I was recently Taylor-expanding ln around (1,0). I noticed that this polynomial will have a range of input that converges between 0 and 2 regardless of Taylor order. I then found an expansion that did not seem to have this issue, namely: lim_(n->+oo)2*sum lim_(i=0)^n ((1)/(2i+1)*(((x-1)/(x+1)))^(2i+1))=ln(x) My question is: how is this formula derived (as in created, not the derivative) from ln?

Jack Ingram

Jack Ingram

Answered question

2022-11-01

Different Polynomial Expansions of Natural Logarithm
I was recently Taylor-expanding ln around ( 1 , 0 ). I noticed that this polynomial will have a range of input that converges between 0 and 2 regardless of Taylor order. I then found an expansion that did not seem to have this issue, namely:
lim n + 2 i = 0 n ( 1 2 i + 1 ( x 1 x + 1 ) 2 i + 1 ) = ln ( x )
My question is: how is this formula derived (as in created, not the derivative) from ln?

Answer & Explanation

Tania Alvarado

Tania Alvarado

Beginner2022-11-02Added 15 answers

If you combine the Taylor series for log ( 1 + x ) and log ( 1 x ) you arrive to a well known series which is
log 1 + x 1 x = 2 i = 0 x 2 i + 1 2 i + 1
So now define y = 1 + x 1 x that is to say x = y 1 y + 1 and you end with
log ( y ) = i = 0 1 2 i + 1 ( y 1 y + 1 ) 2 i + 1
Evelyn Freeman

Evelyn Freeman

Beginner2022-11-03Added 5 answers

You start with the taylor series about x = 0 for log ( 1 + x ) = n ( 1 ) n + 1 n x n for | x | < 1
Then, apply a Binomial/Euler transform to get a series for log x x 1 = n 1 n x n
Then, substitute x = u u 1 to get the desired series after a shift of indices. note that it doesn't converge for all x (exercise: what x does it converge for?)

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?