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

QUESTION

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

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

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

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

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

DISCUSSION

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

$\displaystyle 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

$\displaystyle 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 $\displaystyle c \in (0,x) $? Or is this Maclaurin expansion?

Here $\displaystyle 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 $\displaystyle \le $ 1. All I can think to do is

$\displaystyle 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 $\displaystyle \le $ 1 ??

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

5. I kind of understand this part. the Maclaurin expansion is $\displaystyle f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n $ which we know converges absolutely for $\displaystyle 0 \le x \le 1 $ (f(0) = 0) so it will converge absolutely for$\displaystyle 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