1. ## Maclaurin series estimate

How many term of the MAclaurin series for ln(1+x) do you need to use to estimate ln(1.4) to within 0.001?

I have tried this 10 different ways and I keep getting bigger values.

The MacLaurin series is:

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

2. Note this is an alternating series so you can use |S-Sn| < a(n+1)

replace x with .4

you get (-1)^(n+1) *.4^(n+1)/(n+1)

|S-Sn| < .4^(n+2)/(n+2) < .001

Use a table to find the smallest n to satisfy this inequality

3. For the alternating series, if we stop the interation after the summing of the term $\displaystyle a_{n}$, the [obvious] error criterion is...

$\displaystyle |e_{n}|<|a_{n+1}|$ (1)

Computing the 'infinite sum'...

$\displaystyle \ln (1+x)= - \sum_{n=1}^{\infty}(-1)^{n}\cdot \frac{x^{n}}{n}$ (2)

... the condition $\displaystyle |a_{n+1}|<.001$ is met for n= 5, i.e. five term of (2) are enough and that gives...

$\displaystyle \ln(1.4)\approx .336$ (3)

The 'exact' value is...

$\displaystyle \ln(1.4) =.336472236621\dots$ (4)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. The series expansion...

$\displaystyle \ln (1+z) = -\sum_{n=1}^{\infty} (-1)^{n} \frac{z^{n}}{n}$ (1)

... converges in genearal for complex values of z if |z|<1. It is interesting the computation of (1) for $\displaystyle z=\frac{i}{\sqrt{3}}$. Summing 'only' 44 complex terms of (1) we obtain...

$\displaystyle \ln (1+ \frac{i}{\sqrt{3}}) = \ln \frac{2}{\sqrt{3}} + i\cdot \frac {\pi}{6} = .143841036226\dots + i\cdot .523598775598\dots$

... where both real and imaginary part have an error less than $\displaystyle 10^{-12}$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. n= 4 will work