I have written a program in C which is calculating for me the square root with the heron procedure. x is my number, r is estimated value and steps are steps. I want to output the difference between the exact value and the value obtained by the heron method. But it seems that my function is not correct. For my calculated value I get no value. Can anyone help me?
#include <stdio.h>
#include <math.h>
int heron (x, r, steps)
{
int k = 0;
double xold, xnew;
double rel_error = 1.0;
while(k <= steps && rel_error > 1e-4) {
++k;
xnew = .5 * (xold + x / xold);
rel_error = (xnew - xold) / xnew;
if(rel_error < 0)
rel_error = -rel_error;
xold = xnew;
}
printf("exact value: %.10f\n", sqrt(x));
return (xnew);
}
int main()
{
int x=4, r=10, steps=50;
printf("%f\n", heron(x, r, steps));
return 0;
}
Change int heron (x, r, steps) to double heron(double x, double r, int steps). You need to declare the types of the parameters, and the function works with floating-point values, so it ought to return float or double, not int, and x and r should be double.
Change double xold , xnew; to double xold = r, xnew;. xold must be initialized before it is used.
Change return sqrt(x); to return xold; to return the value that the function calculated.
With this prefix
int heron (x, r, steps)
{
your function is a function that takes an integer x, another integer r, and a third integer steps. Indeed, it also returns an integer.
The algorithm you describe can be implemented in this way:
#include <stdio.h>
#include <math.h>
double heron(double x, double err)
{
double a = x, b = 1.0;
while (a - b > err) {
a = (a + b)/2.0; /* arithmetic mean */
b = x / a; /* approx to geometric mean */
}
return a; /* or b, depending if you want a value in excess or in defect */
}
int main()
{
printf("heron(2.0, 1.0E-10) = %.10f\n", heron(2.0, 1.0E-10));
printf("sqrt(2.0) = %.10f\n", sqrt(2.0));
}
and that will work.
Read about function parameter type definitions in one of the many references of the C programming languages, e.g. "The C programming language" from Brian Kernighan & Dennis Ritchie, for reference.
$ ./heron
heron(2.0, 1.0E-10) = 1.4142135624
sqrt(2.0) = 1.4142135624
$ _
Related
Consider my attempt to implement the Babylonian method in C:
int sqrt3(int x) {
double abs_err = 1.0;
double xold = x;
double xnew = 0;
while(abs_err > 1e-8) {
xnew = (2 * xold + x/(xold* xold))/3;
abs_err= xnew-xold;
if (abs_err < 0) abs_err = -abs_err;
xold=xnew;
}
return xnew;
}
int main() {
int a;
scanf("%d", &a);
printf(" Result is: %f",sqrt3(a));
return 0;
}
Result is for x=27: 0.0000?
Where is my mistake?
While the function returns an int, that value is printed with the wrong format specifier, %f instead of %d.
Change the signature (and the name, if I may) into something like this
double cube_root(double x) { ... }
Or change the format specifier, if you really want an int.
Following the explanation from tutorialspoint, which states, that the basic idea is to implement the Newton Raphson method for solving nonlinear equations, IMHO, the code below displays this fact more clearly. Since there is already an accepted answer, I add this answer just for future reference.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
double rootCube( double a)
{
double x = a;
double y = 1.0;
const double precision = 0.0000001;
while(fabs(x-y) > precision)
{
x = (x + y) / 2.0;
y = a / x / x;
}
return x;
}
int main(int argc, const char* argv[])
{
if(argc > 1)
{
double a =
strtod(argv[1],NULL);
printf("cubeRoot(%f) = %f\n", a, rootCube(a));
}
return 0;
}
Here, in contrast to the original code of the question, it is more obvious, that x and y are the bounds, which are being improved until a sufficiently accurate solution is found.
With modification of the line in the while block, where y is being updated, this code can also be used to solve similar equations. For finding the square root, for example, this line would look like this: y = a / x.
I am trying to use the function that I was given by my professor to calculate the integral of a polynomial function (polynomial such as: ax^2+bx+c). the function is:
double numbericalIntegration(double a ,double b ,double(*func)(double)){
double delta = (b - a)/32;
double sum=0, x;
for(x= a+0.5*delta; x<b ; x+=delta)
{
sum+=(*func)(x);
}
return sum*delta;
}
I changed a lot in order to integrate a polynomial function. but I was get the answer 0. why is that? and I'd appreciate if anybody tried to correct my work. my code is:
double integralPoly(double x, double a, double b, double c){
return (a*pow(x,3))/3 +(b*pow(x,2))/2 + (c*x);
}
double numbericalIntegration(double a ,double b ,double(*func)(double,double,double,double), double firstNum, double secondNum, double thirdNum){
double delta = (b - a)/32;
double sum=0, x;
for(x= a+0.5*delta; x<b ; x+=delta)
{
sum+=(*func)(x, firstNum, secondNum, thirdNum);
}
return sum*delta;
}
int main()
{
double (*func)(double,double,double,double);
func = integralPoly;
double sum = numbericalIntegration(2,4,func,1,1,4);
printf("sum = %d",sum);
return 0;
}
You need to change two things. First your polynomial function doesn't make any sense. You said it needs to be in the form of ax^2+bx+c but in your code polynomial is (ax^3)/3+(bx^2)/2+c*x. Your function should be:
double integralPoly(double x, double a, double b, double c){
return (a*pow(x,2)) +(b*x) + c;
}
Also you need to change your printf. %d is integer type specifier and you need double, so you need to use %f for example:
printf("sum = %f",sum);
Now the output of your program is:
sum = 32.666016
which is correct for your parameters.
When I run my code, for Y I am consistently getting the value -2147483648, regardless of what value y was fed into my equation.
Here is my code.
#define MAX 1000
#define EQ(y) ((2*(pow(y, 4)))+1)
int check(int value);
int main()
{
int i, y, x;
for(y = 1; y < MAX; y++)
{
i = EQ(y);
if(check(i))
printf("combination found: x = %d, y = %d", sqrt(i), y);
}
}
int check(int value)
{
int x = sqrt(value);
if ((x*x) == value)
return 1;
else
{
return 0;
}
}
After reviewing my code, I realized my problem was with my "int x = sqrt(value)". Aside from the problem with the "value" variable being an int, of course, a bogus value was still being returned due to the fact that the purpose of check is to evaluate whether or not (2*(pow(y, 4)))+1) returned a perfect whole square for any given value of y, and this was not possible due to variable x in check(double value) being datatype integer.
UPDATE: I rewrote my code as follows. I still don't get any correct returns
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
/*
* the solution I implemented basically involved dropping x from the equation, solving for y, checking to see if it has a
* perfect square root. if it does, then x = the squareroot of y in the function EQ.
* original problem: for equation x^2 - 2y^4 + 1 = 0, find all possible solutions up to arbitrary maximum
*/
#define MAX 100000
#define EQ(g) (((pow(g, 4.0)))+1)
int check(double value);
int main()
{
int y, x;
double i;
for(y = 1; y < MAX; y++)
{
i = EQ(y);
if(x = check(i) > 0)
printf("combination found: x = %d, y = %d\n", y, x);
}
}
int check(double value)
{
double x = sqrt(value);
int n = (int) x;
printf("%d\n%f\n%f\n", n*n, value, x);
if (n*n == value)
return n*n;
else
return 0;
}
Read the comments are the top of my code, and the purpose for this selection should be pretty obvious.
You don't have a prototype for double pow(double, double); so the compiler implicitly assumes its signature is int pow(int, int);. Not good!
The solution is to #include the appropriate header at the top of your .c file.
#include <math.h>
Make sure you enable warnings, and if they're already enabled, pay attention to them! Your compiler should warn you about the missing prototype. (It should also spit out a similar warning for printf.)
pow() returns double and you are using integer i to store the return value.
Due to type promotion during expression evaluation the expression:
((2*(pow(y, 4)))+1)
will give a double value and you are storing this in integer type which will give unexpected results.
In reference to your updated question, this line:
if(x = check(i) > 0)
needs to be parenthesized:
if((x = check(i)) > 0)
This is the declaration of pow:
double pow(double x, double y)
Which means it operates in double. By using int instead, variable y is overflowing.
I'm new to C, and quite unfamiliar with writing any program larger than a few lines.
I'm trying to write a model for an object in freefall acted upon by gravity and drag. It uses Eulers method to solve two first order differential equations, one for position and one for velocity.
So we have: F = m dv/dt = -mg - k|v|v and dy/dt = v
These are solved by: Vn+1 = Vn - (delta t*(g+(k/m)|Vn|Vn)) and Yn+1 = Yn + (delta t * Vn)
(In this Vn+1 is the n+1th term etc.)
In my program i've tried to have two functions, for position and velocity, which work by passing pointers with Y and V values between them and the main function, and it should then loop until Y=0 and print off the values at each step.
When I run it it comes up with something like this: http://imgur.com/DNHIhHI
Could anyone tell me either what is wrong with this, or if I need to use a different approach completely?
Many Thanks, Code below
#include <stdio.h>
void Velocity(double *ptr, double m, double k, double t);
void Position(double *pst, double *ptr, double t );
int main()
{
double k = 18833.5608;
double t = 0;
double m;
double speed = 0;
double *ptr = &speed;
double y = 1000;
double *pst = &y;
printf("Enter mass of object: \n");
scanf("%f" , &m);
do
{
Velocity( ptr, m, k, t );
printf("Velocity at time %f is: %f\n" , t, speed);
Position( pst, ptr, t);
printf("Position at time %f is: %f\n" , t , y);
t++;
}
while((y>0));
return 0;
}
void Velocity(double *velo, double m, double k, double t)
{
double g = 9.80665;
*velo = *velo - (t*(g+((k/m)*fabs(*velo)**(velo))));
}
void Position(double *Y , double *velo, double t )
{
*Y = *Y+(t*(*velo));
}
When writing programs that do calculations -- in any language, not just C -- try to make the code that does the computation take arguments and return results but not mutate variables. That is, do not write:
void do_calculation( double * result, double x, double y)
{
*result = x + y;
}
...
double r;
do_calculation(&r, 123, 456);
instead write
double do_calculation(double x, double y)
{
return x + y;
}
...
double r = do_calculation(123, 456);
Make sense?
If you want to modify an existing value, again, don't pass it in as a variable to be mutated. Instead of
void do_calculation(double * accumulator, double x, double y)
{
*accumulator = *accumulator + x + y;
}
...
double r = 10;
do_calculation(&r, 123, 456);
instead say
double do_calculation(double original, double x, double y)
{
return original + x + y;
}
...
double r = 10;
r = do_calculation(r, 123, 456);
Now, once you've got your program architected more sensibly, you need to learn how to debug small programs. Some good advice on that subject can be found here:
http://ericlippert.com/2014/03/05/how-to-debug-small-programs/
A misconcept. I believe you're trying to solve the equations by using small increments of time. Nothing wrong with that, just make the time increment as small as possible, and correct the formulas:
#include <stdio.h>
#include <math.h>
void Velocity(double *velocity, double m, double k, double t)
{
double g = 9.80665;
double velo = *(velocity);
velo = velo - (t*(g+((k/m)*abs(velo)*(velo))));
*(velocity)=velo;
}
void Position(double *position , double *velocity, double t )
{
double Y = *(position);
double velo = *(velocity);
Y = Y+(t*(velo));
*(position)=Y;
}
int main()
{
double k = 18833.5608;
double t = 0;
double dt = 0.001; //making a small increment of time
double m=100;
double speed = 0;
double y = 1000;
//printf("Enter mass of object: \n");
//scanf("%f" , &m);
do
{
Velocity( &speed, m, k, dt );
printf("Velocity at time %f is: %f\n" , t, speed);
Position( &y, &speed, dt);
printf("Position at time %f is: %f\n" , t , y);
t+=dt; //increment time by delta t
}
while((y>0));
return 0;
}
Little bit of a 2 parter. First of all im trying to do this in all c. First of all I'll go ahead and post my program
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <omp.h>
#include <string.h>
double f(double x);
void Trap(double a, double b, int n, double* integral_p);
int main(int argc, char* argv[]) {
double integral=0.0; //Integral Result
double a=6, b=10; //Left and Right Points
int n; //Number of Trapezoids (Higher=more accurate)
int degree;
if (argc != 3) {
printf("Error: Invalid Command Line arguements, format:./trapezoid N filename");
exit(0);
}
n = atoi(argv[2]);
FILE *fp = fopen( argv[1], "r" );
# pragma omp parallel
Trap(a, b, n, &integral);
printf("With n = %d trapezoids....\n", n);
printf("of the integral from %f to %f = %.15e\n",a, b, integral);
return 0;
}
double f(double x) {
double return_val;
return_val = pow(3.0*x,5)+pow(2.5*x,4)+pow(-1.5*x,3)+pow(0*x,2)+pow(1.7*x,1)+4;
return return_val;
}
void Trap(double a, double b, int n, double* integral_p) {
double h, x, my_integral;
double local_a, local_b;
int i, local_n;
int my_rank = omp_get_thread_num();
int thread_count = omp_get_num_threads();
h = (b-a)/n;
local_n = n/thread_count;
local_a = a + my_rank*local_n*h;
local_b = local_a + local_n*h;
my_integral = (f(local_a) + f(local_b))/2.0;
for (i = 1; i <= local_n-1; i++) {
x = local_a + i*h;
my_integral += f(x);
}
my_integral = my_integral*h;
# pragma omp critical
*integral_p += my_integral;
}
As you can see, it calculates trapezoidal rule given an interval.
First of all it DOES work, if you hardcode the values and the function. But I need to read from a file in the format of
5
3.0 2.5 -1.5 0.0 1.7 4.0
6 10
Which means:
It is of degree 5 (no more than 50 ever)
3.0x^5 +2.5x^4 −1.5x^3 +1.7x+4 is the polynomial (we skip ^2 since it's 0)
and the Interval is from 6 to 10
My main concern is the f(x) function which I have hardcoded. I have NO IDEA how to make it take up to 50 besides literally typing out 50 POWS and reading in the values to see what they could be.......Anyone else have any ideas perhaps?
Also what would be the best way to read in the file? fgetc? Im not really sure when it comes to reading in C input (especially since everything i read in is an INT, is there some way to convert them?)
For a large degree polynomial, would something like this work?
double f(double x, double coeff[], int nCoeff)
{
double return_val = 0.0;
int exponent = nCoeff-1;
int i;
for(i=0; i<nCoeff-1; ++i, --exponent)
{
return_val = pow(coeff[i]*x, exponent) + return_val;
}
/* add on the final constant, 4, in our example */
return return_val + coeff[nCoeff-1];
}
In your example, you would call it like:
sampleCall()
{
double coefficients[] = {3.0, 2.5, -1.5, 0, 1.7, 4};
/* This expresses 3x^5 + 2.5x^4 + (-1.5x)^3 + 0x^2 + 1.7x + 4 */
my_integral = f(x, coefficients, 6);
}
By passing an array of coefficients (the exponents are assumed), you don't have to deal with variadic arguments. The hardest part is constructing the array, and that is pretty simple.
It should go without saying, if you put the coefficients array and number-of-coefficients into global variables, then the signature of f(x) doesn't need to change:
double f(double x)
{
// access glbl_coeff and glbl_NumOfCoeffs, instead of parameters
}
For you f() function consider making it variadic (varargs is another name)
http://www.gnu.org/s/libc/manual/html_node/Variadic-Functions.html
This way you could pass the function 1 arg telling it how many "pows" you want, with each susequent argument being a double value. Is this what you are asking for with the f() function part of your question?