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Math Help - Why is this trollmath wrong?

  1. #1
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    Why is this trollmath wrong?

    Hi forum,

    I saw this trollmath on youtube. i dont claim its correct, but i cant understand where the mistake is. help

    Consider a geometric series as follows:

    S = 2 + 4+ 8 + 16 + ...... [1]

    2S = 4 + 8+ 16 +  ...... [2]

    [2]-[1]

    S = 2

    Clearly the last line is false, but where is the mistake?
    Last edited by SpringFan25; July 12th 2013 at 07:50 AM.
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  2. #2
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    Re: Why is this trollmath wrong?

    Quote Originally Posted by SpringFan25 View Post
    I saw this trollmath on youtube. i dont claim its correct, but i cant understand where the mistake is. help Consider a geometric series as follows:

    S = 2 + 4+ 8 + 16 + ...... [1]

    2S = 4 + 8+ 16 +  ...... [2]

    [2]-[1]
    S = 2 Clearly the last line is false, but where is the mistake?
    You are only allowed to subtract finite sums. SO

    S_n = 2 + 4+ 8 + 16 + ...+2^n [1]

    2S_n = 4 + 8+ 16 +  ....+2^{n+1} [2]

    2S_n-S_n=2^{n+1}-2
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  3. #3
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    Re: Why is this trollmath wrong?

    That is not exactly correct. You can subtract infinite sums if they converge. For example, 1+ 1/2+ 1/4+ 1/8+ ... (adding 1 over powers of 2) is a "geometric series" that converges to 1/(1- 1/2)= 2. Similarly, 1+ 1/3+ 1/9+ 1/27+ ... (adding 1 over powers of 3) is a geometric series that converges to 1/(1- 1/3)= 3/2. (1+ 1/2+ 1/4+ 1/9+...)- (1+ 1/3+ 1/9+ 1/27+ ...)= (1/2- 1/3)+ (1/4- 1/9)+ (1/8- 1/27)+ ...= 1/6+ 5/36+ 19/216+ ... converges to 2- 3/2= 1/2.

    The problem here is that the series does not converge.
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  4. #4
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    Re: Why is this trollmath wrong?

    Quote Originally Posted by HallsofIvy View Post
    That is not exactly correct. You can subtract infinite sums if they converge.
    Of course that is true. But the point is that to prove convergence one looks at finite sums.
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  5. #5
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    Re: Why is this trollmath wrong?

    Hello, SpringFan25!

    Consider a geometric series as follows:

    . . \begin{array}{ccccc}S &=& 2 + 4 + 8 + 16 + \hdots & {\color{red}[1]} \\ 2S &=& \quad\;\; 4 + 8 + 16 +  \hdots & {\color{red}[2]} \end{array}

    {\color{red}[2 ] - [1]}:\;S = -2

    Clearly the last line is false, but where is the mistake?

    The mistake is in thinking that we can "do" arithmetic with Infinity.

    Clearly: . S \:=\:2 + 4 + 8 + 16 + \hdots \:=\:\infty

    Then: . 2S \:=\:\infty

    Hence: . 2S - S \:=\:\infty-\infty, an indeterminate form.

    That is, S could be any value.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Here is a classic "paradox".

    \text{Evaluate: }\:S \;=\;1 - 1 + 1 - 1 + 1 - 1 + \hdots


    Method 1

    S \;=\;(1-1) + (1-1) + (1-1) + \hdots

    S \;=\;0 + 0 + 0 + \hdots

    \boxed{S \;=\;0}


    Method 2

    S \;=\;1 - (1-1) - (1-1) - (1-1) - \hdots

    S \;=\;1 - 0 - 0 - 0 - \hdots

    \boxed{S \;=\;1}


    Method 3

    S is a geometric series
    . . with first term a = 1 and common ratio r = \text{-}1.

    Hence: . S \;=\;\frac{a}{1-r} \;=\;\frac{1}{1-(\text{-}1)}

    Therefore: . \boxed{S \;=\;\tfrac{1}{2}}
    Thanks from topsquark
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  6. #6
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    Re: Why is this trollmath wrong?

    Quote Originally Posted by SpringFan25 View Post
    Hi forum,

    I saw this trollmath on youtube. i dont claim its correct, but i cant understand where the mistake is. help

    Consider a geometric series as follows:

    S = 2 + 4+ 8 + 16 + ...... [1]

    2S = 4 + 8+ 16 +  ...... [2]

    [2]-[1]

    S = 2

    Clearly the last line is false, but where is the mistake?
    We CAN get the right answer if we are careful

    2S = (2 + 2) + (4 + 4) + (8 + 8) + ....
     \ \ S = \ \ \ \ \ 2 \ \ \ \ + \ \ \ \ 4 \ \  \ + \ \ \  \ 8 \ \ \ \ + ...

    Subtract vertically to get S

    Just because we have the difference of two infinite divergent sums doesn't necessarily put the right answer out of reach , like Soroban said infinity minus infinity could be anything... that includes the right answer. Finding a procedure that produces the right answer may be difficult though.

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