Radius of Convergence: Complex Series Need Not Be Defined Everywhere

[freddyfish]

A complex series need not be defined for all z within the "circle of convergence"?

The (complex) radius of convergence represents the radius of the circle (centered at the center of the series) in which a complex series converges.

Also, a theorem states that a (termwise) differentiated series has the same radius of convergence as the original series.

Now, Ln z is obviously singular (at least) at the negative real axis which is a distance 1 away from the z₀=-1 + i. But the Taylor series of Ln z centered at z₀=-1 + i has a radius of convergence equal to 2^0.5. Thus, the derivative of Ln z is not defined on negative real axis, but according to the theorem it has radius of convergence R=2^0.5.

This implies that a series need not be defined everywhere a distance less than R from the center of the series.

However, the definition of convergence of a complex series is that that the limit of the partial sums converge to some finite value.

How can this contradiction be eliminated?

My own thoughts about this is that this contradiction would not rise from the above definitions and theorems if the derivative of Ln z does not equal the termwise differentiated complex series representation of Ln z. If this is the case, then why?

Thanks
//Freddy

 

[lurflurf]

The trouble logarithms have on the (by convention) negative real axis (apart from the origin) is not a pole, but a branch cut. The series expansions are perfectly well defined, but they will not reflect any branch cuts. For example a series centered at z0=-1 + i defines a function inside a circle of radius sqrt(2) centered at -1 + i but this function only agrees with the usual logarithm (having a branch cut on the negative real axis) up to the branch cut. The series will not have the discontinuity across the cut.
log(-1+0⁺i)-log(-1+0^-i)=2 π i
while
f(-1+0⁺i)-f(-1+0^-i)=0
If f is the series you describe

 

[jackmell]

The (complex) radius of convergence represents the radius of the circle (centered at the center of the series) in which a complex series converges.

Also, a theorem states that a (termwise) differentiated series has the same radius of convergence as the original series.

Now, Ln z is obviously singular (at least) at the negative real axis which is a distance 1 away from the z₀=-1 + i. But the Taylor series of Ln z centered at z₀=-1 + i has a radius of convergence equal to 2^0.5. Thus, the derivative of Ln z is not defined on negative real axis, but according to the theorem it has radius of convergence R=2^0.5.

This implies that a series need not be defined everywhere a distance less than R from the center of the series.

However, the definition of convergence of a complex series is that that the limit of the partial sums converge to some finite value.

How can this contradiction be eliminated?

My own thoughts about this is that this contradiction would not rise from the above definitions and theorems if the derivative of Ln z does not equal the termwise differentiated complex series representation of Ln z. If this is the case, then why?

Thanks
//Freddy

You have a mis-understanding of branch-cuts. They're purely arbitrary and can be placed anywhere to isolate a single-valued component of a multi-valued function. That branch-cut along the negative real axis for the log function is just an arbritrary line of demarcation to isolate a convenient single-valued part of it. But Log is perfectally defined, continuous, and analytic there and in fact everywhere except the origin.

But a single-valued power series is convergent in a disc extending to the nearest singular point of the function. It's convergent for all points inside that disc. So I could just as well center a series for Log(z) at some point along the negative real axis and it will converge to Log(z) for every single point inside a disc the size of which is equal to the distance to the origin and what determination of Log(z) is used to construct the series will determine which single-valued branch of Log(z) the series converges to.

 

[freddyfish]

Thank you guys. That one I should have seen through! What a stupid mistake of me.

Anyway, thanks again :)