numerical methods with c codes

8/2/2019 Numerical Methods With c Codes

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Numerical Methods Algorithm in C++

The scope of these lectures is to study the numerical methods frequently used in

Physics and to write (where possible) their C++ codes. The main topics include.

The Solution of Nonlinear Equations f(x) = 0o Bisection Methodo Newton Raphson Methodo Secant Method

The Solution of Linear Systems AX = Bo Gauss Jordan Methodo Cramers Ruleo Jacobi Methodo Gauss Seidel iterative Methodo Exercise

Numerical Integrationo Riemanns Sumo Trapezoidal Ruleo Simpsons 1/3 Ruleo Simpsons 3/8 Ruleo Monte Carlo -1 Methodo Monte Carlo -2 Method

Ordinary Differential Equationso Euler Methodo Euler Improved Methodo Runge Kutta 4th order Methodo Runge Kutta Fehlberg Method


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The Solution of Nonlinear Equations f(x) = 0

Bisection Method or interval halving method

The bisection method is one of the bracketing methods for finding roots of

equations.

Given a function f(x)=0 and an interval which might contain a root, perform a

predetermined number of iterations using the bisection method. The interval must

be large enough to avoid discontinuity.

The method works like this if [a, b] is the interval in which the root of f(x) lies then

we compute f(a) and f(b). Now we compute f(x) at M=a+b/2 and then calculate

f(M).

Now multiply f(a) with f(M) if the answer of this is negative than the root lies in

[a, M] if not (i.e. positive) then root lies in [M, b].

Example: Consider the non-linear equation

f(x) = x3+x

2-3x-3=0

find the root in the interval [1, 2]

Sol:- Now f(1) = -4

And f(2) = 3

Sign change between [1, 2] guarantees a root in this interval.

Now consider mid point i.e. M = 1.5 than

f(1.5) = -1.475

And f(1)f(1.5) = +ive value

Hence we take the interval [1.5, 2] and repeat the process again and again till we

find that x = 1.731925 is a root to the above equation up to 6 decimal places.


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C++ program for the above example

#include

#include

#include

#include

double f(double x)

{

double value;

value=pow(x,3)+pow(x,2)-3*x-3;

return value;

}

void main()

{

clrscr();

double a,b,M,ans,comp;

int i;

ofstream res("bisectionresults.txt");

couta>>b;

for(i=0;i


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cout


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Newton Raphson Method

The method states that if xn is an approximate to an exact root of the equation f(x)

= 0 then a better approximation is in general given by xn+1 where

xn+1 = xn - f(xn)/f(xn)

Example illustrates the method

Example:

Using Newton Raphson method find the solution of e-x

- sinx = 0 correct up to four

decimal places near x0 = 0.5.

Sol:

f(x) = e-x

sinx

hence

f(x) = -(e-x

+ cosx)

Now

xn+1 = xn - e-x

sinx /-(e-x

+ cosx)

for n = 0 x1 = 0.5885

for n = 1 x2 = 0.586

for n = 2 x3 = 0.5885

hence x = 0.5885 is the root of the equation.


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#include

#include

#include

#include

double f(double x)

{

double value;

value=1.0/exp(x)-sin(x);

return value;

}

double df(double x)

{

double value;

value=-(1.0/exp(x)+cos(x));

return value;

}

void main()

{

clrscr();

double init,Xn1,Xn,ans;

int i;

ofstream res("New_Raph.txt");

cout


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cin>>init;

Xn=init;

for(i=1;i


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Secant Method

The basic disadvantage with Newton Raphson method is that it needs to evaluate

the derivative of f(x). This can be avoided by using the definition of derivatives.

That is we can replace f(x) with the gradient of the secant of f(x) defined as

f'(xn) ~= f(xn) - f(xn-1) / xn xn-1

putting this in

xn+1 = xn - f(xn)/f(xn)

we get

xn+1 = xn f(xn) (xn xn-1) / f(xn) - f(xn-1)

on solving we get

xn+1 = xn-1 f(xn) xn f(xn-1) / f(xn) f(xn-1)

this is known as secant formula. Note that this formula requires two values namely

x0 and x1, between whom the root lays, for the evaluation of the formula.

Example:

Use the secant method to estimate the root of the equation f(x) = sinx 5x + 2 = 0,

given that x0 = 0.4 and x1 = 0.6.

Sol:

For n = 1 the formula becomes

x2 = x0 f(x1) x1 f(x0) / f(x1) f(x0)

putting values and solving we find x2 = 0.494, x3 = 0.4950 and x4 = 0.4950.

hence 0.4950 is the root of the equation.


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#include

#include

#include

#include

double f(double x)

{

double value;

value=sin(x)-5*x+2;

return value;

}

void main()

{

clrscr();

double init,final,Xn1,Xn,Xn_1,ans;

int i;

ofstream res("Secant.txt");

coutinit>>final;

Xn=init;

Xn_1=final;

for(i=1;i


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res


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The Solution of Linear Systems AX = B

Simultaneous linear equations arise in many practical situations. These equations

can be solved either by direct method or by iterative methods. We discuss bothdirect and indirect (iterative) methods.

Consider the system of algebraic equations

a11x1+ a12x2+ a13x3+. a1nxn = c1

a21x1+ a22x2+ a2nxn = c2

..

..

an1x1+ an2x1+ an3x1+ .. annx1 = cn

we can write the above system as

AX = C

Where A is

a11 a12 a13 a1n

a21 a22 a23 a1n

an1 an2 an3 ann

And X is And B is

x1 b1x2 b2

. .

. .

xn bn


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Gauss Jordan Method

In this method we eliminate coefficient of x1 from each equation except from first,

then eliminate x2 from second equation and so on for the next equations. i.e.

making the diagonal elements of the matrix of the coefficients equal to 1 and all

others equal to zero.

Example:

Solve the following system of equation using Gauss Jordan method

2x1+ 2x2+4x3 = 18

1x1+ 3x2+2x3 = 13

3x1+ 1x2+3x3 = 14Sol:

x1 x2 x3 c

2 2 4 181 3 2 13

3 1 3 14

1 1 2 9

1 3 2 13

3 1 3 14

1 1 2 90 2 0 4

0 -2 -3 -13

1 1 2 90 1 0 2

0 -2 -3 -13

1 0 2 7

0 1 0 2

0 0 -3 -9

1 0 2 7

0 1 0 20 0 1 3

1 0 0 1

0 1 0 2

0 0 1 3

Hence x1=1, x2 = 2 and x3 = 3


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#include

#include

#include

#include#include

void main()

{

clrscr();

cout


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//muliplying first element of row 3 with 1 of 1st row

//to make 1st element of row 3 zero

arr[2][0] = -mt2*arr[0][0]+arr[2][0];

arr[2][1] = -mt2*arr[0][1]+arr[2][1];

arr[2][2] = -mt2*arr[0][2]+arr[2][2];

arr[2][3] = -mt2*arr[0][3]+arr[2][3];

//now making 2nd element of 2nd row equal to 1

t=1;

if (arr[1][1] != 1) // start second pivot

{

t = (1/(arr[1][1]));

arr[1][1] = t * arr[1][1];

}

//no need to multi arr[1][0] with t since it is zero

arr[1][2] = t * arr[1][2];arr[1][3] = t * arr[1][3];

//making 2nd element of 1st row zero

float mt3=arr[0][1];

arr[0][1] = -mt3*arr[1][1]+arr[0][1];

arr[0][2] = -mt3*arr[1][2]+arr[0][2];

arr[0][3] = -mt3*arr[1][3]+arr[0][3];

//making second element of 3rd row zero

float mt4 = arr[2][1];

arr[2][1] = -mt4*arr[1][1]+arr[2][1];

arr[2][2] = -mt4*arr[1][2]+arr[2][2];

arr[2][3] = -mt4*arr[1][3]+arr[2][3];

t=1;

if (arr[2][2] != 1) // start third pivot

{

t = (1/(arr[2][2]));

arr[2][2] = t * arr[2][2];

}

//making 3rd element of row three equal to 1

arr[2][3] = t * arr[2][3];//making remaining elements of row 2 and 3 zero


arr[1][2] = -mt5*arr[2][2]+arr[1][2];

arr[1][3] = -mt5*arr[2][3]+arr[1][3];



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arr[0][2] = -mt6*arr[2][2]+arr[0][2];

arr[0][3] = -mt6*arr[2][3]+arr[0][3];

//prining the solved matrix

for(i=0; i


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Cramers Rule

It stated that if AX = C is the system of equations and D is the determinant of A,

D1 is the determinant of A after replacing the 1st

column of A be C, and D2 is the

determinant of A after replacing the 2nd column of A by C and D3 is the

determinant of A after replacing the 3rd

column of A by C, than

x1 = D1/D

x2 = D2/D

x3 = D3/D

Example:

Find the solution of3x1+x2+x3=3x1-3x2+x3=5

x1+x2+4x3=4

Using Cramers Rule

Sol:

D = -38

D1= -38

D2= 38

D3= 38

Hence

x1 = 1

x2 = -1

and

x3 = 1


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#include

#include

#include

#include#include

void main()

{

clrscr();

cout


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float x3 = D3/D;

delay(700);

cout


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Jacobi Method

Given a system of linear equations

AX = C

Where the diagonal elements of A are none zero. The system can be written as

ak+1

= Bxk+b

where the elements of B are

bij = -aij / aii

and bii=0

initial guess is x0

= (0,0,0,0,.)T

and continue the process of iteration for different

values of k, unless we get the required accuracy.For the system

a11x1+a12x2+a13x3 = a14, a21x1+a22x2+a23x3 = a24, a31x1+a32x2+a33x3 = a34

The iterations equations can be written as

x1k+1

= 1/a11(a14-a12x2k-a13x3

k), x2

k+1= 1/a22(a24-a21x1

k-a23x3

k), x1

k+1= 1/a33(a34-a31x1

k-

a33x3k)

Example:

Consider the system, 8x+y-z = 8, x-7y+2z = -4, 2x+y+9z =12 Find sol with x,y and

z starting at zero.

Sol:

Setting the equations as

x = 1/8(8-y+z) -1, y = 1/7(4+x+2z) -2, z = 1/9(12-2x-y) -3

putting x,y and z = 0, we get

x = 1.095, y = 1.095 and z = 1.048

putting these values in 1,2 and 3

and repeating the process enough time we see that the solution

converges to the values.

x = 1.000 , y = 1.000 and z = 1.000


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#include

#include

#include

#include#include

void main()

{

clrscr();

cout


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X2=X2_it;

X3=X3_it;

}

}

delay(700);

cout


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Gauss Seidel iterative Method

The Jacoby method is difficult to implement because we need the equations to be

in such an order that the system becomes diagonally dominant also the

convergence is slow. Another method that is also diagonally dominant (i.e. largest

values must come in the diagonal elements of matrix A) but converges faster that

Jacoby method is Gauss Seidel method. The iterations equations of this method are

slightly different from the Jacoby method as given below.

Consider the system

a11x1+a12x2+a13x3 = a14,

a21x1+a22x2+a23x3 = a24,

a31x1+a32x2+a33x3 = a34

The iterations equations for Gauss Seidel method are

x1k+1

= 1/a11(a14-a12x2k-a13x3

k),

x2k+1

= 1/a22(a24-a21x1k+1

-a23x3k),

x1k+1

= 1/a33(a34-a31x1k+1

-a33x3k+1

)

That is in second and third equations the values already calculated are used.


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#include

#include

#include

#include#include

void main()

{

clrscr();

cout


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X2=X2_it;

X3=X3_it;

}

}

delay(700);

cout


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Excrcise

Consider the network as shown below

Applying Kirchoff s loop rule we obtain

Solve the network #2 for the currents given the following value for the resistors and

battery:

Consider the network

Applying Kirchoff s loop rule we obtain

Solve the network #3 for the currents given the following value for the resistors and

batteries:


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Numerical integration

Riemanns Sum

Let be continuous over the interval , and let

be a partition, then the definite integral is given by

b n

f(x) dx = lim f(xn-1) hna 1

Where xn-1 [x0, xn] and the mesh size of the partition goes to zero in the

"limit," i.e. h 0 as n.Example:

Compute the integral of the function f(x) = x2

over the interval [1, 2] with n=20.

Sol:

Here x0 = 1, x1 = 1+0.05, x2 = 1.2(0.05),x20 = 2, h = 2-1/20 = 0.05

2

Hence x2 dx = f[x0] h+ f[x1] h+ f[x2] h+ f[xn-1] h+ f[xn] h1

= 2.25875

Exact answer is = 2.3333333333


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#include

#include

#include

#include

#include

float f(float x)

{

float value;

value=pow(x,2);

return value;

}

void main()

{

clrscr();

int n;

float a,b,h,Fx,sum=0.0;

ofstream res("Riemann.txt");

couta>>b>>n;

h=(b-a)/n;

clrscr();

//performing itterations


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cout


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Trapezoidal Rule

Consider y = f(x) continuous over [a=x0, b=xn], where b = a + nh. n = 1,2,3 than

by definition trapezoidal rule states that.b

f(x) dx = h/2 [f(x0)+2{f(x1)+ f(x2)+ f(xn-1)}+f(xn)]a

where

h = b-a/n

and

x0 = a, x1 = a+h, x2 = a+2h, x3 = a+3h xn-1 = a+(n-1)h+ xn = b


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C++ program for Trapezoidal Rule

#include

#include

#include

#include

#include

float f(float x)

{

float value;

value=2+2*cos(2*sqrt(x));

return value;

}

void main()

{

clrscr();

int n;

float a,b,h,sum=0.0;

ofstream res("Trapezoidal.txt");

couta>>b>>n;

h=(b-a)/n;

sum=f(a)+f(b);

clrscr();



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cout


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Simpsons 1/3 Rule

Consider y = f(x) continuous over the interval [a, b]. Suppose that the interval is

divided into n subintervals (where n must be an even number) with

h = b-a/n

such that x0 = a, x1 = a+h, x2 = a+2h, x3 = a+3h xn-1 = a+(n-1)h+ xn = b. Than

Simpsons Rule states that

b

f(x) dx = h/3 [f(x0)+2{f(x2)+ f(x4)+ f(xn-2)}+ 4{f(x1)+ f(x3)+ f(xn-1)}+f(xn)]a

is an approximation of the integral of f(x) over the interval [a, b]


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C++ program for Simpsons 1/3 Rule

#include

#include

#include

#include

#include

float f(float x)

{

float value;

value=2+2*cos(2*sqrt(x));

return value;

}

void main()

{

clrscr();

int n;


ofstream res("Simpsons3.txt");

couta>>b>>n;

h=(b-a)/n;

sum=f(a)+f(b);

clrscr();



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for(int k=1;k


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Simpsons 3/8 Rule

Consider y = f(x) continuous over the interval [a, b]. Suppose that the interval is

divided into n subintervals (where n must divisible by 3) with

h = b-a/n

such that x0 = a, x1 = a+h, x2 = a+2h, x3 = a+3h xn-1 = a+(n-1)h+ xn = b. Than

Simpsons 3/8 Rule states that

b

f(x) dx = 3h/8 [f(x0)+3{f(x1)+ f(x2)+ f(xn-1)}+f(xn)]a

Is an approximation of the integral of f(x) over the interval [a, b]


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C++ program for Simpsons 3/8 Rule

#include

#include

#include

#include

#include

float f(float x)

{

float value;

value=3*(1/exp(x))*sin(pow(x,2))+1;

return value;

}

void main()

{

clrscr();

int n;


ofstream res("Simpsons3-8.txt");

couta>>b>>n;

h=(b-a)/n;

sum=(f(a)+f(b));

clrscr();



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for(int k=1;k


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Monte Carlo -1 Method

Monte Carlo methods can be thought of as statistical simulation methods that

utilize sequences of random numbers to perform the simulation. The name "Monte

Carlo'' was coined by Nicholas Constantine Metropolis (1915-1999), because of

the similarity of statistical simulation to games of chance, and because Monte

Carlo is a center for gambling and games of chance.

From previous discussion of trapezoidal rule and Simpsons rule we found out that

the integral of a function f(x) over the interval [a, b] can be written as according to

mid point rule. (Where c is the mid point between a and b)

.

Or equivalently

.

The above equations are for n = 1, i.e we are using the whole interval from a to b

and have not made any subintervals. If we do divide the interval into sub interval

than f(c) will be the average of all the values of f(ck) in the sub interval. In that

case the above relation can be written as

.

using h = (b-a)/n

, where .

With error

, where .


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The Algorithm for the Monte Carlo method goes like this.

Pickn randomly distributed points in the interval .

Determine the average value of the function

.

Compute the approximation to the integral

.

An estimate for the error is

, where

Every time a Monte Carlo simulation is made using the same sample size itwill come up with a slightly different value. Larger values of will

produce more accurate approximations. The values converge very slowly of

the order .


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Following Excel Sheet demonstrate the procedure for Monte Carlo Method

In column A we have used =RAND()*PI()/2 to generate 100random numbers between 0 and Pi.

In column B we have calculated the values if function f(x) = Cosxfrom the values of the random numbers generated in column ausing the function =COS(A#) where # represent the number of

cell just next to the cell of column B.

In cell C2 we calculate the average of the values of column Busing the function =AVERAGE(B2:B101).

In cell D2 we have calculated the numerical answer using(PI()/2)*C2. Which is the required numerical solution.

Next four columns contain the calculations of the error in theroutine. According to the formula given above.


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C++ program for Monte Carlo -1 Method

#include

#include

#include

#include

#include

#include

double f(double x)

{

double value;

value=cos(x);

return value;

}

void main()

{

clrscr();

double a,b;

couta>>b;

int n=1;

coutn;

double random[1000];


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//generating random numbers

for(int j=1;j


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getch();

}Use Monte Carlo Method to find

the numerical solution of

1. f(x) = cos x, with a=0 and b = Pi/2 = 1.570796.Ans : 1.002148

2. Let . Use the Monte Carlomethod to calculate approximations to the

integral . Use 500 random

number.

Ans : 5.2659091

3. Let . Use the MonteCarlo method to calculate approximations to

the integral . Use 500

Random Number

Ans: 1.035687


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Monte Carlo -2 Method

The Monte Carlo method can be used to numerically approximate the value of a

double integral.

The algorithm for Monte Carlo method with two variable or double integrals goes

like this

Pick n randomly distributed points inthe rectangle

Determine the average value of the function

.

Compute the approximation to the integral

.

An estimate for the error is

, where .


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C++ program for Monte Carlo -2 Method

#include

#include

#include

#include

#include

#include

double f(double x,double y)

{

double value;

value=4-pow(x,2)-pow(y,2);

return value;

}

void main()

{

clrscr();

double a,b;

couta>>b;

double c,d;

coutc>>d;

int n=1;

coutn;


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double random1[1000];//For the first variable say x

//generating random numbers

for(int j=1;j


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double ans=avg*(b-a)*(d-c);//computes answer of integral

delay(700);

cout


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Ordinary Differential Equations

Euler Method

To approximate a solution to the equation

y(x) = f(x, y(x))

Where x is independent variable and y is dependent on x.

With initial condition

y(x0) = y0

On an interval Ix = [x0, x], choose a small value for h > 0 and an integer n such that

x0 + nh = x. n =1,2,3,

Compute y1, y2, . . . , ym+1 using the difference equation

ym+1 = ym + hf(xm, ym). m = 0,1,2,n-1

Then ym+1= yx is an approximation for y(x0 + nh) = y(x).


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C++ program for Eulers Method for y(x) = f(x, y(x)) = 0.04x, y(0) =50, Ix = [0, 50]

#include

#include

#include

#include

#include

double f(double a,double b){

double value;

value=(0.04)*b;

return value;

}

void main()

{

again:

clrscr();

int n;

double X0,Y0,X,Y,h;

coutX0;

coutY0;

cout


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cin>>X;

coutn;

if(n


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Euler Improved Method


y(x) = f(x, y(x))



y(x0) = y0


x0 + nh = x. n =1,2,3,

Compute y1, y2, . . . , ym+1 using the difference equationym+1 = ym + h/2(f(xm+h, z)+f(xm,ym)). m = 0,1,2,n-1

where z = ym+1 = ym + h f(xm,ym) m = 0,1,2,n-1



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C++ program for Eulers Improved Method for y(x) = f(x, y(x)) = 0.04x, y(0) =50,

Ix = [0, 50]

#include

#include#include

#include

#include

double f(double a,double b)

{

double value;

value=(0.04)*b;

return value;

}

void main()

{

again:

clrscr();

int n;

double X0,Y0,X,Y,h;

coutX0;coutY0;

coutX;


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coutn;

if(n


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Runge Kutta 4th order Method


y(x) = f(x, y(x))



y(x0) = y0


x0 + nh = x. n =1,2,3,

Compute y1, y2, . . . , ym+1 using the difference equationk1= h* f(xm,ym)

k2 = h* f(xm+h/2,ym+k1/2)

k3 = h* f(xm+h/2,ym+k2/2)

k4 = h* f(xm+h/2,ym+k3)

ym+1 = ym + 1/6(k1+2k2+2k3+k4). m = 0,1,2,n-1



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C++ program for R-K 4th

order Method for y(x) = f(x, y(x)) = 0.04x, y(0) =50, Ix =

[0, 50]

#include

#include#include

#include

#include


{

double value;

value=(0.04)*b;

return value;

}

void main()

{

again:

clrscr();

int n;

double X0,Y0,X,Y,h;

coutX0;

coutY0;

coutX;

cout


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cin>>n;

if(n


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Runge Kutta Fehlberg (RKF45) Method


y(x) = f(x, y(x))



y(x0) = y0


x0 + nh = x. n =1,2,3,

Compute y1, y2, . . . , ym+1 using the difference equationk1= h* f(xm,ym)

k2 = h* f(xm+h/4,ym+k1/4)

k3 = h* f(xm+3*h/8,ym+3*k2/32+9*k2/32)

k4 = h* f(xm+12*h/13,ym+1932*k1/2197-7200*k2/2197+7296*k3/2197)

k5 = h* f(xm+12*h/13,ym+439*k1/216-8*k2+3680*k3/513-845*k4/4104)

k6 = h* f(xm+h/2,ym-8*k1/27+2*k2-3544*k3/2565+1859*k4/4104-11*k5/40)

ym+1 = ym + 16*k1/135+6656*k3/12825+28561*k4/56430-9*k5/50+2*k6/55).

m = 0,1,2,n-1



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C++ program for RKF45 Method for y(x) = f(x, y(x)) = 0.04x, y(0) =50, Ix = [0, 50]

#include

#include

#include

#include

#include


{double value;

value=(0.04)*b;

return value;

}

void main()

{

again:

clrscr();

int n;

double X0,Y0,X,Y,h;

coutX0;

coutY0;

coutX;


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coutn;

if(n


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delay(700);

cout

numerical methods with c codes

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