sum of 1/x^x
 

Hello everybody,

whats the result of 1/1^1 + 1/2^2 + 1/3^3...
My math software can't solve it.
I think it's unknown.

Thanks to everybody who could help me.
nuggetprime

Well, I had no problem doing it with Excel:

I put [FONT="System"]1/(ROW(A1)^ROW(A1))[/FONT] in cell [FONT="Impact"]A1[/FONT] and did the copy drag down a few rows. Then in cell [FONT="Impact"]B1[/FONT] I put [FONT="System"]SUM($A$1:A1)[/FONT] and did the copy drag thing again. The values rapidly go past the precision of Excel, but the value is there to 10+ places.

1.29128599706266
That's the most precise value I can get OpenOffice to show. There's no difference between the 40th and the 140th (shortly after which it gives an error).

EDIT: it gives the number I posted above starting with the 13th iteration, after that the difference is too small for the precision of the program

cancel

[quote=Mini-Geek;104483]1.29128599706266
[/quote]
In QBasic I get a 4 as an additional decimal place

[CODE]1.2912859970626635404072825905956005414986193682745223173100024451369445387652344555588170411294297089849950709248154305484104874192848641975791635559479136964969741568780207997291779482730090256492305507209666381284670120536857459787030012778941292882535517702223833753193457492599677796483008495491110669649755010519757429116210970215616695328976892427890058093908147880940367993055895352006337161104650946386068088649986065310218534124791597373052710686824652246770336860469870234201965831431339685[/CODE]

Arbitrary-precision FTW!

I went to 500, then to 600, and both gave me the same result on the precision I required (500 places)

[QUOTE=fetofs;104489]1.291285997062663540407282590595600541[/QUOTE]

The [url="http://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html"]Inverse Symbolic Calculator[/url] didn't turn up anything, so the sum likely is not a linear combination of elementary transcendental numbers.

Plouffe's Inverter says:
[url]http://bootes.math.uqam.ca/cgi-bin/ipcgi/lookup.pl?Submit=GO+&number=1.291285997062663540407282590595600541498619368274522317310002445136944538765234&lookup_type=simple[/url]
[quote]
Your input of 1.291285997062663540407282590595600541498619368274522317310002445136944538765234 was probably generated by one
the following functions or found in one of the given tables.
Answers are given from shortest to longest description

Mixed constants with 4 operations
1291285997062663 = 1/ln(exp(1))*Powernn

Bases constants, Pi, e, sqrt(2), etc...
1291285997062663 = sum(1/n^n,n=1..infinity)
[/quote]

Chris

I made a monster for you: n=144(20000 digits), after 144 it fails to give a value using my current technique.

[URL]http://kuratkull.pri.ee/var/bignum.txt[/URL]

How many digits you want? ;)

@Chris: what is Powernn?

@kuratkull: You want war? :smile: I went to 800 and got the exact value (not the decimal representation)

[URL="http://www.geocities.com/fetofs/bignum.txt.gz"]http://www.geocities.com/fetofs/bignum.txt.gz[/URL] (read it on an editor that supports arbitrary line sizes or it won't make any sense).

Well I used Perl to solve this.
I used 20000 for accuracy, so it has 20000 decimal places. And after 144 it
gives 0 as the answer for the equasion, but this can't be right.