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Explanation on Taylor expansions needed

  1. Jun 24, 2011 #1
    I have a question about Taylor expanding functions. For both cases I can't get my head around why things are the way they are. I just don't see how one would perform Taylor expansions like that.

    The first:
    The starting point of a symmetry operations is the following expansion:
    [tex]f(r+a) = f(r) + \left(a\cdot\frac{\partial}{\partial r}\right) f(r) + \frac{1}{2} \left(a\cdot\frac{\partial}{\partial r}\right)^2 f(r) + \frac{1}{3!} \left(a\cdot\frac{\partial}{\partial r}\right)^2 f(r) + \ldots[/tex]

    The second:
    This one comes from Landau's book, first chapter, fourth section:
    Given the equation [itex] L = L(v^{\prime 2}) [/itex] where [itex]\boldsymbol{v}^\prime = \boldsymbol{v} + \boldsymbol{\varepsilon}[/itex] and [itex]\boldsymbol{v}^2 = v^2[/itex] s.t. [itex] L(v^{\prime 2}) = L(v^2 + 2\boldsymbol{v}\boldsymbol{\varepsilon} + \boldsymbol{\varepsilon}^2)[/itex]: why does expanding in powers of [itex]\boldsymbol{\varepsilon}[/itex] and neglecting terms above first order lead to:
    L(v^{\prime 2}) = L(v^2)+\frac{\partial L}{\partial v^2} 2\boldsymbol{v}\cdot \boldsymbol{\varepsilon}

    Some insight into why this is would be greatly appreciated. :-)
  2. jcsd
  3. Jun 24, 2011 #2
    Okay, the first and the second one are equivalent if I set [itex]r=v^2[/itex], [itex]a=2\boldsymbol{v}\boldsymbol{\varepsilon}[/itex] and disregard terms [itex]\boldsymbol{\varepsilon}^2[/itex]. I still do not quite see how to get the first equation done properly.

    For a normal Taylor expansion (up to first order), we have:
    f(x) = f(c) + (x-c) \frac{\operatorname{d}}{\operatorname{d}x}f(c) + \ldots

    If I now identify [itex]x=r+a[/itex] and [itex]c=a[/itex] I recover most of the terms I need, with the exception of [itex]\frac{\operatorname{d}}{\operatorname{d}r+a}f(r)[/itex].

  4. Jun 24, 2011 #3
    Hi Stalafin! :smile:

    This is an abuse of notation and an abuse that I don't like at all. Basically, this equation has two kind of r's: an r that represents a value and an r that represents a variable.

    Rewriting your equation gives us
    [tex]f(r+a) = f(r) + a\cdot\frac{\partial}{\partial r}f(r) + \frac{1}{2} a^2\cdot\frac{\partial^2}{\partial r^2}f(r) + \frac{1}{3!} a^3\cdot\frac{\partial^3}{\partial r^3} f(r) + \ldots[/tex]

    Because the notation [itex]\frac{\partial}{\partial f}[/itex] sucks (because r is used as a variable there), we will replace it with a better one:

    [tex]f(r+a) = f(r) + af^\prime(r) + \frac{1}{2} a^2f^{\prime\prime}(r) + \frac{1}{3!} a^3f^{\prime\prime\prime}(r) + \ldots[/tex]

    But this is exactly the Taylor series! So the original formula was correct (but abusive).
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