Liu Hui's π algorithm: Difference between revisions

Revision as of 19:34, 22 June 2017

Revision as of 01:25, 24 June 2017

Line 45:

:Yellow area + green area + red area = <math>A_{2N} + D_{2N}.</math>

Let ''<math>A_{C''}</math> represent the area of the circle. Then

:<math>A_{2N} < A_{C} < A_{2N} + D_{2N}.</math>

If the radius of the circle is taken to be 1, then we have Liu Hui's {{pi}} inequality:

Line 64:

: <math>{} G^2 = r^2 - \left(\tfrac{M}{2}\right)^2</math>

: <math>{} G = \sqrt{r^2- \tfrac{M^2}{4}}</math>

: <math>{} j = r - G = r - \sqrt{r^2- \tfrac{M^2}{4}}</math>

: <math>{} m^2 = \left(\tfrac{M}{2}\right)^2 + j^2</math>

: <math>{} m = \sqrt{\left(\tfrac{M}{2}\right)^2 + j^2}</math>

: <math>{} m = \sqrt{\left(\tfrac{M}{2}\right)^2 + \left(r - G\right)^2}</math>

: <math>{} m = \sqrt{\left(\tfrac{M}{2}\right)^2 + \left(r - \sqrt{r^2- \tfrac{M^2}{4}}\right)^2}</math>

From here, there is now a technique to determine {{math|m}} from {{math|M}}, which gives the side length for a polygon with twice the number of edges. Starting with a [[hexagon]], Liu Hui could determine the side length of a dodecagon using this formula. Then continue repetitively to determine the side length of a 24-gon given the side length of a dodecagon. He could do this recursively as many times as necessary. Knowing how to determine the area of these polygons, Liu Hui could then approximate {{pi}}.

With {{math|''r''}} = 10 units, he obtained

With {{<math|''>r''}} = 10</math> units, he obtained

: area of 48-gon <math>{}A_{96} = 313 {584 \over 625} </math>

Line 79 ⟶ 83:

:from Liu Hui's {{pi}} inequality:

:<math> A_{2N} < A_{C} < A_{2N} + D_{2N}.</math>

:Since {{math|''r''}} = 10, <math>A_{{math|''C''}} = <math> 100 \times \pi</math>

:therefore:

:<math>{}314\frac{64}{625}<100 \times \pi <314 \frac{64}{625} +\frac{105}{625}</math>