Quote:
Originally Posted by plaaynde
Put a square outside and inside the circle, sides and corners touching the circle. First the outside one: the diameter of the circle is one fourth of the circumference of the square. So pi is less than 4. Then the inside one: use the Pythagorean theorem. Pi has to be bigger than 4 divided by the square root of 2 ~= 2.83.
So I'd say pi iz zomewhere between 2.83 and 4.
Here is my best method so far, still working on other ideas. (use Desmos to visualize the next plotting relevant curves
https://www.desmos.com/calculator)
Consider the circle x^2+(y-1)^2=1 and the parabola y=1/2*x^2. Then consider a=((1-3^(1/2)/2))/2)^(1/2) (that is sine of Pi/12 basically ie 15 deg, you can get that from the double angle trigonometric identity formula).
Because the parabola approximates very well the curve of the circle from -a to a (do Taylor expansion of the circle curve to see why), the area of the rectangle with dimensions 2a*1 is equal to the circular section Pi/12 plus the two triangles 2*1/2*a*(1-a^2)^(1/2) plus the area below the parabola from -a to a that is 1/3*a^3.
eg see here all this
https://www.desmos.com/calculator/42s4uakype
zoom
https://www.desmos.com/calculator/smwzcimika
Since one can calculate square roots of eg K easily with the sequence x_(n+1)=1/2(x_n+K/x_n)
one can use the above to get Pi~3.1423 that beats 22/7 by my countryman Archimedes lol. Probably done within less than 2-3 hours without calculator assistance on the square roots.
One can improve on that going to 7.5 degrees even with a=0.130526192 and then get 3.1416 hehe!
Go for 3.75 degrees and it gets to 3.141596. All this doable within 4h of math using the rational approximation of square roots above that converges fast.
He could obtain this result by the way because he knew what the area below a parabola is with other methods than calculus but its too tough to ask for such long time ago in a world without paper and more compact symbols formalism for the algebra.
Last edited by masque de Z; 07-31-2018 at 02:36 AM.