Series Representation

**jules027** · May 20th 2008, 07:15 AM

How would i find a series representation for ln(x) from the series of ln(1+x)? which is: [(-1)^n * x^n] / n

**kalagota** · May 20th 2008, 07:38 AM

Originally Posted by jules027

How would i find a series representation for ln(x) from the series of ln(1+x)? which is: [(-1)^n * x^n] / n

i dont know if this would help..

$\ln x = \ln (1+ (x-1))$
and so $\frac{(-1)^n (x-1)^n}{n} = \frac{(1-x)^n}{n}$

haha.. am i ryt?

**Mathstud28** · May 20th 2008, 12:20 PM

Originally Posted by kalagota

i dont know if this would help..

$\ln x = \ln (1+ (x-1))$
and so $\frac{(-1)^n (x-1)^n}{n} = \frac{(1-x)^n}{n}$

haha.. am i ryt?

$\ln(x+1)=\int\frac{dx}{x+1}=\int\sum_{n=0}^{\infty }(-1)^nx^n=\sum_{n=0}^{\infty}\frac{(-1)^nx^{n+1}}{n+1}$

Now to find ln(x) in terms of this

Kalgotta was correct the first step is to see that $\ln(x)=\ln(1+(x-1))$

So by letting $\lambda=x-1$ we get

$\ln(1+\lambda)=\sum_{n=0}^{\infty}\frac{(-1)^n\lambda^{n+1}}{n+1}$

Substituting back in what u equals we get

$\ln(x)=\sum_{n=0}^{\infty}\frac{(-1)^n(x-1)^{n+1}}{n+1}$

To verify we see that

$\ln(x)=\int\frac{dx}{x}=\int\sum_{n=0}^{\infty}(-1)^n(x-1)^ndx=\sum_{n=0}^{\infty}\frac{(-1)^n(x-1)^{n+1}}{n+1}$

QED

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