Polynomial arithmetic #6 - Questions and Answers

Question 1 The finite arithmetic sequence's term count should be determined. Given: 3 , − 2 , − 7 , … , − 87 3,−2,−7,…,−87 Answer: 𝑎 = a= first term = 3 𝑑 = d= common difference = -5 𝑙 = l= last term = -87 Using the formula: 𝑙 = 𝑎 + ( 𝑛 − 1 ) 𝑑 l=a+(n−1)d Substitute the values: − 87 = 3 + ( 𝑛 − 1 ) ( − 5 ) −87=3+(n−1)(−5) − 87 = 3 − 5 𝑛 + 5 −87=3−5n+5 − 87 = 8 − 5 𝑛 −87=8−5n 5 𝑛 = 95 5n=95 𝑛 = 19 n=19 Thus, there are 19 terms in the given arithmetic sequence. Question 2 The scoring of modern IQ tests is such that Intelligence Quotients (IQs) have a Normal distribution with 𝜇 = 100 μ=100 and 𝜎 = 15 σ=15. a) What percent of people have IQ less than 80? b) What percent of people have IQ greater than 120? c) Mensa International accepts people within the top 2%. What level of IQ qualifies one to be a member of Mensa? Answer: Step 1 Given: 𝜇 = 100 μ=100 (mean) 𝜎 = 15 σ=15 (standard deviation) a) Percent of people with IQ < 80: The Z-score is calculated as: 𝑧 = 𝑥 − 𝜇 𝜎 = 80 − 100 15 = − 1.33 z= σ x−μ ​ = 15 80−100 ​ =−1.33 Using the standard normal table: 𝑃 ( 𝑍 < − 1.33 ) = 0.0918 P(Z<−1.33)=0.0918 Thus, 9.18% of people are expected to have an IQ less than 80. b) Percent of people with IQ > 120: 𝑧 = 120 − 100 15 = 1.33 z= 15 120−100 ​ =1.33 Using the standard normal table: 𝑃 ( 𝑍 > 1.33 ) = 1 − 𝑃 ( 𝑍 < 1.33 ) = 1 − 0.9082 = 0.0918 P(Z>1.33)=1−P(Z<1.33)=1−0.9082=0.0918 Thus, 9.18% of people have an IQ greater than 120. c) Mensa membership (top 2%): The boundary for the top 2% corresponds to a Z-score of approximately 2.05. Using: 𝑥 = 𝜇 + 𝑧 𝜎 = 100 + ( 2.05 ) ( 15 ) = 130.75 x=μ+zσ=100+(2.05)(15)=130.75 Thus, a score of 130.75 or higher qualifies for Mensa. Question 3 This problem deals with polynomials in 𝑍 2 [ 𝑋 ] Z 2 ​ [X]. Remember these have 0–1 coefficients. a) Show that the Hamming weight of an irreducible polynomial (nonzero coefficients) of degree 1 is odd. b) Provide addition and multiplication tables for polynomial arithmetic mod 2. c) List all irreducibles of degree 3 and 4. Answer: a) For any polynomial 𝑝 ( 𝑥 ) p(x) in 𝑍 2 [ 𝑋 ] Z 2 ​ [X]: Hamming weight = the number of nonzero coefficients. Suppose 𝑝 ( 𝑥 ) p(x) has even nonzero coefficients. Then: 𝑝 ( 1 ) p(1) is even, so 𝑝 ( 1 ) = 0 p(1)=0 implies reducibility. Thus, irreducible polynomials in 𝑍 2 [ 𝑋 ] Z 2 ​ [X] must have odd Hamming weight. b) Addition and multiplication tables mod 2: + + 0 𝑥 x 1 + 𝑥 1+x 0 0 𝑥 x 1 + 𝑥 1+x 𝑥 x 𝑥 x 0 1 1 + 𝑥 1+x 1 + 𝑥 1+x 1 0 ∗ ∗ 0 𝑥 x 1 + 𝑥 1+x 0 0 0 0 𝑥 x 0 𝑥 2 x 2 𝑥 + 𝑥 2 x+x 2 1 + 𝑥 1+x 0 𝑥 + 𝑥 2 x+x 2 1 + 𝑥 + 𝑥 2 1+x+x 2 c) Degree 3 irreducibles: 𝑥 3 + 𝑥 2 + 1 ,   𝑥 3 + 𝑥 + 1 ,   𝑥 3 + 𝑥 2 + 𝑥 + 1 x 3 +x 2 +1,x 3 +x+1,x 3 +x 2 +x+1 Degree 4 irreducibles: 𝑥 4 + 𝑥 + 1 ,   𝑥 4 + 𝑥 3 + 1 ,   𝑥 4 + 𝑥 3 + 𝑥 2 + 𝑥 + 1 x 4 +x+1,x 4 +x 3 +1,x 4 +x 3 +x 2 +x+1 Question 4 Find the sum of the finite arithmetic series: 1 + 5 + 9 + ⋯ + 157 1+5+9+⋯+157. Answer: Here: 𝑎 = 1 a=1 (first term) 𝑙 = 157 l=157 (last term) 𝑑 = 4 d=4 (common difference) To find the number of terms ( 𝑛 n): 𝑙 = 𝑎 + ( 𝑛 − 1 ) 𝑑 l=a+(n−1)d 157 = 1 + ( 𝑛 − 1 ) ⋅ 4 157=1+(n−1)⋅4 𝑛 − 1 = 39    ⟹    𝑛 = 40. n−1=39⟹n=40. The sum of 𝑛 n terms is: 𝑆 𝑛 = 𝑛 2 ( 𝑎 + 𝑙 ) . S n ​ = 2 n ​ (a+l). Substitute values: 𝑆 40 = 40 2 ( 1 + 157 ) = 20 ⋅ 158 = 3160. S 40 ​ = 2 40 ​ (1+157)=20⋅158=3160. Thus, the sum is 3160.