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Thread: Continuous is lesser than Discrete?...

  1. #1 Continuous is lesser than Discrete?... 
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    I was reading about Zeno's paradox again, and I was curious to know what the difference was between the summation of and the Integral of . I expected the latter to be larger.

    But the results did not match...





    Why is the discreet version larger? It seems unreasonable...


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    Take the interval (1,2) to start, 1/2^x goes from 1/2 to 1/4. The corresponding term in the sum = 1/2, which dominates. This relationship is true for every unit interval starting at an integer value of x.


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    Quote Originally Posted by mathman View Post
    Take the interval (1,2) to start, 1/2^x goes from 1/2 to 1/4. The corresponding term in the sum = 1/2, which dominates. This relationship is true for every unit interval starting at an integer value of x.
    Can you please explain specifically how this justifies what the OP posted? Please use 1 to infinity for your explanation.

    He is actually asking how you sum up parts of a whole out to infinity and arrive at a number greater than the entire whole out to infinity if you did not get that.
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    Quote Originally Posted by JohnWisp View Post
    Can you please explain specifically how this justifies what the OP posted? Please use 1 to infinity for your explanation.

    He is actually asking how you sum up parts of a whole out to infinity and arrive at a number greater than the entire whole out to infinity if you did not get that.
    No, that's not what he is asking. He may think that's what he's asking, but he's not.

    When approximating an integral by a sum, the latter may (subject to the standard caveats regarding convergence criteria) converge to the former if the successive intervals in the summation are "sufficiently small." With a coarse unit difference in the interval here, one should not expect the two to be equal.

    I don't know why you have a problem with mathman's argument. He shows how you can decompose the comparison problem into a sequence of comparisons, unit interval by unit interval.
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  6. #5  
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    Quote Originally Posted by tk421 View Post
    Quote Originally Posted by JohnWisp View Post
    Can you please explain specifically how this justifies what the OP posted? Please use 1 to infinity for your explanation.

    He is actually asking how you sum up parts of a whole out to infinity and arrive at a number greater than the entire whole out to infinity if you did not get that.
    No, that's not what he is asking. He may think that's what he's asking, but he's not.

    When approximating an integral by a sum, the latter may (subject to the standard caveats regarding convergence criteria) converge to the former if the successive intervals in the summation are "sufficiently small." With a coarse unit difference in the interval here, one should not expect the two to be equal.

    I don't know why you have a problem with mathman's argument. He shows how you can decompose the comparison problem into a sequence of comparisons, unit interval by unit interval.
    That is not the way it works clown.

    If the derivative of a function is negative on the unit interval (a.b), then the integral will be less than the "sum".

    On the other hand, if the derivative of a function is positive on the unit interval (a.b), then the integral will be more than the "sum".

    On the other hand, if the derivative of a function is zero on the unit interval (a.b), then the integral will be equal to the "sum".

    This has nothing to do with your crackpot idea of convergence with I guess sums. I don't even know.
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    Hey John, you're mouth is moving again.

    See to it.
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    Quote Originally Posted by JohnWisp View Post
    That is not the way it works clown.
    That's precisely how it works. And I was kicked out of clown college, so that makes two errors in one sentence.

    If the derivative of a function is negative on the unit interval (a.b), then the integral will be less than the "sum".
    Yup. Blindingly obvious, and irrelevant.

    On the other hand, if the derivative of a function is positive on the unit interval (a.b), then the integral will be more than the "sum".
    Still blinding and irrelevant.

    On the other hand, if the derivative of a function is zero on the unit interval (a.b), then the integral will be equal to the "sum".
    Yes -- you have stated the epoch-shattering result that if a function is constant within an interval, then the sum and integral will be equal.

    This has nothing to do with your crackpot idea of convergence with I guess sums. I don't even know.
    You don't know, because you are painfully ignorant, as is evident from your other posts here.

    What's mildly amusing about your didactic attempt is that you are evidently ignorant of how Riemann sums are used to "sneak up" on an integral. If one approximates the integral with such a sum, the quality of the approximation depends a great deal on the coarseness of the discretization. The OP's question is entirely equivalent to asking why a Riemann sum might not equal the integral. And the answer I gave is wholly correct.

    Once you graduate high school -- if you graduate high school -- you might someday have an opportunity at university to learn about integrals and such things.

    And failing that, there's always Clown College as a fallback option. I think you might have the aptitude I lacked.
    Last edited by tk421; April 20th, 2013 at 10:46 PM.
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    Oh I get it now. If you plotgraph.php.jpg this function continuously, the area is actually bounded more than if you take discrete points ( which looks like a step function ). Am I getting this right?
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  10. #9  
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    Quote Originally Posted by Wise Man View Post
    Oh I get it now. If you plotgraph.php.jpg this function continuously, the area is actually bounded more than if you take discrete points ( which looks like a step function ). Am I getting this right?
    Yes - that is what I was trying to say in the first place.
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