yes
This is the codes for Newton-Raphson method for solving Quadratic equations
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define F(x)(x*x*x)-(4*x)
#define FD(x)(3*x*x)-4
#define MAXIT 20
void main()
{
int count;
float x0,x1,fx,fdx;
clrscr();
printf("NEWTON-RAPHSON METHOD\n");
printf("---------------------\n");
printf("initial value:");
scanf("%f",&x0);
count=1;
begin:
fx=F(x0);
fdx=FD(x0);
x1=(x0-(fx/fdx));
if(fabs((x1-x0)/x1)<0.00001)
{
printf("The root is:%f\n",x1);
printf("Iteration is:%d\n",count);
}
else
{
x0=x1;
count=count+1;
if((count<MAXIT));
goto begin;
}
getch();
}
#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<process.h>
void main()
{
float x0=0,x1=0,a=0,b=0,c=0,f0=0,e=0,df0,slope=.0001;
int i=0,n=0;
clrscr();
printf("\n Enter the interval");
printf("\n Enter the value of x0= ");
scanf("%f",&x0);
printf("\n Enter the error allowed in the root of equation ");
scanf("%f",&e);
printf("\n Enter the cofficients of the equation");
printf("\n Enter the value of a = ");
scanf("%f",&a);
printf("\n Enter the value of b= ");
scanf("%f",&b);
printf("\n Enter the value of c= ");
scanf("%f",&c);
printf("\n Enter the number of iterations= ");
scanf("%d",&n);
for(i=0;i<n;i++)
{
f0=a*x0*x0+b*x0+c;
df0=2*a*x0+b;
if((fabs(df0))<=slope)
{
printf("\n Slope too small so will become divergent from solution");
getch();
exit(0);
}
else
{
x1=x0-(f0/df0);
if(fabs((x1-x0)/x1)<e)
{
printf("\n Root of the equation is %f",x1);
printf("\n No of iterations for calculating the root of the equation is %d",i);
getch();
exit(0);
}
else
{
x0=x1;
}
}
}
printf("\n Solution does not converge in %d iterations",n);
getch();
}
/*programm for modified newton raphson method*/
#include<stdio.h>
#include<math.h>
#define epsilon 1e-6
main()
{
/*This is a programm based on modified
newton raphson method to
find out root of the equation x^3-5*x^2+7*x-3=0*/
double g,g1,v,v1,v2,x,dx;
int converged=0,i;
printf("plz enter the guess value\n");
scanf("%f",&g1);
i=1;
while (converged==0)
{
printf("\n iteration no=%d\n",i);
v=g1*g1*g1-5*g1*g1+7*g1-3;
printf("v=%lf\n",v);
v1=3*g1*g1-10*g1+7;
printf("v1=%lf\n",v1);
v2=6*g1-10;
printf("v2=%lf\n",v2);
x=(v*v1)/(v1*v1-v*v2);
printf("value=%lf\n",x);
g=g1-((v*v1)/(v1*v1-v*v2));
printf("new guess is=%lf\n",g);
dx=((g-g1)/g);
printf("error=%lf\n",dx);
g1=g;
if (fabs(dx)'less than'epsilon)
{converged=1;}
printf("the root of equation is=%lf
i=%d\n",g,i);
i=i+1;
}
}
Function f(x)
x2=x*x
f=x*x2-3*x2+2*x-1
return
end
function f1(x)
f1=3*x*x-6*x+2
return
end
5 write(*,*)'enter initial guess'
read(*,*) x0
write(*,*)'enter tolerrence'
read(*,*) eps
10 x1=x0-f(x0)/f1(x0)
if(abs((x1-x0)/x0).lt.eps) then
write(*,1) x1
1 format('solution=',f10.4)
stop
else
x0=x1
go to 10
endif
end
Ideally, quadratic. Please see the link.
You don't need a flow chart for that; just use the quadratic formula directly; most programming languages have a square root function or method. You would only need to do this in many small steps if you use Assembly programming. The formulae would be something like this: x1 = (-b + sqrt(b^2 - 4*a*c)) / (2 * a) and x2 = (-b - sqrt(b^2 - 4*a*c)) / (2 * a) where a, b, and c are the coefficients of the quadratic equation in standard form, and x1 and x2 are the solutions you want.
An improved root finding scheme is to combine the bisection and Newton-Raphson methods. The bisection method guarantees a root (or singularity) and is used to limit the changes in position estimated by the Newton-Raphson method when the linear assumption is poor. However, Newton-Raphson steps are taken in the nearly linear regime to speed convergence. In other words, if we know that we have a root bracketed between our two bounding points, we first consider the Newton-Raphson step. If that would predict a next point that is outside of our bracketed range, then we do a bisection step instead by choosing the midpoint of the range to be the next point. We then evaluate the function at the next point and, depending on the sign of that evaluation, replace one of the bounding points with the new point. This keeps the root bracketed, while allowing us to benefit from the speed of Newton-Raphson.
It's a method used in Numerical Analysis to find increasingly more accurate solutions to the roots of an equation. x1 = x0 - f(x0)/f'(x0) where f'(x0) is the derivative of f(x0)
You can find this charge by looking online. Many sites can help you to get the chart you need or explain how to make one.
Ideally, quadratic. Please see the link.
how to use gauss programming to find LM unit root test with structural breaks and kpss
Newton and Raphson used ideas of the Calculus to generalize this ancient method to find the zeros of an arbitrary equation.
By knowing how to use the quadratic equation formula.
Start with a quadratic equation in the form � � 2 � � � = 0 ax 2 +bx+c=0, where � a, � b, and � c are constants, and � a is not equal to zero ( � ≠ 0 a =0).
at first the first person to solve the quadratic equation is from the middle kingdom of Egypt. Greeks were also able to solve the quadratic equation but that was on the unproper way. Greeks were able to solve the quadratic equation by geometric method or equlid's method. equlid's method contains only three quadratic equation. dipohantus have also solved the quadratic equations but he have solved by giving only two roots any they both were only of positive signs.After that arbhatya also gave the two formulas for quadratic equation but the bentaguptahave only accepted only one of them after theat some of the Indian mathematican have also solved the quadratic equation who gave the proper definations and formula and in this way quadratic equation have been formed. Prabesh Regmi Kanjirowa National School
using the quadratic formula or the graphics calculator. Yes, you can do it another way, by using a new method, called Diagonal Sum Method, that can quickly and directly give the 2 roots, without having to factor the equation. This method is fast, convenient and is applicable to any quadratic equation in standard form ax^2 +bx + c = 0, whenever it can be factored. It requires fewer permutations than the factoring method does, especially when the constants a, b, and c are large numbers. If this method fails to get answer, then consequently, the quadratic formula must be used to solve the given equation. It is a trial-and-error method, same as the factoring method, that usually takes fewer than 3 trials to solve any quadratic equation. See book titled:" New methods for solving quadratic equations and inequalities" (Trafford Publishing 2009)
Although there is a method for cubics, there are no simple analytical ways.Sometimes you may be able to use the remainder theorem to find one solutions. THen you can divide the original equation using that solution so that you are now searching for an equation of a lower order. If you started off with a cubic you will now have a quadratic and, if all else fails, you can use the quadratic formula.You could use a graphic method. A cubic musthave a solution although that solution need not be rational. A quartic need no have any.Lastly, you could use a numeric method, such as the Newton-Raphson iteration.
It means you are required to "solve" a quadratic equation by factorising the quadratic equation into two binomial expressions. Solving means to find the value(s) of the variable for which the expression equals zero.
I guess you mean the standard quadratic equation, of the form ax^2 + bx + c = 0.There are three main algebraic methods, namely: * Factoring * Completing the square * Using the quadratic formula Since you want five, here are a few more, but they are usually not very convenient to use for this particular type of equation: * Trial and error * Graphic the equation * Diverse iterative methods, such as Newton's method, etc.
In numerical analysis finding the roots of an equation requires taking an equation set to 0 and using iteration techniques to get a value for x that solves the equation. The best method to find roots of polynomials is the Newton-Raphson method, please look at the related question for how it works.
In general, there are two steps in solving a given quadratic equation in standard form ax^2 + bx + c = 0. If a = 1, the process is much simpler. The first step is making sure that the equation can be factored? How? In general, it is hard to know in advance if a quadratic equation is factorable. I suggest that you use first the new Diagonal Sum Method to solve the equation. It is fast and convenient and can directly give the 2 roots in the form of 2 fractions. without having to factor the equation. If this method fails, then you can conclude that the equation is not factorable, and consequently, the quadratic formula must be used. See book titled:" New methods for solving quadratic equations and inequalities" (Trafford Publishing 2009) The second step is solving the equation by the quadratic formula. This book also introduces a new improved quadratic formula, that is easier to remember by relating the formula to the x-intercepts with the parabola graph of the quadratic function.