Ln series

1. Mar 29, 2007

DevoBoy

I'm looking for a series for

$$ln|cos\frac{\Theta}{2}|$$

Using r=1 for my complex variable, if that matters...

Any ideas?

Last edited: Mar 29, 2007
2. Mar 29, 2007

Sourabh N

u may be knowing the series of ln(1+x). Try to convert ur expression in such form.

3. Mar 29, 2007

DevoBoy

Ok, so I expand it like this:

$$ln|cos\frac{\Theta}{2}|=ln|1+(cos\frac{\Theta}{2}-1)|=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(cos\frac{\Theta}{2}-1)^n$$

Ultimatly, I want to use this expansion to prove that

$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=ln2$$

I can't quite see how this will help .... ??

4. Mar 29, 2007

Sourabh N

i think u should multiply 1/2 to ln and convert cos to cos^2(i hope i'm clear) ; then convert half angle to full angle.

5. Mar 29, 2007

DevoBoy

Clever! :)

So now I end up with the relation:

$$ln|2cos\frac{\theta}{2}|=\frac{1}{2}(ln2+ln(1+cos\theta))$$

Thanks!

EDIT: I'm still not able to use this series to show that

$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=ln2$$

Any sugestions?

Last edited: Mar 29, 2007
6. Mar 29, 2007

AlphaNumeric

$$\ln(1+x) = x - \frac{x^{2}}{2} + \ldots$$
$$\ln(1+\cos \theta) = \cos \theta - \frac{\cos^{2}\theta}{2} + \ldots = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{\cos^{n}\theta}{n}$$

Using trig identities you have that $$\ln | \cos \theta/2 | = \frac{1}{2}\ln \left( \cos^{2}\frac{\theta}{2} \right) = \frac{1}{2}\ln (1+\cos \theta) - \frac{1}{2}\ln 2$$

Therefore

$$\ln | \cos \theta/2 | = \frac{1}{2}\sum_{n=1}^{\infty} (-1)^{n+1}\frac{\cos^{n}\theta}{n} - \frac{1}{2}\ln 2$$

$$\theta = 0$$ gives the result.

Last edited: Mar 29, 2007
7. Mar 30, 2007

Gib Z

Or perhaps the most OBVIOUS way would have been to let x=1 in the taylor series for natural log, which looks just a tiny bit similar to the series you want huh?