Can I Combine the Series for ln(x+1) and ln(x-1) to Expand ln((x+1)/(x-1))?

  • #1
brendan
65
0
Hi all,
I am trying to work out a series expansion for ln ((x+1)/(x-1)).


I have got the series expansion for ln(x+1) ie x- (x^2/2) + (x^3/3) - (x^4/4) ...

and for ln(x-1) -x- (x^2/2) - (x^3/3) - (x^4/4) ...

Can I tie these two together to get the series for ln ((x+1)/(x-1)).

Or do I treat this as a new series.

regards
Brendan
 
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  • #2
Welcome to PF!

brendan said:
I am trying to work out a series expansion for ln ((x+1)/(x-1)).

I have got the series expansion for ln(x+1) ie x- (x^2/2) + (x^3/3) - (x^4/4) ...

and for ln(x-1) -x- (x^2/2) - (x^3/3) - (x^4/4) ...

Can I tie these two together to get the series for ln ((x+1)/(x-1)).

Or do I treat this as a new series.

Hi Brendan! Welcome to PF! :smile:

Yes, you can just subtract them …

but do you really mean ln(x-1) … at x = 0, that would be ln(-1) … do you mean ln(1-x)? :smile:
 
  • #3
Thanks a lot I did mean ln(1-x).

Heres the expansion

2x + (2/3)x^3 + (2/5)x^5+ (2/7)x^7

Once again thanks
 

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