3

This is what I've achieved so far:
$$\tan^{-1}1 = \frac{\pi}{4} \Rightarrow \pi = 4\tan^{-1}1$$ $$\tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5}+ \cdots + (-1)^k\frac{x^{2k+1}}{2k+1}$$ $$\pi = 4\tan^{-1}1 = 4\left(1-\frac{1}{3}+\frac{1}{5} +\cdots + (-1)^k\frac{1}{2k+1}\right)$$ What is the next step?

Sajad Rastegar
  • 275
  • 1
    look at http://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80#Inefficiency if you want to increase the rate of convergence of your series – Jam Jul 21 '14 at 19:16
  • 1
    If the question asks how many terms one needs, the error in an alternating series like this (terms decreasing in absolute value) is less than the first "neglected" term. That means you will need an awful lot of terms. If you need to give an actual approximation, you will need a more efficient method. – André Nicolas Jul 21 '14 at 19:16
  • John Machin (1706) gave the formula $\frac{\pi}{4} = \tan^{-1}1 = 4\tan^{-1}\frac{1}{5} - \tan^{-1}\frac{1}{239}$ which allows the power series expansion to be used more effectively (smaller arguments, more rapid convergence). Since then a number of related formulas have been discovered. – hardmath Jul 21 '14 at 19:32
  • formula 49 is very fast. But of course the fastest for your particular question is just the costant sequence $u_n=3.141592653$ :) –  Jul 21 '14 at 20:36

2 Answers2

5

The remainder for an alternating series is given by $$|R_k| \le b_{k+1}.$$

This means you are looking for a $k$ for which $$\frac{1}{2(k+1)+1} < \frac{1}{2\cdot 10^8}.$$

This tells us that the $k$ we are looking for is greater than $$10^8 - \frac{3}{2}$$ which is fairly large. It would be impractical to actually compute $\pi$ with this series. Heuristically, this is a bad series to use, since it is not absolutely convergent. In this sense it converges "slowly."

There are several better methods out there. There is a book by the Borwein brothers "Pi and the AGM" that discusses approximations of $\pi$. For a long time they held the record for the number of digits of $\pi$ computed. I am not sure if they still hold that record.

Joel
  • 16,256
1

You have all that you need. You need to solve this for k : $\frac{1}{2}10^{-8}>\frac{4}{2k+1}$ and then comptute the sum to k.

By solving that inequation, you know that every term you add wont get you further from Pi by more than the error you want.

http://en.wikipedia.org/wiki/Alternating_series See the part "approximating sums"

But your way of approach is bad because minimal k is very big here.

nicomezi
  • 21