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April 4th, 2013, 08:57 AM   #1
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Discontinuity of a power serie on the circle of convergence

Good evening to all members,
Excuse-me my inglish is not very good !

My question:it is well-known that if f(z) is the sum of a power serie "Sum of a_n z^n " of radius 1 which converges for z=1(that is to say "Sum of a_n converges"), then the fonction f is continue in every triangle inclused in the open disk of convergence B(0,1) and which has 1 as a vertex( it comes from the uniform convergence of the serie in this triangle).

I would like to find such a power serie such as the function f is not continue in the open disk of radius 1 U{1}= B(0,1)U {1}. ( obviously it is seems sufficent to find a sequence (z_n) which converges towards 1 "interiously tangently" in the B(0,1) and such that f(z_n) does not converges towards f(1)).

Thank you to everybody who will take time to answer tome

Have a good day

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April 4th, 2013, 11:14 AM   #2
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Re: Discontinuity of a power serie on the circle of converge

[color=#000000] for |z|>1.

And be careful how you express maths,
Quote:
radius 1 U{1}= B(0,1)U {1}
here you mean the unit disc with its border .

[/color]
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April 5th, 2013, 11:13 PM   #3
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Re: Discontinuity of a power serie on the circle of converge

Thank you very mutch ZardoZ,

but the exemple you give me is not a power series of radius 1 (that is to say a function of the kind "Sum of a_n z^n) and your fonction does not exist for z= 1 . Apart of that, for me " B(0,1) U {1}" means the union of the open ball of radius 1 and the single point 1 alone.

Have a good evening and thank you or your help,

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April 6th, 2013, 12:59 AM   #4
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Re: Discontinuity of a power serie on the circle of converge

[color=#000000]There are no balls on the complex plane. Anyway the context of your question is confusing! [/color]
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April 6th, 2013, 12:52 PM   #5
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Re: Discontinuity of a power serie on the circle of converge

Thank you Zardoz,

excuse me i said "ball" but i should have say unit open disk in the complex plane:

By B(0,1) i mean the unit disk in the complex plane that is to say B(0,1) = {z belonging C / IzI < 1}

Good morning to all of you,
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