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#1 2007-02-02 09:26:30

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

about induction

Fibonacci Number(f_(n-1)+f_n=f_(n+1)) , f_0=0 , f_1=1, prove that f_n=f_(k+1)*f_(n-k)+f_k*f(n-k-1)


I want to use induction to prove that , so I need to assume when n=k is true , then move on to prove k+1 is true ,  but in this case , I need to assume k and k+1 are true , but can I assume k and k+1 are true , it's a little wierd assuming too much .


Numbers are the essence of the Universe

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#2 2007-02-02 10:07:36

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: about induction

I want to use induction to prove that , so I need to assume when n=k is true , then move on to prove k+1 is true ,  but in this case , I need to assume k and k+1 are true, but can I assume k and k+1 are true , it's a little wierd assuming too much .

No, if you want to prove k+1 is true, you can't assume k+1 is true.  What you can do however is assume that k and all integers less than k are true.  This is known as "strong" induction, and I believe it's what you want to use.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#3 2007-02-02 14:25:12

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

Re: about induction

Doesn't matter, you can assume k and k+1, but you have to prove k+2 case from them. And you need to verify both k=0 and k=1.


X'(y-Xβ)=0

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#4 2007-02-02 14:27:54

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

Re: about induction

For more on Fibonacci Numbers, Read my post #36 on this page


X'(y-Xβ)=0

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