'√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT]

[Gerson W. Barbosa]

.

[Massimo Gnerucci]

.

[Paul Dale]

\( \sqrt {\sqrt {\sqrt {\sqrt {2π \sqrt{2} \left( 10^3 \left( 10^3+\frac{π}{3^4} \right)+\frac{1}{π \left(π^5+π^2+1 \right)+\frac{1}{\sqrt{2 π}}}\right) }}}}\)

[Gerson W. Barbosa]

(02-18-2020 10:00 PM)Paul Dale Wrote:  \( \sqrt {\sqrt {\sqrt {\sqrt {2π \sqrt{2} \left( 10^3 \left( 10^3+\frac{π}{3^4} \right)+\frac{1}{π \left(π^5+π^2+1 \right)}+\frac{1}{\sqrt{2 π}}\right) }}}}\)

Looks like the algebraic expression has been incorrectly parsed. As it is above, half the correct digits are lost. The HP 50g parses it right, but obviously it will show e to only 12 digits.

-------

\[\sqrt[16]{2\pi\sqrt{2} \left \{ 10^{3}\left ( 10^{3}+\frac{\pi }{3^{4}} \right )+\left [ \pi \left ( \pi ^{5}+\pi ^{2} +1\right )+\frac{1}{\sqrt{2\pi }} \right ]^{-1} \right \}}=e-5.29\times 10^{-17}\]

[Paul Dale]

I mis-parsed the parenthesis placement
Fixed now.

Pauli

[Gerson W. Barbosa]

(02-19-2020 02:51 AM)Paul Dale Wrote:  I mis-parsed the parenthesis placement
Fixed now.

Yes, too many parentheses. Quite easy to get lost, that’s why I used the HP 50g (m48+ emulator) Equation Editor.

——-

\( \sqrt {\sqrt {\sqrt {\sqrt {2π \sqrt{2} \left( 10^3 \left( 10^3+\frac{π}{3^4} \right)+\frac{1}{π \left(π^5+π^2+1 \right)+\frac{1}{\sqrt{2 π}}}\right) }}}}\)

Using only one-digit mantissas, the digits 0 through 5, π, the square root symbol and two pairs of parentheses to yield the first 17 digits of e. If I were mean, I would have posted this as a mini-challenge :-)

[Dave Britten]

It looks like you guys might need to fix your baud rate.

[Gerson W. Barbosa]

Here’s a double pandigital expression:

'9/5-1/(30*672*√√4+8)+√(9/5-1/(30*672*√√4+8))'

→NUM ->

3.14159265362

( From Ramanujan’s '9/5 + √(9/5) = 3.1416407865' )

These are as easy to find as they are useless and futile.

[BruceH]

(02-19-2020 12:52 PM)Dave Britten Wrote:  It looks like you guys might need to fix your baud rate.

Where's the like button? :-)

That took me back. In my head I now have that V.92 connection "be-dang, be-dang" ear-worm going.

[Paul Dale]

(02-19-2020 06:34 PM)Gerson W. Barbosa Wrote:  These are as easy to find as they are useless and futile.

They are still fun

Pauli

[mfleming]

(02-19-2020 09:24 PM)BruceH Wrote:  That took me back. In my head I now have that V.92 connection "be-dang, be-dang" ear-worm going.

I once had the Telebit "moose call" as a ringtone, but yeah, it was just as annoying as you might think!

[Gerson W. Barbosa]

.
How to convert RPL algebraic expressions to LaTeX

Copy the algebraic expression to the HP 50g emulator:

'ƒƒƒƒ(2*‡*ƒ2*K(10^3*K(10^3+‡/3^4)+1/(‡*K(‡^5+‡^2+1)+1/ƒ(2*‡))))'

ƒ = √; ‡ = \(\pi\);  = →

(Place K before every opening parenthesis you want to show up in the formula)

Then on the emulator do

OBJ→ →HP2TX "\cdot" "" SREPL DROP

( "\cdot" "" SREPL DROP only in case you want to remove all dots in the expression )

where

→HP2TX is the last object in the hp2txt directory, by forum member peacecalc here.

That will return the corresponding LaTeX expression:

\sqrt{\sqrt{\sqrt{\sqrt{2\ \pi \ \sqrt{2}\ \left({10}^{3}\ \left({10}^{3}+\frac{\pi }{{3}^{4}}\right)+\frac{1}{\pi \ \left({\pi }^{5}+{\pi }^{2}+1\right)+\frac{1}{\sqrt{2\ \pi }}}\right)}}}}

Append /\[ and/\] to the beginning and end... and voilà!

\[\sqrt{\sqrt{\sqrt{\sqrt{2\ \pi \ \sqrt{2}\ \left({10}^{3}\ \left({10}^{3}+\frac{\pi }{{3}^{4}}\right)+\frac{1}{\pi \ \left({\pi }^{5}+{\pi }^{2}+1\right)+\frac{1}{\sqrt{2\ \pi }}}\right)}}}}\]

"\cdot" "" SREPL should have removed the "\" in "\cdot", but it didn't. Remove them manually if don't want the extra spaces:

\[\sqrt{\sqrt{\sqrt{\sqrt{2 \pi \sqrt{2} \left({10}^{3} \left({10}^{3}+\frac{\pi }{{3}^{4}}\right)+\frac{1}{\pi \left({\pi }^{5}+{\pi }^{2}+1\right)+\frac{1}{\sqrt{2\ \pi }}}\right)}}}}\]