Ln series

  • Thread starter DevoBoy
  • Start date
  • #1
8
0

Main Question or Discussion Point

I'm looking for a series for

[tex]ln|cos\frac{\Theta}{2}|[/tex]

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

Any ideas?
 
Last edited:

Answers and Replies

  • #2
631
0
u may be knowing the series of ln(1+x). Try to convert ur expression in such form.
 
  • #3
8
0
Ok, so I expand it like this:

[tex]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[/tex]

Ultimatly, I want to use this expansion to prove that

[tex]\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=ln2[/tex]

I can't quite see how this will help .... ??
 
  • #4
631
0
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
8
0
Clever! :)

So now I end up with the relation:

[tex]ln|2cos\frac{\theta}{2}|=\frac{1}{2}(ln2+ln(1+cos\theta))[/tex]

Thanks!

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

[tex]\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=ln2[/tex]

Any sugestions?
 
Last edited:
  • #6
289
0
[tex]\ln(1+x) = x - \frac{x^{2}}{2} + \ldots[/tex]
[tex]\ln(1+\cos \theta) = \cos \theta - \frac{\cos^{2}\theta}{2} + \ldots = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{\cos^{n}\theta}{n}[/tex]

Using trig identities you have that [tex]\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[/tex]

Therefore

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

[tex]\theta = 0[/tex] gives the result.
 
Last edited:
  • #7
Gib Z
Homework Helper
3,346
5
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?
 

Related Threads on Ln series

  • Last Post
Replies
2
Views
113K
Replies
3
Views
14K
Replies
2
Views
12K
  • Last Post
Replies
5
Views
10K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
1K
Replies
12
Views
2K
  • Last Post
Replies
7
Views
3K
Replies
2
Views
1K
  • Last Post
Replies
6
Views
5K
Top