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Quotient of infinitesimals indeterminate?

  1. Sep 28, 2010 #1
    In Lectures on the hyperreals: an introduction to nonstandard analysis, pp. 50-51, Goldblatt includes among his hyperreal axioms that the sum of two infinitesimals is infinitesimal, that the product of an infinitesimal and an appreciable (i.e. nonzero real) number is infinitesimal, and that the quotient of two infinitesimals is an indeterminate form.

    Yet Stroyan, p. 50, writes

    [tex]f[x+\delta x]-f[x]=f'[x] \; \delta x + \varepsilon[/tex]

    where [itex]\delta x[/itex] and [itex]\varepsilon[/itex] are infinitesimal. How does this not make

    [tex]\frac{f[x+\delta x]-f[x]}{\delta x}[/tex]

    an indeterminate form in non-standard analysis? (By indeterminate form, I understand something not defined, something not ascribed any meaning, such as x/0.) I thought defining the derivative as the standard part of a quotient of infinitesimals was one of the main motivations for the hyperreal number system. Is the derivative to be understood in nonstandard analysis as an approximation to something undefined, just as in standard analysis?
  2. jcsd
  3. Sep 28, 2010 #2
    It's difficult to ascribe a meaning to quotient of infinitesimals as limits of the type 0/0 can be 0, finite or infinite.
  4. Sep 29, 2010 #3


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    Because division is defined for any pair of numbers, as long as the denominator is nonzero.

    The derivative at a standard value x is, as I've seen it defined, given by:
    [tex]f'(x) = \mathop{std} \frac{f(x + \epsilon) - f(x)}{\epsilon}[/tex]
    if and only if this expression exists and has the same value for all non-zero infinintessimals [itex]\epsilon[/itex]. (And the derivative is extended to non-standard numbers by the transfer principle) (the expression could fail to exist becuase the "standard part" is not defined for infinite numbers)

    Of course, the definition in terms of limits gives the same result.
    Last edited: Sep 29, 2010
  5. Sep 29, 2010 #4
    Thanks, Hurkyl. So is Goldblatt's prohibition on dividing an infinitesimal by an infinitesimal not part of "standard" nonstandard analysis, or have I misunderstood what indeterminate means? Or is the derivative being defined as the standard part of an indeterminate form, if that's even possible?
  6. Sep 29, 2010 #5


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    This, I think. From the information given, one cannot determine whether any of the expressions listed are unlimited, appreciable, or infinitessimal, unlike the forms listed in the previous bullets, so that's the sense I'm sure he means "indeterminate". Of course, there is the close analogy to the concept of "indeterminate form" appearing in elementary calculus; the fact that [itex]\epsilon / \delta[/itex] is indeterminate in the sense I just mentioned implies that 0/0 is an indeterminate form in the sense of elementary calculus.
  7. Sep 29, 2010 #6
    Okay, so indeterminate in that it's not possible to say in general that any infinitesimal divided by any other infinitesimal is always a particular one of these: finite, infinite, infinitesimal or even undefined (as in the case of 0/0); but it may be possible for a given pair of infinitesimal numbers defined by a particular function, depending on the function (and hence on what those numbers are).
  8. Sep 29, 2010 #7


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    Right. For example, if f is a differentiable function e infinitessimal, and x standard, then
    f(x+e) - f(x)​
    is also infinitessimal, and the quotient
    (f(x+e) - f(x))/e​
    is infinitessimally close to f'(x).
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