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Calculus Taylor Approximation Proof

  1. Feb 23, 2016 #1
    • Member warned about deleting the template
    1. The question is. Show that if |nx| <1, the following is exact up to (and including) the x^2 order. The hint giving says to use the Taylor Expansion for both sides of the equation


    2. (1+x)^n = e^n(x-(1/2)x^2) ; the n(x-(1/2)x^2) is all an exponent


    3. My first attempt was to take the taylor series of both sides. The left hand side is pretty easy.

    1 + nx + (n/2)(n-1)x^2 = LHS

    the right hand side gives the same thing though

    f(0) = RHS = 1
    f'(0) = n(1-x)e^n(x-(1/2)x^2) = n
    f''(0) = -n*e^n(x-(1/2)x^2 + n^2 (1+x)^2 * e^n(x-(1/2)x^2) = -n + n^2

    which in the taylor series gives the same as the LHS

    1 + nx + (n/2)(n-1)x^2 = RHS

    but now I'm stuck... how do I show that |nx| < 1
     
    Last edited by a moderator: Feb 24, 2016
  2. jcsd
  3. Feb 24, 2016 #2

    HallsofIvy

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    Then use parentheses to make that clear: (1+ x)^n= e^(n(x- (1/2)x^2)).


    You don't- that is not what is asked. You are asked to show that if |nx|< 1 then that equality is true, not the other way around.
     
    Last edited by a moderator: Feb 24, 2016
  4. Feb 24, 2016 #3
    Any hint on how to start that?
     
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