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Topic: Sequence of entire functions (Read 4100 times) |
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Maria
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Sequence of entire functions
« on: May 15th, 2006, 5:04am » |
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Hi,
I need to construct a sequence (h_j) of entire functions with the property that h_j -> 1 uniformally on compact subsets of the right half plane, but h_j doesn't converge at any point of the open left half plane.
By Runge's theorem I know that such sequence exists, but how do I construct the sequence? I have never constructed sequences of any kind before so I don't know how to approach this problem. Any ideas?
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Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
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Re: Sequence of entire functions
« Reply #1 on: May 15th, 2006, 3:10pm » |
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I have not been familiar with this particular theorem, but the versions I find referenced on the web do not guarantee that the sequence will not converge anywhere on the left half-plane, merely that the rational functions within the sequence will have all their poles there, contained within any set you like.
However, Runge's theorem is not really necessary here, and I would have almost guaranteed the existance of such sequences even without it.
I suggest you consider the behavior of eaz for various real values of a. It's all you need.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Michael Dagg
Senior Riddler
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Re: Sequence of entire functions
« Reply #2 on: May 17th, 2006, 5:33pm » |
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In addition to Icarus' remarks, notice that the graph of the modulus of the exponential function
z --> |e^(x+iy)| \equiv e^x
is like a skateboard ramp, of a sort.
Let h_j(z) = 1 + e^(-j z).
If in your question you meant that you want h_j --> 1 uniformly on compact subsets of
the OPEN right-half plane, then that should do it. If you like, each h_j can be replaced
by a polynomial coming from a Taylor polynomial of the exponential function.
If you meant CLOSED right-half plane, then solutions can be devised from the exponential
function or Taylor polynomials which I'll leave to you (using Icarus' remarks and you have it).
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| « Last Edit: May 17th, 2006, 5:34pm by Michael Dagg » |
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Regards,
Michael Dagg
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