C programming - combining power(x, n) and fact(n) functions together - c

Question

I've been working on a program that calculates sin(x), cos(x), and exp(x) without using math.h and compares them to the library values of their functions. I've been forbidden from actually using the basic power(x, n) and fact(n) functions. The only hint is that I have to do division before doing multiplication when combining the functions into one.
double power(double x, int n)
{
int i;
double prod=1.;
for(i=0;i<n;i++){
prod = prod*x;
}
return prod;
}
double fact(int n)
{
int i;
double prod=1.;
for(i=1;i<=n;i++) {
prod = prod*i;
}
return prod;
}
My idea is to somehow nest the for-loops together, and piecemeal the Taylor Expansion formula for each iteration of the loop, but I haven't had luck actually combining the two.
Any help or hint would be appreciated on how to combine these.
The other aspect of the program that confuses me is that there can only be a single input of X per iteration of the program, and therefore no dynamically defined 'n' for the loops.

Answer 1

Use Taylor series for the exponential:
e^x = 1 + x/1! + x^2/2! + x^3/3!...
and using Euler after that you can calculate sinx and cosx.

Answer 2

The trick is to look at the changes between each successive term in the Taylor series expansion. Let's start with ex:
e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + x^5/5! ...
Notice that each term is x / n times the prior term, where n is the term number. So start with a term of 1, then multiply by the above expression to get the next term.
That gives you the following implementation:
double etox(double x)
{
long double sum = 0;
// term starts at 1
long double term = 1;
// term number
int i = 1;
// continue until the term is below the precision of the current sum
while (sum + term != sum) {
sum += term;
// new term is x/i times the prior term, where i is the term number
term *= (long double)x / i;
i++;
}
return sum;
}
Note that with this implementation, you'll get some degree of error in the least significant digits. If you start adding from a higher term number and work your way back, this can be avoided.
Similarly for sin(x) and cos(x):
sin(x) = x - x^3/3! + x^5/5! - x^7/7! + x^9/9! ...
cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + x^8/8! ...
Each term is - (x*x) / ((2*n)*((2*n)-1)) times the prior term, where n is the term number.
I'll leave the the implementation of these two as an exercise for the reader.

Answer 3

Part of this comes from my answer for doing this in MIPS assembly: Taylor Series in MIPS assembly
You can do Taylor series on the fly without having to call sub-functions. In the series, each term can be calculated from the previous term in a loop. (i.e. no need to call fact and/or pow repeatedly, where each starts from the beginning). See https://en.wikipedia.org/wiki/Taylor_series
Anyway, here's code for sin and cos:
// mipstaylor/mipstaylor -- fast sine/cosine calculation
#include <stdio.h>
#include <math.h>
#define ITERMAX 10
// qcos -- calculate cosine
double
qcos(double x)
{
int iteridx;
double x2;
double cur;
int neg;
double xpow;
double n2m1;
double nfac;
double sum;
// square of x
x2 = x * x;
// values for initial terms where n==0:
xpow = 1.0;
n2m1 = 0.0;
nfac = 1.0;
neg = 1;
sum = 0.0;
iteridx = 0;
// NOTES:
// (1) with the setup above, we can just use the loop without any special
// casing
while (1) {
// calculate current value
cur = xpow / nfac;
// apply it to sum
if (neg < 0)
sum -= cur;
else
sum += cur;
// bug out when done
if (++iteridx >= ITERMAX)
break;
// now calculate intermediate values for _next_ sum term
// get _next_ power term
xpow *= x2;
// go from factorial(2n) to factorial(2n+1)
n2m1 += 1.0;
nfac *= n2m1;
// now get factorial(2n+1+1)
n2m1 += 1.0;
nfac *= n2m1;
// flip sign
neg = -neg;
}
return sum;
}
// qsin -- calculate sine
double
qsin(double x)
{
int iteridx;
double x2;
double cur;
int neg;
double xpow;
double n2m1;
double nfac;
double sum;
// square of x
x2 = x * x;
// values for initial terms where n==0:
xpow = x;
n2m1 = 1.0;
nfac = 1.0;
neg = 1;
sum = 0.0;
iteridx = 0;
// NOTES:
// (1) with the setup above, we can just use the loop without any special
// casing
while (1) {
// calculate current value
cur = xpow / nfac;
// apply it to sum
if (neg < 0)
sum -= cur;
else
sum += cur;
// bug out when done
if (++iteridx >= ITERMAX)
break;
// now calculate intermediate values for _next_ sum term
// get _next_ power term
xpow *= x2;
// go from factorial(2n+1) to factorial(2n+1+1)
n2m1 += 1.0;
nfac *= n2m1;
// now get factorial(2n+1+1+1)
n2m1 += 1.0;
nfac *= n2m1;
// flip sign
neg = -neg;
}
return sum;
}
// testfnc -- test function
void
testfnc(int typ,const char *sym)
{
double (*efnc)(double);
double (*qfnc)(double);
double vale;
double valq;
double x;
double dif;
int iter;
switch (typ) {
case 0:
efnc = cos;
qfnc = qcos;
break;
case 1:
efnc = sin;
qfnc = qsin;
break;
default:
efnc = NULL;
qfnc = NULL;
break;
}
iter = 0;
for (x = 0.0; x <= M_PI_2; x += 0.001, ++iter) {
vale = efnc(x);
valq = qfnc(x);
dif = vale - valq;
dif = fabs(dif);
printf("%s: %d x=%.15f e=%.15f q=%.15f dif=%.15f %s\n",
sym,iter,x,vale,valq,dif,(dif < 1e-14) ? "PASS" : "FAIL");
}
}
// main -- main program
int
main(int argc,char **argv)
{
testfnc(0,"cos");
testfnc(1,"sin");
return 0;
}