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Question about the definition of df

  1. Apr 23, 2014 #1
    I guess I have several definitions of [itex]df[/itex] flying at me, and I am having trouble getting a continuous definition. So in basic Calculus, we are taught [itex]df = f'(x)dx[/itex], and it's taught as sort of a linear approximation of the change of f for small values dx, whch makes sense with the definition of the derivative being a linearization of the change in a function.

    [itex]df = f(x+h)-f(x)[/itex]

    [itex]f(x+h)-f(x)≈f'(x)h[/itex]

    That also makes sense with the higher level definition of a differential being a mapping to the tangent space. I have trouble when I consider a Taylor series based derivation for change in f, it seems to be paradoxical.

    [itex]\displaystyle{f(x) = f(a)+f'(a)(x-a)+\frac{f''(a)(x-a)^{2}}{2!}+\frac{f'''(a)(x-a)^{3}}{3!}...}[/itex]

    sub [itex]\Delta x = x-a[/itex] and [itex]x = a + \Delta x [/itex]

    [itex]\displaystyle{f(\Delta x + a) = f(a)+f'(a)\Delta x+\frac{f''(a)(\Delta x)^{2}}{2!}+\frac{f'''(a)(\Delta x)^{3}}{3!}...}[/itex]

    rearrange and you can see the confusion...

    [itex]\displaystyle{f(\Delta x + a) -f(a) = f'(a)\Delta x+\frac{f''(a)(\Delta x)^{2}}{2!}+\frac{f'''(a)(\Delta x)^{3}}{3!}...}[/itex]

    So now I have different definition of df?? Can anyone explain this to me?
     
  2. jcsd
  3. Apr 23, 2014 #2
    I think the source of you confusion is that you have assumed that ##\Delta f## and ##df## are interchangeable. They are not. In general, ##df\ne\Delta f##.

    In the calculus texts that I've worked with, the differentials ##dx## and ##df## are defined as ##dx=\Delta x## and ##df=f'(x)dx##. And you get that ##df## is a fair approximation of ##\Delta f=f(x+dx)-f(x)## when ##dx## is small. As you've noted, ##df## corresponds specifically to the change in the linear approximation of ##f##; i.e. ##df=\Delta L##, where ##L## is the linear approximation to ##f##. Higher-order Taylor polynomials give different (usually better) approximations for ##f## and, thus, ##\Delta f##. But that doesn't change our definition of ##dx##.

    Note that the ##dx## described above is different from the ##dx## used in integral notation. The ##dx## above is a finite, real number. It is not an infinitesimal. Depending on your approach, the ##dx## of integrals is either an infinitesimal or, more likely, a kind of notational artifact.
     
  4. Apr 25, 2014 #3
    Thanks, I think I'm starting to get the independence of [itex]\Delta f[/itex] and [itex]df[/itex]. So [itex]df = f'(x)dx[/itex], but [itex]\Delta f[/itex] could not necessarily be even "tangent", it could be secant depending on [itex]h[/itex], whereas [itex]df[/itex] is always a tangential approximation, and [itex]df≈f(x+h)-f(x)[/itex] for very small values of [itex]h[/itex].
     
  5. May 16, 2014 #4
    I recently had a breakthrough regarding this topic while I was thinking about something that didn't seem to be directly related. I thought about construction of the integral, and I thought about boxes under the curve. So, We get this.

    [itex]\displaystyle\sum f(x_{i})\Delta x \Leftrightarrow \int f(x) dx[/itex]

    What if, given the sum was taken such that the boxes were always an under approximation, we also used triangles to better approximate the area, so we added those into each term. So the area of the triangle plus the normal rectangle would be

    [itex]f(x_{i})\Delta x + \frac{1}{2} \Delta y \Delta x [/itex].

    Now, as things become infinitesimal, we get the following.

    [itex] f(x)dx + \frac{1}{2} dy dx[/itex]

    We sub [itex]dy=f'(x)dx[/itex], and we get the following

    [itex]f(x)dx + \frac{1}{2} f'(x) dx^{2}[/itex]

    I was very excited to see this, so to me it would seem that there would be higher order terms to better approximate the geometry, as we see in the taylor series I posted previously
     
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