Implementation of pow function in c language with power of -3/2 or -5/2
I am new to C language. I am trying to implement the formula (r to power of (-3/2)) * exp(t*(r-1/r). I am using pow function but when I keep this equation in for loop it continuously implementing the loop saying that "pow domain error". I want the increment in for loop to be of 0.2. Please help me in implementing this formula. The code is:
#include <stdio.h>
#include <conio.h>
#include <math.h>
void main()
{
double R,r;
float f;
clrscr();
for(r=0;r<=5;r+0.2)
{
R=pow(r,-1.5)*exp(2.5*(r-1)/r);
f=r-R;
printf("value of f is:%f",f);
}
getch();
}
You may want to be aware that:
The C Standard, 7.12.1 [ISO/IEC 9899:2011], defines three types of errors that relate specifically to math functions in . Paragraph 2 states
A domain error occurs if an input argument is outside the domain over which the mathematical function is defined.
Paragraph 3 states
A pole error (also known as a singularity or infinitary) occurs if the mathematical function has an exact infinite result as the finite input argument(s) are approached in the limit.
Paragraph 4 states
A range error occurs if the mathematical result of the function cannot be represented in an object of the specified type, due to extreme magnitude.
The available domain for function pow(x,y) is:
(x > 0) || (x == 0 && y > 0) || (x < 0 && y is an integer)
Which means that you cannot use r = 0 as an argument of pow.
1)
pow(r, -1.5) = r^(-3/2) = 1.0/ r^3/2 = 1.0/sqrt(r*r*r);
2) In C language
main function should return int value and takes void not empty bracket ()
3) Incrementing operation can be done via r = r + 0.2 or r += 0.2
4) To avoid division by zero in (r-1)/r close to 0 value has to be used instead.
5) In order to not loose double precision the variable f should be also declared double.
6) The value of f for r=0 is extremely difficult to calculate. The algorithm is numerically unstable for r = 0. We have division of two infinities. The proper way to solve it is to find function limit at point 0+. Numerically, its is enough to set r close to 0.
7) You may consider much more smaller step for r in order to capture a better plot due to the nature of your f function.
#include<stdio.h>
#include<conio.h>
#include<math.h>
#define ZERO_PLUS 1e-14
void calculate_and_print(double r)
{
double f, f1;
f = r - pow(r, -1.5) * exp(2.5*(r-1)/r);
f1 = r - (1.0 /sqrt(r*r*r)) * exp(2.5*(r-1)/r);
printf("Value of f for r= %f is f= %f f1= %f \n", r, f, f1);
}
int main(void)
{
clrscr();
calculate_and_print(ZERO_PLUS);
for (double r = 0.2; r < 5.0; r += 0.2 )
calculate_and_print(r);
calculate_and_print(5.0);
return 0;
}
Test:
Value of f for r= 0.000000 is f= 0.000000 f1= 0.000000
Value of f for r= 0.200000 is f= 0.199492 f1= 0.199492
Value of f for r= 0.400000 is f= 0.307038 f1= 0.307038
Value of f for r= 0.600000 is f= 0.193604 f1= 0.193604
Value of f for r= 0.800000 is f= 0.051949 f1= 0.051949
Value of f for r= 1.000000 is f= 0.000000 f1= 0.000000
Value of f for r= 1.200000 is f= 0.046058 f1= 0.046058
Value of f for r= 1.400000 is f= 0.166843 f1= 0.166843
Value of f for r= 1.600000 is f= 0.338256 f1= 0.338256
Value of f for r= 1.800000 is f= 0.542116 f1= 0.542116
Value of f for r= 2.000000 is f= 0.765977 f1= 0.765977
Value of f for r= 2.200000 is f= 1.001644 f1= 1.001644
Value of f for r= 2.400000 is f= 1.243810 f1= 1.243810
Value of f for r= 2.600000 is f= 1.489073 f1= 1.489073
Value of f for r= 2.800000 is f= 1.735278 f1= 1.735278
Value of f for r= 3.000000 is f= 1.981075 f1= 1.981075
Value of f for r= 3.200000 is f= 2.225642 f1= 2.225642
Value of f for r= 3.400000 is f= 2.468498 f1= 2.468498
Value of f for r= 3.600000 is f= 2.709387 f1= 2.709387
Value of f for r= 3.800000 is f= 2.948194 f1= 2.948194
Value of f for r= 4.000000 is f= 3.184898 f1= 3.184898
Value of f for r= 4.200000 is f= 3.419535 f1= 3.419535
Value of f for r= 4.400000 is f= 3.652177 f1= 3.652177
Value of f for r= 4.600000 is f= 3.882916 f1= 3.882916
Value of f for r= 4.800000 is f= 4.111856 f1= 4.111856
Value of f for r= 5.000000 is f= 4.339103 f1= 4.339103
Let me know if you have more questions.
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