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Derivation of Taylor Series in R^n

  1. Oct 25, 2016 #1
    I was studying the derivation for taylor series in ℝ##^n## on my book and I have some trouble understanding a passage; it's the very beginning actually:

    ##f : A## ⊆ ℝ##^n## → ℝ
    ##f ## ∈ ##C^2(A)##
    ##x_0## ∈ ##A##

    "be ##g_{(t)} = f_{(x_0 + vt)}## where v is a generic versor, then we have:
    ##g_{(t)} = g_{(0)} + g'_{(0)}t + \frac{1}{2}g''_{(\tau)}t^2## where ##\tau ∈ [0, t]##"

    I don't understand why is it ok to do taylor expansion centered in t=0 and then use ##g''_{(\tau)}## instead of ##g''_{(0)}##?
    I'm actually fine with the rest of the demonstration which is quite easy but I'd like to understand what he's doing here; I was pretty sure it could be something coming from Lagrange's theorem(he uses it everywhere) but I can't really see it here.
     
  2. jcsd
  3. Oct 25, 2016 #2

    Ray Vickson

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    The Taylor expansion of first order with remainder for a univariate function is
    $$g(t) = g(0) + t g'(0) + \frac{t^2}{2!} g''(\tau),$$
    where ##\tau## is a value between ##0## and ##t##.

    In general, if ##g \in C^{n+1}## we have
    $$g(t) = g(0) + t g'(0) + \frac{t^2}{2!} g''(0) + \cdots + \frac{t^n}{n!} g^{(n)}(0) + \frac{t^{n+1}}{(n+1)!} g^{(n+1)}(\tau).$$

    See your textbook, or look on-line for "Taylor series with remainder".
     
  4. Oct 25, 2016 #3
    Oh, so it is the remainder in lagrange form for the 1st order expansion.
    Its so obvious now that i see it! I cant believe i got stuck on this ahah thank you ^^
     
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