Re: Print value of p from the infinite series

On Feb 27, 9:17 pm, Lew <l...@xxxxxxxxxxxxxxxxxxxx> wrote:
Lew wrote:
while ( (p - Math.PI) < 1E-6 )
or
for ( int i = 1; (p - Math.PI) < 1E-6; i += 2 )

Use your own values for the target and epsilon.

Exuse me, that condition should be inverted

while ( (p - Math.PI) > 1E-6 )

- Lew


Well, after tried a couple of changes, it doesn't works as expected.
Maybe I mis-understood the key.

Anyway, here is what come after the changes:

Still, it is near but not accurate:

<code>
real numbers: p: counter: > 1E-3
4.000000000000000 4.000 1 0.858407346410207
2.666666666666667 2.667 2 -0.474925986923126
3.466666666666667 3.467 3 0.325074013076874
....
3.140590649833284 3.141 998 -0.001002003756509
3.142593654340044 3.143 999 0.001001000750251
3.140592653839794 3.141 1000 -0.000999999749999
3.142591654339544 3.143 1001 0.000999000749751
</code>

The loops terminate at 1001, since 3.142591654339544 - Math.PI =
0.00099...

So, if I were to count the terms for only 3.141 :

<code>
double p = 0.0;
boolean increment = true;

System.out.printf("%s%12s%17s%12s\n", "real numbers:", "p:",
"counter:",
"> 1E-3");

int count = 0;
int i = 1;
do {
p = (increment) ? (p + (4.0 / i)) : (p - (4.0 / i));
System.out.printf("%.15f %10.3f %12d % 24.15f\n",
p, p, ++count, p - Math.PI);
increment = !increment;
i += 2;
} while ((p - Math.PI) > 1E-3 || (p - Math.PI) < 0);
</code>

Any ideas?

.

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