Optimization of the program. Division by factorial. Taylor series

No need to calculate factorial. You can use the recurrence relation.

Below is a sample program that uses the standard C function exp , which is declared in the header <math.h> , to compare the results. Naturally, this standard function and the corresponding title are not needed for the work of the written function.

 #include <stdio.h> #include <float.h> #include <math.h> double f(double x) { double sum = 1.0; double member = 1.0; for ( int n = 1; !(fabs( member ) < DBL_EPSILON); n++ ) { member = member * -x / n; sum += member; } return sum; } int main( void ) { for (int i = 1; i < 10; i++) { printf( "%lf %lf\n", f(i ), exp(-i)); } } 

The output of the program to the console:

 0.367879 0.367879 0.135335 0.135335 0.049787 0.049787 0.018316 0.018316 0.006738 0.006738 0.002479 0.002479 0.000912 0.000912 0.000335 0.000335 0.000123 0.000123 

As can be seen from the output, the results for the shown function and the standard C function of exp coincide.

Most likely the teacher had in mind that the next member of the series for a given function can be calculated recurrently.

Write out the members of the series:

 a_{0} = 1 a_{1} = - x/1 a_{2} = x^2/(1*2) a_{3} = -x^3/(1*2*3) a_{4} = x^4/(1*2*3*4) 

This shows that the i -th member of the series can be calculated based on the i-1 1th member and the value of i :

 a_{i} = a_{i-1}*(-x/i), при i = 1,2... 

With this approach, the inner loop can be removed.

To make it clearer, I'll sign for math more:

Well, or even more :) -

So as a result, the series is considered a simple function.

 double exp_neg(double x, double eps) { double sum = 1.0, term = 1.0; for(int n = 1; term > eps || term < -eps; ++n) { sum += term *= -x/n; } return sum; }