Pandigital RPL algebraic pi approximation

[Gerson W. Barbosa]

'2*√LN(ALOG(1)+LN(6)+3*SQ(9-0!)/(ALOG(8)-SQ(75/4)))'

[Gerson W. Barbosa]

Or, in an expression WolframAlpha can understand:

2*√ln(alog(1)+ln(6)+3*sq(9-0!)/(alog(8)-sq(75/4)))

[EdS2]

wow that's a lot of digits of correct approximation! well done!

[naddy]

What's "alog"?

I thought anti-logarithm to base 10, so 10^, but that doesn't make sense.

Edit: Never mind, it works out with alog as 10^. I was lost among the parentheses.

[Maximilian Hohmann]

Hello!

(10-16-2024 02:20 PM)naddy Wrote:  What's "alog"?

I thought anti-logarithm to base 10, so 10^, but that doesn't make sense.

Wolfram Alpha interprets it as natural logarithm. But like you, I have never seen that written as "alog". And why square root of nine minus factorial of zero?

Regards
Max

[klesl]

if alog is natural logarithm, what is ln in the expression?

[AnnoyedOne]

(10-16-2024 02:32 PM)Maximilian Hohmann Wrote:  I have never seen that written as "alog". And why square root of nine minus factorial of zero?

Yeah. I'd write "alog" as "e^x" and "9-0!" as "8" (9-1) but that's just me.

Ln is fine (natural log = log base e where e = 2.718 approx).

A1

[Gerson W. Barbosa]

(10-16-2024 02:32 PM)Maximilian Hohmann Wrote:  Hello!

(10-16-2024 02:20 PM)naddy Wrote:  What's "alog"?

I thought anti-logarithm to base 10, so 10^, but that doesn't make sense.

Wolfram Alpha interprets it as natural logarithm. But like you, I have never seen that written as "alog". And why square root of nine minus factorial of zero?

ALOG is RPL for 10^x, likewise SQ stands for x squared (x^2). WolframAlpha interprets ALOG as log, but gets alog in lowercase right.
I used (9 - 0!)^2 instead of simply 8^2 because 8 had already been used elsewhere. Pandigital expressions use all ten decimal digits only once. They are somewhat difficult to write. Resorting to SQ and ALOG instead of x^2 and 10^x is kind of cheating, but that makes the task a bit easier.

Best,

Gerson.

[Gerson W. Barbosa]

(10-16-2024 01:51 PM)EdS2 Wrote:  wow that's a lot of digits of correct approximation! well done!

Thanks, Ed!

Yes, nineteen correct digits from the nine significant digits in this more conventional equivalent expression:

\(2\sqrt{\ln\left({10+\frac{3\times32^2}{200^4-75^2}+\ln\left({6}\right)}\right)}\)

[KeithB]

8 digits if you go for pi/2. 8^)