Questions and Answers #17 Polynomials

Questions and Answers Sheet 17 Polynomials Question #1 Determine whether the following polynomials u, v, w in P(t) are linearly dependent or independent: 𝑢 = 𝑡 3 − 4𝑡 2 + 3𝑡 + 3, 𝑣 = 𝑡 3 + 2𝑡 2 + 4𝑡 − 1, 𝑤 = 2𝑡 3 − 𝑡 2 − 3𝑡 + 5 Answer: Let 𝑐1 , 𝑐2 , 𝑐3 be the constants. So 𝑢𝑐1 + 𝑣𝑐2 + 𝑤𝑐3 = 0 gives 𝑐1 (𝑡 3 − 4𝑡 2 + 3𝑡 + 3) + 𝑐2 (𝑡 3 + 2𝑡 2 + 4𝑡 − 1) + 𝑐3 (2𝑡 3 − 𝑡 2 = 3𝑡 + 5) = 0 𝑡 3 (𝑐1 + 𝑐2 + 2𝑐3 ) + 𝑡 2 (−4𝑐1 + 2𝑐2 − 𝑐3 ) + 𝑡(3𝑐1 + 4𝑐2 − 3𝑐3 ) + (3𝑥1 − 𝑐2 + 5𝑐3 ) = 0 Comparing coefficients of 𝑡 3 , 𝑡 2 , 𝑡 and constant term both sides we get, 𝑐1 + 𝑐2 + 2𝑐3 = 0(1) −4𝑐1 + 2𝑐2 − 𝑐3 = 0(2) 3𝑐1 + 4𝑐2 − 3𝑐3 = (3) and 3𝑐1 − 𝑐2 + 5𝑐3 = 0(4) Subtracting equation (4) from equation (3), we get 8𝑐3 = 5𝑐2 𝑐3 = (5/8)𝑐2 (5) Substituting the value of c3 in equation (1), we get 𝑐1 + 𝑐2 + 2(5/8)𝑐2 = 0 𝑐1 = (−9/4)𝑐2 (6) Substituting values from (5) & (6) in equation (1), we get (−9/4)𝑐2 + 𝑐2 + (5/8)𝑐2 = 0 So, 𝑐2 = 0 And so from (5) and (6) 𝑐1 = 𝑐2 = 𝑐3 = 0 Therefore, polynomials 𝑢 = 𝑡 3 − 4𝑡 2 + 3𝑡 + 3, 𝑣 = 𝑡 3 + 2𝑡 2 + 4𝑡 − 1, 𝑤 = 2𝑡 3 − 𝑡 2 − 3𝑡 + 5 are linearly independent. Question #2 Determine whether the following polynomials u, v, w in P(t) are linearly dependent or independent: 𝑢 = 𝑡 3 − 5𝑡 2 − 2𝑡 + 3, 𝑣 = 𝑡 3 − 4𝑡 2 − 3𝑡 + 4, 𝑤 = 2𝑡 3 − 17𝑡 2 − 7𝑡 + 9 Answer: 𝑢𝑐1 + 𝑣𝑐2 + 𝑤𝑐3 = 0 gives 𝑐1 (𝑡 3 − 5𝑡 2 − 2𝑡 + 3) + 𝑐2 (𝑡 3 − 4𝑡 2 − 3𝑡 + 4) + 𝑐3 (2𝑡 3 − 17𝑡 2 − 7𝑡 + 9) = 0 𝑡 3 (𝑐1 + 𝑐2 + 2𝑐3 ) + 𝑡 2 (−5𝑐1 − 4𝑐2 − 17𝑐3 ) + 𝑡(−2𝑐1 − 3𝑐2 − 7𝑐3 ) + (3𝑐1 + 4𝑐2 + 9𝑐3 ) = 0 Comparing coefficients of 𝑡 3 , 𝑡 2 , and constant term from both sides we get, 𝑐1 + 𝑐2 + 2𝑐3 = 0(1) −5𝑐1 − 4𝑐2 − 17𝑐3 = 0 𝑠𝑜, 5𝑐1 + 4𝑐2 + 17𝑐3 = 0(2) −2𝑐1 − 3𝑐2 − 7𝑐3 = 0(3) and 3𝑐1 + 4𝑐2 + 9𝑐3 = 0(4) Subtracting equation (4) from equation (2), we get 2𝑐1 = 8𝑐3 𝑐1 = (−4)𝑐3 (5) Substituting the value of 𝑐1 in equation (1), we get (−4)𝑐3 + 𝑐2 + 2𝑐3 = 0 𝑐2 = 2𝑐3 (6) So, 𝑐3 = 0 And so from (5) and (6), 𝑐1 = 𝑐2 = 𝑐3 = 0 Therefore, polynomials 𝑢 = 𝑡 3 − 5𝑡 2 − 2𝑡 + 3, 𝑣 = 𝑡 3 − 4𝑡 2 − 3𝑡 + 4, 𝑤 = 2𝑡 3 − 17𝑡 2 − 7𝑡 + 9 are linearly independent. Question #3 Multiply the polynomials. (6𝑚 + 5)(8𝑚2 − 13𝑚 + 3) Answer: We use = 𝑚𝑎 ⋅ 𝑚𝑏 = 𝑚(𝑎+𝑏) given 𝐹 = (6𝑚 + 5)(8𝑚2 − 13𝑚 + 3) (6𝑚 + 5)(8𝑚2 − 13𝑚 + 3) = 6𝑚 ⋅ 8𝑚2 + 6𝑚 ⋅ (−13𝑚) + 6𝑚 ⋅ 3 + 5 ⋅ 8𝑚2 + 5 ⋅ (−13𝑚) + 5 ⋅ 3 = 48𝑚3 − 78𝑚2 + 18𝑚 + 40𝑚2 − 65𝑚 + 15 = 48𝑚3 + (−78 + 40)𝑚2 + (18 − 65)𝑚 + 15 = 48𝑚3 − 38𝑚2 − 47𝑚 + 15 Question #4 Simplify the polynomials: 2 3𝑥 2 (5𝑥) (− 𝑥 3 ) 3 Answer: 2 Given polynomials is 𝑝(𝑥) = 3𝑥 2 (5𝑥) (− 𝑥 3 ) 3 2 2 3𝑥 2 (5𝑥) (− 𝑥 3 ) = 3 × 5 × 𝑥 2 × 𝑥 × (− 𝑥 3 ) 3 3 2 = 15𝑥 2+1 × (− 𝑥 3 ) 3 3 2 3 = 15𝑥 (− 𝑥 ) 3 2 = 15 × (− ) × 𝑥 3 × 𝑥 3 3 5 × (−2) × 𝑥 3+3 = −10𝑥 6 Question #5 Show that factorial polynomial corresponding to 𝑃𝑘 = 𝑘 5 − 4𝑘 4 + 2𝑘 3 − 𝑘 2 + 𝑘 + 10, to prove 𝑃𝑘 = 𝑘 (5) + 11𝑘 (4) − 𝑘 (3) − 9𝑘 (2) − 𝑘 (1) + 10 Answer: Given: The given polynomials is 𝑃𝑘 = 𝑘 5 − 4𝑘 4 + 2𝑘 3 − 𝑘 2 + 𝑘 + 10, to prove 𝑃𝑘 = 𝑘 (5) + 11𝑘 (4) − 𝑘 (3) − 9𝑘 (2) − 𝑘 (1) + 10 Calculation: Consider the polynomial are 𝑃𝑘1 = 𝑘 5 − 4𝑘 4 + 2𝑘 3 − 𝑘 2 + 𝑘 + 10 𝑃𝑘2 = 𝑘 (5) + 11𝑘 (4) − 𝑘 (3) − 9𝑘 (2) + 𝑘 (1) + 10 𝑃𝑘 = 𝑘 5 − 4𝑘 4 + 2𝑘 3 − 𝑘 2 + 𝑘 + 10 = 𝐴𝑘 (5) + 𝐵𝑘 (4) + 𝐶𝑘 (3) + 𝐷𝑘 (2) + 𝐸𝑘 (1) + 𝐹(1) 𝑃𝑘 = 𝑘 5 − 4𝑘 4 + 2𝑘 3 − 𝑘 2 + 𝑘 + 10 = 𝐴𝑘(𝑘 − 1)(𝑘 − 2)(𝑘 − 3)(𝑘 − 4) + 𝐵𝑘(𝑘 − 1)(𝑘 − 2)(𝑘 − 3) + 𝐶𝑘(𝑘 − 1)(𝑘 − 2) + 𝐷𝑘(𝑘 − 1) + 𝐸𝑘 + 𝐹 Put k=0 on both sides F=10 Now put k=1 on both sides, 1 − 4 + 2 − 1 + 1 + 10 = 𝐸 + 𝐹 9 = 𝐸 + 10 𝐸 = 9 − 10 = −1, ∴ 𝐸 = −1 Now put k=2 on both sides, 32 − 64 + 16 − 4 + 2 + 10 = 𝐷2(2 − 1) + 2𝐸 + 𝐹 -8=2D-2+10 -8=2D+8 2𝐷 = −16 ⇒ 𝐷 = −8 𝑘 5 − 4𝑘 4 + 2𝑘 3 − 𝑘 2 + 𝑘 + 10 = 𝐴𝑘(𝑘 − 1)(𝑘 − 2)(𝑘 − 3)(𝑘 − 4) + 𝐵𝑘(𝑘 − 1)(𝑘 − 2)(𝑘 − 3) + 𝐶𝑘(𝑘 − 1)(𝑘 − 2) + 𝐷𝑘(𝑘 − 1) + 𝐸𝑘 + 𝐹 Now put k=3 on both sides, −23 = 3𝐶(3 − 1)(3 − 2) + 3𝐷(3 − 1) + 3𝐸 + 𝐹 −23 = 6𝐶 − 48 − 3 + 10 6𝐶 = 18 ⇒ 𝐶 = 3 Now put k=4 on both sides, 126 = 24𝐵 + 24𝐶 + 12𝐷 + 4𝐸 + 𝐹 126 = 24𝐵 − 18 144 = 24𝐵 𝐵=6 𝑘 5 − 4𝑘 4 + 2𝑘 3 − 𝑘 2 + 𝑘 + 10 = 𝐴𝑘(𝑘 − 1)(𝑘 − 2)(𝑘 − 3)(𝑘 − 4) + 𝐵𝑘(𝑘 − 1)(𝑘 − 2)(𝑘 − 3) + 𝐶𝑘(𝑘 − 1)(𝑘 − 2) + 𝐷𝑘(𝑘 − 1) + 𝐸𝑘 + 𝐹 Both sides compare the coefficient of 𝑘 5 1=A Pu all values of in equation (1) 𝑘 (5) + 6𝑘 (4) + 3𝑘 (3) − 8𝑘 (2) − 𝑘 (1) + 10 Hence it is not proved. Question #6 Let 𝑃2 be the vector space of all polynomials of degree 2 or less, and let H be the subspace spanned by −(4𝑥 2 + 5𝑥 + 4), 1 − (6𝑥 2 + 17𝑥) and −(𝑥 2 + 2). The dimension of the subspace H is. Is {−(4𝑥 2 + 5𝑥 + 4), 1 − (6𝑥 2 + 17𝑥), −(𝑥 2 + 2)} a basis for 𝑃2 ? choose Be sure you can explain and justify your answer. A basis for the subspace H is { }. Enter a polynomial or a comma separated list of polynomials Answer: Given that H is a subspace of the vector space 𝑃2 of all polynomials of degree 2 or less spanned by the vectors −(4𝑥 2 + 5𝑥 + 4), 1 − (6𝑥 2 + 17𝑥) and −(𝑥 2 + 2). We know that the dimension of 𝑃𝑛 is n+1 and hence the dimension of 𝑃2 is 3 Since H is subspace of 𝑃2 its dimension is less than or equal to dimension of 𝑃2 = 3 A basis for vector space V is a subset of V which spans V and the set of vectors in the subset are linearly independent. Here, the set {−(4𝑥 2 + 5𝑥 + 4), 1 − (6𝑥 2 + 17𝑥), −(𝑥 2 + 2)} spans H. Next, we shall check whether the above vectors are linearly independent or not. If the linear combination 𝑎(−(4𝑥 2 + 5𝑥 + 4)) + 𝑏(1 − (6𝑥 2 + 17𝑥)) + 𝑐(−(𝑥 2 + 2)) = 0 only for a=b=c=0 , then they are linearly independent. Otherwise they are linearly dependent. Now, 𝑎(−(4𝑥 2 + 5𝑥 + 4)) + 𝑏(1 − (6𝑥 2 + 17𝑥)) + 𝑐(−(𝑥 2 + 2)) = 0 ⇒ (4𝑎 − 6𝑏 − 𝑐)𝑥 2 + (−5𝑎 − 17𝑏)𝑥 + (−4𝑎 + 𝑐 − 2𝑐) = 0 ⇒ −4𝑎 − 6𝑏 − 𝑐 = 0, −5𝑎 − 17𝑏 = 0, −4𝑎 + 𝑏 − 2𝑐 = 0 5 ⇒ −4𝑎 − 6𝑏 − 𝑐 = 0, 𝑏 = − 𝑎, −4𝑎 + 𝑏 − 2𝑐 = 0 17 5 ⇒ −4𝑎 − 6 (− 𝑎) − 𝑐 = 0, −4𝑎 + 𝑏 − 2𝑐 = 0 17 30 5 ⇒ −4𝑎 + 𝑎 − 𝑐 = 0, −4𝑎 − 𝑎 − 2𝑐 = 0 17 17 38 73 ⇒ − 𝑎 − 𝑐 = 0, − 𝑎 − 2𝑐 = 0 17 ⇒𝑐=− 38 17 𝑎, − 73 17 17 𝑎 − 2𝑐 = 0 ⇒− ⇒− ⇒ 3 17 73 17 73 17 𝑎 − 2 (− 𝑎+ 76 17 38 17 𝑎) = 0 𝑎=0 𝑎=0 ⇒𝑎=0 5 5 38 38 𝑎=− × 0 = 0, ⇒ 𝑐 = − 𝑎 = − ×0=0 17 17 17 17 Thus, all a=b=c=0 Hence, the set of vectors {−(4𝑥 2 + 5𝑥 + 4), 1 − (6𝑥 2 + 17𝑥), −(𝑥6 =2 + 2)} are linearly independent. Thus, the subset {−(4𝑥 2 + 5𝑥 + 4), 1 − (6𝑥 2 + 17𝑥), −(𝑥 2 + 2)} spans H as well as linearly independent. Therefore, the set {−(4𝑥 2 + 5𝑥 + 4), 1 − (6𝑥 2 + 17𝑥), −(𝑥 2 + 2)} forms a basis for H. Since the basis for H has 3 elements, the dimension of the subspace H is 3. We know that if B is a subset of a vector space V of dimension n such B has exactly n linearly independent vectors, then B is a basis for V. Here, {−(4𝑥 2 + 5𝑥 + 4), 1 − (6𝑥 2 + 17𝑥), −(𝑥 2 + 2)} is a subset of a vector space 𝑃2 of dimension 3 such this subset has exactly 3 linearly independent vectors. Hence, {−(4𝑥 2 + 5𝑥 + 4), 1 − (6𝑥 2 + 17𝑥), −(𝑥 2 + 2)} is a basis for 𝑃2 . ⇒𝑏=− Question #7 Use coordinate vectors to show that (4 − 𝑡)3 , (3 − 𝑡)2, 2 − 𝑡 and 1 + 5𝑡 − 6𝑡 2 + 𝑡 3 form a basis for 𝑃3 . Then find the coordinate vector for 1 − 2𝑡 + 3𝑡 2 − 4𝑡 3 relative 𝛽. Find q in 𝑃3 given that 1  −3  [ q ]B =    −7    5 Answer: To check that the polynomials (4 − 𝑡)3 , (3 − 𝑡)2, 2 − 𝑡, 1 + 5𝑡 − 6𝑡 2 + 𝑡 3 form a basis for 𝑃3 We will write the coordinate vector of each polynomial with respect to the standard basis 𝑆 = {1, 𝑡, 𝑡 2 , 𝑡 3 } Since, 𝑝1 (𝑡) = (4 − 𝑡)3 = 64 − 48𝑡 + 12𝑡 2 − 𝑡 3 . So coordinator vector of 𝑝1 (𝑡) w.r.t is  64   −48  p1 (t ) =   12     −1  Similarly, 𝑝2 (𝑡) = (3 − 𝑡)2 = 9 − 6𝑡 + 𝑡 2 + 0𝑡 3 . So coordinator vector of 𝑝2 (𝑡) w.r.t. S is 9  −6  p2 (t ) =   1   0 𝑝3 (𝑡) = (2 − 𝑡) = 2 − 𝑡 + 0𝑡 2 + 0𝑡 3 . So coordinator vector of 𝑝3 (𝑡) w.r.t. S is 2  −1 [ p3 ]S =   0   0 and 𝑝4 (𝑡) = 1 + 5𝑡 − 6𝑡 2 + 𝑡 3 . So coordinator vector of 𝑝4 (𝑡) w.r.t. S is 1 5 [ p4 ]S =    −6    1 So, the matrix generated by these coordinate vectors is  64 9 2 1   −48 −6 −1 5   A=  12 1 0 −6     −1 0 0 1  Since the determinant of the matrix A is not zero. So, matrix A is invertible, i.e. the columns of the matrix A are linearly independent. So, we can say that the given polynomials form a basis for 𝑃3 Coordinator vector of 𝑝(𝑡) = 1 − 2𝑡 + 3𝑡 2 − 4𝑡 3 relative to B is [𝑝]𝐵 = 𝐴−1 [𝑝]𝑆 2  1 − 3 − 3  4  2  [ p]B =  7 11  3  3  1 2 −  3  3 7  22   3 1  3     7 20     −65 −2 44    = 170  1 −  3    3    3  −4 4     10  −1 −    3  3   64 9 2 1   1   28   −48 −6 −1 5   −3  2    =   Since, [q ]S = A[q ]B =   12 1 0 −6   −7   −21       −1 0 0 1   5   4  −1 − Therefore, q in 𝑃3 is 28 + 2𝑡 − 21𝑡 2 + 4𝑡 3 . Question #8 A polynomial f(x) with real coefficients and leading coefficient 1 has the given zero(s) and degree. Express f(x) as a product of linear and quadratic polynomials with real coefficients that are irreducible over 𝑅. 2,-2-5i ; degree 3 Answer: At the given condition: Polynomial with real coefficients , each complex root (non-real root) goes in pair with its conjugate. Thus, if: -2-5i is one root, then -2+5i is the other root. Hence, We have three roots: One real root and other two complex roots: 2, -2-5i and -2+5i To determine: The polynomial f(x): So, the polynomial can be written as: 𝑓(𝑥) = (𝑥 − 2)(𝑥 − (−2 − 5𝑖))(𝑥 − (−2 + 5𝑖)) 𝑓(𝑥) = (𝑥 − 2)(𝑥 + 2 − 5𝑖)(𝑥 + 2 + 5𝑖) 𝑓(𝑥) = (𝑥 − 2)((𝑥 + 2)2 − (5𝑖)2 ) 𝑓(𝑥) = (𝑥 − 2)((𝑥 + 2)2 − 25𝑖 2 Since: 𝑖 2 = −1 𝑓(𝑥) = (𝑥 − 2)((𝑥 + 2)2 + 25) 𝑓(𝑥) = (𝑥 − 2)(𝑥 2 + 4 + 4𝑥 + 25) 𝑓(𝑥) = (𝑥 − 2)(𝑥 2 + 4𝑥 + 29) Hence, f(x) is a product of linear and quadratic polynomials with real coefficients that are irreducible over 𝑅 Question #9 Divide the following polynomials 18𝑥 5 +6𝑥 4 +9𝑥 3 3𝑥 2 Answer: To divide the given polynomial. Given information: 18𝑥 5 +6𝑥 4 +9𝑥 3 3𝑥 2 Calculation: Write the denominator separately for each term. 18𝑥 5 +6𝑥 4 +9𝑥 3 = = 3𝑥 2 18𝑥 5 3𝑥 2 6𝑥 5 𝑥2 + + 6𝑥 4 3𝑥 2 2𝑥 4 5−2 𝑥2 + + 9𝑥 3 3𝑥 2 3𝑥 3 𝑥2 = 6𝑥 + 2𝑥 4−2 + 3𝑥 3−2 3 = 6𝑥 + 2𝑥 2 + 3𝑥 1 = 6𝑥 3 + 2𝑥 2 + 3𝑥 Answer: 6𝑥 3 + 2𝑥 2 + 3𝑥 Question #10 The set of first four Hermite polynomials is 𝑆 = {1,2𝑡, −2 + 4𝑡 2 , −12𝑡 + 8𝑡 3 }. These polynomials have a wide variety of applications in physics and engineering. 1) Show that the set S of first four Hermite polynomials form a basis for 𝑃3 2) Find the coordinate vector of the polynomial 𝑝(𝑡) = −1 − 𝐴𝑡 + 8𝑡 2 + 𝐵𝑡 3 relative to S. Answer: Given the set of first four Hermite polynomials is 𝑆 = {1,2𝑡, −2 + 4𝑡 2 , −12𝑡 + 8𝑡 3 }. 1) To prove S is a basis for 𝑃3 First, show that S is linearly independent. Consider 𝑐1 ⋅ 1 + 𝑐2 ⋅ 2𝑡 + 𝑐3 ⋅ (−2 + 4𝑡 2 ) + 𝑐4 ⋅ (−12𝑡 + 8𝑡 3 ) = 0 𝑐1 + 2𝑐2 𝑡 − 2𝑐3 + 4𝑐3 𝑡 2 − 12𝑐4 𝑡 + 8𝑐4 𝑡 3 = 0 (𝑐1 − 2𝑐3 ) + (2𝑐2 − 12𝑐4 )𝑡 + 4𝑐3 𝑡 2 + 8𝑐4 𝑡 3 = 0 Equating the coefficients gives: 𝑐1 − 2𝑐3 = 0 ⇒ 𝑐1 = 2𝑐3 = 0 2𝑐2 − 12𝑐4 = 0 ⇒ 𝑐2 = 6𝑐4 = 0 4𝑐3 = 0 ⇒ 𝑐3 = 0 8𝑐4 = 0 ⇒ 𝑐4 = 0 Therefore, the vectors are linearly independent. Using the theorem, ''Let V be a vector space of dimension n. Any n linearly independent vectors of V form a basis of V'', S is a basis of 𝑃3 since 𝑃3 has dimension 4. 2) Given the polynomial 𝑝(𝑡) = −1 − 38𝑡 + 8𝑡 2 + 84𝑡 3 . Find the coordinate vector of S relative to S. Consider: 𝑐1 ⋅ 1 + 𝑐2 ⋅ 2𝑡 + 𝑐3 ⋅ (−2 + 4𝑡 2 ) + 𝑐4 ⋅ (−12𝑡 + 8𝑡 3 ) = −1 − 38𝑡 + 8𝑡 2 + 84𝑡 3 𝑐1 + 2𝑐2 𝑡 − 2𝑐3 + 4𝑐3 𝑡 2 − 12𝑐4 𝑡 + 8𝑐4 𝑡 3 = −1 − 38𝑡 + 8𝑡 2 + 84𝑡 3 (𝑐1 − 2𝑐3 ) + (2𝑐2 − 12𝑐4 )𝑡 + 4𝑐3 𝑡 2 + 8𝑐4 𝑡 3 = −1 − 38𝑡 + 8𝑡 2 + 84𝑡 3 Equate the coefficients. 𝑐1 − 2𝑐3 = −1 ⇒ 𝑐1 = −5 2𝑐2 − 12𝑐4 = −38 ⇒ 𝑐2 = 44 4𝑐3 = 8 ⇒ 𝑐3 = 2 21 8𝑐4 = 84 ⇒ 𝑐4 = 2 Therefore, 21 −5 ⋅ 1 + 44 ⋅ 2𝑡 + 2 ⋅ (−2 + 4𝑡 2 ) + ⋅ (−12𝑡 + 8𝑡 3 ) = −1 − 38𝑡 + 8𝑡 2 + 84𝑡 3 2  −5    44   Hence the coordinate vector of p(t) relative to S is  −2 + 4t 2   3  −12t + 8t 