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+--- Thread: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] (/thread-14532.html)
'√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - Gerson W. Barbosa - 02-18-2020 06:49 PM
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≈e [NT] - Massimo Gnerucci - 02-18-2020 07:38 PM
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RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - Paul Dale - 02-18-2020 10:00 PM
\( \sqrt {\sqrt {\sqrt {\sqrt {2π \sqrt{2} \left( 10^3 \left( 10^3+\frac{π}{3^4} \right)+\frac{1}{π \left(π^5+π^2+1 \right)+\frac{1}{\sqrt{2 π}}}\right) }}}}\)
RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - Gerson W. Barbosa - 02-18-2020 10:13 PM
(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.
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\[\sqrt[16]{2\pi\sqrt{2} \left \{ 10^{3}\left ( 10^{3}+\frac{\pi }{3^{4}} \right )+\left [ \pi \left ( \pi ^{5}+\pi ^{2} +1\right )+\frac{1}{\sqrt{2\pi }} \right ]^{-1} \right \}}=e-5.29\times 10^{-17}\]
RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - Paul Dale - 02-19-2020 02:51 AM
I mis-parsed the parenthesis placement
Fixed now.
Pauli
RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - Gerson W. Barbosa - 02-19-2020 04:36 AM
(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.
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\( \sqrt {\sqrt {\sqrt {\sqrt {2π \sqrt{2} \left( 10^3 \left( 10^3+\frac{π}{3^4} \right)+\frac{1}{π \left(π^5+π^2+1 \right)+\frac{1}{\sqrt{2 π}}}\right) }}}}\)
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 :-)
RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - Dave Britten - 02-19-2020 12:52 PM
It looks like you guys might need to fix your baud rate.
RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - Gerson W. Barbosa - 02-19-2020 06:34 PM
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.
RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - BruceH - 02-19-2020 09:24 PM
(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.
RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - Paul Dale - 02-20-2020 12:36 AM
(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
RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - mfleming - 02-20-2020 01:36 AM
(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!
RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT] - Gerson W. Barbosa - 02-20-2020 07:29 PM
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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:
\[\sqrt{\sqrt{\sqrt{\sqrt{2 \pi \sqrt{2} \left({10}^{3} \left({10}^{3}+\frac{\pi }{{3}^{4}}\right)+\frac{1}{\pi \left({\pi }^{5}+{\pi }^{2}+1\right)+\frac{1}{\sqrt{2\ \pi }}}\right)}}}}\]