# How to prove that the natural logarithm function is real analytic on (0, infinity)?

• Nov 10th 2010, 05:22 AM
CSM
[SOLVED]Proving that ln(x) is real analytic on (0, infinity)
I already know that it is real analytic in 1, with radius of convergence 1 and power series expansion:

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

for $\displaystyle x \in (0,2)$

Now what?

I think that I can show that ln is real analytic for all points of (0,2). But I'd like to prove it for $\displaystyle (0, + \infty)$.
Some thoughts?
• Nov 10th 2010, 07:43 AM
chisigma
Setting $\displaystyle z=x + i\ y$ is...

$\displaystyle \displaystyle f(z)= \ln z = \ln \sqrt{x^{2} + y^{2}} + i\ \tan^{-1} \frac{y}{x} = u(x,y) + i\ v(x,y)$ (1)

Now necessary condition for f(*) analytic is done by the Cauchy-Riemann relations...

$\displaystyle \displaystyle \frac{\partial{u}}{\partial{x}}= \frac{\partial{v}}{\partial{y}}$

$\displaystyle \displaystyle \frac{\partial{u}}{\partial{y}}= - \frac{\partial{v}}{\partial{x}}$ (2)

What are the values of z [if any...] that satisfy (2)?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Nov 10th 2010, 07:52 AM
Drexel28
Quote:

Originally Posted by CSM
I already know that it is real analytic in 1, with radius of convergence 1 and power series expansion:

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

for $\displaystyle x \in (0,2)$

Now what?

I think that I can show that ln is real analytic for all points of (0,2). But I'd like to prove it for $\displaystyle (0, + \infty)$.
Some thoughts?

In fact, it's real analytic on $\displaystyle (0,2]$. Which criteria do you know? The clearest possible one is that $\displaystyle \ln$ can be extended holomorphically to an open set in $\displaystyle \mathbb{C}$ containing $\displaystyle \mathbb{R}$
• Nov 10th 2010, 10:30 AM
CSM
Thanks for your replies. But I cannot use any knowledge of complex numbers. It's only real analysis, no complex numbers.
I use the definition from Tao's Analysis II:
Let E be a subset of R and let f:E--> R be a function. If a is an interior point of E, we say that f is real analytic at a if there exists an open interval (a-r,a+r) in E for some r>0 such that there exists a power series $\displaystyle \Sigma_{n=o}^\infty c_n(x-a)^n$centered at a which has a radius of convergence greater than or equal to r, and which converges to f on (a-r,a+r). If E is an open set, and f is real analytic at every point a of E, we say that f is real analytic on E.
• Nov 10th 2010, 10:42 AM
Also sprach Zarathustra
I think you should try to prove the following:

The function f(x)=ln(x) is an infinitely differentiable function such that the Taylor series at any point x0=1 in its domain.
• Nov 10th 2010, 10:43 AM
Drexel28
Quote:

Originally Posted by Also sprach Zarathustra
I think you should try to prove the following:

The function f(x)=ln(x) is an infinitely differentiable function such that the Taylor series at any point x0=1 in its domain.

Infinite differentiability does not imply real analyticity. Take $\displaystyle f(x)=\begin{cases}e^{\frac{-1}{x^2}} & \mbox{if}\quad x\ne 0\\ 0 & \mbox{if}\quad x=0\end{cases}$
• Nov 14th 2010, 12:04 PM
CSM
Any ideas?
How do you in general proof a function to be analytic?
• Nov 14th 2010, 04:55 PM
Jose27
$\displaystyle f\in C^{\infty}(U,\mathbb{R})$ (where $\displaystyle U\subset \mathbb{R}$ open) is analytic in $\displaystyle U$ if and only if for all $\displaystyle x\in U$ there exist $\displaystyle x\in J\subset U$ an interval, and constants $\displaystyle R,C>0$ with the property that $\displaystyle |f^{(k)}(y)|\leq C \frac{k!}{R^k}$ for all $\displaystyle y\in J$. The proof is not difficult, the key is using an appropiate form of the residue in Taylor's formula.

From this, and noting that $\displaystyle (\ln(x))^{(k)}=\frac{(-1)^{k+1}(k-1)!}{x^k}$ the result follows.
• Nov 14th 2010, 05:37 PM
Bruno J.
Quote:

Originally Posted by Drexel28
In fact, it's real analytic on $\displaystyle (0,2]$. Which criteria do you know? The clearest possible one is that $\displaystyle \ln$ can be extended holomorphically to an open set in $\displaystyle \mathbb{C}$ containing $\displaystyle \mathbb{R}$

That's not true... the logarithm has an essential singularity at the origin, of the worst possible kind on top of that! The largest possible domain on which a fixed branch of a logarithm can live comfortably is the plane cut along some curve joining $\displaystyle 0$ and $\displaystyle \infty$. But you're right in the sense that the existence of a holomorphic function on $\displaystyle \mathbb{C}-\mathbb{R}_{\leq 0}$ which coincides with the usual (real) logarithm on the real line immediately implies that the usual logarithm is analytic there (and everywhere else on $\displaystyle \mathbb{C}-\mathbb{R}_{\leq 0}$).
• Nov 14th 2010, 05:45 PM
Drexel28
Quote:

Originally Posted by Bruno J.
That's not true... the logarithm has an essential singularity at the origin, of the worst possible kind on top of that! The largest possible domain on which a fixed branch of a logarithm can live comfortably is the plane cut along some curve joining $\displaystyle 0$ and $\displaystyle \infty$. But you're right in the sense that the existence of a holomorphic function on $\displaystyle \mathbb{C}-\mathbb{R}_{\leq 0}$ which coincides with the usual (real) logarithm on the real line immediately implies that the usual logarithm is analytic there (and everywhere else on $\displaystyle \mathbb{C}-\mathbb{R}_{\leq 0}$).

Ah, of course. It must have been late.
• Nov 14th 2010, 07:23 PM
chisigma
Quote:

Originally Posted by Bruno J.
That's not true... the logarithm has an essential singularity at the origin, of the worst possible kind on top of that!...

That is not fully exact!... the function $\displaystyle \ln z$ is a multivalued function that in $\displaystyle z=0$ has a singularity classified as branch point. Effectively such type of singularity is, in a certain sense, 'the worst possible' because for an $\displaystyle f(z)$ with an essential singularity in $\displaystyle z=0$ [like for example $\displaystyle f(z)= e^{\frac{1}{z}}$...] it is possible to obtain the Laurent expansion around $\displaystyle z=0$, but for $\displaystyle f(z)= \ln z$ that is impossible...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Nov 14th 2010, 07:33 PM
chisigma
For more information about the singularity of $\displaystyle \ln z$ in $\displaystyle z=0$ see...

Logarithmic Singularity -- from Wolfram MathWorld

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$