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Prove that its a linear operator

  1. Mar 4, 2009 #1
    prove that a linear operator..
    [tex]
    T(f):=\frac{\mathrm{d^2f} }{\mathrm{d} x^2}+2\frac{\mathrm{df} }{\mathrm{d} x}
    [/tex]

    T(kf)=kT(f) part:
    [tex]
    T(kf):=k\frac{\mathrm{d^2f} }{\mathrm{d} x^2}+2k\frac{\mathrm{df} }{\mathrm{d} x}=k(\frac{\mathrm{d^2f} }{\mathrm{d} x^2}+2\frac{\mathrm{df} }{\mathrm{d} x})=kT(f)\\
    [/tex]

    is it correct??
     
  2. jcsd
  3. Mar 4, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes it is. Now what is
    [tex]\frac{d(f(x)+ g(x))}{dx}[/tex]
     
  4. Mar 4, 2009 #3
    Yes, it's correct, but you skipped a skip in the derivation, if you want to be explicit. It should be:

    [tex]T(kf)= \frac{d^2 (kf)}{dx^2} + 2\frac{d(kf)}{dx^2} = k\frac{\mathrm{d^2f} }{\mathrm{d} x^2}+2k\frac{\mathrm{df} }{\mathrm{d} x}=k(\frac{\mathrm{d^2f} }{\mathrm{d} x^2}+2\frac{\mathrm{df} }{\mathrm{d} x})=kT(f)\\ [/tex]

    Since the entire proof relies on this step, it is important to include it. Now to finish the proof you need to show [tex]T(f+g) = Tf + Tf[/tex].
     
  5. Mar 4, 2009 #4
    " T(f + g) = T(f) + T(f) "

    Actually, what phreak meant was

    " T(f + g) = T(f) + T(g) "
     
  6. Mar 4, 2009 #5
    i think its
    [tex]
    \frac{d(f(x)+ g(x))}{dx}=\frac{d(f(x)+d(g(x)}{dx}
    [/tex]
     
  7. Mar 4, 2009 #6
    af ter that
    if i got a derivative of a sum and there is dx in the demoniator
    then i just brake it into two peaces
    [tex]
    T(f):=\frac{\mathrm{d^2f} }{\mathrm{d} x^2}+2\frac{\mathrm{df} }{\mathrm{d} x}\\
    [/tex]
    [tex]
    T(f+g):=\frac{\mathrm{d^2(f+g)} }{\mathrm{d} x^2}+2\frac{\mathrm{d(f+g)} }{\mathrm{d} x}=\frac{\mathrm{d^2f} }{\mathrm{d} x^2}+2\frac{\mathrm{df} }{\mathrm{d} x}+\frac{\mathrm{d^2g} }{\mathrm{d} x^2}+2\frac{\mathrm{dg} }{\mathrm{d} x}
    =T(f)+T(g)
    [/tex]
     
  8. Mar 4, 2009 #7
    Yep, that's right.
     
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