Not really understanding Taylor/Maclaurin Theorems

**slevvio** · November 23rd 2009, 02:28 PM

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

**slevvio** · November 24th 2009, 01:27 AM

sorry - giant erorr in part 3, there shouldnt be a factorial term

part 2 is wrong too - ive answered my own question here.

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