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What's the point in using differentials (in integrals)?

  1. Dec 23, 2007 #1
    I don't get why we need to use differentials and why they are the way they are.

    For example: [tex]dy=f'(x)dx[/tex] vs. the derivative [tex]\frac{dy}{dx}=f'(x)[/tex]

    Why are they equivalent? Why are integrals written in the differential form? I don't get the purpose of it. (other than to be used as an estimation for error propagation)

    It seems weird how you can just move the dx over...

    I get how if the change in (delta)x is small then (delta)y can be approximated by dy where dy is the change in y of the tangent line.
  2. jcsd
  3. Dec 23, 2007 #2


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    The d is part of the integral symbol, as it's usually first taught:

    [tex]\int \underline{\quad \quad} \, d \underline{\ \ } [/tex]

    is the symbol for the indefinite integration operator. When I say that, I mean the entire thing is just one giant symbol; never mind the fact that it looks like it consists of two separate parts.

    The grammar for this operator is that you declare a dummy variable by placing it in the second blank, and then you place the expression to be integrated in the first blank. (the expression will be interpreted as a function of the dummy variable)

    When you're using this integration operator, expressions like dx are really just shorthand -- e.g. when making the substitution y = f(x), the notation dy = f'(x) dx is really just shorthand for the theorem that says

    [tex]\forall g: \int g(f(x)) \, f'(x) dx = \int g(y) \, dy[/tex]

    Eventually, it becomes convenient to adopt the more sophisticated approach of differential geometry, and you define things like line bundles, vector fields, and differential n-forms. In this setting, you can define a new integration operator that takes an n-dimensional region and a differential n-form, and produces a number.

    It just so happens that you can rewrite any in your "simple" definite integral as one of these more "sophisticated" integrals, and you get the same answer either way. So, one usually notates the differential geometry in a way such that you can do this translation in a mostly transparent way.

    e.g. if I choose x to be a coordinate function on the line R, (and f is an invertible function) I can define a new coordinate function on R by

    y = f(x)

    and then we have the identity that the differential form f'(x) dx is equal to the differential form dy. Consequently, for any interval I and scalar field g, we have

    [tex]\int_I g \, dy = \int_I g f'(x) \, dx.[/tex]
  4. Dec 24, 2007 #3
    The way I like to think of an integral is as a sum of slices.

    For example

    [tex]\int_0^5 x^2 dx[/tex]

    You are finding the area under the curve x^2 from 0 to 5. This is done by finding the sum of a bunch of infinitesimal vertical slices.

    Think of dx as the infinitesimal width of each slice. x^2 is the height of each slice. The integral sign is summation operator.

    (This logic doesn't work so cleanly with indefinite integrals, but this is the idea)
  5. Dec 24, 2007 #4


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    Dearly Missed

    However, chickendude:
    What is an infinitesemal other than the ghost of a departed quantity?
  6. Dec 24, 2007 #5

    D H

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    Just to elaborate a bit: Some people put the d right after the integral sign to emphasize that it is indeed one giant symbol:
    [tex]\int d \underline{\ \ } \; \underline{\quad \quad} [/tex]
    The advantage of this form is that it makes it a bit clearer that the integration variable is a dummy variable, particularly so when it comes to definite integrals. The disadvantage of this form is that it is not clear where the thing to be integrated ends.
  7. Dec 24, 2007 #6


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    From the point of view of notation, there are two important things that the differential at the end of the integral provides:
    The variable of integration:
    [tex]\int x\sin (y) dx[/tex]
    [tex]\int x\sin (y) dy[/tex]

    The close of the integral:
    [tex]\int x+2x dx[/tex]
    [tex]\int x dx + 2x[/tex]

    As things get more involved those two are very important.
  8. Dec 24, 2007 #7
    For example, when you are finding the area under a curve, each slice is basically a vertical line whose width is approaching zero.

    Each slice has height f(x) and width dx. The dx here represents that tiny "ghost of a departed quantity", meaning the width that has shrunk toward zero.

    The integral is summing an infinite amount of these infinitesimal slices.
  9. Dec 25, 2007 #8

    Gib Z

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    If the width is dx for every slice, and dx is an infinitesimal, does that not imply the width of each partition must be the same? That is definitely not required, it is only required that the width of the largest partition tends to zero as the number of partitions grows infinitely large.
  10. Dec 25, 2007 #9
    I think what arildno was implying is that we can easily say things like summing an infinite amount of these infinitesimal slices, but really.....what does that even mean?

    Show me how to add an infinite amount of quantities...then show me how to do it when those quantities are infinitesimally small.

    Now where did I leave my infinitesimal calculator?:smile:

  11. Dec 25, 2007 #10

    Gib Z

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    Well I can help you on the first point =]

    [tex]\sum_{n=1}^{\infty} \frac{1}{2^n} = 1[/tex] =]
  12. Dec 25, 2007 #11
    Thanks Gib:rofl:
  13. Dec 25, 2007 #12


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    None of those numbers are infinitessimally small!
  14. Dec 25, 2007 #13

    Gib Z

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    I never claimed they were =] I only said I could do something about the first point (infinite amount of quantities), not the second (infinitely small).
  15. Dec 26, 2007 #14
    Actually that is not a differential form, it's only notation. Even it may to be written like this:

    [tex]\int f = g[/tex]

    Where [tex]f = g'[/tex] and being [tex]f[/tex] and [tex]g[/tex] functions of real variables.

    Well, [tex]dy=f'(x)dx[/tex] is a "cheating" form. However it's used to simplify the things and make them more intuitive (and because the calculus was formulated in that way in the era of Newton, but nobody knew what "differentials" were rigorously).

    The equation: [tex]\frac{dy}{dx}=y[/tex] may be solved without using strange manipulations:



    Integrating with respecto to x:

    [tex]\int\frac{1}{y}\frac{dy}{dx}dx=\int 1dx[/tex]

    As a result for the chain rule:
    [tex]\int\frac{1}{y}{dy}=\int 1dx[/tex]

    For the definition of [tex]\int [/tex] (see up):
    [tex]\ln y=x+C_1[/tex]


    [tex]y=Ce^x[/tex] (where [tex]C=e^{C_1}[/tex] )

    But really differentiales are a cheating form? why it works "well" when it's used like tiny numbers? The answer is in the non-standard analysis. Please, check out this: http://en.wikipedia.org/wiki/Infinitesimal.

    I hope that this has clarified something.

    Good night!
    Last edited: Dec 26, 2007
  16. Dec 26, 2007 #15


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    While I agree that non-standard analysis is more intuitive -- you have to realize that it doesn't make all of these issues vanish! Even in NSA, the notion of an infinitessimal difference is not synonymous with the notion of a differential Liekwise, the notion of a derivative is not synonymous with the notion of a ratio of two infintiessimal differences!

    And even in standard analysis, dy = f'(x) dx is not a 'cheating form'; it is a perfectly good expression in the language of differential geometry.
  17. Dec 26, 2007 #16
    I don't say that they are synonymous. Using this theory can be demonstrated rigorously many manipulations in which are used differentials.

    I named it 'cheating form' because many times it is used without clarify its properties and proof them (for example can I do this? dx/dy=1/f'(x) ). Even though, I don't think so that expression has very much sense in standard analysis.

    Sure, but in the rigor mathematician, it has very little to do with the concept of diffential that it's used in some calculus of one variable books .

    Good good nights!
    Last edited: Dec 26, 2007
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