1=0.999... infinities and box of chocolates..Phliosophy of Math...

Discussion in 'General Philosophy' started by Quantum Quack, Nov 2, 2013.

  1. Tach Banned Banned

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  3. Quantum Quack Life's a tease... Valued Senior Member

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    @Billy T
    I presume that there is a distinction between these two terms and is not a contradiction as Tach has suggested:
    Convergent infinite series
    &
    Convergent geometric series


    Am I correct?
     
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  5. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    I don't care to whom your post was addressed. I would like its location so I can reconsider it.

    I'll be very surprised if it is not some version of: "I can make calculations with any specific n in a^(-n) and see they become smaller the larger n becomes. Furthermore, that given any small number, d, but not zero, I can find an n and calculate with it the value of a^(-n) such that that value is less than d, the "target" number, so I therefore I guess the true value of
    a^(-n) is zero as n becomes larger than any specific number I can write and calculate with. However, I don't want to "just guess" so I define a^(-n) to be zero as n becomes larger than any specific number I can write and calculate with."
     
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  7. Tach Banned Banned

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    Here. Your incorrect retort follows a couple of posts below.

    Why do you persist and you refuse to dust off your calculus books? The proof is standard, it is one of the easier proofs in introduction to limits. It was introduced in 11-th grade, in high school.
     
  8. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    no. In your post 19 your realized I made a typo with "than" I quickly corrected it to "that" but not before tach had quoted my dyslectic version. Both your guesses as to what I intended *then & therefore) don't correct my dyslectic version. What follows the intended that is a qualification, not a consequence.
     
  9. Tach Banned Banned

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    This is not as bad as before but it is still bad. The limit of any convergent series is \(\lim_{n \to \infty}S_n\).
    You are just inventing your own dichotomy. It doesn't exist in any calculus books.
     
  10. Tach Banned Banned

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    Yes, the "Quack" made it even worse, with his "corrections". I guess, this is what quacks do.
     
  11. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    Yes I admit that "proof of calculability" is not exactly the one I just guessed in post 23, but it is defective (as a "proof of calculability") until you show that you can calculate natural logs.
    I think you can calculate only quite accurate approximations of them.*

    here was your "proof of calculability"
    * I don't mean to be crude or offensive, but converting problem of showing calculability of limits into one of showing calculability of logs is just like a religious person's explaining the where / how the universe came to exist by "God made it." - Swaps one problem for another just a hard.
     
  12. Tach Banned Banned

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    Come on, you are now just grasping at straws.

    Doesn't matter, you only need to find \(n>N(\epsilon)\) for which \(|a^n|<\epsilon\). This gives you the lower limit at \(N(\epsilon)=\frac{ln \epsilon}{ln |a|}\). Have you tried finding your calculus books?


    Yes, you can find it in most mainstream calculus books <shrug>
     
  13. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    I already noted that logs are among the many things in math that can not actually be calculated, 0nly very good approximations* of them can be:
    As for calculus books I sold then two decades ago, just before moving to Brazil.

    * we usually know the upper bound the approximations are approaching as the approximations become more refined, so DEFINE that as the value we can not actually calculate.
     
  14. Tach Banned Banned

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    But this is not a valid argument, the books teach you that you only need to find \(n>N(\epsilon)\) such that \(S_n<\epsilon\). \(N(\epsilon)\) doesn't have to be exact. As a matter of fact, in the example I gave you, you need to round up \(\frac{ln \epsilon}{ln a}\) in order to get the next natural number. All this is textbook stuff, if you no longer have the books , you could certainly go to the nearest library.



    Repeating the same fringe ideas doesn't make them right, I am quite sure that the Brazilian libraries have a few good calculus books. Do yourself a favor and read the appropriate chapter.
     
  15. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    It certainly is if one is discussing what is, and what is not, "calculable." - Beats appeal to authority all to hell.

    Your post 31 is another attempt to switch the discussion. I have ALWAYS agreed that one can find n such that the calculated value is less than epsilon different from the defined value for any value of epsilon (not zero) you like.

    I will now even go further: Yes you can use approximations to find an n as f(epsilon) and even use N= n + 1 million just to be sure, but can not calculate the limit, say zero if that is the limit, which has been defined.
     
  16. Tach Banned Banned

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    Find a book. Take a class. Go talk with a professor. Read a link from a math professor (I know, I gave you this one before).
     
  17. Fednis48 Registered Senior Member

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    I think you're confused about the definition of a limit. The limit of f(n) as n goes to infinity is defined as the number x such that for any nonzero epsilon, one can choose an N such that f(n>N) is within epsilon of x. You're saying that the whole "within epsilon" method is just an approximation, but in fact that's literally what the word "limit" means in mathematics.
     
  18. Yazata Valued Senior Member

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    I haven't read that discussion, nor do I know what Tac and BillyT are arguing about in this thread.

    I don't want to agree that's correct. To make it correct, you would probably need to follow Newton and Leibniz and say that

    0.999... + an infinitesimal = 1

    where an infinitesimal is an infinitely small number such that

    1 - 0.999... = an infinitesimal

    ...or else follow most of contemporary mathematics and say that endlessly adding 9's to the butt-end of 0.999... produces a growing number that approaches closer and closer to 1 as its limit, without ever quite getting there.

    I'm not a mathematician and I'm not really qualified to discuss this stuff, but I do know that these kind of ideas have been a source of no end of controversy and consternation in the history and philosophy of calculus.

    See here:

    http://www.sjsu.edu/faculty/watkins/infincalc.htm
     
  19. gmilam Valued Senior Member

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    It's a simple "glitch" in our decimal representation of fractions.

    A cool example is 1/7 = .(142857)
    6/7 = .(857142)

    1/7 + 6/7 = .(142857) + .(857142)
    7/7 = .(9)
     
  20. Pete It's not rocket surgery Registered Senior Member

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    Billy, are you just making a point about your preferred definition of "calculate", or are you saying something about the numeric value of 0.999...?
    If a value for an infinite sum can't be calculated, by your definition, then so what?
    What does that tell us about the actual result of the infinite sum?
    Are you claiming that results determined by calculation are better in some sense than results determined through wider algorithms?

    If you can't calculate \(\sqrt{2}\), then so what?
    Does that mean that \(\sqrt{2}\) doesn't have an actual value?


    On the definition of a limit:
    Do you think that the definition of a limit is arbitrary? Could it be meaningfully defined in any other way, so as to give different results?
    I don't think it could be. And that suggests to me that the definition is representative of something just and real and meaningful as calculation.
     
  21. Tach Banned Banned

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    This is very cool, indeed!
     
  22. someguy1 Registered Senior Member

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    Hi, I haven't read through the rest of the discussion. But I think you are getting at something meaningful. A calculation is defined as an execution of some algorithm implemented in the physical world. If we make that definition, then a calculation can never evaluate an infinite series.

    In fact we could look at this from the point of view of the hardware, and observe that a calculation takes a certain amount of energy in terms of the electrons that have to flow through the circuits. In a universe with finite total energy, there is literally a boundary between things that can be calculated and things that can't.

    Is that what you are getting at? I agree that we can never calculate the result of an infinitary process.

    But that's what so clever about what mathematicians have done over the last few hundred years. They've worked out a logically rigorous theory, starting from some simple axioms of set theory, that allows us to define the results of infinitary processes.

    So we can make a definition of "the sum of an infinite series" and we can prove that .999... = 1. It's a triumph of human intellectual history, from Archimedes to Newton to Cauchy to Cantor and so many others. We are privileged to live in the age in which humanity knows how to symbolically deal with the infinite.

    But no infinitary process can ever be carried out in the physical world. And if that's what you mean by a calculation, you are correct.

    Am I understanding your point about calculation?
     
  23. Quantum Quack Life's a tease... Valued Senior Member

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    that it is and has been
     

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