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Relationship between differential and the operator

  1. Mar 6, 2012 #1
    Is there some sort of relationship between the differential

    [tex]dx[/tex]

    and the differential operator which means to take the derivative [tex]d/dx[/tex]

    if x is a dependent variable? My prof said that [tex] dx * d/dx = 1[/tex] but that doesn't seem to work out in the case I'm looking at, so I must be missing something.
     
  2. jcsd
  3. Mar 6, 2012 #2

    Char. Limit

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    [itex]dx \frac{d}{dx}[/itex] is just an operator multiplied by a differential. If you apply it to a function f(x), it'll give the differential df, however. So you could call it the function differential operator, I suppose.
     
  4. Mar 6, 2012 #3
    So then can you take [tex]dx = 1/(d/dx)[/tex]? As in the differential is one over the operator? Or does that not work? I was having trouble making that work out in one of my problems.
     
  5. Mar 6, 2012 #4

    Char. Limit

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    No, that wouldn't work. I really don't believe it's possible to say that dx d/dx = 1, anyway. If you're really trying to assign it a value, it would be just d, or a differential operator. That is, you're basically saying "the derivative operator with respect to a variable, multiplied by the differential of that variable, is the differential operator".
     
  6. Mar 6, 2012 #5

    chiro

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    You might want to revert back to the limit definition for the derivative for understanding cases like this.
     
  7. Mar 6, 2012 #6
    On a multivariable function, [itex]\frac{\partial}{\partial x}[/itex] and [itex]\partial x[/itex] can be thought of as unit vectors. I am still not totally comfortable with the idea, but perhaps this is what your prof was getting at?
     
  8. Mar 6, 2012 #7

    mathwonk

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    have you had linear algebra? df is a field of (co)vectors, and dx is a field of standard basis covectors. In this basis, d/dx is the operator that sends f to its field of coefficients in terms of the basis dx. I.e. at every point p, df(p) = df/dx(p) dx.

    I.e. at every point df/dx(p) is a number, and at every point df(p) is a covector.

    At every point dx(p) is also a covector, and we have at every point, that the number df/dx(p) times the covector dx(p), equals the covector df(p).

    so df/dx . dx = df, an equation that holds at every point p.


    for instance if f(x) = x^2, then df(p) = 2p. dx(p) = df/dx(p) dx(p).
     
  9. Mar 6, 2012 #8
    Your professor is wrong. [tex] dx * d/dx = d[/tex]. In words, the change [of something] per change in x times the change in x is the change [in the thing].
     
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