :

Here f(2)=6.1837>0 and f(2.5)=-1.1138<0

∴ Now, Root lies between 2 and 2.5

x1=2+2.52=2.25

f(x1)=f(2.25)=-9â‹…(2.25-sin(2.25))+16=2.7527>0

3rd iteration :

Here f(2.25)=2.7527>0 and f(2.5)=-1.1138<0

∴ Now, Root lies between 2.25 and 2.5

x2=2.25+2.52=2.375

f(x2)=f(2.375)=-9â‹…(2.375-sin(2.375))+16=0.8682>0

4th iteration :

Here f(2.375)=0.8682>0 and f(2.5)=-1.1138<0

∴ Now, Root lies between 2.375 and 2.5

x3=2.375+2.52=2.4375

 

The real root accurate= 2.4375

2)     Fixed Point Iteration Method:

Method-1

Let f(x)=-9⋅(x-sin(x))+16

Here 16-9(x-sinx)=0

∴-9⋅(x-sin(x))+16=0

∴-9⋅(x-sin(x))=-16

∴x=-16

∴ϕ(x)=-16

Here f(2)=6.1837>0 and f(3)=-9.7299<0

∴ Root lies between 2 and 3

3)     Method of False Position

Here 16-9(x-sinx)=0

∴-9⋅(x-sin(x))+16=0

Let f(x)=-9⋅(x-sin(x))+16

 iteration :

Here f(2)=6.1837>0 and f(3)=-9.7299<0

∴ Now, Root lies between x0=2 and x1=3

x2=x0-f(x0)â‹…x1-x0f(x1)-f(x0)

x2=2-6.1837â‹…3-2-9.7299-6.1837

x2=2.4375

4)     Newton-Raphson Method

Here 16-9(x-sinx)=0

∴-9⋅(x-sin(x))+16=0

Let f(x)=-9⋅(x-sin(x))+16

ddx(-9â‹…(x-sin(x))+16)=-9+9cos(x)

∴f′(x)=-9+9cos(x)

x0=2

1st iteration :

f(x0)=f(2)=-9â‹…(2-sin(2))+16=6.1837

f′(x0)=f′(2)=-9+9cos(2)=-12.7453

x1=x0-f(x0)f′(x0)

x1=2-6.1837-12.7453

x1=2.4852

2nd iteration :

f(x1)=f(2.4852)=-9â‹…(2.4852-sin(2.4852))+16=-0.874

f′(x1)=f′(2.4852)=-9+9cos(2.4852)=-16.1296

x2=x1-f(x1)f′(x1)

x2=2.4852--0.874-16.1296

x2=2.431

3rd iteration :

f(x2)=f(2.431)=-9â‹…(2.431-sin(2.431))+16=-0.0083

f′(x2)=f′(2.431)=-9+9cos(2.431)=-15.8217

x3=x2-f(x2)f′(x2)

 x3=2.431--0.0083-15.8217

x3=2.4305

4th iteration :

f(x3)=f(2.4305)=-9â‹…(2.4305-sin(2.4305))+16=0

f′(x3)=f′(2.4305)=-9+9cos(2.4305)=-15.8186

x4=x3-f(x3)f′(x3)

x4=2.4305-0-15.8186

x4=2.4305