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How fundamental are functions, really?

  1. Nov 17, 2013 #1
    I'm sorry if this is in the wrong place but I believe the answer is geometric in nature somehow.

    So as a young physics student it is banged into your head that vectors/tensors are genuine geometric and algebraic "objects" and the coordinates by which you label them are metaphysical, so to speak, and not at all fundamental. The same object has many different labels depending on the basis.

    So this got me thinking about functions. Dirac seems to want us to imagine functions (say over just the real numbers) as an uncountable set of coordinates, analogous to vectors, one for each element of R. So just like we can think geometrically/algebraically about vectors, is there a way of thinking about functions without projecting them into any coordinate space?

    I suppose the thrust of my point is this: if it is non-sensical to say "(1,2,3) is a vector" then ought it be equally cavalier to say "f(x)=x^3" is a function?"
     
  2. jcsd
  3. Nov 22, 2013 #2
    I don't quite see the connection with vectors and functions but here is some relevant comments which might help.

    From a mathematical point of view [itex]v = (1, 2, 3)[/itex] is indeed a vector as an element of the vector space [itex]\mathbb{R}^3[/itex] over [itex]\mathbb{R}[/itex] with the usual operations. Furthermore [itex]v[/itex] is independent of coordinates. When you pick a basis, then you get a column matrix of [itex]v[/itex] in terms of the basis. Of course if you pick the standard basis then [itex]v[/itex] looks like the same thing but as a column matrix.

    I guess from a physics perspective a vector would be some geometric object (an arrow in space with no coordinates) and you would not have any quantitative representation of it until you pick a basis and measure the components relative to that basis.

    For functions, its not exactly the same thing. In the example given [itex]f[/itex] is the function. It is something which is somewhat abstract and you cannot directly give [itex]f[/itex] but indirectly by saying what its value is at each point in the domain. In your example, you have to define [itex]f[/itex] by saying the value is [itex]f(x) = x^3[/itex] for each [itex]x[/itex] in [itex]\mathbb{R}[/itex] (or whatever the domain is).
     
  4. Nov 22, 2013 #3
    Well I was motivated to think about this since in Dirac notation, we take abstract elements of a Hilbert Space and project them into coordinates to get "wavefunctions." Like the ket Y_lm projected into spherical coordinates on the 2-sphere. It seems to be that these wavefunctions are just abstract elements, that take uncountably many values when projected into coordinate space. This is the connection I had in mind.

    I agree that (1,2,3) is a vector algebraically in the vector space R^3 over R but how does the coordinate independence that is so algebraic carry over when you try to imagine (1,2,3) geometrically in R^3?

    As for the last part...so you are indeed saying that functions cannot be described without explicitly giving their values over the domain? If so, this is what I've been confused about! So given this, the spherical harmonics, are "functions" on the 2-sphere, but then what are they before the projection onto the sphere? Thanks PSarkar
     
  5. Nov 23, 2013 #4
    I guess the difference is that in math the coordinates (basis) we are picking are not some real physical thing whereas in physics, the coordinates are physical. So [itex](1, 2, 3)[/itex] is coordinate independent in math but in physics a vector is something attached to a physical space and so it depends on the physical coordinates. In other words, the coordinates (basis) in math are not exactly the same thing as coordinates in physics.

    About functions ... I realized that defining functions by giving the value at each point is not the only way to define a function. For example you can also define a function as a solution to some differential equation (for some differential equation with a unique solution).

    Unfortunately, I don't know about wavefunctions in quantum mechanics yet so may be someone else can give a better answer.
     
  6. Nov 23, 2013 #5

    lurflurf

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    Do you know anything about calculus or linear algebra?

    We might say v=(1,2,3) is an element of R^3
    Where do the one two and three come from?
    We might pick three elements from the set and take inner products
    1=v.i
    2=v.j
    3=v.k

    All that is arbitrary. v in no way depends upon what three basis vectors we choose or how we define the inner product. Someone may say v=(37,potato,v). v has not changed, but it is represented in a different way.

    It is the same for functions. The subject that mixes calculus and linear algebra is called functional analysis. There are some tricky things that come up there. We might say a function (R->R) belong to R^R. Often we need to restrict the functions we consider to avoid a number of problems. For example we might consider polynomials and one basis could be 1,x,x^2,x^3,...
    then f=(0,0,0,1,0,0,0,...)
    but we might just as well have a basis 1,1+x,1+3x+3x^2,1+6x+15x^2+15x^3,1+10x+45x^2+105x^3+105x^4+...
    then f=(-1/3,3/5,-1/3,1/15,0,0,0,...
    Just as for R^3 the basis is not important

    hope that helps
     
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