07Ma6aSol1

07Ma6aSol1 - Math 6a Solution Set 1 Problem 1 Liu 1.30(d...

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Unformatted text preview: Math 6a - Solution Set 1 Problem # 1: Liu 1.30 (d) Determine and prove a general formula that includes the results in (a), (b) and (c) as special cases. Solution. The general formula, which can easily be inferred by observing the formulae in the previous parts, is n X i =1 1 ( ki- k + 1)( ki + 1) = n kn + 1 . We prove this is true for all natural numbers n via induction. First, note that if n = 1 , the left hand side of this equation is 1 ( k- k +1)( k +1) = 1 k +1 which is equal to the right hand side when n = 1 , so the base case is proven. So now we assume it is true for n = m and consider the case when n = m + 1 . Then, using the induction hypothesis, we have: m +1 X i =1 1 ( ki- k + 1)( ki + 1) = m X i =1 1 ( ki- k + 1)( ki + 1) + 1 ( k ( m + 1)- k + 1)( k ( m + 1) + 1) n +1 X i =1 1 ( ki- k + 1)( ki + 1) = m km + 1 + 1 ( k ( m + 1)- k + 1)( k ( m + 1) + 1) n +1 X i =1 1 ( ki- k + 1)( ki + 1) = m km + 1 + 1 ( km + 1)( km + k + 1) = m ( km + k + 1) + 1 ( km + 1)( km + k + 1) n +1 X i =1 1 ( ki- k + 1)( ki + 1) = km 2 + km + m + 1 ( km + 1)( km + k + 1) = km ( m + 1) + ( m + 1) ( km + 1)( km + k + 1) n +1 X i =1 1 ( ki- k + 1)( ki + 1) = ( km + 1)( m + 1) ( km + 1)( km + k + 1) = m + 1 km + k + 1 = m + 1 k ( m + 1) + 1 . This is the general form of the statement for n = m + 1 and thus we have, by induction, the truth of the statement for all natural numbers n. Alternate Solution. Note that we have 1 ki- k + 1- 1 ki + 1 = ki + 1- ( ki- k + 1) ( ki- k + 1)( ki + 1) = k ( ki- k + 1)( ki + 1) . 1 2 Thus, we have 1 ( ki- k + 1)( ki + 1) = 1 k ( 1 ki- k + 1- 1 ki + 1 ) which means that n X i =1 1 ( ki- k + 1)( ki + 1) = n X i =1 1 k ( 1 ki- k + 1- 1 ki + 1 ) = 1 k n X i =1 ( 1 ki- k + 1- 1 ki + 1 ) . But note that n X i =1 ( 1 ki- k + 1- 1 ki + 1 ) = n X i =1 ( 1 k ( i- 1) + 1- 1 ki + 1 ) is clearly a telescoping series, and thus we have n X i =1 ( 1 k ( i- 1) + 1- 1 ki + 1 ) = 1- 1 kn + 1 = kn + 1- 1 kn + 1 = kn kn + 1 and thus we have n X i =1 1 ( ki- k + 1)( ki + 1) = 1 k n X i =1 ( 1 ki- k + 1- 1 ki + 1 ) = ( 1 k )( kn kn + 1 ) = n kn + 1 which is what we needed to prove. Problem # 2: Liu 1.48 (a) Give a verbal description of the sentences in the language specified by the grammar: { S → aSb, S → ab } (b) Give a verbal description of the sentences in the language specified by the grammar: { S → aB, S → bA, A → a, A → aS, A → bAA, B → b, B → bS, B → aBB } (c) Give a verbal description of the sentences in the language specified by the grammar: { S → ABx, A → Ay, yB → zB, A → w, xB → u } (d) Construct a grammar for a language whose sentences are of the form { a n b 2 n | n ≥ 1 } . (e) Construct a grammar for a language whose sentences are of the form { αabab | α ∈ { a, b } * } . 3 Solution....
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07Ma6aSol1 - Math 6a Solution Set 1 Problem 1 Liu 1.30(d...

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