[SOLVED] Logarithm real analytic

**bram kierkels** · November 16th 2009, 01:26 AM

I want to proof that the natural logarithm is real analytic.
For $x \in (0,2)$ I have $ln(x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(x-1)^n$

Is it satisfied when I say when $x \in (2,\infty)$ ,
I have $ln(a)=-ln(\frac{1}{a})$
Because $\frac{1}{a} \in (0,2)$ it is also real analytic when $x \in (2,\infty)$ ?
Thanks

**bram kierkels** · November 16th 2009, 01:27 AM

Definition Real Analytic;

F is real analytic at a if there excist an open interval (a-r, a+r) in E for some r>0 such that there exist a power series $\sum_{n=0}^{\infty} c_n (x-a)^n$ centered at a which has radius of convergence greater than or equal to r, and which converges to f on (a-r, a+r).

**chisigma** · November 16th 2009, 06:29 AM

The 'natural logarithm' function is analytic in $x=1$ so that the Taylor expansion around this point exists and it is...

$\ln x = -\sum_{n=1}^{\infty} \frac{(1-x)^{n}}{n}$ , $|1-x|<1$ (1)

If You want to 'expand' the range of definition of the function, there are two possibilities...

a) because is $\ln \frac{1}{x}= - \ln x$ , writing $\frac{1}{x}$ instead to $x$ in the second term of (1) and changing sign You obtain...

$\ln x = \sum_{n=1}^{\infty} \frac{(1-\frac{1}{x})^{n}}{n}$ , $|1-\frac{1}{x}|<1$ (2)

b) because is $\ln \frac{x}{a} = \ln x - \ln a$ , writing $\frac{x}{a}$ instead of $x$ in the second term of (1) You obtain…

$\ln x = \ln a - \sum_{n=1}^{\infty} \frac{(1-\frac{x}{a})^{n}}{n}$ , $|1-\frac{x}{a}|<1$ (3)

Kind regards

$\chi$ $\sigma$

**bram kierkels** · November 16th 2009, 07:59 AM

Originally Posted by chisigma

The 'natural logarithm' function is analytic in $x=1$ so that the Taylor expansion around this point exists and it is...

$\ln x = -\sum_{n=1}^{\infty} \frac{(1-x)^{n}}{n}$ , $|1-x|<1$ (1)

If You want to 'expand' the range of definition of the function, there are two possibilities...

a) because is $\ln \frac{1}{x}= - \ln x$ , writing $\frac{1}{x}$ instead to $x$ in the second term of (1) and changing sign You obtain...

$\ln x = \sum_{n=1}^{\infty} \frac{(1-\frac{1}{x})^{n}}{n}$ , $|1-\frac{1}{x}|<1$ (2)

b) because is $\ln \frac{x}{a} = \ln x - \ln a$ , writing $\frac{x}{a}$ instead of $x$ in the second term of (1) You obtain…

$\ln x = \ln a - \sum_{n=1}^{\infty} \frac{(1-\frac{x}{a})^{n}}{n}$ , $|1-\frac{x}{a}|<1$ (3)

Kind regards

$\chi$ $\sigma$

Thanks a lot Chisigma.
The Taylor expansion is $\ln x = -\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x-1)^n$
so if i'm right the answer should be something like this;
$\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(\frac{1}{x}-1)^n$
Or am I missing something?

**chisigma** · November 16th 2009, 08:21 AM

Originally Posted by bram kierkels

... the Taylor expansion is $\ln x = -\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x-1)^n$
so if i'm right the answer should be something like this;
$\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(\frac{1}{x}-1)^n$

All right!... because is...

$(x-1)^{n}= (-1)^{n} \cdot (1-x)^{n}$ (1)

... we have...

$\ln x = - \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}\cdot (x-1)^{n} = - \sum_{n=1}^{\infty} \frac{(1-x)^{n}}{n} = \sum_{n=1}^{\infty} \frac{(1-\frac{1}{x})^{n}}{n}$ (2)

Kind regards

$\chi$ $\sigma$

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