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B Multiterm Taylor expansion

  1. Jul 15, 2017 #1
    So in the book it says expend function ƒ in ε to get following.

    ƒ=√ (1 + (α + βε)2) = √ (1 + α2) + (αβε)/√ (1 + α2) + (β2ε2)/2 (1 + α2)3/2 + O(e3)

    When I expend I get(keeping ε = 0)
    ƒ(0) = √ (1 + α2) -->first term
    ƒ'(0) = (αβ)/√ (1 + α2) --> sec term with gets multiplied by ε

    for third term
    I used
    d/de (αβ + εβ2 ) * √ (1 + (α + βε)2) + d/de √ (1 + (α + βε)2) * (αβ + εβ2)


    at most I got

    ƒ''(0) = β2 / √ (1 + (α + βε)2) + -1/2 (α + βε) β(αβ + εβ2) /3√ (1 + ( α + βε)2)

    = β2 / √ (1 + (α )2) + -1/2 (α + β0) β(αβ + 0β2) /3√ (1 + α 2)

    Why I am not getting the term as in the book which is
    2ε2)/2 (1 + α2)3/2
     
  2. jcsd
  3. Jul 15, 2017 #2

    andrewkirk

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    The error is in the differentiation to get the third term.

    The function being expanded is
    $$f(x)=\sqrt{1+x^2}$$
    and it is being expanded at ##x=\alpha##.

    The first derivative is correct:
    $$\frac d{dx}\sqrt{1+x^2} = x(1+x^2)^\frac12$$

    The second derivative is
    $$\frac d{dx}\left(x(1+x^2)^\frac12\right) = (1+x^2)^{-\frac12} - x^2(1+x^2)^{-\frac32}$$
     
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