Calculating the value of arctan(x) in C++ - c++

Question

I have to calculate the value of arctan(x) . I have calculated the value of this by evaluating the following series :
Arctan (x) = x – x^3/3 + x^5/5 – x^7/7 + x^9/9 - …
But the following code can not calculate the actual value. For example, calculate_angle(1) returns 38.34 . Why?
const double DEGREES_PER_RADIAN = 57.296;
double calculate_angle(double x)
{
int power=5,count=3;
double term,result,prev_res;
prev_res = x;
result= x-pow(x,3)/3;
while(abs(result-prev_res)<1e-10)
{
term = pow(x,power)/power;
if(count%2==0)
term = term*(-1);
prev_res=result;
result=result+term;
++count;
power+=2;
// if(count=99)
// break;
}
return result*DEGREES_PER_RADIAN;
}

Answer 1

EDIT: I found the culprit. You forgot to include stdlib.h, where the function abs resides. You must have ignored the warning about abs being implicitly declared. I checked that removing the include yields the result 38.19 and including it yields the result ~45.
The compiler is not required to stop compilation when an undeclared function is being used (in this case abs). Instead, it is allowed to make assumptions on how the function is declared (in this case, wrong one.)
Besides, like other posters already stated, your use of abs is inappropriate as it returns an int, not a double or float. The condition in the while should be >1e-100 not <1e-100. The 1e-100 is also too small.
--
You forgot to increase count and power after calculating the first two summands:
prev_res = x;
result= x-pow(x,3)/3;
count = 4; <<<<<<
power = 5; <<<<<<
while(abs(result-prev_res)<1e-100)
{
term = pow(x,power)/power;
if(count%2==1)
term = term*(-1);
Also I consider your use of the count variable counterintuitive: it is intialized with 3 like if it denotes the last used power; but then, loop iterations increase it by 1 instead of 2 and you decide the sign by count%2 == 1 as opposed to power%4 == 3

Answer 2

The series converges to tan^{-1} x, but not very fast. Consider the series when x=1:
1 - 1/3 + 1/5 - 1/7 + 1/9 - ...
What is the error when truncating at the 1/9 term? It's around 1/9. To get 10^{-100} accuracy, you would need to have 10^{100} terms. The universe would end before you'd get that. And, catastrophic round-off error and truncation error would make the answer utterly unreliable. You only have 14 digits to play with for doubles.
Look at reference works like Abramowitz and Stegun [AMS 55] or the new NIST Digital Library of Mathematical Functions at http://dlmf.nist.gov to see how these are done in practice. Often, one uses Padé approximants instead of Taylor series. Even when you stick with Taylor series, you often use Chebyshev approximation to cut down on the total error.
I also recommend Numerical Methods that [Usually] Work, by Forman Acton. Or the Numerical Recipes in ... series.

Answer 3

Your sign is the wrong way around after the first two terms. It should be:
if(count%2==0)
term = term*(-1);
Your comparison is the wrong way around in the while condition. Also, you're expecting an unrealistically high level of precision. I would suggest something more like this:
while(fabs(result-prev_res)>1e-8)
Finally, you'll get a more accurate result with a better value for DEGREES_PER_RADIAN. Why not something like this:
const double DEGREES_PER_RADIAN = 180/M_PI;