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Rudin 5.18 (Trick guidance)

  1. Mar 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose f is a real function on [a, b], n is a positive integer, and [tex]\f^{(n-1)}[/tex]
    exists for every t in [a, b]. Let [tex]\alpha,\beta [/tex], and P be as in Taylor’s theorem
    (5.15). Define

    [tex]\ Q(t) = \frac{f(t)-f(\beta)}{t-\beta}[/tex]

    for [tex]\ t \in [a, b], t \neq \beta [/tex],

    differentiate

    [tex]\ f(t)-f(\beta)=(t-\beta)Q(t)[/tex]

    n − 1 times at [tex]\ t = \alpha[/tex], and derive an alternate Taylor’s theorem:

    [tex]\ f(\beta)=P(\beta)+\frac{Q^{(n-1)}(\alpha)}{(n-1)!}(\beta-\alpha)^{n}[/tex] (I had to put this here to make the above expression stay on one line)



    2. Relevant equations



    3. The attempt at a solution

    So first I did the differentiation n-1 times, you notice a pattern and since f(beta) is constant, you get

    [tex]\ f^{(n}}(t)= nQ^{(n-1)}+(t-\beta)^nQ^{(n)}(t)[/tex]

    Then from Taylor's theorem we know that

    [tex]\ f(\beta) = P(\beta) + \frac{f^{(n)}(x)}{n!}(\beta - \alpha)^{n}[/tex]

    Just plugging in that expression into Taylor's theorem is real damn close to the result I need. How do I get rid of the extra Q^{(n)} in the numerator? (or is my differentiation wrong and I'm not catching it?)

    Thanks a mil guys.
     
    Last edited: Mar 4, 2009
  2. jcsd
  3. Mar 4, 2009 #2

    Avodyne

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    Science Advisor

    Both your equations in part (3) are wrong.
     
  4. Mar 4, 2009 #3
    The second equation is Rudin 24... I don't see how it could be wrong -_-.

    The first is just differentiating the given expression. We'd use the product rule (n-1) (I think there I used it n times), and that would be the result yes?
     
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