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Showing a twice differentiable function is a vector space

  1. Mar 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Show that the set of twice differentiable functions f: R→R satisfying the differential equation

    sin(x)f"(x)+[itex]x^{2}[/itex]f(x)=0


    is a vector space with respect to the usual operations of addition of functions and multiplication by scalars. Here, f" denotes the second derivative of f.

    2. Relevant equations

    sin(x)f"(x)+[itex]x^{2}[/itex]f(x)=0


    3. The attempt at a solution

    I have no idea how to start this question.
    I know that to prove it is a vector space, proving that it is a subspace is enough?
    I flipped through the notes for my course and followed an example done in class. I got :

    0(x) = sin(x)0"(x) + [itex]x^{2}[/itex]0(x) = 0
     
  2. jcsd
  3. Mar 25, 2014 #2

    Mark44

    Staff: Mentor

    If your set was contained within an encompassing vector space. Can you establish that the space that your subset belongs to is a vector space? If you can't do that, you'll need to verify that all of the vector space axioms are satisfied.
    This is pretty sketchy. What are you trying to say here?
     
  4. Mar 25, 2014 #3
    Hi, thanks for the quick reply.

    I copied that from another example that my lecturer did in class.

    I know that I need to prove the axioms but I don't know how to?

    For example, to prove u+v is in V, how do I do it? what's u and what's v?

    Do I say something like:
    sin(u)f"(u)+[itex]u^{2}[/itex]f(u)=0
    sin(v)f"(v)+[itex]v^{2}[/itex]f(v)=0

    since [itex]u+v\in R[/itex]
    sin(u+v)f"(u+v)+[itex](u+v)^{2}[/itex]f(u+v)=0


    thanks
     
  5. Mar 25, 2014 #4


    So I googled how to prove the u + v is an element of V thing.

    In this context, if I do: sin(u+v)f"(u+v) + [itex](u+v)^{2}[/itex]f(u+v) = o
    I want to get it to become something like

    sin(u)f"(u) + [itex]u^{2}[/itex]f(u) + sin(v)f"(v)+[itex]v^{2}[/itex]f(v) = 0
    so 0 + 0 = 0 right?

    but how can I expand them? I mean with sin(u+v) I know I can at least get sin(u)cos(v) + sin(v)cos(u) but what about f(u+v) and f"(u+v)??

    Thanks a bunch
    and hope I make sense
     
    Last edited: Mar 25, 2014
  6. Mar 25, 2014 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No. The point is NOT to replace "x" with u+ v. You want to prove that the set of all solutions to this (linear, homogeneous) differential equation forms a vector space. So need to show that if f(x) and g(x) both satisfy the equation then so does af(x)+ bg(x) for any numbers a and b.
     
  7. Mar 25, 2014 #6

    Hi there,
    Thanks for the reply,

    So to show f(x) and g(x) both satisfy the equation, do I just write

    sin(x)f"(x)+[itex]x^{2}[/itex]f(x)=0
    sin(x)g"(x)+[itex]x^{2}[/itex]g(X)=0


    Thus af(x) + bg(x)
    =a[sin(x)f"(x)+[itex]x^{2}[/itex]f(x)] + b[sin(x)g"(x)+[itex]x^{2}[/itex]g(X)]
    =a(0) +b(0)
    =0


    Sorry I'm a bit slow at this
     
  8. Mar 25, 2014 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, that shows that applying the differential operator to both af and bg separately gives 0.
    But what you want to show is that
    [itex]sin(x)(af+ bg)''+ x^2(af+ bg)= 0[/itex]

    Can you do that? (It is important that this is a "linear, homogeneous" differential equation and that differentiation is a "linear" operation.)

     
  9. Mar 25, 2014 #8

    So I can just expand it?

    Can I say, showing that this is a vector space via the addition of functions:
    [itex]sin(x)(af+ bg)''+ x^2(af+ bg)= 0[/itex]

    LHS = sin(x)(af+ bg)''+ [itex]x^{2}[/itex](af+ bg)
    = sin(x)(af)"+sin(x)(bg)" + [itex]x^{2}[/itex] (af) + [itex]x^{2}[/itex] (bg)
    = sin(x)af"(x) + [itex]x^{2}[/itex]af(x) + sin(x) bg"(x) + [itex]x^{2}[/itex] bg(x)
    = a[sin(x)f"(x) + [itex]x^{2}[/itex]f(x)] +b[sin(x) g"(x) + [itex]x^{2}[/itex] g(x)]
    = 0 + 0
    = 0
    = RHS
     
  10. Mar 25, 2014 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, that's it.
     
  11. Mar 25, 2014 #10
    So that is the additional of functions right?

    How about multiplication? is it included in the equation as well?
     
  12. Mar 25, 2014 #11

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes. If you take the case a=1 and b=1, it tests addition on f+g. If you take the case g=0 it tests multiplication on af. Your test does both at once.
     
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