If [math] \sum\limits_{n=1}^{|\mathbb{N}|} 2^{-n} = 1 [/math], then what does [math] \sum\limits_{n=1}^{|2^\mathbb{N}|} 2^{-n} [/math] equal?
Can you define what it means to sum a series over uncountably many terms?
But it doesn't equal 1. It's asymptotic.
>>8169027
You add unaccountably many terms together.
>>8169036
Addition is a linear operation that requires enumeration. So you can't.
>>8168992
I am not used to your notation
why |N|? and on the left N starts at 1 and on the right at 0?
so I will interpret it my own way
the one on the left is obvious
the one on the right is [math]2^{-2}+2^{-4}+...[/math]
this equals [math]\sum_{i=1}^{\infty}2^{-2i}[/math]
very easy to solve... 0.25 is half of 0.5 and so on. you can immediately recognize the proportions, that means
[math]\sum_{i=1}^{\infty}2^{-2i}=\frac{1}{3}[/math]
>>8169046
>on the left N starts at 1 and on the right at 0?
N doesn't start anywhere. n starts at one in both cases.
>>8169046
|N| is the number of natural number = infinity
>>8169046
ooops did I misread the OP?
it's 2^N, not 2N
so you mean [math]2^{-1}+2^{-2}+2^{-4}+2^{-8}+... = \sum_{i=1}^{\infty}2^{-2^{i}}[/math]
sooo yeah that will be a bit smaller than 1/3
I dont know how to solve this precisely but wolframalpha says 0.316422
if N includes 0 then the answer will obviously be 0.816422
>>8169051
well the equasion is wrong but you get the point
I cant remember the last time I did math so I am pretty shit at it
