1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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],


    [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


    User Avatar
    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook