# Thread: MacLaurin series error estimation

1. ## MacLaurin series error estimation

Hey everyone, quick question:

Approximate ln(0.6) by using the MacLaurin series expansion for ln(1-x), with an error guaranteed to be less than 0.01. Show your error estimate.
My workings:

Since we want ln(0.6), but make use of the series expansion for ln(1-x), x = 0.4

$\displaystyle \ln(1-x) = - \left(x + \frac{x^2}{2} + \frac{x^3}{3} + ... \right)$

$\displaystyle \ln(1-x) = - \sum ^{\infty} _{n=0} (-1) \frac{x^{n+1}}{n+1}$

Let $\displaystyle e_n$ be the error estimate.

$\displaystyle | e_1 | = \frac{0.4^{1 + 1}}{1+1} = 0.08$

$\displaystyle | e_2 | = 0.02$

$\displaystyle | e_3 | = 0.0064$

So therefore $\displaystyle \ln(0.6) = \sum ^{2} _{n=0} (-1) \frac{0.4^{n+1}}{n+1} = -1,101333...$
However, my calculator gives me -0.510826.

Where did I go wrong?

2. Originally Posted by janvdl
Hey everyone, quick question:

My workings:

However, my calculator gives me -0.510826.

Where did I go wrong?
Your first series doesn't alternate - your second does. They are not the same.

3. Originally Posted by Danny
Your first series doesn't alternate - your second does. They are not the same.
Sorry, could you please elaborate on this?

4. Originally Posted by janvdl
Sorry, could you please elaborate on this?
Sorry, it doesn't. Why are you writing the series

$\displaystyle -\sum_{n=0}^{\infty} (-1) \frac{x^{n+1}}{n+1}?$

and further, how are you getting your error bounds?