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Can one define higher order infinitesimals?

  1. Jul 4, 2015 #1
    Say ##A##, ##B##, ##C##,... are finite numbers; real, complex, quarternians, tensors, or what have you.

    "First Order" infinitesimals are finite variables prepended with the letter ##d##.
    Infinitesimal of any order, are prepended with ##d^n## where ##n## is the infinitesimal "order".

    Finite numbers may be notated as ##d^0 B##. ##B## is any finite number.

    Each transfinite number of would be the reciprocal of a infinitesimal. ##d^{-n}B = \frac{1}{d^{n}B}= \partial_{(n)}B##.

    Examples: ##d^nC = d^mD d E^{m-n}##. ##n## and ##m## range over the integers, and ##d^n(d^m C) = d^{n+m}C##.

    ##d^mX<d^nY## for all ##X## and ##Y## where ##n<m## and ##X## and ##Y## are positive.

    Is this a consistent system, or could there be cases where ##d^nX>d^mY## is true and ##d^nU>d^mV## is false?
     
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  3. Jul 5, 2015 #2

    mfb

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    Transfinite numbers have a specific, different meaning already.
    Ordinal numbers have a structure that could match your question. Note that you have to define ">" on the number system anew, and there is a consistent way to do that.
     
  4. Jul 5, 2015 #3

    micromass

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  5. Jul 5, 2015 #4

    micromass

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    This is a perfectly consistent system, and a lot of number systems have these characteristics. For example, the Laurent series. https://en.wikipedia.org/wiki/Infinitesimal#Laurent_series
     
  6. Jul 5, 2015 #5

    WWGD

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    An interesting question ,somewhat related, is that of doing Calulus/Analysis in powers of the Hyperreals. We don't have an actual metric, but a hyperreal-valued one.
     
  7. Jul 5, 2015 #6

    micromass

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    Well, the issue with the hyperreals here is that you cannot cleanly divide them like the OP wants to. The OP wants to have an infinitesimal ##\varepsilon##, and then divide all the infinitesimals into classes of the form ##\{a\varepsilon^n~\vert~a\in \mathbb{R}\}## for ##n\in \mathbb{Z}##. This does not exhaust the hyperreals however, since there are hyperreals of the form ##\sqrt{\varepsilon}## which are not included in those classes. Even taking ##n\in \mathbb{Q}## would not exhaust the hyperreals, since you can have stuff like ##e^{-1/\varepsilon}## which is smaller than each ##\varepsilon^n##. This of course is related to the existence of a smooth function with vanishing Taylor series. So the hyperreals are far too rich to do this.

    In constructive analysis and smooth differential geometry, we sometimes do subdivide the infinitesimals into classes. This can even be made to work in standard mathematics using some adjustments. http://www.emis.de/journals/AMUC/_vol-73/_no_2/_giordano/giordano.pdf
     
  8. Jul 5, 2015 #7
    Thanks for all that. I think it may be the surreals that you first referenced as well as the surcomplex that I'm looking for, though I don't know enough yet about the other number systems.

    Are the surreals a subset of the hyperreals? I would need to be able to take the square root of ##\epsilon##.
     
    Last edited: Jul 5, 2015
  9. Jul 5, 2015 #8

    mfb

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    They other way round, the hyperreal numbers are a subset of the surreal numbers.
    From Wikipedia: "the surreal numbers are the largest possible ordered field; all other ordered fields, such as the rationals, the reals, the rational functions, the Levi-Civita field, the superreal numbers, and the hyperreal numbers, can be realized as subfields of the surreals".
     
  10. Jul 5, 2015 #9

    WWGD

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    I think this is true modulo choice of set theoretic-axioms ( although the surreals are a class, not a set ).
     
  11. Jul 5, 2015 #10

    micromass

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    Yes, you can take square roots in the surreal number system. But this is dangerous territory: not just all algebraic operations make sense in the surreal numbers. Unlike the hyperreals where you can essentially do every operation that you can do in ##\mathbb{R}#.
     
  12. Jul 5, 2015 #11
    This question came up in the context of nonstandard analysis, were I am not happy throwing away all but the finite parts.

    ##f'(x) = \frac{f(x+dx)-f(x)}{(x+dx)-(x)}##

    For ##y=f(x)=x^2##, ##f'(x)= st(2x + dx)=2x##. The ##dx## is thrown away.

    I would rather like ## f'(x)= \{ \frac{f(x+dx)-f(x)}{(x+dx)-(x) }, \frac{f(x)-f(x-dx)}{(x)-(x-dx)} \} = \{ 2x+dx, 2x-dx \} ##, and the differential ##dy={2xdx+d^2 x,2xdx-d^2 x}##.

    In the case where ##f(x)=x^3##, ## f'(x) = { 3x^2+3xdx + (dx)^2, ...}##, (This, unfortunately, doesn't correspond well with my notation in post#1.) So the question comes up: Is ##(dx)^2 ## smaller than any infinitesimal ##du##, where u is finite.

    micromass, we've cross posted, so this post is out of sequence.
     
  13. Jul 5, 2015 #12
    It seems that any function with a Taylor series expansion would make sense. Do you have an example of what doesn't work?
     
  14. Jul 5, 2015 #13

    micromass

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    But then your derivative ##f'(x)## will depend on the specific ##dx## you take.

    What does ##du## with ##u## finite mean? It is true that ##(dx)^2 < dx##.
     
  15. Jul 5, 2015 #14

    micromass

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  16. Jul 5, 2015 #15
    I'm not sure what you have in mind. I would think that, if ##x## is an independent variable, it's differential ##dx##, is an independent variable so it doesn't have a specific value until the value of ##x## is specified.

    ##u## is any finite number, and ##du## is it's infinitesimal, but I've noticed, in defining the differential ##dy##, I have an inconsistency. It's supposed to be order 1, infinitesimal, but, in terms of ##x##, it is mixed order. It may be fatal.
     
  17. Jul 6, 2015 #16
    That's too far outside my knowledge base. I'll learn more about surreal numbers, hyperreal numbers, etc. before I tackle this again. Thanks for the help! And everyone has been helpful.
     
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