Euler method: Difference between revisions

Line 2,819:

<lang ruby>func euler_method(t0, t1, k, step_size) {

var results = [[0, t0]];

var results = [[0, t0]]

step_size.~~to(100)~~.by(step_size)~~.each~~ { ~~|s|~~

for s in (step_size..100 -> by(step_size)) {

t0 -= ((t0 - t1) * k * step_size);

t0 -= ((t0 - t1) * k * step_size)

results << [s, t0];

results << [s, t0]

}

return results;

Line 2,828:

func analytical(t0, t1, k, time) {

(t0 - t1) * exp(-time * k) + t1;

(t0 - t1) * exp(-time * k) + t1

}

var (T0, T1, k) = (100, 20, .07);

var (T0, T1, k) = (100, 20, .07)

var r2 = euler_method(T0, T1, k, 2).grep { _[0] %% 10 };

var r2 = euler_method(T0, T1, k, 2).grep { _[0] %% 10 }

var r5 = euler_method(T0, T1, k, 5).grep { _[0] %% 10 };

var r5 = euler_method(T0, T1, k, 5).grep { _[0] %% 10 }

var r10 = euler_method(T0, T1, k, 10).grep { _[0] %% 10 };

var r10 = euler_method(T0, T1, k, 10).grep { _[0] %% 10 }

say "Time\t 2 err(%) 5 err(%) 10 err(%) Analytic";

say "Time\t 2 err(%) 5 err(%) 10 err(%) Analytic"

say "-"*76;

say "-"*76

r2.range.each { |i|

var an = analytical(T0, T1, k, r2[i][0]);

var an = analytical(T0, T1, k, r2[i][0])

printf("%4d\t#{'%9.3f' * 7}\n",

r2[i][0],

r2[i][1], (r2[i][1] / an) * 100 - 100,

r2[i][1], ( r2[i][1] / an) * 100 - 100,

r5[i][1], (r5[i][1] / an) * 100 - 100,

r5[i][1], ( r5[i][1] / an) * 100 - 100,

r10[i][1], (r10[i][1] / an) * 100 - 100,

an);

an)

}</lang>