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1 + 2 + 4 + 8 + ... = -1

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  • 12-10-2012 2:06pm
    #1
    Closed Accounts Posts: 49


    S = 1 + 2 + 4 + 8 + ...

    = 1 + 2(1 + 2 + 4 + ...)

    = 1 + 2S

    S = 1 + 2S

    S = -1

    Right?


Comments

  • Registered Users Posts: 607 ✭✭✭BroLo


    Not really. 0 =/= 1 + 2(S), so you cannot say S = -1


  • Closed Accounts Posts: 49 Snotzenfartz


    Wat? Where's the error in the reasoning?


  • Closed Accounts Posts: 49 Snotzenfartz


    S = 1 + 2S

    So S = -1, right?


  • Registered Users Posts: 607 ✭✭✭BroLo


    If S* = 1+2+4+8... (is this continued to inifinity?)

    Then 1 + 2(S*) = 1+1+2+4+8...

    As S is already determined, you cannot then say 0 = 1+2(S*). You can say 0 = 1+2(S) but they are two separate equations. S* (in the first equations) does not equal S.

    EDIT
    Sorry. I see what you mean now. Yes I guess you are correct, my mistake. I didn't realise the equation was continuing on, I misread the OP completely.

    Yes you are correct, in a way.

    You should edit the OP and add S to the every left side of the equation, and put a plus before the "..." to show it continues to infinity.


  • Closed Accounts Posts: 5,064 ✭✭✭Gurgle


    Ah, I see...
    S = 1 + 2 + 4 + 8 ...
    = 1 + 2(1 + 2 + 4 ...)

    You can only get to the second line if S = infinity, in which case 2S = S and the rest doesn't work
    If S isn't infinity, then the sequence 1+2+4+8... is finite, so you will have one less term inside the brackets on the second line. Then you can't substitute S for whats in the brackets.


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  • Registered Users Posts: 607 ✭✭✭BroLo


    Gurgle wrote: »
    Ah, I see...


    You can only get to the second line if S = infinity.

    I think he meant that with the "..." he should have put a "+" before it to make it clear though. :)


  • Registered Users Posts: 5,141 ✭✭✭Yakuza


    The sum S doesn't converge, so you're essentially saying an one aleph-null infinity is the same as another aleph-null infinity. Big deal.


  • Closed Accounts Posts: 5,064 ✭✭✭Gurgle


    Yakuza wrote: »
    The sum S doesn't converge, so you're essentially saying an one aleph-null infinity is the same as another aleph-null infinity. Big deal.

    Jaysus, why can't I remember words like 'converge' when I need them.


  • Registered Users Posts: 3,745 ✭✭✭Eliot Rosewater


    It does converge using the "p-adic norm" - i.e., if you change the way in which you measure the difference between two numbers.

    If you take a rational number n and write it in the form
    [LATEX]\displaystyle n=\frac{2^k r}{s}[/LATEX], where r and s are odd integers, and where k is a positive or negative integer, then the 2-adic norm of n is [LATEX]\|n\|_2 =2^{-k}[/LATEX]. The norm is a measure of how big a number is.

    So, for instance [LATEX]\displaystyle \| 3 \|_2 = 1; \|4\|_2 = \frac{1}{2}; \left\|\frac{7}{12}\right\|_2 = 4[/LATEX]. It's a strange notion: 3 is a "bigger" number than 4, and both are "smaller" than 7/12!

    We can show that the sum in the OP converges in the 2-adic norm to -1. We look at the size of the different between the first n terms in the sum, and the supposed limit -1:
    [LATEX] \displaystyle
    \| S_n - (-1) \| = \left\| \sum_{k=0}^n2^k -(-1) \right\| = \| 2^{n+1} -1 +1 \| = \| 2^{n+1} \| = \frac{1}{2^{n+1}} \rightarrow 0
    [/LATEX] as [LATEX]n \rightarrow \infty [/LATEX]. So in the 2-adic norm, [LATEX]\displaystyle \lim_{n\rightarrow \infty} \sum_{k=0}^n 2^k =-1[/LATEX] . :)


  • Registered Users Posts: 5,141 ✭✭✭Yakuza


    Well shut my mouth. I'd never heard of that!
    I thought once you started equating infinite divergent sums with each other, all bets were off.

    edit: Who'da thunk it - Wiki already has an article on it.


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  • Registered Users Posts: 3,745 ✭✭✭Eliot Rosewater


    The p-adic norm actually has some pretty nice "applications" - it's not a triviality! Naturally, the Wikipedia article doesn't mention these...but I've seen some, and they are nice.

    Last year at the Irish Student Math Conference a professor from NUI Maynooth proved a theorem about binomial co-efficients using p-adic norms. Fix a rational number a. Then if the denominator of the binomial coefficient [LATEX]\displaystyle {a \choose k}= \frac{a(a-1)...(a-n+1)}{k!}[/LATEX] (for k an integer) has a given prime p in it, then a also has the same p in its denominator.

    The approach was to show that the function [LATEX]\displaystyle a \mapsto {a \choose k}[/LATEX] is continuous with respect to the p-adic norm, and that hence the inverse image of some closed set was closed, and then something about the inverse image only containing numbers with primes p in their denominator. It was a nice mix of arithmetic/algebra/kind of number theory with analysis.


  • Closed Accounts Posts: 49 Snotzenfartz


    If we add in negative powers as well it sums to 0 since 1/2 + 1/4 + 1/8 + ... = 1.

    What we get is,

    ... + 1/8 + 1/4 + 1/2 + 1 + 2 + 4 + 8 + ... = 0.

    All the powers of 2 add up to zero. WTF is going on? :eek:


  • Closed Accounts Posts: 49 Snotzenfartz


    I just realised the same is true for any number and here's why.

    Define

    [latex]S = ... 1/a^2 + 1/a + 1 + a + a^2 + ...[/latex]

    Now multiply through by [latex]a[/latex], to give,

    [latex]aS = ... 1/a + 1 + a + a^2 + a^3 ...[/latex]

    The whole series has been shifted left one step which doesn't change its value because it has no beginning or end. It's the same series that we started with, it's just viewed from a different point.

    So, [latex]aS = S[/latex].

    Which means [latex]S = 0.[/latex]


  • Closed Accounts Posts: 11 Splagmata


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  • Closed Accounts Posts: 11 Splagmata


    1+1+1+... = -1/2

    ...+1+1+1 = 1/2

    ...1+1+1+... = 0

    trufax.


  • Moderators, Recreation & Hobbies Moderators, Science, Health & Environment Moderators, Technology & Internet Moderators Posts: 90,855 Mod ✭✭✭✭Capt'n Midnight


    S = 1 + 2 + 4 + 8 + ...

    = 1 + 2(1 + 2 + 4 + ...)

    Right?
    You have just said
    S = X

    = 1 + 2(X)


    Even if infinity is the same as (infinity +1) it's not the same as twice infinity and neither are the same as infinite infinities. It doesn't behave as a normal number.



    You've probably seen this
    http://scidiv.bellevuecollege.edu/math/infinitehotel.html


  • Closed Accounts Posts: 11 Splagmata


    1 + 2 + 3 + 4 + ... = -1/12

    tru story.


  • Registered Users Posts: 5,063 ✭✭✭Greenmachine


    Wow


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