MacLaurin series error estimation

**janvdl** · October 15th 2009, 12:20 PM

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

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

$\ln(1-x) = - \sum ^{\infty} _{n=0} (-1) \frac{x^{n+1}}{n+1}<br />$

Let $e_n$ be the error estimate.

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

$| e_2 | = 0.02$

$| e_3 | = 0.0064$

So therefore $\ln(0.6) = \sum ^{2} _{n=0} (-1) \frac{0.4^{n+1}}{n+1} = -1,101333...$

However, my calculator gives me -0.510826.

Where did I go wrong?

**Danny** · October 15th 2009, 12:59 PM

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.

**janvdl** · October 15th 2009, 01:07 PM

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?

**Danny** · October 15th 2009, 01:13 PM

Originally Posted by janvdl

Sorry, could you please elaborate on this?

Sorry, it doesn't. Why are you writing the series

$<br /> -\sum_{n=0}^{\infty} (-1) \frac{x^{n+1}}{n+1}?<br />$

and further, how are you getting your error bounds?

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