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Uchiha Sasuke

21/04/2017 at 07:39
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A function f is difined for non - negative integers n and k as follows

f (0,n) = n+1 , f (k,0) = f (k-1,1) , f (k+1,n+1) = f (k,f (k+1,n))

Evaluate f (2,2)

Help me please! I will give you a like if you answer ithihiok

 



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    FC Alan Walker 06/05/2018 at 09:06

    Put : \(f\left(0,n\right)=n+1\)(1)

            \(f\left(k,0\right)=f\left(k-1,1\right)\)(2)

            \(f\left(k+1,n+1\right)=f\left(k,f\left(k+1,n\right)\right)\)(3)

    After that we have:

    From (3) \(\Rightarrow f\left(2,2\right)=f\left(1,f\left(2,1\right)\right)=f\left(1,f\left(1,f\left(1,1\right)\right)\right)=f\left(1,f\left(1,f\left(0,f\left(1,0\right)\right)\right)\right)\)

    From (2) \(\Rightarrow f\left(1,f\left(1,f\left(0,f\left(1,0\right)\right)\right)\right)=f\left(1,f\left(1,f\left(0,f\left(0,1\right)\right)\right)\right)\)

    From (1) \(\Rightarrow f\left(1,f\left(1,f\left(0,f\left(0,1\right)\right)\right)\right)=f\left(1,f\left(1,f\left(0,2\right)\right)\right)=f\left(1,f\left(1,3\right)\right)\)

    From (3) \(\Rightarrow f\left(1,f\left(1,3\right)\right)=f\left(1,f\left(0,f\left(1,2\right)\right)\right)=f\left(1,f\left(0,f\left(0,f\left(1,1\right)\right)\right)\right)=f\left(1,f\left(0,f\left(0,f\left(0,f\left(1,0\right)\right)\right)\right)\right)\)

    From (2) \(\Rightarrow f\left(1,f\left(0,f\left(0,f\left(0,f\left(1,0\right)\right)\right)\right)\right)=f\left(1,f\left(0,f\left(0,f\left(0,f\left(0,1\right)\right)\right)\right)\right)\)

    From (1) \(\Rightarrow f\left(1,f\left(0,f\left(0,f\left(0,f\left(0,1\right)\right)\right)\right)\right)=f\left(1,f\left(0,f\left(0,f\left(0,2\right)\right)\right)\right)=f\left(1,f\left(0,f\left(0,3\right)\right)\right)=f\left(1,f\left(0,4\right)\right)=f\left(1,5\right)\)

    From (3) \(\Rightarrow f\left(1,5\right)=f\left(0,f\left(1,4\right)\right)=f\left(0,f\left(0,f\left(1,3\right)\right)\right)=f\left(0,f\left(0,f\left(0,f\left(1,2\right)\right)\right)\right)=f\left(0,f\left(0,f\left(0,f\left(0,f\left(1,1\right)\right)\right)\right)\right)=f\left(0,f\left(0,f\left(0,f\left(0,f\left(0,f\left(1,0\right)\right)\right)\right)\right)\right)\)

    From (2) \(\Rightarrow f\left(0,f\left(0,f\left(0,f\left(0,f\left(0,f\left(1,0\right)\right)\right)\right)\right)\right)=f\left(0,f\left(0,f\left(0,f\left(0,f\left(0,f\left(0,1\right)\right)\right)\right)\right)\right)\)

    From (1) \(\Rightarrow f\left(0,f\left(0,f\left(0,f\left(0,f\left(0,f\left(0,1\right)\right)\right)\right)\right)\right)=f\left(0,f\left(0,f\left(0,f\left(0,f\left(0,2\right)\right)\right)\right)\right)=f\left(0,f\left(0,f\left(0,f\left(0,3\right)\right)\right)\right)=f\left(0,f\left(0,f\left(0,4\right)\right)\right)=f\left(0,f\left(0,5\right)\right)=f\left(0,6\right)=7\)

    So f(2,2)=7



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