Your favourite Paradox

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  1. cuckoo99's Avatar
    • Benevolent Member
    • Location: Brighton
    • Posts: 705
    Re: Your favourite Paradox
    (Original post by MinpoloD)
    The Fermi paradox



    Also: Achilles and the Tortoise
    The Grandfather Paradox
    read the top comment on that achilles and the tortoise paradox
  2. Lord of the Flies's Avatar
    • The foul fiend Flibbertigibbet
    • Location: Paris, France
    • Warning points: 2
    Re: Your favourite Paradox
    (Original post by Aeonstorm)
    I dunno about the first few, but the last one is evidently fallacious as it uses the limiting sum formula of a/(1-r) to give 1/(1-2) = -1.

    However, it fails to recognise that the limiting sum formula actually contains the inherent condition that -1 < r < 1 as the formula is derived from the geometric sum a(r^n - 1)/(r-1) where n approaches infinity, hence equalling
    ar^infinity + a/(1-r)
    which only equals a/(1-r) if -1<r<1
    I *think* that equality is obtained by analytical continuation, something I do not know much about... But basically I think the idea is:

    You have the analytical function f(z)=\displaystyle\sum_{i=0}^{ \infty}z^i and the meromorphic function h(z)=\dfrac{1}{1-z}
    Since f(z)=h(z) over a non-empty subset of the complex plane, h(z) is a unique analytical continuation of f(z) into \mathbb{C}/\{1\}. So h(2)=f(2)=-1

    ... something along these lines anyway. I am not entirely sure why the radius of convergence of h can be extended though...

    On a side note, I just found this fairly basic reasoning:

    y=1+2+4+8+\,...
    y=1+2y
    y=-1

    Not sure this sort of manipulation holds up on its own though.
  3. Aeonstorm's Avatar
    • Exalted Member
    • Location: Sydney
    • Posts: 312
    Re: Your favourite Paradox
    (Original post by Lord of the Flies)
    I *think* that equality is obtained by analytical continuation, something I do not know much about... But basically I think the idea is:

    You have the analytical function f(z)=\displaystyle\sum_{i=0}^{ \infty}z^i and the meromorphic function h(z)=\dfrac{1}{1-z}
    Since f(z)=h(z) over a non-empty subset of the complex plane, h(z) is a unique analytical continuation of f(z) into \mathbb{C}/\{1\}. So h(2)=f(2)=-1

    ... something along these lines anyway. I am not entirely sure why the radius of convergence of h can be extended though...

    On a side note, I just found this fairly basic reasoning:

    y=1+2+4+8+\,...
    y=1+2y
    y=-1

    Not sure this sort of manipulation holds up on its own though.
    I don't pretend to understand the analytical continuation stuff. However, do you know why your basic reasoning is wrong? I thought it was cos 2y and the (2 + 4 + 8...) part of y are different types of infinity. However, I then realised that this is the same principle involved in proving that 0.9 recurring equals 1:

    x = 0.999...
    10x = 9.999...
    so 9x = 9
    so x = 1.

    Isn't this essentially the same logic?
  4. Lord of the Flies's Avatar
    • The foul fiend Flibbertigibbet
    • Location: Paris, France
    • Warning points: 2
    Re: Your favourite Paradox
    (Original post by Aeonstorm)
    I don't pretend to understand the analytical continuation stuff. However, do you know why your basic reasoning is wrong? I thought it was cos 2y and the (2 + 4 + 8...) part of y are different types of infinity. However, I then realised that this is the same principle involved in proving that 0.9 recurring equals 1:

    x = 0.999...
    10x = 9.999...
    so 9x = 9
    so x = 1.

    Isn't this essentially the same logic?
    After a quick search, I am fairly convinced the more basic approach is not wrong at all. According to this as long as your summation method for the divergent series has linearity and stability, you can legally perform simple algebraic manipulations on it. Basically, you can do the following as long as t is different to 1:

    s(t)=\displaystyle\sum_{i=0}^{ \infty}t^i

    s(t)=t^0+\displaystyle\sum_{i=0}  ^{ \infty}t^{i+1}\quad\Rightarrow \text{stability}

    s(t)=t^0+t\displaystyle\sum_{i=0  }^{ \infty}t^{i}\quad\Rightarrow \text{linearity}

    s(t)=1+ts(t)


    t=2\Rightarrow s(2)=-1

    As far as 0.(9) goes, I don't think the logic is entirely the same: the infinity of the decimals is not countably infinite, whereas that of the power series is.
  5. Aeonstorm's Avatar
    • Exalted Member
    • Location: Sydney
    • Posts: 312
    Re: Your favourite Paradox
    (Original post by Lord of the Flies)
    After a quick search, I am fairly convinced the more basic approach is not wrong at all. According to this as long as your summation method for the divergent series has linearity and stability, you can legally perform simple algebraic manipulations on it. Basically, you can do the following as long as t is different to 1:

    s(t)=\displaystyle\sum_{i=0}^{ \infty}t^i

    s(t)=t^0+\displaystyle\sum_{i=0}  ^{ \infty}t^{i+1}\quad\Rightarrow \text{stability}

    s(t)=t^0+t\displaystyle\sum_{i=0  }^{ \infty}t^{i}\quad\Rightarrow \text{linearity}

    s(t)=1+ts(t)


    t=2\Rightarrow s(2)=-1

    As far as 0.(9) goes, I don't think the logic is entirely the same: the infinity of the decimals is not countably infinite, whereas that of the power series is.
    Haha what does countably infinite mean? Does it mean that that one less term in the power series matters more because that term will be infinitely large? As opposed to the last decimal place being infinitely small.
  6. Lord of the Flies's Avatar
    • The foul fiend Flibbertigibbet
    • Location: Paris, France
    • Warning points: 2
    Re: Your favourite Paradox
    (Original post by Aeonstorm)
    Haha what does countably infinite mean? Does it mean that that one less term in the power series matters more because that term will be infinitely large? As opposed to the last decimal place being infinitely small.
    Not really. A set is countably infinite if it contains an infinity of countable sets. A set is uncountably infinite if the set is infinite but not countable. An example will explain this better:

    Countably infinite: \mathbb{N}, or \mathbb{Z} (you can "count" in these sets, and they are infinite)

    Uncountably infinite: \mathbb{R} (you cannot "count" in R, and it is infinite)

    For instance, the set \{0,2,4,8\,...\} is countably infinite. Whereas the set \{[0;1]\} (real numbers between 0 and 1 included) is not countable, therefore it is uncountably infinite.

    The "countability" of a set has to do with its cardinality (number of elements within the set). If the cardinality of a set is more than \mathbb{N}, it is uncountable.
  7. Aeonstorm's Avatar
    • Exalted Member
    • Location: Sydney
    • Posts: 312
    Re: Your favourite Paradox
    (Original post by Lord of the Flies)
    Not really. A set is countably infinite if it contains an infinity of countable sets. A set is uncountably infinite if the set is infinite but not countable. An example will explain this better:

    Countably infinite: \mathbb{N}, or \mathbb{Z} (you can "count" in these sets, and they are infinite)

    Uncountably infinite: \mathbb{R} (you cannot "count" in R, and it is infinite)

    For instance, the set \{0,2,4,8\,...\} is countably infinite. Whereas the set \{[0;1]\} (real numbers between 0 and 1 included) is not countable, therefore it is uncountably infinite.

    The "countability" of a set has to do with its cardinality (number of elements within the set). If the cardinality of a set is more than \mathbb{N}, it is uncountable.
    Ah cheers for the explanation, it makes sense now.
  8. FO12DY's Avatar
    • Adored and Respected Member
    • Location: Dorset
    • Posts: 590
    Re: Your favourite Paradox
    (Original post by Aeonstorm)
    I then realised that this is the same principle involved in proving that 0.9 recurring equals 1:

    x = 0.999...
    10x = 9.999...
    so 9x = 9
    so x = 1.
    I've never liked that proof much - it seems more convoluted than it needs to be.

    0.333........ * 3 = 0.999.......
    1/3 = 0.333........
    1/3 * 3 = 1
    0.999......... = 1
  9. Hugues*'s Avatar
    • Full Member
    • Location: Cornwall | Posts: ♋
    Re: Your favourite Paradox
    Probably been posted before but I'll say it anyway as it's my favourite:

    If Pinocchio says "My nose will grow now!", what will happen?
  10. FO12DY's Avatar
    • Adored and Respected Member
    • Location: Dorset
    • Posts: 590
    Re: Your favourite Paradox
    (Original post by Hugues*)
    If Pinocchio says "My nose will grow now!", what will happen?
    Holy ****.

    Madness would ensue.
  11. Hugues*'s Avatar
    • Full Member
    • Location: Cornwall | Posts: ♋
    Re: Your favourite Paradox
    Any takers on solving it?
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