Mixing
creating a Sin formula in the c programming language
Hello everyone I'm trying to implement a programm like the fomular sin
The program will compile but when running it i am not getting the correct values back from my inputs. I'm still getting a negativ value.
can someone help me please ? I had a look at the other posts but that didn't help me :(.
my code is:
#include <stdio.h>
#include <stdlib.h>
int fac (int a) { // fac. => factorial and i is for the loop
int i,fac;
fac=1;
for (i=1; i<=a; i++){
fac=fac*i;
}
return fac;
}
int power_func(int x,int y) // x is exponent and y is the number that would be multiplied by itself.
{
int i;//i is for the loop
int ret = 1;
for(i=1;i<=x;i++)
{
ret *= y;
}
return ret;
}
int main()
{
int num,denom,i;//num. is numerator and denom. is denominator
int sin,x,result=0;
printf("Enter the number of x n");
scanf("%d",&x);
for(i=0;i<x;i++)
{
num= power_func(2*i+1,x);
denom=fac((2*i+1));
sin=power_func(i,-1)*num/denom;
result =result+sin;
printf("%d n",result);
}
return 0;
}
c visual-c++ sin
asked Nov 20 '18 at 20:51
AmerovAmerov
397
|
show 9 more comments
Hello everyone I'm trying to implement a programm like the fomular sin
The program will compile but when running it i am not getting the correct values back from my inputs. I'm still getting a negativ value.
can someone help me please ? I had a look at the other posts but that didn't help me :(.
my code is:
#include <stdio.h>
#include <stdlib.h>
int fac (int a) { // fac. => factorial and i is for the loop
int i,fac;
fac=1;
for (i=1; i<=a; i++){
fac=fac*i;
}
return fac;
}
int power_func(int x,int y) // x is exponent and y is the number that would be multiplied by itself.
{
int i;//i is for the loop
int ret = 1;
for(i=1;i<=x;i++)
{
ret *= y;
}
return ret;
}
int main()
{
int num,denom,i;//num. is numerator and denom. is denominator
int sin,x,result=0;
printf("Enter the number of x n");
scanf("%d",&x);
for(i=0;i<x;i++)
{
num= power_func(2*i+1,x);
denom=fac((2*i+1));
sin=power_func(i,-1)*num/denom;
result =result+sin;
printf("%d n",result);
}
return 0;
}
c visual-c++ sin
asked Nov 20 '18 at 20:51
AmerovAmerov
397
Yourpower_funcis not correct. It computesy^(2^x).
– lurker
Nov 20 '18 at 20:54
@lurker I have changed this line to num= power_func(x, 2*i+1); but I still get a wong answer like : Enter the number of x 3( ,-1.000000 ,1092.000000 ,4347.000000) the result
– Amerov
Nov 20 '18 at 20:58
See my initial comment about it being incorrectly calculated. Sorry, I was mistaken about the backward part.
– lurker
Nov 20 '18 at 20:59
ohh mmm it's ok @lurker
– Amerov
Nov 20 '18 at 21:04
1
More specifically,y = y*yis a problem...
– lurker
Nov 20 '18 at 21:07
|
show 9 more comments
Hello everyone I'm trying to implement a programm like the fomular sin
The program will compile but when running it i am not getting the correct values back from my inputs. I'm still getting a negativ value.
can someone help me please ? I had a look at the other posts but that didn't help me :(.
my code is:
#include <stdio.h>
#include <stdlib.h>
int fac (int a) { // fac. => factorial and i is for the loop
int i,fac;
fac=1;
for (i=1; i<=a; i++){
fac=fac*i;
}
return fac;
}
int power_func(int x,int y) // x is exponent and y is the number that would be multiplied by itself.
{
int i;//i is for the loop
int ret = 1;
for(i=1;i<=x;i++)
{
ret *= y;
}
return ret;
}
int main()
{
int num,denom,i;//num. is numerator and denom. is denominator
int sin,x,result=0;
printf("Enter the number of x n");
scanf("%d",&x);
for(i=0;i<x;i++)
{
num= power_func(2*i+1,x);
denom=fac((2*i+1));
sin=power_func(i,-1)*num/denom;
result =result+sin;
printf("%d n",result);
}
return 0;
}
c visual-c++ sin
asked Nov 20 '18 at 20:51
AmerovAmerov
397
Hello everyone I'm trying to implement a programm like the fomular sin
The program will compile but when running it i am not getting the correct values back from my inputs. I'm still getting a negativ value.
can someone help me please ? I had a look at the other posts but that didn't help me :(.
my code is:
#include <stdio.h>
#include <stdlib.h>
int fac (int a) { // fac. => factorial and i is for the loop
int i,fac;
fac=1;
for (i=1; i<=a; i++){
fac=fac*i;
}
return fac;
}
int power_func(int x,int y) // x is exponent and y is the number that would be multiplied by itself.
{
int i;//i is for the loop
int ret = 1;
for(i=1;i<=x;i++)
{
ret *= y;
}
return ret;
}
int main()
{
int num,denom,i;//num. is numerator and denom. is denominator
int sin,x,result=0;
printf("Enter the number of x n");
scanf("%d",&x);
for(i=0;i<x;i++)
{
num= power_func(2*i+1,x);
denom=fac((2*i+1));
sin=power_func(i,-1)*num/denom;
result =result+sin;
printf("%d n",result);
}
return 0;
}
c visual-c++ sin
c visual-c++ sin
asked Nov 20 '18 at 20:51
AmerovAmerov
397
asked Nov 20 '18 at 20:51
AmerovAmerov
397
edited Nov 20 '18 at 21:19
Amerov
asked Nov 20 '18 at 20:51
AmerovAmerov
397
asked Nov 20 '18 at 20:51
AmerovAmerov
397
asked Nov 20 '18 at 20:51
AmerovAmerov
397
397
Yourpower_funcis not correct. It computesy^(2^x).
– lurker
Nov 20 '18 at 20:54
@lurker I have changed this line to num= power_func(x, 2*i+1); but I still get a wong answer like : Enter the number of x 3( ,-1.000000 ,1092.000000 ,4347.000000) the result
– Amerov
Nov 20 '18 at 20:58
See my initial comment about it being incorrectly calculated. Sorry, I was mistaken about the backward part.
– lurker
Nov 20 '18 at 20:59
ohh mmm it's ok @lurker
– Amerov
Nov 20 '18 at 21:04
1
More specifically,y = y*yis a problem...
– lurker
Nov 20 '18 at 21:07
|
show 9 more comments
Yourpower_funcis not correct. It computesy^(2^x).
– lurker
Nov 20 '18 at 20:54
@lurker I have changed this line to num= power_func(x, 2*i+1); but I still get a wong answer like : Enter the number of x 3( ,-1.000000 ,1092.000000 ,4347.000000) the result
– Amerov
Nov 20 '18 at 20:58
See my initial comment about it being incorrectly calculated. Sorry, I was mistaken about the backward part.
– lurker
Nov 20 '18 at 20:59
ohh mmm it's ok @lurker
– Amerov
Nov 20 '18 at 21:04
1
More specifically,y = y*yis a problem...
– lurker
Nov 20 '18 at 21:07
Your
power_func is not correct. It computes y^(2^x).– lurker
Nov 20 '18 at 20:54
Your
power_func is not correct. It computes y^(2^x).– lurker
Nov 20 '18 at 20:54
@lurker I have changed this line to num= power_func(x, 2*i+1); but I still get a wong answer like : Enter the number of x 3( ,-1.000000 ,1092.000000 ,4347.000000) the result
– Amerov
Nov 20 '18 at 20:58
@lurker I have changed this line to num= power_func(x, 2*i+1); but I still get a wong answer like : Enter the number of x 3( ,-1.000000 ,1092.000000 ,4347.000000) the result
– Amerov
Nov 20 '18 at 20:58
See my initial comment about it being incorrectly calculated. Sorry, I was mistaken about the backward part.
– lurker
Nov 20 '18 at 20:59
See my initial comment about it being incorrectly calculated. Sorry, I was mistaken about the backward part.
– lurker
Nov 20 '18 at 20:59
ohh mmm it's ok @lurker
– Amerov
Nov 20 '18 at 21:04
ohh mmm it's ok @lurker
– Amerov
Nov 20 '18 at 21:04
1
1
More specifically,
y = y*y is a problem...– lurker
Nov 20 '18 at 21:07
More specifically,
y = y*y is a problem...– lurker
Nov 20 '18 at 21:07
|
show 9 more comments
1 Answer
1
active
oldest
votes
You have various misconceptions about your code. First, let's look at the formula you have provided:
sin(x) = sum((−1)^k * x^(2*k + 1) / (2*k + 1)! for x ∈ R; k = 0, ..., infinity
The sine function takes a real and returns a real. Therefore, you should use a floating-point type for x and sin(x). Use a double. Let's also write a function that emulates sin from <math.h>:
double my_sin(double x);
The above series is accurate when there are infinitely many terms. We can't calculate that many, of course, and it would be a waste of time, too, because the terms are getting ever smaller until they can no longer be represented by a double. So let's chose a maximum number of terms, say
enum {
nTerms = 8
};
Factorials grow big fast. A regular 32-bit int can hold 12! = 479,001,600. A 64-bit int can hold 20! = 2,432,902,008,176,640,000. Since we are going to use these factorials in a double calculation, we can just as well use double here. That will even allow us to represent 22! = 1,124,000,727,777,607,680,000 accurately.
Your power function should also have a double base. The exponent is integer. (But please use the more natural order power(base, exp).
Finally, (−1)^k is just an alternating sign. It is positive when k is even and odd otherwise.
Putting all this together:
double fact(int n)
{
double result = 1.0;
while (n > 0) {
result *= n;
n--;
}
return result;
}
double power(double a, int n)
{
double result = 1.0;
while (n > 0) {
result *= a;
n--;
}
return result;
}
enum {
nTerms = 8
};
double my_sin(double x)
{
double result = 0.0;
double sign = 1.0;
for(int k = 0; k < nTerms; k++)
{
double num = power(x, 2*k + 1);
double denom = fact(2*k + 1);
double term = sign * num / denom;
result = result + term;
sign = -sign;
}
return result;
}
If we write a driver program to print out some test values and compare them with the standard math library's implementation of sin:
int main(void)
{
for (int i = 0; i < 15; i++) {
double x = 0.1 * i;
double m = my_sin(x); // series approximation
double s = sin(x); // <math.h> implementation
printf("%16g%16g%16g%16gn", x, m, s, m - s);
}
return 0;
}
we can see that we're not doing so badly:
x my_sin(x) sin(x) difference
-------- ------------ ------------ ------------
0 0 0 0
0.1 0.0998334 0.0998334 1.38778e-17
0.2 0.198669 0.198669 2.77556e-17
0.3 0.29552 0.29552 0
0.4 0.389418 0.389418 -5.55112e-17
0.5 0.479426 0.479426 0
0.6 0.564642 0.564642 0
0.7 0.644218 0.644218 0
0.8 0.717356 0.717356 0
0.9 0.783327 0.783327 -4.44089e-16
1 0.841471 0.841471 -2.77556e-15
1.1 0.891207 0.891207 -1.43219e-14
1.2 0.932039 0.932039 -6.20615e-14
1.3 0.963558 0.963558 -2.42029e-13
1.4 0.98545 0.98545 -8.52318e-13
(But it gets worse the farther we go from zero. Try other values for nTerms.)
I've said in a comment above that you don't need to calculate factorials and powers and that's true. If you look at the terms of the series, you will see that:
s[n] = -1 * s[n - 1] * x^2 / (2*n * (2*n +1))
s[0] = x
s[1] = x^3 / (1 * 2 * 3) = x * x^2 / (2 * 3)
s[2] = x^5 / (1 * 2 * 3 * 4 * 5) = x^3 / (1 * 2 * 3) * x^2 / (4 * 5)
s[3] = ...
Here's a function that implements that.It calculates terms until adding them to the sum doesn't change it, because they are too small:
double sin_r(double x)
{
double sum = x;
double a = x;
int n;
for (n = 1; ; n++) {
double was = sum;
a = -a * x*x / (2*n) / (2*n + 1);
sum += a;
if (was == sum) break;
}
return sum;
}
The addition still loses some precision by summing the first terms first, but it has the benefit that it doesn't have to calculate factorials and powers. You don't even need <math.h>.
answered Nov 21 '18 at 10:00
M OehmM Oehm
21.4k31832
add a comment |
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1 Answer
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1 Answer
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You have various misconceptions about your code. First, let's look at the formula you have provided:
sin(x) = sum((−1)^k * x^(2*k + 1) / (2*k + 1)! for x ∈ R; k = 0, ..., infinity
The sine function takes a real and returns a real. Therefore, you should use a floating-point type for x and sin(x). Use a double. Let's also write a function that emulates sin from <math.h>:
double my_sin(double x);
The above series is accurate when there are infinitely many terms. We can't calculate that many, of course, and it would be a waste of time, too, because the terms are getting ever smaller until they can no longer be represented by a double. So let's chose a maximum number of terms, say
enum {
nTerms = 8
};
Factorials grow big fast. A regular 32-bit int can hold 12! = 479,001,600. A 64-bit int can hold 20! = 2,432,902,008,176,640,000. Since we are going to use these factorials in a double calculation, we can just as well use double here. That will even allow us to represent 22! = 1,124,000,727,777,607,680,000 accurately.
Your power function should also have a double base. The exponent is integer. (But please use the more natural order power(base, exp).
Finally, (−1)^k is just an alternating sign. It is positive when k is even and odd otherwise.
Putting all this together:
double fact(int n)
{
double result = 1.0;
while (n > 0) {
result *= n;
n--;
}
return result;
}
double power(double a, int n)
{
double result = 1.0;
while (n > 0) {
result *= a;
n--;
}
return result;
}
enum {
nTerms = 8
};
double my_sin(double x)
{
double result = 0.0;
double sign = 1.0;
for(int k = 0; k < nTerms; k++)
{
double num = power(x, 2*k + 1);
double denom = fact(2*k + 1);
double term = sign * num / denom;
result = result + term;
sign = -sign;
}
return result;
}
If we write a driver program to print out some test values and compare them with the standard math library's implementation of sin:
int main(void)
{
for (int i = 0; i < 15; i++) {
double x = 0.1 * i;
double m = my_sin(x); // series approximation
double s = sin(x); // <math.h> implementation
printf("%16g%16g%16g%16gn", x, m, s, m - s);
}
return 0;
}
we can see that we're not doing so badly:
x my_sin(x) sin(x) difference
-------- ------------ ------------ ------------
0 0 0 0
0.1 0.0998334 0.0998334 1.38778e-17
0.2 0.198669 0.198669 2.77556e-17
0.3 0.29552 0.29552 0
0.4 0.389418 0.389418 -5.55112e-17
0.5 0.479426 0.479426 0
0.6 0.564642 0.564642 0
0.7 0.644218 0.644218 0
0.8 0.717356 0.717356 0
0.9 0.783327 0.783327 -4.44089e-16
1 0.841471 0.841471 -2.77556e-15
1.1 0.891207 0.891207 -1.43219e-14
1.2 0.932039 0.932039 -6.20615e-14
1.3 0.963558 0.963558 -2.42029e-13
1.4 0.98545 0.98545 -8.52318e-13
(But it gets worse the farther we go from zero. Try other values for nTerms.)
I've said in a comment above that you don't need to calculate factorials and powers and that's true. If you look at the terms of the series, you will see that:
s[n] = -1 * s[n - 1] * x^2 / (2*n * (2*n +1))
s[0] = x
s[1] = x^3 / (1 * 2 * 3) = x * x^2 / (2 * 3)
s[2] = x^5 / (1 * 2 * 3 * 4 * 5) = x^3 / (1 * 2 * 3) * x^2 / (4 * 5)
s[3] = ...
Here's a function that implements that.It calculates terms until adding them to the sum doesn't change it, because they are too small:
double sin_r(double x)
{
double sum = x;
double a = x;
int n;
for (n = 1; ; n++) {
double was = sum;
a = -a * x*x / (2*n) / (2*n + 1);
sum += a;
if (was == sum) break;
}
return sum;
}
The addition still loses some precision by summing the first terms first, but it has the benefit that it doesn't have to calculate factorials and powers. You don't even need <math.h>.
answered Nov 21 '18 at 10:00
M OehmM Oehm
21.4k31832
add a comment |
You have various misconceptions about your code. First, let's look at the formula you have provided:
sin(x) = sum((−1)^k * x^(2*k + 1) / (2*k + 1)! for x ∈ R; k = 0, ..., infinity
The sine function takes a real and returns a real. Therefore, you should use a floating-point type for x and sin(x). Use a double. Let's also write a function that emulates sin from <math.h>:
double my_sin(double x);
The above series is accurate when there are infinitely many terms. We can't calculate that many, of course, and it would be a waste of time, too, because the terms are getting ever smaller until they can no longer be represented by a double. So let's chose a maximum number of terms, say
enum {
nTerms = 8
};
Factorials grow big fast. A regular 32-bit int can hold 12! = 479,001,600. A 64-bit int can hold 20! = 2,432,902,008,176,640,000. Since we are going to use these factorials in a double calculation, we can just as well use double here. That will even allow us to represent 22! = 1,124,000,727,777,607,680,000 accurately.
Your power function should also have a double base. The exponent is integer. (But please use the more natural order power(base, exp).
Finally, (−1)^k is just an alternating sign. It is positive when k is even and odd otherwise.
Putting all this together:
double fact(int n)
{
double result = 1.0;
while (n > 0) {
result *= n;
n--;
}
return result;
}
double power(double a, int n)
{
double result = 1.0;
while (n > 0) {
result *= a;
n--;
}
return result;
}
enum {
nTerms = 8
};
double my_sin(double x)
{
double result = 0.0;
double sign = 1.0;
for(int k = 0; k < nTerms; k++)
{
double num = power(x, 2*k + 1);
double denom = fact(2*k + 1);
double term = sign * num / denom;
result = result + term;
sign = -sign;
}
return result;
}
If we write a driver program to print out some test values and compare them with the standard math library's implementation of sin:
int main(void)
{
for (int i = 0; i < 15; i++) {
double x = 0.1 * i;
double m = my_sin(x); // series approximation
double s = sin(x); // <math.h> implementation
printf("%16g%16g%16g%16gn", x, m, s, m - s);
}
return 0;
}
we can see that we're not doing so badly:
x my_sin(x) sin(x) difference
-------- ------------ ------------ ------------
0 0 0 0
0.1 0.0998334 0.0998334 1.38778e-17
0.2 0.198669 0.198669 2.77556e-17
0.3 0.29552 0.29552 0
0.4 0.389418 0.389418 -5.55112e-17
0.5 0.479426 0.479426 0
0.6 0.564642 0.564642 0
0.7 0.644218 0.644218 0
0.8 0.717356 0.717356 0
0.9 0.783327 0.783327 -4.44089e-16
1 0.841471 0.841471 -2.77556e-15
1.1 0.891207 0.891207 -1.43219e-14
1.2 0.932039 0.932039 -6.20615e-14
1.3 0.963558 0.963558 -2.42029e-13
1.4 0.98545 0.98545 -8.52318e-13
(But it gets worse the farther we go from zero. Try other values for nTerms.)
I've said in a comment above that you don't need to calculate factorials and powers and that's true. If you look at the terms of the series, you will see that:
s[n] = -1 * s[n - 1] * x^2 / (2*n * (2*n +1))
s[0] = x
s[1] = x^3 / (1 * 2 * 3) = x * x^2 / (2 * 3)
s[2] = x^5 / (1 * 2 * 3 * 4 * 5) = x^3 / (1 * 2 * 3) * x^2 / (4 * 5)
s[3] = ...
Here's a function that implements that.It calculates terms until adding them to the sum doesn't change it, because they are too small:
double sin_r(double x)
{
double sum = x;
double a = x;
int n;
for (n = 1; ; n++) {
double was = sum;
a = -a * x*x / (2*n) / (2*n + 1);
sum += a;
if (was == sum) break;
}
return sum;
}
The addition still loses some precision by summing the first terms first, but it has the benefit that it doesn't have to calculate factorials and powers. You don't even need <math.h>.
answered Nov 21 '18 at 10:00
M OehmM Oehm
21.4k31832
add a comment |
You have various misconceptions about your code. First, let's look at the formula you have provided:
sin(x) = sum((−1)^k * x^(2*k + 1) / (2*k + 1)! for x ∈ R; k = 0, ..., infinity
The sine function takes a real and returns a real. Therefore, you should use a floating-point type for x and sin(x). Use a double. Let's also write a function that emulates sin from <math.h>:
double my_sin(double x);
The above series is accurate when there are infinitely many terms. We can't calculate that many, of course, and it would be a waste of time, too, because the terms are getting ever smaller until they can no longer be represented by a double. So let's chose a maximum number of terms, say
enum {
nTerms = 8
};
Factorials grow big fast. A regular 32-bit int can hold 12! = 479,001,600. A 64-bit int can hold 20! = 2,432,902,008,176,640,000. Since we are going to use these factorials in a double calculation, we can just as well use double here. That will even allow us to represent 22! = 1,124,000,727,777,607,680,000 accurately.
Your power function should also have a double base. The exponent is integer. (But please use the more natural order power(base, exp).
Finally, (−1)^k is just an alternating sign. It is positive when k is even and odd otherwise.
Putting all this together:
double fact(int n)
{
double result = 1.0;
while (n > 0) {
result *= n;
n--;
}
return result;
}
double power(double a, int n)
{
double result = 1.0;
while (n > 0) {
result *= a;
n--;
}
return result;
}
enum {
nTerms = 8
};
double my_sin(double x)
{
double result = 0.0;
double sign = 1.0;
for(int k = 0; k < nTerms; k++)
{
double num = power(x, 2*k + 1);
double denom = fact(2*k + 1);
double term = sign * num / denom;
result = result + term;
sign = -sign;
}
return result;
}
If we write a driver program to print out some test values and compare them with the standard math library's implementation of sin:
int main(void)
{
for (int i = 0; i < 15; i++) {
double x = 0.1 * i;
double m = my_sin(x); // series approximation
double s = sin(x); // <math.h> implementation
printf("%16g%16g%16g%16gn", x, m, s, m - s);
}
return 0;
}
we can see that we're not doing so badly:
x my_sin(x) sin(x) difference
-------- ------------ ------------ ------------
0 0 0 0
0.1 0.0998334 0.0998334 1.38778e-17
0.2 0.198669 0.198669 2.77556e-17
0.3 0.29552 0.29552 0
0.4 0.389418 0.389418 -5.55112e-17
0.5 0.479426 0.479426 0
0.6 0.564642 0.564642 0
0.7 0.644218 0.644218 0
0.8 0.717356 0.717356 0
0.9 0.783327 0.783327 -4.44089e-16
1 0.841471 0.841471 -2.77556e-15
1.1 0.891207 0.891207 -1.43219e-14
1.2 0.932039 0.932039 -6.20615e-14
1.3 0.963558 0.963558 -2.42029e-13
1.4 0.98545 0.98545 -8.52318e-13
(But it gets worse the farther we go from zero. Try other values for nTerms.)
I've said in a comment above that you don't need to calculate factorials and powers and that's true. If you look at the terms of the series, you will see that:
s[n] = -1 * s[n - 1] * x^2 / (2*n * (2*n +1))
s[0] = x
s[1] = x^3 / (1 * 2 * 3) = x * x^2 / (2 * 3)
s[2] = x^5 / (1 * 2 * 3 * 4 * 5) = x^3 / (1 * 2 * 3) * x^2 / (4 * 5)
s[3] = ...
Here's a function that implements that.It calculates terms until adding them to the sum doesn't change it, because they are too small:
double sin_r(double x)
{
double sum = x;
double a = x;
int n;
for (n = 1; ; n++) {
double was = sum;
a = -a * x*x / (2*n) / (2*n + 1);
sum += a;
if (was == sum) break;
}
return sum;
}
The addition still loses some precision by summing the first terms first, but it has the benefit that it doesn't have to calculate factorials and powers. You don't even need <math.h>.
answered Nov 21 '18 at 10:00
M OehmM Oehm
21.4k31832
You have various misconceptions about your code. First, let's look at the formula you have provided:
sin(x) = sum((−1)^k * x^(2*k + 1) / (2*k + 1)! for x ∈ R; k = 0, ..., infinity
The sine function takes a real and returns a real. Therefore, you should use a floating-point type for x and sin(x). Use a double. Let's also write a function that emulates sin from <math.h>:
double my_sin(double x);
The above series is accurate when there are infinitely many terms. We can't calculate that many, of course, and it would be a waste of time, too, because the terms are getting ever smaller until they can no longer be represented by a double. So let's chose a maximum number of terms, say
enum {
nTerms = 8
};
Factorials grow big fast. A regular 32-bit int can hold 12! = 479,001,600. A 64-bit int can hold 20! = 2,432,902,008,176,640,000. Since we are going to use these factorials in a double calculation, we can just as well use double here. That will even allow us to represent 22! = 1,124,000,727,777,607,680,000 accurately.
Your power function should also have a double base. The exponent is integer. (But please use the more natural order power(base, exp).
Finally, (−1)^k is just an alternating sign. It is positive when k is even and odd otherwise.
Putting all this together:
double fact(int n)
{
double result = 1.0;
while (n > 0) {
result *= n;
n--;
}
return result;
}
double power(double a, int n)
{
double result = 1.0;
while (n > 0) {
result *= a;
n--;
}
return result;
}
enum {
nTerms = 8
};
double my_sin(double x)
{
double result = 0.0;
double sign = 1.0;
for(int k = 0; k < nTerms; k++)
{
double num = power(x, 2*k + 1);
double denom = fact(2*k + 1);
double term = sign * num / denom;
result = result + term;
sign = -sign;
}
return result;
}
If we write a driver program to print out some test values and compare them with the standard math library's implementation of sin:
int main(void)
{
for (int i = 0; i < 15; i++) {
double x = 0.1 * i;
double m = my_sin(x); // series approximation
double s = sin(x); // <math.h> implementation
printf("%16g%16g%16g%16gn", x, m, s, m - s);
}
return 0;
}
we can see that we're not doing so badly:
x my_sin(x) sin(x) difference
-------- ------------ ------------ ------------
0 0 0 0
0.1 0.0998334 0.0998334 1.38778e-17
0.2 0.198669 0.198669 2.77556e-17
0.3 0.29552 0.29552 0
0.4 0.389418 0.389418 -5.55112e-17
0.5 0.479426 0.479426 0
0.6 0.564642 0.564642 0
0.7 0.644218 0.644218 0
0.8 0.717356 0.717356 0
0.9 0.783327 0.783327 -4.44089e-16
1 0.841471 0.841471 -2.77556e-15
1.1 0.891207 0.891207 -1.43219e-14
1.2 0.932039 0.932039 -6.20615e-14
1.3 0.963558 0.963558 -2.42029e-13
1.4 0.98545 0.98545 -8.52318e-13
(But it gets worse the farther we go from zero. Try other values for nTerms.)
I've said in a comment above that you don't need to calculate factorials and powers and that's true. If you look at the terms of the series, you will see that:
s[n] = -1 * s[n - 1] * x^2 / (2*n * (2*n +1))
s[0] = x
s[1] = x^3 / (1 * 2 * 3) = x * x^2 / (2 * 3)
s[2] = x^5 / (1 * 2 * 3 * 4 * 5) = x^3 / (1 * 2 * 3) * x^2 / (4 * 5)
s[3] = ...
Here's a function that implements that.It calculates terms until adding them to the sum doesn't change it, because they are too small:
double sin_r(double x)
{
double sum = x;
double a = x;
int n;
for (n = 1; ; n++) {
double was = sum;
a = -a * x*x / (2*n) / (2*n + 1);
sum += a;
if (was == sum) break;
}
return sum;
}
The addition still loses some precision by summing the first terms first, but it has the benefit that it doesn't have to calculate factorials and powers. You don't even need <math.h>.
answered Nov 21 '18 at 10:00
M OehmM Oehm
21.4k31832
answered Nov 21 '18 at 10:00
M OehmM Oehm
21.4k31832
answered Nov 21 '18 at 10:00
M OehmM Oehm
21.4k31832
answered Nov 21 '18 at 10:00
M OehmM Oehm
21.4k31832
21.4k31832
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Your
power_funcis not correct. It computesy^(2^x).– lurker
Nov 20 '18 at 20:54
@lurker I have changed this line to num= power_func(x, 2*i+1); but I still get a wong answer like : Enter the number of x 3( ,-1.000000 ,1092.000000 ,4347.000000) the result
– Amerov
Nov 20 '18 at 20:58
See my initial comment about it being incorrectly calculated. Sorry, I was mistaken about the backward part.
– lurker
Nov 20 '18 at 20:59
ohh mmm it's ok @lurker
– Amerov
Nov 20 '18 at 21:04
1
More specifically,
y = y*yis a problem...– lurker
Nov 20 '18 at 21:07