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

[michael.wes]

Homework Statement

Let f_n(z)=\sum_{k=0}^n\frac{1}{k!}z^n. Show that for sufficiently large n the polynomial f_n(z) has no roots in D_0(100), 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!

 

[Dick]

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

 

[michael.wes]

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!

 

[Dick]

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.