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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!