3
$\begingroup$

For Gregory–Leibniz series, wikipedia has - "after 500,000 terms, it produces only five correct decimal digits of π.". But how do you know that those five decimal values are correct when you reach 500,000?

What if during any random calculation (not considering pi) the number is 2.82999 and we were to add 0.00001 to it. The result will be 2.83000 which changes the second, third, fourth and fifth digit after decimal. How do you know the number of digits that will not change?

$\endgroup$
  • 1
    $\begingroup$ its called a limit bro.. $\endgroup$ – Matthew Levy Oct 19 '14 at 16:11
  • 5
    $\begingroup$ It's an alternating series, whose terms are decreasing. Those have the property that the sum is always in between any two consecutive partial sums. As in, $1$ is too big, $1-\frac13$ is too small, $1-\frac13+\frac15$ is too big, etc., so $\pi$ has to be in between any two consecutive partial sums. If two partial sums agree on the first five digits, then, those digits have to be correct. $\endgroup$ – Akiva Weinberger Oct 19 '14 at 16:12
  • 3
    $\begingroup$ @columbus8myhw: I think that comment should be an answer $\endgroup$ – Ben Millwood Oct 19 '14 at 16:16
  • 1
    $\begingroup$ @PIMan: I gave the question a title I think is more informative. Let me know if you disagree. $\endgroup$ – Ben Millwood Oct 19 '14 at 16:19
0
$\begingroup$

The Leibniz series is an alternating series (similar to the Wallis Product), whose terms are decreasing toward $\frac{\pi}{4}$. Thus, those terms must have the property that the sum is always in between any two consecutive partial sums. About $0.785398...$ (i.e $\frac{\pi}{4}$) is the goal, so $1$ is too big, $1−\frac{1}{3}$ ( $0.\overline{6}$ ) is too small, 1−$\frac{1}{3}+\frac{1}{5}$ ( $0.8\overline{6}$ ) is too big, (etc.). So, $\pi$ has to be in between any two consecutive partial sums. If two partial sums agree on the first five digits, then, those digits have to be correct.


Also, another thing you might be interested in: $13$ trillion digits of $\pi$ (no joke! click this link)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.