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Multiplying by dx

  1. Jul 31, 2009 #1
    When solving an ODE, physcists sometimes multiply the differential dx.

    Example

    df/dx = x/2

    df = (x/2)dx

    f = x^2 + c

    I've always been told that this mathematically unsound, but I've never been told why. Is it because the necessary constant c may not necessarily be additive? Are there any specific counter-examples where this doesn't work?
     
  2. jcsd
  3. Jul 31, 2009 #2
    It works fine. The objection is that, in Leibnitz's notation, df/dx is not the quotient of something df by something dx. The calculation is conveniently done that way, and eaisiy checked at the end, so even if someone "questions" the method, you still get the right answer.
     
  4. Jul 31, 2009 #3

    arildno

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    1. It is mathematically unsound because "dx" is not a number, so you can't "multiply" with it.

    2. Doing so "works" in the very limited case of an ordinary first derivative, because it can then be coupled to the rigorous mathematical theory concerning differential forms.

    It does NOT work, in general, for higher order derivatives, or partial derivatives, not even first-order ones.
     
  5. Aug 1, 2009 #4
    Wonderful, two opposite answers. I think I know who's the physicist and who's the mathematician.
     
  6. Aug 1, 2009 #5

    Hurkyl

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    How about a third answer?

    The differential (a.k.a. covector field) df is, in fact, the scalar product of the differential dx by the function df/dx.

    But that fact has nothing to do with the related property of real numbers. (at least, not directly)

    (okay, I lied -- arildno already said this)
     
  7. Aug 1, 2009 #6

    arildno

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    How are the two "opposite"??

    All g_edgar did was to pre-gurgitate my point 2.

    (Is pregurgitate an English word, by the way?)
     
  8. Aug 2, 2009 #7

    HallsofIvy

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    Hey, English is flexible. If it wasn't, it is now!

    My look at it:
    [tex]\frac{dy}{dx}= \lim_{h\rightarrow 0}\frac{f(x+h)- f(x)}{h}[/tex]

    Writing "dy/dx= f(x) so dy= f(x)dx" by "multiplying both sides by dx" is "mathematically unsound" because dy/dx is the limit of a fraction, not a fraction.

    But you can use that equation: "dy= f(x)dx" (though not the justification "multiplying both sides by dx") because it is the limit of a fraction!

    Basically, you can justify treating dy/dx as if it were a fraction (note the use of subjunctive mode?) by going back before the limit, using the fact that (f(x+h)- f(x))/h is a fraction, then taking the limit again. Take any Calculus text, and look at how the "chain rule" (which could be symbolized by
    [tex]\frac{dy}{dt}= \frac{dy}{dx}\frac{dx}{dt}[/tex]
    and note that they don't "cancel" the "dx"s.


    In order to make use that, we define the "differentials": "dx" is an "undefined ("primitive") term and then we define "dy" by "dy= f'(x)dx". (I intentionally used the notation "y' " rather than "dy/dx" to avoid anyone thinking I was "cancelling"!)

    Although used a lot in Calculus and Differential Equations, differentials really come into there own in Differential Geometry.
     
  9. Aug 2, 2009 #8

    arildno

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    "Writing "dy/dx= f(x) so dy= f(x)dx" by "multiplying both sides by dx" is "mathematically unsound" because dy/dx is the limit of a fraction, not a fraction."

    This was, as it happens, also Newton's view:

    He made the analogy that if we take the fraction 2x/x and let x pass into infinity, the expression still equals 2, but that nobody would say that in the limit, we had a fraction between two numbers.

    Ps:
    Thanks, Halls, I thought pregurgitate was a nice word myself! :smile:
     
  10. Aug 3, 2009 #9

    HallsofIvy

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    Yes, it was Newton's view, but he was using "infinitesmals" and it has taken us almost 500 years to define those properly! Anyway, in "non-standard" analysis, I would have had to tell the OP that his first question made no sense!
     
  11. Aug 3, 2009 #10
    Leibniz notation is unsound, but miraculously produces the correct answer most of the time anyway (at least for first-year calculus).

    As it's been said before, the derivative and integral are defined by limits. You can't break the limit into two halves. In other words, it is blatantly false that [tex]\lim \frac{a}{b} = \frac{\lim a}{\lim b}[/tex] or variation like that. You simply can't split a limit. The "dx" can't exist on its own like that. It's not a real number. It's not an infinitesimal. It's nonsense.

    But it's nonsense that produces the correct answer (in most cases). The reason for this probably has to do with a generalized notion of a limit. If you had a limit operator that acted on an entire equation (as opposed to an expression -- ie: on both sides of the = sign, not just on the left or right), then you can do a few more tricks. In that case, the definition of a derivative would be

    [tex]\lim_{h->0} (f'(x) = \frac{f(x+h) - f(x)}{h})[/tex]

    The parentheses are there to show that the limit applies to the whole equation. The important difference is that the variable h is allowed to live anywhere inside the equation (as opposed to just on the left or right side). That means, we can rearrange the equation like this:

    [tex]\lim_{h->0} (f'(x) h = f(x+h) - f(x))[/tex]

    If someone learned calculus with a concentration in physics, they will recognize this form immediately. The h plays the role of the dx. The limit suggests it plays the role of an infinitesimal. If you use the first equation, you can reduce the limit equation to a regular one, and in the process, you learn the derivative. In the second equation, (the one right above here), the equation reduces to 0 = 0, which (while useless) is still true.

    So these bastard "dx"'s can be used to do simple calculus. If you can solve problems using this method, that's cool. But you should understand that you're really working with a more powerful framework. The dx's are NOT real numbers. They are NOT functions. They are just the shadows of more powerful tools.

    These methods tend to break down when you get to more difficult problems. When there are more than one function or you are working with a function of more than one parameter, it would not be difficult to run into problems. Especially in multivariable calculus, when learning to do multiple integrals, surface, or line integrals, the equation is more of a summary of the problem than the whole problem.
     
  12. Aug 5, 2009 #11
    I know people have already tried to give an explanation, but I would recommend you to browser Thomas's Calculus book and look for section 3.8.

    I am glad I looked it up, I had the same question for ages. Its no fun solving problems when you know you don't understand the first principles.
     
  13. Aug 5, 2009 #12
    In nonstandard analysis the derivative is a perfectly good fraction and things like df = (x/2)dx describe numbers and make sense.
     
  14. Aug 6, 2009 #13

    arildno

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    Sure, but then again, neither division or multiplication in non-standard analysis are the same operations as within "standard" maths.
     
  15. Aug 6, 2009 #14
    This is perhaps misleading. In nonstandard analysis, the quotient of infinitesimals is not equal to the derivative, it is just infinitely close to the derivative.
     
  16. Aug 6, 2009 #15
    dy = a dx is just a different (and, especially in physics, often more natural) way of writing dy/dx = a.
     
  17. Aug 6, 2009 #16

    arildno

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    Nope, it is not.
     
  18. Aug 6, 2009 #17
    I'm not sure what you mean. There are things that have no counterpart in real numbers, like

    [tex]\frac{y}{dx} .[/tex]

    Can you supply an example?
     
  19. Aug 6, 2009 #18

    jgens

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    Not quite! In non-standard analysis the derivative of a function is not defined as the quotient of two infinitesimal numbers, but rather their "standard part." Hence, we arrive at the definition:

    [tex]\frac{dy}{dx} = \mathrm{st}\left (\frac{f(x + \epsilon) - f(x)}{\epsilon} \right )[/tex]

    Note: If [tex]f[/tex] is differentiable, then [tex]f(x + \epsilon) - f(x)[/tex] is infinitesimal if [tex]\epsilon[/tex] is infinitesimal. Also, if [tex]\Delta{y}[/tex] and [tex]\Delta{x}[/tex] are infinitesimal, then their quotient is infinitely close to the function's derivative.
     
  20. Aug 6, 2009 #19
    To be honest, it was an intentionally provocative claim, and I do appreciate all the criticism, but it really depends on how one defines a fraction--a quotient of reals or a quotient that may include infinitesimals in quotient or denominator, doesn't it?
     
    Last edited: Aug 6, 2009
  21. Aug 7, 2009 #20

    Hurkyl

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    The phrase "the derivative is a fraction" is most commonly to mean the assertion
    [tex]f'(x) = \frac{f(x+\epsilon) - \f(x)}{\epsilon}[/itex]
    for... well... people usually aren't clear on what they really think [itex]\epsilon[/itex] means here.

    Of course, one can find all sorts of equations that assert the derivative is equal to a fraction. For example, if
    [tex]f(x) = \frac{1}{1 + x^2}[/tex]​
    then
    [tex]f'(x) = \frac{-2x}{(1 + x^2)^2}[/tex]​
    Voila! The derivative is a fraction!


    Even classicaly, I might be inspired to define, for any univariate function f, a binary function (which I will boldly call df) defined by
    [tex]d_\epsilon f(x) = \epsilon f'(x)[/tex]​
    and whose domain is nonzero [itex]\epsilon[/itex] and those x for which f is differentiable

    Then, we would have the equation
    [tex]\frac{d_\epsilon f(x)}{d_\epsilon x} = f'(x)[/tex]​

    but this is very different from what the phrase "the derivative is a fraction" is most commonly used to mean.


    (Incidentally, the function df I write above is very closely related both to differential forms and to what one might find in a nonstandard analysis textbook such as Keisler's online book)
     
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