Exploring the Upper Bound of f^{n+1}(x) for x in [-1/2, 1/2]

[Punkyc7]

Let f = ln([itex]\frac{1}{1-x}[/itex])


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


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



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


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


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.