[VA] SRC #016 - Pi Day 2024 Special
|
03-19-2024, 09:57 PM
(This post was last modified: 03-19-2024 09:59 PM by Gerson W. Barbosa.)
Post: #24
|
|||
|
|||
RE: [VA] SRC #016 - Pi Day 2024 Special
7th:
When trying to simplify the polynomial equation x³ - 6x² + 4x - 2 = 0, I noticed the change of variable x = y + 2 reduces it to the incomplete cubic equation y³ - 8y - 10 = 0, of the type y³ + p y + q = 0, which can be solved by Cardano’s formula. The real root (only one in this case) is given by y = (-q/2 + √d)^(1/3) + (-q/2 - √d)^(1/3), where d = (p/3)^3 + (q/2)^2 and p = -8; q = -10 Thus y = (5+√(163/27))^(1/3)+(5-√(163/27))^(1/3)) and x = 2 + (5+√(163/27))^(1/3)+(5-√(163/27))^(1/3)) = 5.318628217750185659109680153318022 By replacing x in the given expression we get, on Free 42, log((2+(5+√(163/27))^(1/3)+(5-√(163/27))^(1/3))^24-24)/√163 = 3.141592653589793238462643383279503 which matches exactly all the 34 digits of the built-in constant. The difference to \(\pi\) is about 3.14e-34 in excess, which makes this approximation quite suitable for Free42. The value of the real root of the second polynomial equation can also be given by a more compact expression (see Valentin’s HP-15C Mini-Challenge: Impossibly Short !? for details), which I have used to evaluate the full expression on Free42: y = 4 √(2/3) cosh(1/3 acosh((15 √(3/2))/16)) 00 { 50-Byte Prgm } 01▸LBL "A7" 02 4 03 2 04 3 05 ÷ 06 SQRT 07 × 08 LASTX 09 1/X 10 15 11 × 12 16 13 ÷ 14 ACOSH 15 3 16 ÷ 17 COSH 18 × 19 2 20 + 21 24 22 Y↑X 23 RCL- ST L 24 LN 25 163 26 SQRT 27 ÷ 28 END XEQ A7 -> 3.141592653589793238462643383279503 |
|||
« Next Oldest | Next Newest »
|
User(s) browsing this thread: 1 Guest(s)