Questions and Answers #3 Polynomials

Questions and Answers Sheet 3 Polynomials Question #1 Consider the following polynomials in 𝑃3 : 𝑞(𝑥) = −𝑥 3 + 3𝑥 2 − 𝑥 + 5 𝑟(𝑥) = −4𝑥 3 + 7𝑥 2 − 𝑥 + 10 𝑢(𝑥) = −5𝑥 3 + 8𝑥 2 + 10 For each of the following polynomials, determine whether it is in span {q,r,u}. If so, ecpress it as a lineare combination of the polynomials above. Use e.g. q rather than q(x) to represebt the polynomials in your linear combination. 𝑝1 (𝑥) = −3𝑥 2 − 3𝑥 + 10 Answer: Given the polynomials 𝑞(𝑥) = −𝑥 3 + 3𝑥 2 − 𝑥 + 5, 𝑟(𝑥) = −4𝑥 3 + 7𝑥 2 − 𝑥 + 10 and 𝑢(𝑥) = −5𝑥 3 + 8𝑥 2 . Now, a polynomial p(x) is in span of 𝑞, 𝑟, 𝑢 if there exist nonzero constants a,b and c such that 𝑝(𝑥) = 𝑎𝑞 + 𝑏𝑟 + 𝑐𝑢 The polynomial is 𝑝1 (𝑥) = −3𝑥 2 − 3𝑥 + 10 Then, 𝑝1 = 𝑎𝑞 + 𝑏𝑟 + 𝑐𝑢 −3𝑥 2 − 3𝑥 + 10 = 𝑎(−𝑥 3 + 3𝑥 2 − 𝑥 + 5) + 𝑏(−4𝑥 3 + 7𝑥 2 − 𝑥 + 10) + 𝑐(−5𝑥 3 + 8𝑥 2 + 10) − 3𝑥 2 − 3𝑥 + 10 = (𝑎 − 4𝑏 − 5𝑐)𝑥 3 + (3𝑎 + 7𝑏 + 8𝑐)𝑥 2 + (−𝑎 − 𝑏)𝑥 + 5𝑎 + 10𝑏 + 10𝑐 Equate the coefficients of both sides and find the values of a, b and c. 𝑎 − 4𝑏 − 5𝑐 = 0 (1) 3𝑎 + 7𝑏 + 8𝑐 = −3 (2) −𝑎 − 𝑏 = −3 (3) 5𝑎 + 10𝑏 + 10𝑐 = 10 (4) From equation (3), 𝑎 = −3 + 𝑏 𝑎 =3−𝑏 Substitute 𝑏 = 3 − 𝑎 in equation (1) and equation (2) and (3) −(3 − 𝑏) − 4𝑏 − 5𝑐 = 0 (4) 3(3 − 𝑏) + 7𝑏 + 8𝑐 = −3 (5) 5(3 − 𝑏) + 10𝑏 + 10𝑐 = 10 (6) From equation (4), −3 + 𝑏 − 4𝑏 − 5𝑐 = 0 −3 − 3𝑏 − 5𝑐 = 0 −3𝑏 = 5𝑐 + 3 5𝑐 + 3 𝑏=− 3 5𝑐+3 Substitute 𝑏 = − and in equation (5) and (6) 3 −9 + 4 (− 15 + 5 (− 5𝑐+3 3 5𝑐+3 3 ) + 8𝑐 = −3 (7) ) + 10𝑐 = 10 (8) Find the value of c from euation (7), 5𝑐+3 −9 + 4 (− ) + 8𝑐 = −3 3 5𝑐 + 3 4 (− ) + 8𝑐 = −12 3 4𝑐 − 4 = −12 3 4𝑐 − 12 = −36 𝑐 = −6 Substitute 𝑐 = −6 in equation (8), 15 + 5 (− 5𝑐 + 3 ) + 10𝑐 = 10 3 5𝑐 − 15 + 15 = 10 3 5(−6) − 15 + 15 = 10 3 0 = 10 which leads to a false statement. Therefore, the equation has no solution. Thus, the polynomial 𝑝1 (𝑥) is not a linear combination of the given polynomials. Therefore, 𝑝1 (𝑥) is not in span of 𝑞, 𝑟, 𝑢. Question #2 1 Consider the function ∫0 (1 + 𝑥) ln(1 − 𝑥) 𝑑𝑥 Find the function’s Taylor polynomials of degrees 1, 2, and 3 centered at 𝑐 = 0. Answer: Given: The function is: 𝑓(𝑥) = (1 + 𝑥) ln(1 − 𝑥) Now we have to find the function’s Taylor polynomials of degrees 1, 2, and 3 centred at 𝑐 = 0. Taylor's Polynomial formula: 𝑓 ′′ (𝑐) 𝑓 ′′′ (𝑐) (𝑥 − 𝑐)2 + (𝑥 − 𝑐)3 + ⋯ 𝑓(𝑥) = 𝑓(𝑐) + 𝑓 ′ (𝑐)(𝑥 − 𝑐) + 2! 3! ∞ 𝑓 (𝑛) (𝑥 − 𝑐)𝑛 =∑ 𝑛! 𝑛=0 Find the derivateves: 𝑓 ′ (𝑥) = ln(1 − 𝑥) + (1 + 𝑥) 1 (−1) (1 − 𝑥) (1 + 𝑥) (1 − 𝑥) (1 − 𝑥) − (1 + 𝑥)(−1) −1 𝑓 ′′ (𝑥) = −[ ] (1 − 𝑥)2 1−𝑥 −1 1−𝑥+1+𝑥 = −[ ] (1 − 𝑥)2 1−𝑥 −1 2 = − 1 − 𝑥 (1 − 𝑥)2 −1 −4 𝑓 ′′′ (𝑥) = + (1 − 𝑥)2 (1 − 𝑥)3 For Taylor polynomials of degrees 1 centred at 𝑐 = 0. 𝑓(𝑥) = 𝑓(0) = 𝑓 ′ (0)(𝑥 − 0) = 1 + (−1)𝑥 = 1−𝑥 For Taylor polynomials of degrees 2 centred at c = 0. = ln(1 − 𝑥) − 𝑓(𝑥) = 𝑓(0 + 𝑓 ′ (0))(𝑥 − 0) + (−3) 2 = 1 + (−1)𝑥 + 𝑥 2! 3 2 =1−𝑥− 𝑥 2 𝑓 ′′(0) 2! (𝑥 − 0)2 For Taylor polynomials of degrees 3 centred at 𝑐 = 0. 𝑓 ′′ (0) 𝑓 ′′′ (0) (𝑥 − 0)2 + (𝑥 − 0)3 𝑓(𝑥) = 𝑓(0) + 𝑓 ′ (0)(𝑥 − 0) + 2! 3! (−5) 3 3 = 1 − 𝑥 − 𝑥2 + 𝑥 2 6 3 5 = 1 − 𝑥 − 𝑥2 − 𝑥3 2 6 Question #3 Add the given polynomials. 2 12𝑎 𝑏 − 9𝑎𝑏 and 5𝑎2 𝑏 2 + 4𝑎𝑏 Answer: We have to add the given polynomials: 12𝑎2 𝑏 2 − 9𝑎𝑏 and 5𝑎2 𝑏 2 + 4𝑎𝑏 Adding the given polynomial, (12𝑎2 𝑏 2 − 9𝑎𝑏) + (5𝑎2 𝑏 2 + 4𝑎𝑏) Combing the similar terms, (12𝑎2 𝑏 2 − 9𝑎𝑏) + (5𝑎2 𝑏 2 + 4𝑎𝑏) = 12𝑎2 𝑏 2 − 9𝑎𝑏 + 5𝑎2 𝑏 2 + 4𝑎𝑏 = 12𝑎2 𝑏 2 + 5𝑎2 𝑏 2 − 9𝑎𝑏 + 4𝑎𝑏 = 17𝑎2 𝑏 2 − 5𝑎𝑏 Hence, addition of polynomial is 17𝑎2 𝑏 2 − 5𝑎𝑏 Question #4 Add the given polynomials. 6𝑥 2 + 8𝑥 + 4 and −7𝑥 2 − 7𝑥 − 10 Answer: We have to add the given polynomials: 6𝑥 2 + 8𝑥 + 4 and −7𝑥 2 − 7𝑥 − 10 Adding the given polynomial, (6𝑥 2 + 8𝑥 + 4) + (−7𝑥 2 − 7𝑥 − 10) Removing the parentheses, we get (6𝑥 2 + 8𝑥 + 4) + (−7𝑥 2 − 7𝑥 − 10) = 6𝑥 2 + 8𝑥 + 4 − 7𝑥 2 − 7𝑥 − 10 Simplifying further by combining similar terms, we get 6𝑥 2 + 8𝑥 + 4 − 7𝑥 2 − 10 = 6𝑥 2 − 7𝑥 2 + 8𝑥 − 7𝑥 + 4 − 10 = −𝑥 2 + 𝑥 − 6 Hence, required answer is −𝑥 2 + 𝑥 − 6 Question #5 Express 𝑓(𝑥) = 4𝑥 3 + 6𝑥 2 + 7𝑥 + 2 in term of Legendre Polynomials. Answer: The Legendre equation is, (1 − 𝑥 2 ) 𝑑2𝑦 𝑑𝑥 2 − 2𝑥 𝑑𝑦 𝑑𝑥 + 𝑛(𝑛 + 1)𝑦 = 0 The polynomial denoted by 𝑃𝑛 (𝑥) is called the Legendre polynomial of degree n The first few polynomials are as follows. 𝑃0 (𝑥) = 1 𝑃1 (𝑥) = 𝑥 3𝑥 2 1 𝑃2 (𝑥) = − 2 2 3 5𝑥 3𝑥 𝑃3 (𝑥) = − 2 2 35𝑥 4 30𝑥 2 3 𝑃4 (𝑥) = − + 8 8 8 5 3 63𝑥 70𝑥 15𝑥 𝑃5 (𝑥) = − + 8 8 8 The general form of a Legendre polynomial of order n is given by the sum, 𝑚 𝑃𝑛 (𝑥) = ∑𝑀 𝑚=0(−1) (2𝑛−2𝑚)! 2𝑛𝑚!(𝑛−𝑚)!(𝑛−2𝑚)! 𝑥 𝑛−2𝑚 n for even degree equation  2 , where M =   n − 1 , for odd degree equation  2 Given, 𝑓(𝑥) = 4𝑥 3 + 6𝑥 2 + 7𝑥 + 2 Since the given polynomial is of degree 3. So, 𝑓(𝑥) = 𝑎𝑃0 (𝑥) + 𝑏𝑃1 (𝑥) + 𝑐𝑃2 (𝑥) + 𝑑𝑃3 (𝑥) where the Legendre polynomials are P0 ( x) = 1, P1 ( x) = x, P2 ( x) = 3x 2 1 5 x3 3x − , P3 ( x) = − 2 2 2 2 On comparing, 4𝑥 3 + 6𝑥 2 + 7𝑥 + 2 = 𝑎𝑃0 (𝑥) + 𝑏𝑃1 (𝑥) + 𝑐𝑃2 (𝑥) + 𝑑𝑃3 (𝑥) 3𝑥 2 − 1 5𝑥 3 − 3𝑥 = 𝑎(1) + 𝑏(𝑥) + 𝑐 ( )+𝑑( ) 2 2 3𝑐 𝑐 5𝑑 3 3𝑑 = 𝑎 + 𝑏𝑥 + 𝑥 2 − + 𝑥 − 𝑥 2 2 2 2 𝑐 3𝑑 3𝑐 2 5𝑑 3 = (𝑎 − ) + (𝑏 − ) 𝑥 + 𝑥 + 𝑥 2 2 2 2 Now equating the coefficients of the like powers x in the above equation, On comparing coefficient of 𝑥 3 5𝑑 8 =4⇒𝑑= 2 5 On comparing coefficient of 𝑥 2 3𝑐 =6⇒𝑐=4 2 On comparing coefficient of x 3𝑑 𝑏− =7 2 Putting the value of d in this equation, 3 8 12 47 𝑏− ( )=7⇒𝑏 =7+ = 2 5 5 5 47 ⇒𝑏= 5 And, 𝑐 𝑎− =2 2 Putting the value of c in this equation, 4 𝑎− =2 ⇒𝑎 =2+2 =4 2 ⇒𝑎=4 Putting the values of a, b, c and d in equation 47 8 4𝑥 3 + 6𝑥 2 + 7𝑥 + 2 = 4𝑃0 (𝑥) + 𝑃1 (𝑥) + 4𝑃2 (𝑥) + 𝑃3 (𝑥) 5 5 or 47 8 𝑓(𝑥) = 4𝑃0 (𝑥) + 𝑃1 (𝑥) + 4𝑃2 (𝑥) + 𝑃3 (𝑥) 5 5 This is the expression of f(x) in terms of Legendre polynomials. 47 8 Answer: 𝑓(𝑥) = 4𝑃0 (𝑥) + 𝑃1 (𝑥) + 4𝑃2 (𝑥) + 𝑃3 (𝑥) 5 5 Question #6 If the sum of the squares zeros of the quadratic polynomial 𝑓(𝑥) = 𝑥 2 − 8𝑥 + 𝑘 . Find k. Answer: Consider the quadratic polynomials, 𝑓(𝑥) = 𝑥 2 − 8𝑥 + 𝑘 Comparing the given quadratic polynomias with standart polynomials 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 After comparing, 𝑎 = 1, 𝑏 = −8 and 𝑐 = 𝑘 Let the zeros of the polynomials is 𝛼 and 𝛽 According to the question, sum of the square of the quadratic Polynomials 𝛼 2 + 𝛽 2 = 40 Sum of zeros of the polynomials is, 𝑏 𝛼+𝛽 =− 𝑎 −8 𝛼+𝛽 = 1 𝛼+𝛽 =8 The product of the standard quadratic polynomials, 𝑐 𝛼⋅𝛽 = 𝑎 𝑘 𝛼⋅𝛽 = 1 𝛼⋅𝛽 =𝑘 To calculate the value of ''k'' From equation 𝛼 2 + 𝛽 2 = 40 (𝛼 + 𝛽)2 − 2𝛼𝛽 = 40 (8)2 − 2𝑘 = 40 64 − 2𝑘 = 40 2𝑘 = 64 − 40 2𝑘 = 24 𝑘 = 12 Hence the value of k is 12. Question #7 Add the given polynomials. −𝑥 2 − 𝑥 − 4,2𝑥 2 − 7𝑥 + 9, and −3𝑥 2 + 6𝑥 − 10 Answer: Given: −𝑥 2 − 𝑥 − 4,2𝑥 2 − 7𝑥 + 9 and −3𝑥 2 + 6𝑥 − 10 To add: The given above polynomials. Explanation: As per the question, the expression are, −𝑥 2 − 𝑥 − 4,2𝑥 2 − 7𝑥 + 9 and −3𝑥 2 + 6𝑥 − 10 As we have to add the above expression, so we can write as, = (−𝑥 2 − 𝑥 − 4) + (2𝑥 2 − 7𝑥 + 9) + (−3𝑥 2 + 6𝑥 − 10) = −𝑥 2 − 𝑥 − 4 + 2𝑥 2 − 7𝑥 + 9 − 3𝑥 2 + 6𝑥 − 10 −2𝑥 2 − 2𝑥 − 5 So, the addition of the above expression is −2𝑥 2 − 2𝑥 − 5 Answer: Hence, the addition of the expression −𝑥 2 − 𝑥 − 4,2𝑥 2 − 7𝑥 + 9 and −3𝑥 2 + 6𝑥 − 10 is −2𝑥 2 − 2𝑥 − 5. Question #8 Find the zeros of the following quadratic functions. a) 𝑓(𝑥) = 4𝑥 2 − 4𝑥 + 1 b) 𝑓(𝑥) = 12𝑥 2 − 3𝑥 + 5 Answer: Consider the quadratic polynomials, 𝑓(𝑥) = 4𝑥 2 − 4𝑥 + 1 To calculate the zero's of the given polynomials, Substittute, 𝑓(𝑥) = 0 4𝑥 2 − 4𝑥 + 1 = 0 4𝑥 2 − 2𝑥 − 2𝑥 + 1 = 0 2𝑥(2𝑥 − 1) − 1(2𝑥 − 1) = 0 (2𝑥 − 1)(2𝑥1 ) = 0 if 2𝑥 − 1 = 0 2𝑥 = 1 𝑥= 1 2 Again 2𝑥 − 1 = 0 2𝑥 = 1 1 𝑥= 2 1 1 Hence zeros of the polynomials is 𝑥 = , 2 2 b) Consider the polynomials, 𝑓(𝑥) = 12𝑥 2 − 3𝑥 + 5 To calculate the zeros of the polynomials, Substitute, 𝑓(𝑥) = 0 12𝑥 2 − 3𝑥 + 5 = 0 (−3) ± √(−3)2 − 4(12)(5) 𝑥= 2(12) 3 ± √9 − 240 = 24 3 ± 𝑖√231 = 24 3 + 𝑖√231 3 − 𝑖√231 = , 24 24 1 √231 1 √231 𝑥1,2 = + 𝑖 , −𝑖 8 24 8 24 Question #9 Find the sum of the polynomials: 8𝑥 3 − 2𝑥 2 + 6𝑥 − 2and 3𝑥 4 − 2𝑥 3 + 𝑥 2 + 𝑥 Answer: To find the sum of polynomials: 8𝑥 3 − 2𝑥 2 + 6𝑥 − 2 and 3𝑥 4 − 2𝑥 3 + 𝑥 2 + 𝑥 Solution: (8𝑥 3 − 2𝑥 2 + 6𝑥 − 2) + (3𝑥 4 − 2𝑥 3 + 𝑥 2 + 𝑥) On simplifying further we get: (8𝑥 3 − 2𝑥 2 + 6𝑥 − 2) + (3𝑥 4 − 2𝑥 3 + 𝑥 2 + 𝑥) = 3𝑥 4 + 8𝑥 3 − 2𝑥 3 − 2𝑥 2 + 𝑥 2 + 6𝑥 + 𝑥 − 2 ⇒ (8𝑥 3 − 2𝑥 2 + 6𝑥 − 2) + (3𝑥 4 − 2𝑥 3 + 𝑥 2 + 𝑥) = 3𝑥 4 + (8𝑥 3 − 2𝑥 3 ) − (2𝑥 2 − 𝑥 2 ) + (6𝑥 + 𝑥) − 2 ⇒ (8𝑥 3 − 2𝑥 2 + 6𝑥 − 2) + (3𝑥 4 − 2𝑥 3 + 𝑥 2 + 𝑥) = 3𝑥 4 + 6𝑥 3 − 𝑥 2 + 7𝑥 − 2 Result: Required expression is: 3𝑥 4 + 6𝑥 3 − 𝑥 2 + 7𝑥 − 2 Question #10 Let 𝑃(𝑥) = 8𝑥 4 + 6𝑥 2 − 3𝑥 + 1 and 𝐷(𝑥) = 2𝑥 2 − 𝑥 + 2 . Find polynomials Q(x) and R(x) such that 𝑃(𝑥) = 𝐷(𝑥) ⋅ 𝑄(𝑥) + 𝑅(𝑥) . Answer: The polynomials: 𝑃(𝑥) = 8𝑥 4 + 6𝑥 2 − 3𝑥 + 1, 𝐷(𝑥) = 2𝑥 2 − 𝑥 + 2 To determine: The polynomials Q(x) and R(x) such that 𝑃(𝑥) = 𝐷(𝑥)𝑄(𝑥) + 𝑅(𝑥) Solution: The long division of polynomial 𝑃(𝑥) = 8𝑥 4 + 6𝑥 2 − 3𝑥 + 1 by 𝐷(𝑥) = 2𝑥 2 − 𝑥 + 2 is given by: According to division algorithm, we have, 8𝑥 4 + 6𝑥 2 − 3𝑥 + 1 = (2𝑥 2 − 𝑥 + 2)(4𝑥 2 + 2𝑥) + (−7𝑥 + 1) So, 𝑄(𝑥) = 4𝑥 2 + 2𝑥 and 𝑅(𝑥) = −7𝑥 + 1 Conclusion: Hence, 𝑄(𝑥) = 4𝑥 2 + 2𝑥 and 𝑅(𝑥) = −7𝑥 + 1