Bisection method

1. Using bisection method find the positive root

of the following function F(x)=x3-4x-9

solution; here we can conclude one

root of f(x) lie between 2 & 3 i.e;

f(2)=-9 (-ve) and f(3)=6 (+ve)

If xr represents the root mean of the function

by bisection methods

I. Xr=(2+3)/2=2.5

f(xr)=f(2.5)=-3.375 is –ve

error=(2.5-3)/2.5*100=20%

II. Xr=(2.5+2.75)/2=2.75

f(xr)=f(2.75)=0.796875 is +ve

III. Xr=(2.5+2.75)/2=2.625

f(xr)=f(2.625)=-1.412109 is –ve

Continuous up two required iteration

x F(x)

0 -9

1 -12

2 -9

3 6

Secant method

xi+1=xi-[f(xi)(xi-1-xi)]/[f(xi-1)-f(xi)]

1. Use the secant method to estimate the root of

f(x)=x3-x-1 with initial estimates of 1&2 having

three iteration

Solution;

xi-1=1 and f(2)=5

I. X1=2-[f(2)(1-2)]/[f(1)-f(2)]=2-[5(-1)]/[-1-5] = 1.166667

Now xi-1=2 and xi=1.166667

error= [1.166667-2]/1.166667*100= 71.4286%

II. F(2)=5 and f(1.166667)= -0.5787

X2=1.166667-[-0.5787(2-1.166667)]/[5-(-0.5787)] = 1.25311

Now xi-1 = 1.166667 and xi = 1.25311

III. F(1.166667)=-0.5787 and f(1.25311)= -0.28537

X3 = 1.25311-[-0.28537(1.16667-1.25311)]/[-0.5787-(-0.28537)]

X3= 1.3372 f(1.3372) = 0.0538854

Lease square regression method

1. Find the equation of straight line which best fits

n = 7 ,

Ẋ = ∑xi/n

Ý= ∑yi/n ,

a1=[n∑xiyi-∑yi]/[n∑xi2-(∑xi)2

ao = Ý-a1Ẋ

therefore the line which fits the given tabular

points is y= ao+a1x

x 10 12 13 16 17 20 25 y 10 22 24 27 29 33 37

Xi Yi XiYi xi2 10 10 100 100 12 22 264 144 13 24 312 169 16 27 432 256 17 29 493 289 20 33 660 400 25 37 925 625 Sum 113 182 3186 1983

Fourier approximation

1. The PH in a reactor varies sinusoidally as the couse

of a day use Fourier approximation Time(h) 0 2 4 5 7 8.5 12 15 20 22 24

PH 7.3 7 7.1 6.4 7.2 8.9 8.8 8.9 7.9 7.9 7

Where f=1/t and t=24hr , f=1/24= 0.042hr-1 ,n = 11

Wo=2πf=0.261799388

Ao=∑y/n, A1 = 2/n[∑ycos(wot) ,

B1 = 2/n[∑ysin(wot)

The equation will be y= Ao+A1cos(wot)+B1sin(wot)

time(t) PH(y) ycos(wot) ysin(wot)

0 7.3 7.3 0

2 7 6.062177825 3.500000002

4 7.1 3.549999995 6.14878037

5 6.4 1.656441882 6.18192529

7 7.4 -1.915260944 7.147851112

8.5 7.2 -4.383082299 5.712144043

12 8.9 -8.9 -2.14508E-08

15 8.8 -6.222539656 -6.222539693

20 8.9 4.450000031 -7.707626076

22 7.9 6.841600707 -3.94999997

24 7 7 3.37429E-08

Sum 83.9 15.43933754 10.81053509

Linear interpolation

The general formula for linear interpolation is

𝑓1(𝑥)−𝑓(𝑋𝑜)

𝑋−𝑋𝑂=

f(X1)−f(Xo)

𝑋1−𝑋0 from the above we get the

general formula for linear interpolation

f1(x) = 𝑓(𝑥𝑜) +𝑓(𝑥1) − 𝑓(𝑥𝑜)

𝑥1 − 𝑥𝑜(𝑥 − 𝑥𝑜)

Example find the value of ln(2) if ln(1)=0 and ln(4)=1.386294 using linear interpolation

Solution 1st we make a table

Given f(xo)=0, xo=1

f(x1)=1.386294, x1=4

f1(x)=?, x=2

By using linear interpolation formula

f1(x) = 𝑓(𝑥𝑜) +𝑓(𝑥1) − 𝑓(𝑥𝑜)

𝑥1 − 𝑥𝑜(𝑥 − 𝑥𝑜)

f1(x) = 0 +1.386294 − 0

4 − 1(2 − 1)

f1(x) =1.386294

3= 0.4621 𝑎𝑛𝑠𝑤𝑒𝑟

x ln(x) 1 0

2 ? 4 1.386294

The gauss seidel method

Rule

|a11|>|a12|+|a13|,

|a22|>|a21|+|a23|, and

|a33|>|a31|+|a32|

Iteration 1, y = z =0, xi=?,

x = xi, z = 0 , yi = ?

x = xi, y = yi, zi = ?

Iteration 2,

using yi & zi we get xii = ?

xii & yi we get yii = ?

xii & yii we get zii = ?

continue this to the required iteration

a11 a12 a13 a21 a22 a23 a31 a32 a33

e f g

The naï ve gauss elimination method

Example; given 20x+y+4z=25----------------eq(1)

8x+13y+2z=23--------------eq(2)

4x-11y+21z=14-------------eq(3)

with the value of x=y=z=1

Solution: by multiplying eq(1) by 8/20 and subtract

from eq(2) and also multiply eq(1) by 4/20 and

subtract from eq(3) we get the following set of equations

20x+y+4z=25--------------------eq(1)

12.6y+0.4z=13------------------eq(4)

-11.2y+20.2z=9-------------------eq(5) by proceeding to eliminate we find a single variable z by multiplying

eq(4) by -11.2/12.6 and subtract from eq(5) we get

20.5556z=20.5556----------------eq(6)

finally we rearrange the equation and written as

20x+y+4z=25 -----------------eq(1)

12.6y+0.4z=13 ----------------eq(4)

20.5556z=20.5556-----eq(6)

now by using back substitution we get the unknowns x,y&z

from eq(6) we get z=1 ,

from eq(4) we get y=1,

from eq(1) we get x=1

LU decomposition

© to find the LU decomposition first we find f21=

a21/a11 f31= a31/a11 f32 = a’32/a’22

And the multiplying the first row by f21 and

then subtract this value from second row

Multiplying the first row by f31 and then

subtract this value from third row

Now we find f32 from the matrix we get above

Now multiplying the new matrix second row

by f32 and then subtract this value from the

new matrix row three

Finally we get [U] = [L] =

a11 a12 a13

a21 a22 a23

a31 a32 a33

a11 a12 a13

0 a22 a23

0 0 a33

1 0 0

f21 1 0

f31 f32 1

Matrix inverse using LU decomposition

We find the LU decomposition of the given matrix as the

above form and then we find the inverse of the matrix

Procedure one; [L] =

to find the value of d1,d2,d3 by using forward

substitution

d1+0+0=1 we get d1

F21d1+d2+0=0 we get d2

F31d1+F32d2+d3=0 we get d3

the vector can be then used as the right hand side for

the upper triangular matrix

[U] =

let us find the value of x1,x2 and x3 using backward

substitution

0+0+a33x3=d3 from this we get x3

0+a22x2+a23x3=d2 from this we get x2

a11x1+a12x2+a13x3=d1 from this we get x1

d1 d2 d3

1 0 0

d1 d2 d3

x1 x2 x3

x1,x2 and x3 are the first column of the inverse matrix

As like as procedure one we get all the columns of the

inverse matrix by only changing

To and using in procedure 2 and 3

respectively

finally the inverse of the matrix we get is

[A] -1 =

1 0 0

0 1 0

0 0 1

X1 X1 X1

X2 X2 X2

X3 X3 X3

Trapezoidal rule

Examples;

∫06 1

1 𝑥 dx given strip 6,

Solution; given a = 0 and b = 6

Interval = b-a = 6 and

width of strip = h = interval/no of strip = 6/6 = 1

integration =

h/2 [ (sum of the first and last ordinate)

+ 2(sum of remaining ordinates)]

integration=

½[(1+0.142857)+2(0.5+0.3333+0.25+0.2+0.166667)]

integration = 2.021429

actual value = 1.945910 integrate the above equation

therefore the error =

[(actual value – new value)/actual value]*100%

Error = =3.880%

x 0 1 2 3 4 5 6

f(x) 1 0.5 0.333333 0.25 0.2 0.166667 0.142857

Simpson’s 1/3 rule

To find Simpson’s 1/3 rule we use the following

equation

integration = h/3 [ (sum of the first and last

ordinates) + 2(sum of even ordinates) + 4(sum of

odd ordinates) ]

Example; ∫06 1

1 𝑥 dx given strip 6,

Solution; given a = 0 and b = 6

Interval = b-a = 6 and

width of strip = h = interval/no of strip = 6/6 = 1

Integration =

1/3[(1+0.142857)+2(0.3333+0.2)+4(0.5+0.25+0.166667) ]

integration = 1.95873

actual = 1.945910

Error=[(actual value – new value)/actual value]*100%

Error = = 1.3553%

y0 y1 y2 y3 y4 y5 y6

x 0 1 2 3 4 5 6

f(x) 1 0.5 0.333333 0.25 0.2 0.166667 0.142857

Romberg integration

Example1; ∫12(𝑥 +

1

)2dx and actual value is 4.83333

Solution; a=1 and b=2 f(x) = (𝑥 +1

)2

Step I, f(1) = 4 and f(2) = 6.25

A. n1 = 1 , h1=(b-a)/n1= (2-1)/1 = 1

I1 = h1*[f(1)+f(2)]/2 = 1*[4+6.25]/2

I1 = 5.125

error=[(actual value – new value)/actual value]*100%

Error = 6.0418%

B. n2 =2 , h2 = (b-a )/n2 = 0.5 is width of strip

as a result of h2 = 0.5

we get f(1)=4, f(1.5)=4.6944 and f(2)=6.25

I2 = h2 * [ (f(1)+f(1.5))/2] + h2 * [(f(1.5)+f(2))/2]

I2 = 4.909722

error = [(actual value – new value)/actual value]*100%

Error = 1.58047%

C. n3 = 4, h3 = (b-a)/n3 =0.25 is width of strip

as a result of h3 = 0.25 we get

f(1)=4, f(1.25) , f(1.5)=4.6944, f(1.75), and

f(2)=6.25

I3 = h3/2 [ f(1) + 2f(1.25) + 2f(1.5) + 2f(1.75) + f(2)]

I3 = 4.852744

error = [(actual value – new value)/actual value]*100%

Error = 0.4016%

Step II,

Step III,

I4 = 4/3 (I2) – 1/3 (I1) = 4.837963

Error = 0.09579%

I5 = 4/3 (I3) – 1/3 (I2) = 4.833751

Error = 0.00865%

Step IV,

I6 = 16/15 (I5) – 1/15 (I4) = 4.83347

Error = 0.0028%

Iteration Segment h Integral Error(%)

1 1 1 5.125 6.041795986

2 2 0.5 4.90972222 1.580466776

3 4 0.25 4.85274376 0.401602045

Euler method The formula for Euler method is

Yi + = yi + f(xi,yi) h

Examples solve dy/dx = -2x3+12x2-20x+8.5

for x=0, step size = 0.5 up to x = 4

Solution; x=0,0.5,1,1.5,2,2.5,3,3.5,4

dy/dx = -2x3+12x2-20x+8.5 integrate both sides

y = -1/2x4+4x3-10x2+8.5x+c , f(0) = c = 1

y = -1/2x4+4x3-10x2+8.5x+1 true solution

the initial condition at x = 0 is y = 1 thus xi = 0& yi = 1

y(0.5) = y(0) + f(0,1) *step size = 5.25 predicted value

The true solution at x = 0.5 is

y = −0.5(0.5)4 + 4(0.5)3 − 10(0.5)2 + 8.5(0.5) + 1 = 3.21875

Error =[ (tv-pv)/tv]*100% = [3.218-5.25)/3.218]*100%=63.1%

From the above we get xi = 0.5 and yi = 5.25 then

y(1) = yi + f(xi,yi) h = y(0.5) + f(0.5,5.25)*0.5 = 5.875 is

euler value or predicted value

The true solution at x = 1 is

y(1) = −0.5(1)4 + 4(1)3 − 10(1)2 + 8.5(1) + 1 = 3

Error = [ (tv-pv)/tv]*100% = [(3-5.785)/3]*100% = 95.833%

From the above we get xi = 1 and yi = 5.875

This is continue up to y(4)

Therefore the table will be formatted as

x Y true Y euler Error 0 1 1 0.5 3.218 5.25 63.1% 1 3 5.875 95.833%

Example ; solve y’ = x + y for y(0) = 1 find value of y at

x = 0,0.2,0.4,0.6,0.8,1

Solution by integrating both sides the problem

becomes y = x2/2 + xy + c, solving for y(0) = 1, c = 1

and substitute y = x2/2 + xy + 1 this is a true solution

h = step size = 0.4-0.2 = 0.2 and xi = 0 & yi =1

y(0.2) = y(0) + f(0,1)h = 1.2 is Euler value

The true solution at x = 0.2 is

y(0.2) = (0.2)2/2 + 0.2y +1

y = 0.2 + 0.2y + 1 = 1.5

Error = [(tv – euler value)/tv]*100% = 20%

from the above we get xi = 0.2 and yi = 1.2 it is

continuous up to the given x value y(1)

Leibmann method The general formula for leibmann method is

Ti,jnew = ለTi,jnew + (1-ለ)Ti,jold

Examples solve the following problems with leibmann

method with ለ = 1.25 and Ea(error) < 1% for 1st element

100oc

750c 500c

00C

1st iteration

T1,1 = [right + left + top + bottom] / 4 = [R+L+T+B]/4

T1,1 = [0+75+0+0] / 4 = 18.75

T1,1new = 1.25*18.75+(1-1.25) *0= 23.44

Error=[(tv-pv)/tv]*100% =[(23.4-0)/23.4]*100%=0%

From the 1st iteration we use Ti,jold = 0& error = 0%

It is continuous up to T3,3new

(1,3) (2,3) (3,3)

(1,2) (3,2)

(1,1) (2,1) (3,1)

(2,2)

2nd iteration

T1,1 =[ R+L+T+B]/4 = [7.324+75+30.37+0]/4 = 28.26

T1,1new = 1.25*28.26+(1-1.25)*23.44 = 29.47

Error=[(current iteration–previous iteration)/ci]*100

Error = [(29.47-23.47)/29.47]*100% = 20.5%

At the 2nd iteration we use the formula

Ti,jnew = ለTi,jnew + (1-ለ)Ti,jold

Where

Ti,jnew = the value we get in the second

iteration of Ti,j

Ti,jold = is the value we get in the first iteration of

Ti,j

Then it is continuous up to T3,3new

The table then set us the following form

element 1st iteration 2nd iteration

T 0c Error % T 0c Error %

T1,1 0 100 29.47 20.5

T2,1 0 100 16.64 56.2

T3,1 0 100 24.1 25.7

T1,2 0 100 48.26 36.9

T2,2 0 100 42.46 71.99

T3,2 0 100 52.64 52.7

T1,3 0 100 75.05 9.5

T2,3 0 100 75.4 26.9

T3,3 0 100 68.9 4.3

The fourth order Ruge Kutta method

The formula for the fourth order ruge kutta method is

yi+1 = yi + ɸ(xi,yi,h)*h

ɸ = a1k1+a2k2+a3k3+…+ankn

yi+1 = yi + 1/6 [k1+2k2+2k3+k4]*h

where

k1 = f(xi,yi)

k2 = f(xi+h/2, yi+k1*h/2)

k3 = f(xi+h/2,yi+k2*h/2)

k4 = f(xi+h, yi+k3*h)

Examples

1, solve f(x,y)=-2x3+12x2-20x+8.5,with h=0.5 & y(0)=1

solution from the given equation at xi=0 , yi=1

f(x) = )=-2x3+12x2-20x+8.5 then we find the value of k

k1 = 8.5, k2 = 4.21875, k3 =4.21875 and k4 = 1.25

y(0.5) = y(0) + 1/6 [k1+2k2+2k3+k4]*h

y(0.5) = 3.4875

2, f(x,y)=4e0.8x -0.5y, h = 0.5 with y(0) = 2 from x=0 to x=0.5

Solution

find the value of 1st slope xi=0, yi=2, f(x)= 4e0.8x -0.5y

K1 = f(xi,yi) = f(0,2) = 3

This value is used to compute the value of ‘y’ and slope at

mid-point y(0.25) = 2 + 3(0.25) = 2.75, xi=0 and yi=2

k2 = f(xi+h/2, yi+k1*h/2) = f(0.25,2.75)=3.51

This slope in form is used to compute another slope of

mid-point

y(0.25) = 2 + 3.510611(0.25) = 2.877653 ,xi=0 and yi=2

k3 = f(xi+h/2,yi+k2*h/2) = f(0.25,2.8776)=3.412

This value used to compute the value of y and slope at

the end of the interval

y(0.5) = 2 + 3.071785(0.5) = 3.723392, xi=0 & yi=2

k4 = f(xi+h, yi+k3*h)=f(0.5,3.7233)= 41.056

Let’s find an average slope

yi+1 = yi + 1/6 [k1+2k2+2k3+k4]*h

yi+1 = 3.7515

3, find the value of y at x=0.1,0.2 y’=-y, y(0)=1,h=0.1

f(x,y) = -y

Solution let’s find k’s where xi=0.1 and yi=1

k1 = f(xi,yi) = f(0.1,1) = -1

k2 = f(xi+h/2, yi+k1*h/2)= f(0.15,0.95) = -0.95

k3 = f(xi+h/2,yi+k2*h/2)= f(0.15,0.9525)=-0.9525

k4 = f(xi+h, yi+k3*h) = f(0.2,0.9047) = -0.9047

then yi+1 = yi + 1/6 [k1+2k2+2k3+k4]*h

yi+1 = 0.9048374

then to find another value of y put yi=0.9048374, and

use xi = 0.2, h=0.1, we get the same individual slope of

k1= -1

k2 = -0.95

k3 = -0.9525 and

k4 = -0.9047 hence

yi+1 = yi + 1/6 [k1+2k2+2k3+k4]*h

yi+1 = 0.8097 note that for other questions the

values of k’s may be difference