algorithm - Proving the Horner Function (Polynomial Evaluation) -

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i completed exercise 1-8 in the algorithm design manual, second edition, steven s. skiena:

enter image description here

is convincing?

typically in induction proofs separate steps each other. induction step implicit taste. i'd this:

1) n = 1 horner([a0], x) = a0

2) horner([a0,...,a(n+1)], x) = x * horner([a1,...,a(n+1)], x) + a0 = x * horner([b0,...,bn], x) + a0, bn = a(n+1)

3) having horner n, horner n+1 can calculated 2)

your proof fine, i've said - induction step should accented - how solution n+1 follows solution n (m in case), in separate proof step.


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