# Thread: Maclaurin Series for ln((1+x)/(1-x))

1. ## Maclaurin Series for ln((1+x)/(1-x))

This is an assignment question

So I know the Maclaurin series for ln (1+x), and it is

x - (x^2 )/2 + (x^3 ) /3 - (x^4 )/4....

And after doing some working, I think I found the Maclaurin series for ln (1-x)

-x - (x^2 )/2 - (x^3 )/3 - x(^4 )/4... (all negatives)

So ln(1+x)/(l-x) = ln(1+x)-ln(1-x)

which, when I subtract the series from one another, I get

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

But this doesn't seem correct, as neither approximation is even remotely close to ln(1.5) (0.406 3dp)

Using five terms from ln(1+x) = 1.753
five terms from my series ln((1+x)/(1-x)) - it just gets continually larger...

Maybe I've got the correct answer and just haven't explained why ln(1+x) is more accurate. Any help?
Also, how do I find where it converges?

2. ## Re: Maclaurin Series for ln((1+x)/(1-x))

Originally Posted by lukasaurus
So I know the Maclaurin series for ln (1+x), and it is

x - (x^2 )/2 + (x^3 ) /3 - (x^4 )/4....

And after doing some working, I think I found the Maclaurin series for ln (1-x)

-x - (x^2 )/2 - (x^3 )/3 - x(^4 )/4... (all negatives)
Um... how did you find the maclaurin series for ln(1-x)?

You know that $\displaystyle \ln(1+x) = x + \dfrac{x^2}{2} + \dfrac{x^3}{3} + \dfrac{x^4}{4} ...$

Now if you replace x with -x you get...

You know that $\displaystyle \ln(1 -x) = -x + \dfrac{x^2}{2} - \dfrac{x^3}{3} + \dfrac{x^4}{4} ...$

3. ## Re: Maclaurin Series for ln((1+x)/(1-x))

That's what I thought too, but when I worked it out using the Taylor series formula, it comes out as what I got, and online calculators come out as what I got as well.

eg for f(x) = ln(1-x) ==> f(0) = 0, f'(0) = -1, f''(0) = -1, f'''(0) = -2, f''''(0

You can check on any online calculator.

4. ## Re: Maclaurin Series for ln((1+x)/(1-x))

Originally Posted by Educated
Um... how did you find the maclaurin series for ln(1-x)?

You know that $\displaystyle \ln(1+x) = x + \dfrac{x^2}{2} + \dfrac{x^3}{3} + \dfrac{x^4}{4} ...$

Now if you replace x with -x you get...

You know that $\displaystyle \ln(1 -x) = -x + \dfrac{x^2}{2} - \dfrac{x^3}{3} + \dfrac{x^4}{4} ...$
Also, I think you have your ln(1+x) incorrect. I've got a list here in stewarts calc and it is showing what I have got...

5. ## Re: Maclaurin Series for ln((1+x)/(1-x))

Sorry, I wasn't paying attention before...

Okay I'ma start from the beginning,

$\displaystyle \dfrac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + x^5 ...$
(Geometric sequence)

Integrating both sides, I get:

$\displaystyle \displaystyle \int \dfrac{1}{1-x}dx = \int (1 + x + x^2 + x^3 + x^4 + x^5 ...)dx$

$\displaystyle -\ln(1-x) = x + \dfrac{x^2}{2} + \dfrac{x^3}{3} + \dfrac{x^4}{4} + \dfrac{x^5}{5} + \dfrac{x^6}{6} ...$

$\displaystyle \ln(1-x) = -x - \dfrac{x^2}{2} - \dfrac{x^3}{3} - \dfrac{x^4}{4} - \dfrac{x^5}{5} - \dfrac{x^6}{6} ...$

Replacing x with -x we get:

$\displaystyle \ln(1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \dfrac{x^5}{5} - \dfrac{x^6}{6} ...$

Also, here's a plot of the 2 graphs:
Wolfram|Alpha, plots
Do you see why the approximation is bad?

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