Show that the partial sums of a power series have no roots in a disk as n->infty

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Homework Help Overview

The problem involves analyzing the partial sums of a power series defined by \( f_n(z) = \sum_{k=0}^n \frac{1}{k!} z^k \) and demonstrating that these polynomials do not have roots within a specified disk as \( n \) approaches infinity. The context is rooted in complex analysis, particularly focusing on properties of analytic functions and convergence.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss applying the analytic convergence property and Rouche's theorem to establish the relationship between the roots of \( f_n(z) \) and \( e^z \). There are attempts to justify the use of bounds on the functions involved, and some participants express uncertainty about the validity of their arguments.

Discussion Status

The discussion is ongoing, with participants seeking clarification on previous arguments and hints for further progress. Some guidance has been offered regarding the use of bounds and the properties of compact sets, indicating a productive direction for exploration.

Contextual Notes

Participants note the importance of the disk's boundary and the implications of uniform convergence on closed disks. There is an acknowledgment of potential flaws in earlier reasoning, prompting a reevaluation of the approach taken.

michael.wes
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Homework Statement


Let [tex]f_n(z)=\sum_{k=0}^n\frac{1}{k!}z^n[/tex]. Show that for sufficiently large n the polynomial f_n(z) has no roots in [tex]D_0(100)[/tex], i.e. the disk of radius 100 centered at 0.


Homework Equations



This is a sequence of analytic functions which converges uniformly to e^z on C.

The Attempt at a Solution



I want to apply the analytic convergence property, and a corollary to Rouche's theorem, which says that if I have two analytic functions on a region, a closed path gamma with interior, homologous to 0 in the region, and |g(z)-f(z)|<|f(z)| for all z in the image of gamma, then the number of roots of g in the interior of gamma is the same as the number of roots of f in the interior of gamma, counting multiplicities.

That looks like a handful, but I think I've basically got it..

Let epsilon = 1. We know that for sufficiently large n,

|f_n(z)-e^z|_im(gamma) <= ||f_n(z)-e^z||_(whole disk) < 1 (arbitrary constant).

But 1 is certainly less that ||e^z||_im(gamma), since ||e^z||_im(gamma) >= |e^100| >> 1.

So f_n and e^z have the same number of roots on the interior, that is, none.

I would appreciate it if someone could check my work, and let me know if there are any holes in the argument. Thanks!
 
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z=(-100) is on the boundary of D_0(100). e^(-100) certainly isn't greater than e^(100)!
 
I can't seem to come up with much after you pointed out the flaw in this argument. Could you give me a hint on how to proceed? The assignment itself says: 'hint: circles are compact sets'. The only use I can think of for this is that f_n converges on closed disks to e^z, and hence uniformly and absolutely to e^z on C, but other than that I'm stuck.

Thanks!
 
michael.wes said:
I can't seem to come up with much after you pointed out the flaw in this argument. Could you give me a hint on how to proceed? The assignment itself says: 'hint: circles are compact sets'. The only use I can think of for this is that f_n converges on closed disks to e^z, and hence uniformly and absolutely to e^z on C, but other than that I'm stuck.

Thanks!

Try to find a much better bound for e^z on the circle of radius 100. You've got the right idea, it's just that your bounds are way off.
 

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