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The jacobian matrix of partial derivatives?

  1. Dec 2, 2009 #1
    In differential geometry what does df mean as in


    [tex]

    \mbox{f} : \mathbb{R}^m \mbox{ to } \mathbb{R}^n [/tex]

    Then df is what? the jacobian matrix of partial derivatives?
     
  2. jcsd
  3. Dec 2, 2009 #2

    quasar987

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    Re: df

    More generally, suppose the map f is a differentiable map is defined on an open set U of R^m:
    [tex]f:U\rightarrow\mathbb{R}^n[/tex]

    In differential geometry, we define on each point x of U the tangent space to U at x as the vector space [itex]T_x\mathbb{R}^m[/itex] consisting of pairs (x,v), where [itex]v\in\mathbb{R}^m[/itex] and where addition and scalar multiplication are defined by (x,v) + (x,w) = (x,v+w) and a(x,v) = (x,av). A vector [itex](x,v)\in T_x\mathbb{R}^m[/itex] is to be interpreted as "the vector v "at" x". Then we form the tangent bundle of U
    [tex]
    TU:=\bigcup_{x\in U}T_xU
    [/tex]
    Then df is defined as the map [itex]df:TU\rightarrow T\mathbb{R}^n[/itex] that associates to [itex](x,v)\in T_xU[/itex] the element [itex](f(x),Df(x)v)\in T_{f(x)}\mathbb{R}^n[/itex], where Df(x) is the usual derivative of f at x in the sense of calculus or analysis.
     
  4. Dec 3, 2009 #3

    HallsofIvy

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    Re: df

    Strictly speaking, df is the linear transformation that best approximates f (in the same way that y= mx+ b best approximates f(x) at x= a when m= f'(a)). Given the standard bases for Rm and Rn, that is represented by the Jacobian matrix.
     
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