# Quadrature of a circle solved

Discussion in 'Pseudoscience' started by Jason.Marshall, Jan 7, 2015.

1. ### James RJust this guy, you know?Staff Member

Messages:
29,990
Nonsense. The second two numbers are not $\pi$.

Why do you think that
$\pi = \frac{14-\sqrt{2}}{4}$?
How did you derive that result? (Just to emphasise: it is incorrect.)

Why do you think that
$\pi=\frac{4}{(1+\sqrt{5})/2}$?
How did you derive that result? (Just to emphasise: it is incorrect.)

What is the relevance of this?

What?!

There's only one "modern pi value". Why 4 or 5 decimal places?

No. There is no debate. Only one of the values you have cited approximates pi.

exchemist likes this.

3. ### rpennerFully WiredStaff Member

Messages:
4,827
Actually, it was $\frac{4}{\sqrt{\frac{1 + \sqrt{5}}{2}}} = 4 \sqrt{\frac{2}{1 + \sqrt{5}}} = 2 \sqrt{ 2 ( \sqrt{5} - 1 ) }$

Not "almost" -- 100% accuracy is never found in any manufacturing technique. It is also impossible in Euclidean geometry to construct circles and squares of equal area. Likewise it is impossible to form an exact expression in terms of just the four arithmetic operators and roots or to find a polynomial equation with integer coefficients that gives pi as a root.

But the approximate forms you give are not very impressive as intellectual achievements.

$\frac{14 - \sqrt{2}}{4}$ gets the first 3 digits right. If a circle had a diameter of 12 million meters (about the diameter of the Earth) the error in the circumference calculated with this approximation would be about 58,247 meters too long.
$\frac{4}{\sqrt{\frac{1 + \sqrt{5}}{2}}}$ gets the first 3 digits right, but is better than the above.
3.14 gets the first 3 digits right, but is better than the above. If a circle had a diameter of 12 million meters (about the diameter of the Earth) the error in the circumference calculated with this approximation would be about 19,112 meters too short.
22/7 gets the first 3 digits right, but is better than the above. If a circle had a diameter of 12 million meters (about the diameter of the Earth) the error in the circumference calculated with this approximation would be about 15,178 meters too long.
$\sqrt{ \frac{40}{3} - \sqrt{12}}$ (Kochanski's Approximation) gets the first 5 digits right.
355/113 gets the first 7 digits right. If a circle had a diameter of 12 million meters (about the diameter of the Earth) the error in the circumference calculated with this approximation would be about 3.2 meters too long. That's likely as good as one would ever need for civil engineering.
$\frac{99^2}{2206 \sqrt{2}}$ (Ramanujan) gets the first 7 digits right.
$\frac{66 \sqrt{2}}{ 33 \sqrt{29} - 148 }$ (Borwein and Bailey) gets the first 8 digits right.
3.141592653589793 is a simple truncation of first 16 digits of the infinite decimal sequence for pi.
Due to design considerations, most PC and Mac programs are coded with IEEE double precision numbers which forces them to best approximate pi as $\frac{884279719003555}{2^{48} }$ which is a bit better than the above decimal truncation. This makes me suspect you cribbed the above from the computer without any heavy lifting on your part.
21053343141/6701487259 gets the first 21 digits right.
$\frac{ \ln ( (640320^3 + 744)^2 - 393768 ) }{ 2 \sqrt{163}}$ (Warda) gets the first 46 digits right. But this is using the natural logarithm which is outside of the rules above.
But if we are allowing that, then we could use this: $\frac{ \ln ( -1 ) }{ \sqrt{-1} } = \pi$ and that is exact.

5. ### Confused2Registered Senior Member

Messages:
360
I spy e^(iπ)=-1 EDIT: aka Euler's formula ...

A proof (for example) here:-