
Citação
Originally posted by d_h
less than 1 dollar? I forget how to calculate the limes of the euler-thing/stuff....
Citação
Originally posted by ilex011
HA!
This post has been edited 1 times, last edit by "ilex011" (26/05/2008, 05:51)
Citação
Originally posted by ilex011
Mala Strana, Praha.
Miss it so much. Favorite city in the world. Haven't been back since 1991 though. Yes, I know I'm in for a shock. But I need to go there again. Must be my Czech blood, drawing me back home... the only place I can see my own last name in a phone book! I should have ripped out the page, but that wouldn't have been very nice as guests in other countries shouldn't do things like that. But I was tempted...


+(1/16)+(1/32)+(1/64)+......

Citação
Originally posted by ilex011
I tried calculating it too but I can't read the very fine print to determine the numbers... is that 2(pi) + sum of infinity and something- can't read it... ? I was always curious to see if it actually did come out to anything.
still, I :tongue: :tongue: :tongue: when I first saw this, heh!
Citação
Originally posted by d_h
less than 1 dollar? I forget how to calculate the limes of the euler-thing/stuff....
Citação
Originally posted by ilex011
HA!
Citação
Originally posted by LePand
Citação
Originally posted by ilex011
Mala Strana, Praha.
Miss it so much. Favorite city in the world. Haven't been back since 1991 though. Yes, I know I'm in for a shock. But I need to go there again. Must be my Czech blood, drawing me back home... the only place I can see my own last name in a phone book! I should have ripped out the page, but that wouldn't have been very nice as guests in other countries shouldn't do things like that. But I was tempted...
- Frankfurt
- Köln
- Amsterdam
- Berlin
- Prague
- Milan
wanna join??![]()
Citação
Originally posted by LePand
- Frankfurt
- Köln
- Amsterdam
- Berlin
- Prague
- Milan
wanna join??![]()
)
Citação
Originally posted by d_h
I found the euler-thing with help of the script of my prof.....it is used by complex calculatings.....
1)
e^(i*(pi))=-1 (see script of my prof.)
2)
Sum(1/2^(n))=(1/(2^1))+(1/(2^2))+(1/(2^3))+(1/(2^4))+(1/(2^5))+(1/(2^6))+........
=(1/2)+(1/4)+(1/+(1/16)+(1/32)+(1/64)+......
=0.5+0.25+0.125+0.0625+0.03125+0.015625+.....<1 (it never reach 1 ....I am able only to guess that the guy used a high performance pc... and he gets as result something like = 0.99
3)
0.002
calculating.... -1+0.998+0.002=0 (Zero)



This post has been edited 2 times, last edit by "ilex011" (30/05/2008, 05:39)
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