Memorize just that e^(pi*sqrt(163)) is almost a perfect cube, and its difference from a perfect cube is almost an integer. The other constants can be derived.
First calculate the cube root of the "almost integer". Note exp(x/3) = cbrt(e^x).
? exp(Pi*sqrt(163)/3)
640320.00000000060486373504901603947175
Then find the offset from the cube. Note that many scientific calculators do not have enough internal precision to compute this correctly.
? exp(Pi*sqrt(163))-640320^3
743.99999999999925007258948526705708004
Then you have recovered the necessary ingredients for the formula, which yields 30 correct digits of pi. One did not have to memorize 640320 or +744.
? \p50
realprecision = 57 significant digits (50 digits displayed)
? log(640320^3+744)/sqrt(163)-Pi
2.23735150380481508416601318272292512584 E-31
A similar memorization: sqrt(58) ... perfect 4th power yields 18 digits.
We can get 46 digits with a little more work, memorizing 2*Pi*sqrt(163) and perfect square:
? exp(Pi*sqrt(163))
262537412640768743.99999999999925007259719818568888
? exp(Pi*2*sqrt(163))-262537412640768744^2
-393767.999999999836261355790597560029392
Note 393768/2 = 196884 is Monstrous Moonshine. Skip the next 2 steps if one already found 640320 and 744 from above.
? sqrt(exp(Pi*2*sqrt(163))+393768)^(1/3)
640320.00000000060486373504901664915438949934751465
? sqrt(exp(Pi*2*sqrt(163))+393768)-640320^3
744.00000000000000000000000000031183868722255543956
? log((640320^3+744)^2-393768)/(2*sqrt(163))-Pi
-9.303470618546941055 E-47
Of course, it is a little bit silly to be using a high-precision value of pi to compute a lower-precision approximation of pi. Furthermore, if one is memorizing a formula for pi, easiest would be to memorize 4 atan 1 which gives an infinite number of correct digits.
No comments :
Post a Comment