Use strong induction to prove piecewise function H 0 </msub> = 0 ,

Desirae Washington 2022-07-04 Answered
Use strong induction to prove piecewise function
H 0 = 0 , H 1 = 1 , H 2 = 1, for all n N where n 3:
Prove for all n N ,
H n = H n 1 + H n 2 H n 3 .
H n = { n 2 , if  n  is even n + 1 2 , if  n  is odd
I don't know how the inductive step k + 1 in a strong induction would go for piecewise function like this. I think I'll have to show the proposition hold when k + 1 is even and odd, but I don't know how to continue the proof.
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Answers (1)

grubijanebb
Answered 2022-07-05 Author has 10 answers
Step 1
You would just do both cases and see if each of them is correct. Since this case distinction was exhaustive, the statement is correct. I'll present the inductive step:
Suppose the explicit form of H n holds for k 3 , k 2 , k 1. Now we only have two possible cases:
1. k is even. Note that k 2 is also even, but k 1 and k 3 are odd. We have
H k = H k 1 + H k 2 H k 3 = ( k 1 ) + 1 2 + k 2 2 ( k 3 ) + 1 2 = k 2 ,
which satisfies the given expression for H n .
2. k is odd. Note that k 2 is also odd, but k 1 and k 3 are even. We have
H k = H k 1 + H k 2 H k 3 = k 1 2 + ( k 2 ) + 1 2 k 3 2 = k + 1 2 ,
which also satisfies the given expression for H n .
Step 2
In conclusion, the above expression for H n is correct in every possible case, so it is true in general.
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