Maclaurin Series

[boneill3]

HI Guys,
Just a quick questoin.
I have worked out a series expansion of

ln(x+1)/ln(x-1) which equals 2x+2/3x^3+2/5x^5........

I have been asked to find the series expansion for 1/2 ln(x+1)/ln(x-1)

Do I multiply my result with the scaler 1/2 or do I work out the Taylor again using

1/2 ln(x+1) devide 1/2ln(x-1)

as the functions instead
regards
Brendan

[jemidiah]

Shorter answer: just multiply your original result by 1/2. It'll be the same as working out the Maclaurin series for "1/2*ln(x+1)/ln(x-1)". It may have been a typo but you wouldn't work out the series for "1/2 ln(x+1) devide 1/2ln(x-1)" since this is 1/2*ln(x+1)/[1/2*ln(x-1)] = 2/2*ln(x+1)/ln(x-1) = ln(x+1)/ln(x-1).

Longer answer: one of the results of the study of infinite series is that, if a series converges, then so does that series with each term multiplied by some scalar. That is, given a series {x_n} = {x₁, x₂, ...} where x_n goes to some number X as n goes to infinity, we know that the series {ax_n} = {ax₁, ax₂, ...} goes to aX. Let's say f(n) = the sum of the first n terms of the Maclaurin series for ln(x+1)/ln(x-1), and that f(x) = ln(x+1)/ln(x-1). By construction, as n goes to infinity f(n) goes to f(x). Thus 1/2*f(n) goes to 1/2*f(x), and so the Maclaurin series for 1/2*f(x) can be written as the sum of 1/2 times the terms you already had.

You can also observe that the derivative allows you to take the 1/2 out of each term when finding terms of the Maclaurin series, and note that you can then factor all of the 1/2's out to get 1/2 times your original answer.


Edit: oh yeah, I wanted to say I didn't check your derivation of the first series since you didn't ask for it.

[boneill3]

Thanks alot for that.
The line

"That is, given a series {xn} = {x1, x2, ...} where xn goes to some number X as n goes to infinity, we know that the series {axn} = {ax1, ax2, ...} goes to aX."

really helped me a lot.

regards
Brendan

[jemidiah]

Glad to hear it