Infinite Derivative of the Zeta Function

[seanhbailey]

Homework Statement

When extended to the complex plain, does the zeta function approach zero as the number of derivatives of it approaches infinity?

Homework Equations

The Attempt at a Solution

 

[seanhbailey]

Could someone please help? I really need an answer.

 

[Petek]

I don't understand the question. When extended to the entire complex plane, the zeta function is analytic everywhere, except at 1. Being analytic implies that it's infinitely differentiable everywhere (again, except at 1). What do you mean by "does the zeta function approach zero as the number of derivatives of it approaches infinity?"

Petek

 

[seanhbailey]

In other words, does the infinith derivative of the zeta function equal 0?

 

[Petek]

I think you're saying that, suppose s [itex]\neq[/itex] 1. Then, does the sequence [itex](\zeta^{(n)}(s))[/itex] converge to zero? If that's what you mean, I'll have think about it. Let me know if you mean something else.

Petek

 

[seanhbailey]

Yes, that is what I mean.

 

[Petek]

Sorry, can't figure it out. The only result I could get was the trivial observation that

[tex]\frac{\zeta^{(n)}(s)}{n!}[/tex]

converges to zero for all s [itex]\neq[/itex] 1. That just follows from the fact that zeta is analytic at those points, and hence has a Taylor series whose coefficients are as above. If this is a problem from a book, you might want to check if the previous problem provides a clue. Sometimes authors will do that.

Petek