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[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
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RE: [VA] SRC #016 - Pi Day 2024 Special - Gerson W. Barbosa - 03-19-2024 09:57 PM



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