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Topic: sequence (Read 2548 times) |
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Alex
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Please help me with this problem Let {a_n} and {b_n} be sequence of complex numbers. Assume that {a_n} has no accumulation point. Prove that there exists a holomorphic function f:C->C such that f(a_n)=b_n Thank you.
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Icarus
wu::riddles Moderator Uberpuzzler
    
 Boldly going where even angels fear to tread.
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Re: sequence
« Reply #1 on: Jun 23rd, 2004, 7:28pm » |
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Start with a finite version: Find a polynomial P1 of minimal degree such that P(a1) = b1. The answer is trivial: P1(x)=b1. Now find P2, which works for a1 and a2: A trick for this one is two start off with A(x - a1) + B(x - a2). At a1, the first term is zero, so all contribution comes from the second term - yielding B(a1 - a2). Choosing B to be b1/(a1 - a2) yields the value we want. Similarly, A is chosen to be b2/(a2-a1). So P2(x) = (b1(x - a2) - b2(x - a1))/(a1-a2) If we let Q(x) = (x - a1)...(x - ak), and let Qi = Q(x)/(x - ai), then Pk(x) = [sum]i biQi(x)/Qi(ai) Now let i [to] [infty] ...
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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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puzzlecracker
Senior Riddler
   
 Men have become the tools of their tools
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Re: sequence
« Reply #2 on: Nov 29th, 2004, 10:10pm » |
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Alex are you Russian and where do you study?
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While we are postponing, life speeds by
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Icarus
wu::riddles Moderator Uberpuzzler
    
 Boldly going where even angels fear to tread.
Gender: 
Posts: 4863
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Re: sequence
« Reply #3 on: Dec 5th, 2004, 7:07pm » |
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Out of curiousity, why do you Alex might be Russian?
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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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