How Does the Upper Bound of f^{n+1}(x) Relate to 2^{n + 1} * n! on [-1/2, 1/2]?

[Punkyc7]

Let f = ln(\frac{1}{1-x})


show that if x \in [-1/2 , 1/2] then


|f^{n+1}(x)| <= 2^{n + 1} * n!



I am having a hard time seeing how 2^{n + 1} * n! comes into play.


I have that the taylor series for f is \Sigma \frac{x^n}{n}


If a take a derivative it becomes x^(n-1) and if I plug anything on the interval it is less than one. I am thinking that I did this wrong because of how big that upper bound is/.

 

[jbunniii]

You're currently using $n$ for two different purposes: the order of the derivative and the index of the sum. I strongly suggest using a different letter for one of these. For example, $f(x) = \sum_{k=1}^{\infty} x^k/k$. And you want to show that $|f^{(n+1)}(x)| \leq 2^{n+1}n!$. (I assume your exponent $n+1$ means the $n+1$'st derivative.)

Try starting with $n=0$. Can you show that the inequality is true in that case?

 

[brmath]

No one said you have to use the Taylor's series. Why don't you take 2 or 3 of derivatives directly and see if a pattern emerges. That could lead you to a proof by induction.