math: p_exp: Implement exp function using series

src/math/p_exp.c

@@ -3,23 +3,66 @@
 /**
  *
  * Calculate exponent (e^a), where e is the base of the natural logarithm
- * (2.71828.)
+ * (2.71828.) Using series expansion of function
  *
  * @param a     Pointer to input vector
  *
  * @param c     Pointer to output vector
  *
  * @param n     Size of 'a' and 'c' vector.
  *
+ * @param p     Number of processor to use (task parallelism)
+ *
+ * @param team  Team to work with 
+ *
  * @return      None
  *
  */
-#include <math.h>
+static const float precision = 0.1e-6;
+
+float inverse(float a) //function to calucalte inverse without division operator using Newton Raphson method
+{
+	float result = precision;
+	while(result * a > 1 + precision || result * a < 1 - precision)
+	{
+		result *= (2 - a * result);
+	}
+	return result;
+}
+
+float single_exp(float b)
+{
+	float a = b;
+	if(b < 0) //if exponent is negative then calculate e^b and return 1/e^b
+	{
+		a = -a;
+	}
+	float result = 1;
+	float counter = 1;
+	float next_term = 1;
+	float old_result;
+	float error;
+	do
+	{
+		old_result = result;
+		next_term = next_term * a * inverse(counter);
+		result += next_term;
+		counter += 1;
+		error = (old_result - result) / result;
+		if(error < 0)
+		{
+			error = -error;
+		}
+	}while(error > precision);
+	if(b < 0)
+		return inverse(result);
+	return result;
+}
 void p_exp_f32(const float *a, float *c, int n)
 {
 
     int i;
     for (i = 0; i < n; i++) {
-        *(c + i) = expf(*(a + i));
+        *(c + i) = single_exp(*(a + i));
     }
 }