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

In summary, the conversation discusses the problem of showing that for sufficiently large n, the polynomial f_n(z) has no roots in D_0(100), which is the disk of radius 100 centered at 0. The conversation mentions using the analytic convergence property and a corollary to Rouche's theorem to prove this. However, there is a flaw in the argument as pointed out. The conversation then suggests finding a better bound for e^z on the circle of radius 100 to proceed with the proof.
  • #1
michael.wes
Gold Member
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0

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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  • #2
z=(-100) is on the boundary of D_0(100). e^(-100) certainly isn't greater than e^(100)!
 
  • #3
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!
 
  • #4
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.
 

What is a power series?

A power series is a series of the form ∑n=0∞ an(x-c)n, where an are coefficients, x is a variable, and c is a constant. It is used to represent functions as an infinite sum of terms.

What does it mean for a power series to have roots?

A root of a power series is a value of x that makes the series equal to 0. In other words, it is a solution to the equation ∑n=0∞ an(x-c)n = 0.

Why is it important to show that the partial sums of a power series have no roots in a disk as n->∞?

This result is important because it shows that as the number of terms in the series increases, there are no values of x that make the series equal to 0 within a certain range or "disk". This allows us to determine the convergence of the series and the behavior of the function it represents.

How can we prove that the partial sums of a power series have no roots in a disk as n->∞?

This can be proven using the Ratio Test or the Root Test, which are tests for determining the convergence of infinite series. These tests involve taking the limit of the ratio or the nth root of the terms in the series, respectively, as n->∞. If the limit is less than 1, the series converges and there are no roots in the disk.

Are there any exceptions to this result?

Yes, there are some power series that do have roots within a disk, even as n->∞. These are known as singular power series and they have a specific radius of convergence. However, for most power series, the partial sums will have no roots in a disk as n->∞.

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