How to derive the following series expansion for \ln (

airesex2

airesex2

Answered question

2022-04-24

How to derive the following series expansion for ln(x+1x1)
How to prove :
ln(x+1x1)=2[1x+13x3+15x5+]
where |x|>1
I am not able to get how to proceed on this one ?

Answer & Explanation

Makayla Santiago

Makayla Santiago

Beginner2022-04-25Added 19 answers

Note that x+1x1=1+1x11x
Note that the series for log(1+x) converges absolutely when |x|<1. Therefore it is permissible to rearrange the series in any way you like.
Now note that log(1+1x11x)=log(1+1x)log(11x) Expand both logarithmic series and group the terms.
veceritzpzg

veceritzpzg

Beginner2022-04-26Added 16 answers

If |x>1 then 1|x|<1, so:
ln(x+1x1)=ln(x(1+1x)x(11x))=ln(1+1x11x)=ln(1+t)ln(1t)
With t=1x<1, then |t|<1
Expanding :
ln(1+t)=tt22+t33t44+
ln(1t)=tt22t33t44
Then:
ln(1+t)ln(1t)=(tt22+t33t44+)(tt22t33t44)=
2(t+t33+t55)
Now t=1x;
ln(x+1x1)=2(1x+13x3+15x5+17x7)

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