I have a problem in expressing ln (1 + x) as a convergent series. Can people show me the steps of doing this? thanks in advance!!!

Printable View

- Sep 30th 2010, 06:59 AMReal9999Help needed. Writing ln(1 + x) as a convergent series
I have a problem in expressing ln (1 + x) as a convergent series. Can people show me the steps of doing this? thanks in advance!!!

- Sep 30th 2010, 07:22 AMHappyJoe
You could expand ln(1+x) in its Taylor series around x=0.

For this, you need all the derivative of ln(1+x) evaluated at x=0.

Let f(x)=ln(1+x). (note first that f(0) = 0).

Then

f'(x) = 1/(1+x).

f''(x) = -1/(1+x)^2.

f'''(x) = 2/(1+x)^3.

f''''(x) = -6/(1+x)^4

and so on.

In general, we have that the n'th derivative of f is $\displaystyle (-1)^{n+1}\frac{(n-1)!}{(1+x)^n}$ (which you may prove by induction). Then

$\displaystyle f^{(n)}(0) = (-1)^{n+1}(n-1)!,$

and so the Taylor series is:

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n,$

which converges for $\displaystyle |x|\leq 1$ (except for $\displaystyle x=-1$). - Sep 30th 2010, 09:38 AMReal9999
Thanks! Your response is extremely lucid and helpful! It does not only answer the question perfectly, but it also gives me some very good intuition to understand Taylor series expansion! I have become more confident in real analysis now!

Thank you very much(Bow)(Bow) - Sep 30th 2010, 04:48 PMskeeter
$\displaystyle \displaystyle \frac{1}{1+x} = \frac{1}{1-(-x)} = 1 - x + x^2 - x^3 + ...

$

integrate ...

$\displaystyle \displaystyle \ln(1+x) = C + x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...$

$\displaystyle x = 0 \implies C = 0$

$\displaystyle \displaystyle \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ... = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}$