Infinity defeated
Name:
Anonymous
2007-08-13 20:37
ID:CXng0JnT
let
s = 1 + 2 + 4 + 8 + ...
s = 1 + 2(1 + 2 + 4 + ...)
s = 1 + 2s
s = -1
1 + 2 + 4 + 8 + ... = -1
discuss
Name:
Anonymous
2007-08-13 21:23
ID:SLUEk+Yd
infinite sums do not work that way
Name:
Anonymous
2007-08-13 21:36
ID:Heaven
S = sum from n = 1 to inf of 2^n
S = 1 + sum from n = 2 to inf of 2^n
S = 1 + 2(sum from n = 2 to inf of 2^(n-1))
Name:
Anonymous
2007-08-13 21:50
ID:VVA53Gt9
Infinite summation is not regular addition.
See also: conditionally convergent series.
Name:
Anonymous
2007-08-13 21:54
ID:DszhSKOh
Infinite sums can quite easily be rearranged to equal different numbers. Riemann did some theorem on this.
Name:
4tran
2007-08-13 23:26
ID:Heaven
s = 1 + 2 + 4 + 8 + ...
s = 1 + 0 + 2 + 0 + 4 + 0 + 8 + 0 + ...
s = 1 + (1 - 1) + 2 + (1 - 1) + 4 + (1 - 1) + 8 + (1 - 1) + ...
s = (1 + 1) (- 1 + 2 + 1) (- 1 + 4 + 1) (- 1 + 8 + 1) + ...
s = 1 + 1 + 2 + 4 + 8 + ...
s = 1 + s
0 = 1 ZOMG!!
Inconsistent'd
Name:
4tran
2007-08-13 23:27
ID:Heaven
Name:
Anonymous
2007-08-13 23:32
ID:CXng0JnT
Name:
Anonymous
2007-08-14 21:41
ID:ZXl4NguV
ur a genius
Name:
Anonymous
2007-08-14 21:51
ID:hPFCLLRm
>>7
It's not a conditionally convergent series though.
I was going to post the same thing though to be honest.
Name:
Anonymous
2007-08-14 22:30
ID:wrglwV15
>>6
YOU assume that s IS A REAL NUMBER. you do not use MICHAEL SPIVAK'S EXPERT ANALYSIS METHOD, described in HIS BOOK
Name:
4tran
2007-08-15 0:21
ID:Heaven
>>10
I know, but I referenced the theorem
>>5 was referring to just in case somebody was interested, even if it didn't apply to this particular case.