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

for

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 .

Some thoughts?

- November 10th 2010, 06:22 AMCSM[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:

for

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 .

Some thoughts? - November 10th 2010, 08:43 AMchisigma
Setting is...

(1)

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

(2)

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

Kind regards

- November 10th 2010, 08:52 AMDrexel28
- November 10th 2010, 11:30 AMCSM
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 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. - November 10th 2010, 11:42 AMAlso 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. - November 10th 2010, 11:43 AMDrexel28
- November 14th 2010, 01:04 PMCSM
Any ideas?

How do you in general proof a function to be analytic? - November 14th 2010, 05:55 PMJose27
(where open) is analytic in if and only if for all there exist an interval, and constants with the property that for all . The proof is not difficult, the key is using an appropiate form of the residue in Taylor's formula.

From this, and noting that the result follows. - November 14th 2010, 06:37 PMBruno 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 and . But you're right in the sense that the existence of a holomorphic function on which coincides with the usual (real) logarithm on the real line immediately implies that the usual logarithm is analytic there (and everywhere else on ).

- November 14th 2010, 06:45 PMDrexel28
- November 14th 2010, 08:23 PMchisigma
That is not fully exact!... the function is a multivalued function that in has a singularity classified as

*branch point*. Effectively such type of singularity is, in a certain sense, 'the worst possible' because for an with an essential singularity in [like for example ...] it is possible to obtain the Laurent expansion around , but for that is impossible...

Kind regards

- November 14th 2010, 08:33 PMchisigma
For more information about the singularity of in see...

Logarithmic Singularity -- from Wolfram MathWorld

Kind regards