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?
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.
Sorry, I wasn't paying attention before...
Okay I'ma start from the beginning,
Integrating both sides, I get:
Replacing x with -x we get:
That should be about right.
So you were right with your first answer.
Also, here's a plot of the 2 graphs:
Do you see why the approximation is bad?