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What is the meaning of dx?

  1. Mar 18, 2015 #1
    Apologies if this isn't quite the right forum to post this in, but I was unsure between this and the calculus forum.

    Something that has always bothered me since first learning calculus is how to interpret [itex]dx[/itex], essentially, what does it "mean"? I understand that it doesn't make sense to consider it as an infinitesimal change in [itex]x[/itex] (in a rigorous sense) as the idea of an infinitesimal cannot be formulated rigorously (at least in standard analysis), but can any sense of this notion be retained?

    I read in Spivak's book that we can not consider quantities such as [itex]df[/itex] in the classical sense, but to overcome this we can replace this notion of infinitesimals by promoting the quantities [itex]df[/itex] to functions of infinitesimal changes along particular directions, i.e. functions of tangent vectors. In this sense the function [itex]df: T_{p}M\rightarrow\mathbb{R}[/itex] contains all information about the infinitesimal changes in the function [itex]f[/itex] as it moves in particular directions (i.e. along particular directions). Thus we can consider the functions [itex]dx^{i}: T_{p}M\rightarrow\mathbb{R}[/itex] as containing all information about the infinitesimal change in the coordinate functions in particular directions. I have paraphrased what is written in the book and tried to reformulate it in the way that I can understand it; would what I put be correct?

    Also, is there a way to formulate the idea of a differential in elementary calculus (without resorting to non-standard analysis)? Is it correct to say that one can consider the rate of change in a function at a point, [itex]f'(x_{0})[/itex] which is the gradient of the tangent line [itex]y[/itex] to the function [itex]f[/itex] at this point. From this we can construct a new function [itex]df[/itex] which is dependent on the point [itex]x[/itex] and this change in its value [itex]\Delta x[/itex], such that [tex]df(x_{0},\Delta x)=f'(x_{0})\Delta x[/tex] Thus, the (finite) change in the function near a point [itex]x_{0}[/itex], [itex]\Delta f[/itex] can be expressed as the following [tex]\Delta f=f'(x_{0})\Delta x +\varepsilon =df+\varepsilon[/tex] where [itex]\varepsilon[/itex] is some error function. We note that [tex]dx(x_{0},\Delta x)=\Delta x[/tex] and so [tex]\Delta f=f'(x_{0})dx +\varepsilon =df+\varepsilon \Rightarrow df=f'(x_{0})dx[/tex] where [itex]df=f'(x_{0})dx[/itex] represents a (finite) change along the tangent line to the function [itex]f[/itex] at the point [itex]x_{0}[/itex]. I'm unsure how to proceed from here though?!
     
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  3. Mar 18, 2015 #2

    jedishrfu

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    I found this discussion on dx that may help although you are most likely beyond this point:

    http://mathforum.org/library/drmath/view/60949.html

    As I read your post, it got me thinking about it too. For me from a differential geometry perspective, the dx represents a vector in the x direction vs dy and dz representing vectors along the y and z directions so that df = f'dx + f''dy + f'dz represents a vector that is in the same direction as the the gradient of the function f(x,y,z).

    http://en.wikipedia.org/wiki/Gradient

    http://en.wikipedia.org/wiki/Differential_form
     
  4. Mar 18, 2015 #3
    Thanks for the comments. I understand this bit, but have gotten myself into a bit of a "maths spiral" trying to somehow relate it to the classical (non-rigorous) notion of an infinitesimal quantity?!
     
  5. Mar 18, 2015 #4

    jedishrfu

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    There's no escpaingthe infinitesimal quantity notion as it comes up in limit proofs.

    Is your confusion related to the notion of plotting the function f on x, y vs having a one dimensional geometry where the f provides a value for each x?

    As an example, f could represent the temperature at any point along a one dimensional line.
     
  6. Mar 18, 2015 #5
    Yeah, I think it really arises from the fact that I've seen the differential written as [itex]df=\lim_{\Delta x\rightarrow 0}\Delta f[/itex], but to me this would imply that [tex]df=\lim_{\Delta x\rightarrow 0}f'(x)\Delta x =\lim_{\Delta x\rightarrow 0}f'(x)\lim_{\Delta x\rightarrow 0}\Delta x[/tex] which doesn't seem to make any sense, as clearly [itex]\lim_{\Delta x\rightarrow 0}\Delta x =0[/itex]?!
     
  7. Mar 18, 2015 #6

    jedishrfu

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    Yeah, I see your confusion. I don't think you can do the limit distribution trick with dx though, but I couldn't find an exact reference:

    http://math.oregonstate.edu/home/pr...estStudyGuides/SandS/lHopital/limit_laws.html

    As an example:

    lim y = lim (y)*(x/x) = lim (y/x) lim (x) which leads to a nonsensical answer that all limits approach zero as lim x approaches zero.

    Perhaps reviewing the use of L'Hospital's rule will help resolve your confusion:

    http://math.oregonstate.edu/home/pr...uestStudyGuides/SandS/lHopital/statement.html

    http://en.wikipedia.org/wiki/L'Hôpital's_rule

    @Mark44 would have a much better answer though.
     
    Last edited: Mar 18, 2015
  8. Mar 18, 2015 #7

    dx

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    The idea of dx in single variable calculus is the same as in several dimensions. If you have a function f(x), then

    df = f'(x)dx

    is a 1-form. If you have a small change in x, represented by

    ζ = Δx (∂/∂x)

    Then the change in f is

    Δf = (f'(x)dx)⋅ζ

    = f'(x)Δx

    The interpretation of this is that Δf is the change in f when ζ is very small, or more precisely, it is the first order change in f. You cannot write

    Δf = df + ε

    because Δf is a number and df is a 1-form.
     
  9. Mar 18, 2015 #8

    dx

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    The point of the classical notion of an "infinitesimal quantity" is simply to focus on linear changes and linear approximations. f'(x) tells you the linear approximation of f at the point x. If you change x by a very small amount, then the change in f(x) is mainly coming from its linear behavior, i.e. the second derivatives become unimportant, as x goes to 0.

    In modern calculus, instead of developing calculus on the idea of a derivative (which is a representation of the linear behavior of f, since it is the slope of the tangent line to f), we base calculus on the linear behavior itself, which is df, defined as a linear operator. If you have a curve ξ : λ → M, then

    df( d/dλ )

    tells you the linear part of change in f as you move along the curve.
     
    Last edited: Mar 18, 2015
  10. Mar 18, 2015 #9
    So would it be correct to say that [itex] df[/itex] encodes all the information about the incremental change in [itex] f[/itex] at a particular point due to an infinitesimal change along a particular coordinate direction (itself encoded in the partial derivative [itex] \frac{\partial} {\partial x^{i}} [/itex])?!
     
  11. Mar 18, 2015 #10

    dx

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    Yes. It encodes how f is changing in all directions, not just a particular direction. More precisely, it encodes the linear changes in f in all directions, which is another way of saying that it encodes how f changes with infinitesimal changes in x.
     
  12. Mar 18, 2015 #11

    lavinia

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    I am going to give this modern point of view of differentials because it is the one that one must use when doing calculus on manifolds.

    In modern Mathematics differentials are linear maps between tangent spaces. They codify the idea of the Jacobian matrix of a function.

    From this point of view, dx can be thought of as the identity linear transformation on the tangent space to R. That is dx(∂/∂x) = ∂/∂x. In this context, it is the differential of the identity map, x-> x of R into itself.

    Aside: Usually, since every tangent vector to R is of the form A(x)∂/∂x at some point,x,one often writes dx(A(x)∂/∂x) = A(x) and because of this one can define dx as having values in R. So dx or the differential of any function may be defined as a 1-form that maps the tangent space into R. But here in this way of looking at it dx is being treated as the identity linear map with values in the tangent bundle to R.

    In general if f maps an n-manifold into an m-manifold, df is a linear map of the n dimensional tangent spaces into the m dimensional tangent spaces.

    To each smooth function between two smooth manifolds

    f: M ->N

    the differential, df, is a smooth vector bundle morphism of the tangent bundles,. This means that df: TM -> TN such that df: TM##_{p}## -> TN##_{f(p)}## is linear on each fiber, TM##_{p}##, and df is a smooth function between the two smooth manifolds TM and TN.

    If f and g are two smooth maps

    f: M -> N and g: N -> W then

    d(gof) = dgodf

    This is the general form of the Chain Rule.

    Note also that if

    f: M -> M = id##_{M}## then

    df is the identity linear transformation on each tangent space to M. The case of dx is the special case of the identity map on R.

    So for illustration, take the case of a function from R to R.
    df is a linear map from the tangent bundle of R into itself that is linear on fibers.

    This means that df(A(x)∂/∂x) = A(x)df(∂/∂x) is a tangent vector at the point, f(x).

    Exercise: Figure out what vector df(∂/∂x) is.

    Observation: Suppose φ,ψ: U -> R##^n## are two coordinate charts from an open set,U, in a manifold,M, and let v be a tangent vector at a point in U. Then

    w = dφ(v) and u = dψ(v) are two different tangent vectors at different points in R##^n##.

    Therefore, the inverse differentials dφ##^{-1}## and dψ##^{-1}## map w and u back to v.
    So one can think of v as the vectors w and u identified as the same vector. In fact the identification mapping is

    odψ##^{-1}## since dφodψ##^{-1}##(u) = w ( by the Chain Rule).
    This is how one defines tangent spaces on manifolds in terms of coordinate charts.
     
    Last edited: Mar 19, 2015
  13. Mar 19, 2015 #12
    Thanks for your explanations. I'm really stuck trying to understand what exactly [itex]df[/itex] and [itex]dx[/itex] actually "represent" in a modern context, I was trying to parse Spivak's discussion (that I paraphrased in my first post), but I'm struggling if I'm honest. Is the point that we wish [itex]df[/itex] to describe the infinitesimal change in [itex]f[/itex] due to an infinitesimal change in the point it is evaluated at. To do so it must be a function of this infinitesimal change, i.e. it must be a function of the tangent vectors (in the tangent space at that point) as they themselves describe all the possible directions (and rates of change) that the function [itex]f[/itex] could pass through that point. In this sense, if [itex]f=x^{i}[/itex] (where [itex]dx^{i}[/itex] is a given coordinate function), then [itex]dx^{i}[/itex] describes the infinitesimal change in the coordinate function [itex]x^{i}[/itex] as a function of all possible directions that it could pass through that point. Obviously, as a coordinate function only changes along itself, and at unit "speed" with respect to itself we require that [tex]dx^{i}\left(\frac{\partial}{\partial x^{i}}\right)=1[/tex] or, more generally, [tex]dx^{i}\left(\frac{\partial}{\partial x^{j}}\right)=\delta^{i}_{\; j}[/tex] In this sense the functional [itex]df[/itex] evaluated along the rate of change of a particular coordinate function should describe the rate of change in the function [itex]f[/itex] at a particular point, i.e.
    [tex]df\left(\frac{\partial}{\partial x^{i}}\right)=\frac{\partial f}{\partial x^{i}}[/tex]
    Would this be correct at all?
     
    Last edited: Mar 19, 2015
  14. Mar 19, 2015 #13

    lavinia

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    That is correct. dx is the dual form to ∂/∂x in a coordinate system.

    Whenever one has a vector space, one has a dual space. In the case of coordinate tangent vectors. one has dual coordinate tangent vectors. The concept of a dual basis is general and applies to any basis of any vector space.

    I strongly recommend that you try to understand this stuff from the point of view that I described above.
    Take a look at Milnor's "Topology from the Differentiable Viewpoint". He uses this method I from the beginning.

    As far as infinitesimals go, I think you are getting a little off track. You should think of differentials as linear maps between tangent spaces or in the case of functions - like coordinate functions - you can think of them as 1 forms - i.e. as dual vectors to the tangent space.

    Whenever one has a vector space, one has the dual vector space of linear maps of the vector space into the base field. the differential of a function may be thought of as a dual vector because given any tangent vector at a point, df(v) = v.f is a linear map into R. That is given any linear combination of tangent vectors av + bw

    df(av + bw) = a(v.f) + b(w.f).

    This of course is true for coordinate functions.
     
    Last edited: Mar 19, 2015
  15. Mar 19, 2015 #14
    So is the definition [tex]df(p)(v_{p})=v_{p}[f](p)[/tex] where [itex]v_{p}\in T_{p}M[/itex] given this way because we wish [itex]df[/itex], evaluated at a particular point [itex]p\in M[/itex], to describe the infinitesimal change in [itex]f[/itex] as it passes through [itex]p[/itex] along the direction described by [itex]v_{p}[/itex], and as such this must equal the directional derivative of the function along [itex]v_{p}[/itex]?
     
  16. Mar 19, 2015 #15

    lavinia

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    yes
     
  17. Mar 19, 2015 #16
    Ah, ok. Thank you very much for your help.
     
  18. Mar 19, 2015 #17
    Not a mathematician, but I think ##dx## only has any meaning if it appears in an expression containing another differential such as ##dy##, kind of like how open parenthesis only has meaning when there is a close parenthesis in the same expression. It tells how two or more variables scale with each other, so one differential by itself means nothing. They need to be arranged in an equation such that they can be rearranged into a division, like ##dy/dx## or ##dx/dy##, and the derivatives have their usual epsilon limit definition. But the differential notation is nice because it shows the symmetry between the two variables.

    A multivariable differential like
    ##dz = x dx + y dy##
    means the same thing as
    ##\partial z/\partial x = x##
    ##\partial z/\partial y = y##
    ##\partial x/\partial z = 1/x##
    ##\partial x/\partial y = y/x##
    ##\partial y/\partial z = 1/y##
    ##\partial y/\partial x = x/y##
    with the implication that z is a function of x and y, and vice versa
    but written more elegantly and symmetrically.
    sorry if I'm stating the obvious
     
  19. Mar 19, 2015 #18

    mathwonk

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    df is a function of 2 variables, x and h. for each x, df is the homogeneous linear approximation to the function of h defined by f(x+h)-f(x). Thus if you imagine the graph of y = f(x), and its family of tangent lines, these tangent lines are the graphs of the functions df + f(x), for each value a of x. I.e. for each a, the tangent line at (a,f(a)) is the graph of df(x-a)+f(a). Taking f(x) = x, you get the definition of dx. I hope this is right. I'm watching the Harvard UNC game.

    I.e. df is really a section of the cotangent bundle to the x axis, whose value at x is the linear function of h with value f'(x).h. but then i have to define cotangent bundle.
     
  20. Mar 20, 2015 #19
    If I've understood it correctly, in order to quantify how a coordinate function [itex]x^{i}[/itex] is changing at a particular point [itex]p\in M[/itex] one needs to know how it can change as it passes through that point. The set of tangent vectors in the vector space at that point describe all the possible directions in which a function can pass through that point and all the possible "speeds" (rates of change) that it can do so at. These vectors quantify the instantaneous changes along all possible paths through that point and so we can consider a function [itex]dx^{i}[/itex], that is dependent on the tangent vectors in the tangent space to that point, which quantifies the instantaneous (or infinitesimal) change in the coordinate function along a chosen direction at that point. Obviously, [itex]x^{i}[/itex] will only change along the coordinate curve that it defines, and thus, its differential change along any given vector [itex]v_{p}\in T_{p}M[/itex] at the point [itex]p\in M[/itex] should be equal to the component of that vector along that coordinate curve, i.e. [tex]dx^{i}(v_{p})=v_{p}^{i}[/tex] as [itex]v_{p}^{i}[/itex] quantifies how much change occurs along the direction of the basis vector [itex]\frac{\partial} {\partial x^{i}}[/itex] at that point. Hence, for consistency the change in [itex]x^{i}[/itex] at that point along the basis vector [itex]\frac{\partial} {\partial x^{i}}[/itex] should be 1, as it is a measure of how it changes with respect to itself at that point, i.e. [itex]dx^{i}(\frac{\partial} {\partial x^{i}})=1[/itex].
    Moving on to general (differentiable) functions [itex]f[/itex] we note that all the possible paths that [itex]f[/itex] could pass through a given point [itex]p\in M[/itex], and how "fast" it can do so, are encoded in the tangent vectors [itex]v_{p}\in T_{p}M[/itex] and thus the infinitesimal change in the function, [itex]df[/itex] as it passes through that point should depend on which path it takes, i.e. which vector it "points" along at that point. Thus, for [itex]df[/itex] to quantify the instantaneous change in [itex]f[/itex], when evaluated along a particular vector [itex]v_{p}\in T_{p}M[/itex] it should be equal to the directional derivative of [itex]f[/itex] along [itex]v_{p}[/itex], i.e. [tex]df(v_{p})=v_{p}(f)[/tex]
    Apologies for the long-windedness of this post, just trying to express in words what I think I've understood from this discussion.
     
  21. Mar 20, 2015 #20

    dx

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    It is not true that xi changes only along ∂/∂xi

    Draw a picture and it will be clear. For example, in 2 dimensions, with coordinates x and y, x changes in all directions except ∂/∂y
     
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