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January 22nd, 2013, 05:31 AM   #1
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Is this proof valid

Let: x,u,y,v:C?R be functions in the complex variable s=a+ib defined by convergente series. I have this implication:

If x(s)=u(s),y(s)=v(s) , then after some steps on these series one get an equation of the form: h(s)=0 which give one value of s as s=1+ib for some b .

I call this result (s=1+ib ) as "(P)".

The function h here depends on the given functions x,u,y,v . Its formula is not important.

let A=s?C(s)=u(s),y(s)=v(s) and B=s?C(s)=u(s)=y(s)=v(s)=0 is a proper subset of A .

I know that if s?A then (P) holds true.

My question is as follow: What happen if s?B . Is (P) true or false!.

We remark here that if s?B then we cannot construct the function h and hence we cannot deduce (P), i.e., if x(s)=u(s)=y(s)=v(s)=0 then I cannot do anything and the proof stoped here.
zeraoulia is offline  
January 22nd, 2013, 01:51 PM   #2
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Re: Is this proof valid

I'll start by saying I have no idea what you are trying to do.

However you say that something is true for (all?) s in A. Since B is a subset of A, it will be true for s in B.
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January 28th, 2013, 03:37 AM   #3
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Re: Is this proof valid

could you post original version of problem???
tahir.iman is offline  

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proof, valid

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