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Geometric Tangent Vector

  1. Mar 3, 2013 #1
    So let [itex]ℝ^{n}_{a}[/itex]={(a,v) : a [itex]\in[/itex] [itex]ℝ^{n}[/itex], v [itex]\in[/itex] [itex]ℝ^{n}[/itex]}

    so any geometric tangent vector, which is an element of [itex]ℝ^{n}_{a}[/itex] yields a map

    Dv|af = Dvf(a) = [itex]\frac{d}{dt}|_{t=0}[/itex]f(a+tv)

    this operation is linear over ℝ and satisfies the product rule

    Dv|a(fg) = f(a)Dvg + g(a)Dvf

    if v|a = [itex]\sum_{i=1}^n[/itex] viei|a, then by the chain rule
    Dv|af can be written as:

    Dv|af [itex]\sum_{i=1}^n[/itex] vi [itex]\frac{∂f}{∂x_{i}}(a)[/itex]

    not seeing how the chain rule applies and how the result as such.
  2. jcsd
  3. Mar 3, 2013 #2


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    xi(t) = ai + tvi

    (d/dt) f(a + tv) = (d/dt) f(xi(t))

    = Σ (∂f/∂xi) (dxi/dt)

    = Σ vi(∂f/∂xi)
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