I recently stumbled across the formula:


developed by Euler, for approximating pi. I evaluated it to several thousand decimal places and up to that point, it accurately represented pi.

When does this equation break down in its representation of pi?

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    $\begingroup$ It doesn't. It's an exact identity. $\endgroup$ Jan 11 '12 at 4:02
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    $\begingroup$ A derivation of the identity and related ones can be found here $\endgroup$
    – t.b.
    Jan 11 '12 at 4:43
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    $\begingroup$ Quick proof: The complex number $(7 + i)^{20} (79 + 3 i)^8 = -149011611938476562500000000000000$ lies on the negative real axis, so $20\arctan\frac{1}{7}+8\arctan\frac{3}{79} = \pi + 2\pi n$. Consideration of the sizes of the angles involved shows that $n$ must be zero. $\endgroup$ Jan 11 '12 at 7:46
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    $\begingroup$ See this as well. $\endgroup$ Jan 12 '12 at 9:17
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    $\begingroup$ $-149011611938476562500000000000000 = 2^{14}5^{40}$, just for info. $\endgroup$ Oct 5 '13 at 2:55

That formula doesn't break down. It can be proven that the right-hand side is equal to $\pi$. One reason it comes up in connection with approximation is that the arctangent terms can be approximated well by adding terms from the Taylor expansion of the arctangent function,


This is valid for $|x|\leq 1$, but the series converges more rapidly the closer $x$ is to $0$, so with the relatively small $\frac{1}{7}$ and $\frac{3}{79}$ as inputs it gives a good approximation without having to take too many terms. This can be contrasted with the formula $$\pi=4\arctan(1)=4\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\cdots\right),$$ where the series converges much more slowly.

In 1950 H.C. Schepler published a 3 part "Chronology of pi" in Mathematics Magazine, available to those with access to JSTOR here, here, and here. In the second part there is the following excerpt indicating that Hutton may have suggested using the formula before Euler:

enter image description here

Schepler's chronology can also be found in L. Berggren, J. Borwein, and P. Borwein's Pi: A Source Book.

  • $\begingroup$ Thanks for explaining why it's associated with an 'approximation'. That had me a bit confused $\endgroup$
    – Chris A
    Jan 11 '12 at 12:44

The Soviet 8th grade students usually solve this problem as follows. First, divide by 4. Then a portable arc tangent in the opposite direction. Then calculate the tangent of the right and left part using very simple formula tg(A+B).

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    $\begingroup$ "A portable arc tangent" ... not a correct translation into English of what you want to say. $\endgroup$
    – GEdgar
    Jan 11 '12 at 14:05
  • $\begingroup$ What is a portable arc tangent? $\endgroup$ Apr 7 '13 at 19:52

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