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Non-standard calculus (infinitesimals)

  1. Jan 24, 2005 #1
    Compute the standard part of this, please:
    [tex]\frac{ \sqrt{H+1}}{ \sqrt{2H} + \sqrt{H-1}}[/tex], where H is positive infinite.

    It probably should be some algebra trick I'm not familar with.
     
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  3. Jan 24, 2005 #2
    infinite can't divided by infinite... your question doesn't make sense at all
    I think what you meant was.....
    [tex] \lim_{H \rightarrow \infty} \frac{\sqrt{H+1}}{\sqrt{2H}+\sqrt{H-1}} [/tex]

    hints:
    for a very large [itex] H [/itex], you can assume [itex] \sqrt{H+1}= \sqrt{H} [/itex]
     
  4. Jan 24, 2005 #3

    HallsofIvy

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    No, vincent, he meant what he said. As the title said, this is "non-standard analysis" in which we have both infinite numbers and infinitesmals. (H is a positive infinite number.)

    However, the result will be exactly the same as the lim as H-> infinity.

    And, yes, it is true that for "infinite" H, H+1= H, H-1= H. This is exactly the same as
    [tex]\frac{\sqrt{H}}{\sqrt{2H}+\sqrt{H}}= \frac{\sqrt{H}}{(\sqrt{2}+1)\sqrt{H}}= \frac{1}{\sqrt{2}+1}[/tex]
     
    Last edited: Jan 24, 2005
  5. Jan 24, 2005 #4

    Hurkyl

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    One algebra trick, it so happens, is exactly the same as standard analysis: divide the numerator and denominator by the "highest power" of H... in this case, it's 1/2.



    That's incorrect: H+1 is never equal to H for any hyperreal, even the unlimited ones. (I think unlimited is the preferred term, over infinite or transfinite)

    The unlimited hyperreals don't act like cardinal numbers -- they act like real numbers, in a very real sense. (Pun intended)


    However, H/(H+1) would be a limited (aka finite) number with standard part 1, so that could be used fruitfully in this example, by replacing (H+1) with H * ((H+1) / H).
     
  6. Jan 25, 2005 #5
    Ahh I see! It's really just a beginner calculus book, although it used infinitesimals.
     
  7. Jan 25, 2005 #6

    HallsofIvy

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    Ah! Thanks for clearing up my error, Hurkyl. An expert on non-standard analysis I'm not. (I'm barely competent on STANDARD analysis!)
     
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