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Math Help - Not really understanding Taylor/Maclaurin Theorems

  1. #1
    Senior Member slevvio's Avatar
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    Not really understanding Taylor/Maclaurin Theorems

    Let  f: (-1,\infty) \rightarrow \mathbb{R} be given by  f(x) = log(1+x)

    QUESTION

    1. Write down  f^{(n)}(x).

    2. Write down the n-th stage Taylor's expansion for f, giving a clear statement about the remainder  R_n (x).

    3. For which  x \in (0, \infty) do we have  R_n (x) \rightarrow 0 as  n \rightarrow \infty ?

    4. What does this say about the range of validity of the Maclaurin series for log(1+x) on  (0,\infty)

    5. What is the range of validity of the Maclaurin series for log(1+x) on  (-1, \infty) ?

    DISCUSSION

    1.  f^(n) = \frac{ (-1)^{n+1} (n-1)!}{(1+x)^n}

    2. What does this even mean? I have this statement in my notes

    Suppose  f:[a,b] \rightarrow \mathbb{R} and  \forall n \in \mathbb{N}, f, f^{(1)}, f^{(2)}, f^{(3)}, ... , f^{(n-1)} all exist and are continuous on [a,b] (using left and right derivatives at a and b) and  f^{(n)} exists on (a,b). Then  \exists c \in (a,b) such that

    f(b) = f(a) + f^{(1)}(a)(b-a) + \frac{f^{(2)}(a)(b-a)^2}{2} + ... + \frac{f^{(n-1)}(a)(b-a)^{n-1}}{(n-1)!} + \frac{f^{(n)}(c)}{n!} (b-a)^n

    Am I supposed to set a to 0 and set b to x? That would give me this thing

     f(x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ... + \frac{(-1)^n(n-2)!}{(1+x)^{n-1}} + \frac{(-1)^{n+1}(n-1)!}{(1+c)^n } x^n for  c \in (0,x) ? Or is this Maclaurin expansion?

    Here  R_n (x) = \frac{(-1)^{n+1}(n-1)!}{(1+c)^n } x^n

    3. I have no idea how to do this bit but I knowthe answer is x  \le 1. All I can think to do is

     0 \le |R_n (x) | = | \frac{(n-1)!x^n}{(1+c)^n}| = (n-1)! |\frac{x^n}{(1+c)^n} | but why on earth does this tend to 0 as n tends to infinity, if x  \le 1 ??

    4. The Maclaurin series is valid for 0 < x  \le 1.

    5. I kind of understand this part. the Maclaurin expansion is  f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n which we know converges absolutely for  0 \le x \le 1 (f(0) = 0) so it will converge absolutely for  0 \ge x > - 1 so has radius of convergence at least 1. The maclaurin expansion is valid on (-1, 1]. Not at -1 because then log stops being defined. Is this right?

    Any help with this would be very appreciated
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  2. #2
    Senior Member slevvio's Avatar
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    sorry - giant erorr in part 3, there shouldnt be a factorial term

    part 2 is wrong too - ive answered my own question here.
    Last edited by slevvio; November 24th 2009 at 02:44 AM.
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