For x varying from a to b with a step (ba) / k, where (k = 10), calculate the function f (x) using its expansion in a power series in three cases:

  • a) for a given n;
  • b) for a given accuracy e (e = 0.0001);
  • c) for the “exact” value (according to the analytical formula).

For comparison, find the relative error in calculating the function function value o_pogr = ABS ((exact_value - approximate_value) / exact_value)

  • function: y = -ln | 2sinx / 2 |
  • range of argument variation: pi / 5 <= x <= 9pi / 5
  • n = 40

  • sum: S = cosx + cos2x / 2 + ... + cosnx / n

    package laboratornaya3; import static java.lang.Math.*; public class Laboratornaya3 { public static void main(String[] args) { double a = PI / 5, b = 9 * PI / 5, E = 0.0001, pSN, pSE; int n = 40; double dx=(ba)/10; //Шапка System.out.printf("%-15s %-30s %-30s %-30s %-30s %-30s", "x", "y", "SN", "SE", "pSN", "pSE"); System.out.println(); for (double x = a; x <= b; x += dx) { // Точное значение Y double y = -log(abs(2*sin(x/2))); // SN double SN = 0; for (int i = 1; i <= n; i++) { SN += cos(i*x)/i; } // SE double SE = 0; int i = 1; double step; do { step = cos(i*x)/i; SE+=step; i++; } while (abs(step)>E); // погрешности pSN = abs((y - SN) / y); pSE = abs((y - SE) / y); //вывод System.out.printf("%-15s %-30s %-30s %-30s %-30s %-30s", String.format("%.5f", x), y, SN, SE, pSN, pSE); System.out.println(); } } }
  • what do you need here? error increase? - Saidolim
  • No, on the contrary, to reduce - Madina
  • The large error in calculations with n (pSN) is not very exciting, but the calculation with a given accuracy (pSE) raises questions. Observed observation of the step (step) and for some reason it increases by moments as if it does not have time to count in the specified range, so I think that there is an error in the exit condition from the while, but what condition to set? C mathematics, unfortunately, is mediocre, so maybe something I do not understand in the expansion of the function in a power series - Madina

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