Questions and Answers #15 Polynomials

Questions and Answers Sheet 15 Polynomials Question #1 Find f+g, f-g, fg and 𝑓 𝑔 . Then determine the domain for each function. 2 𝑓(𝑥) = 3𝑥 − 14𝑥 + 16, 𝑔(𝑥) = 𝑥 − 2 Answer: Given, 𝑓(𝑥) = 3𝑥 2 − 14𝑥 + 16, 𝑔(𝑥) = 𝑥 − 2 To find f+g, f-g, fg and 𝑓 𝑔 and domain of each function. 𝑓 + 𝑔(𝑥) = 𝑓(𝑥) + 𝑔(𝑥) = 3𝑥 2 − 14𝑥 + 16 + 𝑥 − 2 = 3𝑥 2 − 13𝑥 + 14 𝑓 − 𝑔(𝑥) = 𝑓(𝑥) − 𝑔(𝑥) = 3𝑥 2 − 14𝑥 + 16 − 𝑥 + 2 = 3𝑥 2 − 15𝑥 + 18 𝑓𝑔(𝑥) = 𝑓(𝑥)𝑔(𝑥) = (3𝑥 2 − 14𝑥 + 16)(𝑥 − 2) = 3𝑥 3 − 6𝑥 2 − 14𝑥 2 + 28𝑥 + 16𝑥 − 32 = 3𝑥 2 − 20𝑥 2 + 44𝑥 − 32 𝑓 𝑔 (𝑥) = 𝑓(𝑥) 𝑔(𝑥) = 3𝑥 2 −14𝑥+16 𝑥−2 ,𝑥 ≠ 2 Since all the functions f+g, f-g and fg are polynomials. So their domain is set of all real numbers 𝑅 However, 𝑓 𝑔 is defined for all x except where its denominator vanishes. It vanishes when x=2 Thus the domain of 𝑓 𝑔 is (−∞, 2) ∪ (2, ∞). Question #2 Simplify each rational expression. You will need to use factoring by grouping. 𝑥 2 −2𝑥+𝑎𝑥−2𝑎 𝑥 2 −2𝑥+3𝑎𝑥−6𝑎 Answer: A polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables. A rational expression is simply a quotient of two polynomials. The given expression is 𝑥 2 −2𝑥+𝑎𝑥−2𝑎 𝑥 2 −2𝑥+3𝑎𝑥−6𝑎 . Simplify the given expression by taking out the common terms and then grouping the common terms as follows: 𝑥 2 −2𝑥+𝑎𝑥−2𝑎 𝑥 2 −2𝑥+3𝑎𝑥−6𝑎 (𝑥+𝑎)(𝑥−2) = (𝑥(𝑥−2)+𝑎(𝑥−2)) (𝑥(𝑥−2)+3𝑎(𝑥−2)) - Take the common terms out = (𝑥+3𝑎)(𝑥−2) - Group the terms (𝑥+𝑎) = (𝑥+3𝑎) - Cancel the common terms in the numerator and the denominator. = (𝑥 + 𝑎)/(𝑥 + 3𝑎) Question #3 Add or subtract the rational expressions as indicated. Be sure to express your answers in simplest form. 3𝑥/(2𝑥 + 5) + 1 Answer: An expression is a sentence with a minimum of two numbers and at least one math operation. This math operation can be addition, subtraction, multiplication, and division. A rational expression is simply a quotient of two polynomials. The given expression is 3𝑥/(2𝑥 + 5) + 1. The LCM of the denominators of the expression (2x+5) and 1 is (2x+5). So multiply with corresponding values to make the denominators equal to (2x+5) and then simplify as follows: 3𝑥 3𝑥/(2𝑥 + 5) + 1 = (2𝑥+5) + 1 (2𝑥+5) 3𝑥 = (2𝑥+5) + 1 × (2𝑥+5) - Make the denominators equal to (2x+5). = 3𝑥+(2𝑥+5) (2𝑥+5) (5𝑥+5) - Combine the terms. = (2𝑥+5) 5(𝑥+1) = (2𝑥+5) = 5(𝑥 + 1)/(2𝑥 + 5) The simplified form of given rational expression is 3𝑥/(2𝑥 + 5) + 1 = 5(𝑥 + 1)/(2𝑥 + 5) Question #4 Use the sum-of-two-cubes or the difference-of-two-cubes pattern to factor each of the following. 1 − 27𝑎3 Answer: A polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables. Factoring polynomials involves breaking up a polynomial into simpler terms (the factors) such that when the terms are multiplied together they equal the original polynomial. The given expression is 1 − 27𝑎3 . Factor the given expression by simplifying and then using sum of two cubes pattern as follows: 1 − 27𝑎3 = 13 − 33 𝑎3 (1)3 + (−3𝑎)3 = (1 + (−3𝑎))(12 − (1)(−3𝑎) + (−3𝑎)2 ) Use sum of two cubes pattern 𝑎3 + 𝑏 3 = (𝑎 + 𝑏)(𝑎2 − 𝑎𝑏 + 𝑏 2 ) = (1 − 3𝑎)(1 + 3𝑎 + 9𝑎2 ) = (1 − 3𝑎)(9𝑎2 + 3𝑎 + 1) - Factored form. Hence, the factored form of given expression is equal to 1 − 27𝑎3 = (1 − 3𝑎)(9𝑎2 + 3𝑎 + 1). Question #5 Find (a) , (b) ||p||, (c) ||q||, and (d) d(p,q) for the polynomials in 𝑃2 using the inner product =𝑎0 𝑏0 + 𝑎1 𝑏1 + 𝑎2 𝑏2 . 𝑝(𝑥) = 1 − 3𝑥 + 𝑥 2 , 𝑞(𝑥) = −𝑥 + 2𝑥 2 Answer: We have given that; 𝑝(𝑥) = 1 − 3𝑥 + 𝑥 2 , 𝑞(𝑥) = −𝑥 + 2𝑥 2 a) The inner product of the polynomials will be: 𝑝(𝑥) = 1 − 3𝑥 + 𝑥 2 , 𝑞(𝑥) = −𝑥 + 2𝑥 2 =1 × 0 + (−3)(−1) + 1(2) =3+2 =5 b) ||𝑝|| = √12 + (−3)2 + 12 ||𝑝|| = √1 + 9 + 1 ||𝑝|| = √11 c) ||𝑞|| = √(−1)2 + 22 ||𝑞|| = √1 + 4 ||𝑞|| = √5 d) 𝑑(𝑝, 𝑞) = ||𝑝 − 𝑞|| 𝑑(𝑝, 𝑞) = ||1 − 3𝑥 + 𝑥 2 − (−𝑥 + 2𝑥 2 )|| 𝑑(𝑝, 𝑞) = ||1 − 3𝑥 + 𝑥 2 + 𝑥 − 2𝑥 2 || 𝑑(𝑝, 𝑞) = ||1 − 2𝑥 − 𝑥 2 || 𝑑(𝑝, 𝑞) = √12 + (−2)2 + (−1)2 𝑑(𝑝, 𝑞) = √1 + 4 + 1 𝑑(𝑝, 𝑞) = √6 Question #6 Using generating function for Legendre polynomials, prove that 1−𝑡 2 ∑𝑙0(2𝑙 + 1)𝑃𝑙 (𝑥)𝑡 𝑙 = (1−2𝑥𝑡+𝑡 2 )3/2 Answer: Introduction The legendre polynomial denoted by 𝑃𝑛 (𝑥). The solution of physical problems is the class of function called legendre polynomials. Explanation To prove, 𝑥𝑡−𝑡 2+𝑥𝑡−𝑡 2 ∑𝑙0 𝑃𝑙 (𝑥)𝑡 𝑙 (𝑙 + 𝑙 + 1) = 3 (1−2𝑥𝑡+𝑡 2 )2 + 1 1 (1−2𝑥𝑡+𝑡 2 )2 Use generating function for legendre polynomial, 1 𝑛 2 − ∑∞ 𝑥=0 𝑃𝑛 (𝑥)𝑡 = (1 − 2𝑥𝑡 + 𝑡 ) 2 Consider the limit 0 to l 1 ∑𝑙𝑛=0 𝑃𝑙 (𝑥)𝑡 𝑙 = (1 − 2𝑥𝑡 + 𝑡 2 )−2 Differentiating with respect to t, 3 1 ∑𝑙0 𝑃𝑙 (𝑥)𝑙𝑡 𝑙−1 = (− ) (1 − 2𝑥𝑡 + 𝑡 2 )−2 (−2𝑥 + 2𝑡) ∑𝑙0 𝑃𝑙 (𝑥)𝑙𝑡 𝑙−1 = 2 (𝑥−𝑡) 3 (1−2𝑥𝑡+𝑡 2 )2 Multiplying t on both sides we get, ∑𝑙0 𝑃𝑙 (𝑥)𝑙𝑡 𝑙−1 𝑡 = ∑𝑙0 𝑃𝑙 (𝑥)𝑙𝑡 𝑙 = (𝑥−𝑡)𝑡 3 (1−2𝑥𝑡+𝑡 2 )2 𝑥𝑡−𝑡 2 3 (1−2𝑥𝑡+𝑡 2 )2 Let us consider, ∑𝑙0 𝑃𝑙 (𝑥)𝑙𝑡 𝑙 + ∑𝑙0 𝑃𝑙 (𝑥)𝑙𝑡 𝑙 + ∑𝑙0 𝑃𝑙 (𝑥)𝑡 𝑙 = ∑𝑙0 𝑃𝑙 (𝑥)𝑡 𝑙 (𝑙 + 𝑙 + 𝑙) = ∑𝑙0(2𝑙 + 𝑙)𝑃𝑙 (𝑥)𝑡 𝑙 = ∑𝑙0(2𝑙 + 𝑙)𝑃𝑙 (𝑥)𝑡 𝑙 = 𝑥𝑡−𝑡 2 +𝑥𝑡−𝑡 2 + 𝑥𝑡−𝑡 2 3 (1−2𝑥𝑡+𝑡 2)2 + 1 1 1 (1−2𝑥𝑡+𝑡 2 )2 (1−2𝑥𝑡+𝑡 2 )2 2 2 2𝑥𝑡−2𝑡 +1−2𝑥𝑡+𝑡 3 (1−2𝑥𝑡+𝑡 2 )2 1−𝑡 2 3 (1−2𝑥𝑡+𝑡 2)2 Question #7 Multiply the following polynomials: (4𝑥 + 4)(4𝑥 + 4) Answer: Our Aim is to multiply the polynomial given below: (4x+4)(4x+4) Considering equation, we have: 𝑥𝑡−𝑡 2 3 (1−2𝑥𝑡+𝑡 2 )2 + 1 1 (1−2𝑥𝑡+𝑡 2 )2 (4𝑥 + 4)(4𝑥 + 4) = (4𝑥)(4𝑥) + (4𝑥)(4) + (4)(4𝑥) + (4)(4) ⇒ (4𝑥 + 4)(4𝑥 + 4) = 16𝑥 2 + 16𝑥 + 16𝑥 + 16 ⇒ (4𝑥 + 4)(4𝑥 + 4) = 16𝑥 2 + 32𝑥 + 16 is the required answer obtained by multiplying the polynomials. Question #8 Let 𝑓(𝑥) = cos(2𝑥) + 7. Compute the following Taylor polynomials of f. For any approximations, you should use around 6 decimals. Answer: From the given problem: 𝑓(𝑥) = cos(2𝑥) + 7 Let centered at a=0 From the Taylor series: 𝑓 ′ (𝑎) 𝑓 ′′ (𝑎) 𝑓 ′′′ (𝑎) (𝑥 − 𝑎) + (𝑥 − 𝑎)2 + (𝑥 − 𝑎)3 + ⋯ (1) 𝑓(𝑥) = 𝑓(𝑎) + 1! 2! 3! 𝑑(cos(2𝑥)+7) 𝑑 𝑑 So, 𝑓 ′ (𝑥) = = (cos(2𝑥)) + (7) = −2 sin(2𝑥) 𝑑𝑥 𝑓 ′′ = 𝑑(−2 sin(2𝑥)) 𝑓 ′′′ (𝑥) = 𝑓 4 (𝑥) = 𝑑𝑥 𝑑(−4 cos(2𝑥)) 𝑑𝑥 𝑑(8 sin(2𝑥)) 𝑑𝑥 𝑑𝑥 𝑑 = cos(2𝑥) 𝑑𝑥 𝑑𝑥 (2𝑥) = −4 cos(2𝑥) = − sin(2𝑥) = − cos(2𝑥) 𝑑 𝑑𝑥 𝑑 𝑑𝑥 (2𝑥) = 8 sin(2𝑥) (2𝑥) = 16 cos(2𝑥) At a=0 𝑓(0) = cos(2 × 0) + 7 = 8 𝑓 ′ (0) = −2 sin(2 × 0) = 0 𝑓 ′′ (0) = −4 cos(2 × 0) = −4 𝑓 ′′′ (0) = 8 sin(2 × 0) = 0 𝑓 4 (0) = 16 cos(2 × 0) = 16 Now substitute these values in equation (1). 0 −4 2 0 3 16 4 cos(2𝑥) + 7 = 8 + 𝑥 + 𝑥 + 𝑥 + 𝑥 +⋯ 1! 2! 3! 4! 2 = 8 − 2𝑥 2 + 𝑥 4 + ⋯ 3 Therefore the Taylor polynomials of the function are: 2 cos(2𝑥) = 8 − 2𝑥 2 + 𝑥 4 + ⋯ 3 Question #9 Find the Taylor polynomials of orders 0,1,2, and 3 generated by 𝑓(𝑥) = ln(9 + 𝑥) at x=4 Answer: We have to find the Taylor polynomial of order 0,1,2, and 3 generated by 𝑓(𝑥) = ln(9 + 𝑥) at x=4 We begin with the formula for a Taylor polynomial generated at x=a: ∑∞ 𝑘=0 𝑓 𝑘 (𝑎) 𝑘! (𝑥 − 𝑎)𝑘 = 𝑓(𝑎) + 𝑓 ′ (𝑎) 1! (𝑥 − 𝑎)1 + 𝑓 ′′ (𝑎) 2! (𝑥 − 𝑎)2 + ⋯ + 𝑓 𝑛(𝑎) 𝑛! (𝑥 − 𝑎)𝑛 + ⋯ The Taylor polynomial of order n refers to the first n+1 (to the nth derivative) terms of this Taylor polynomial. We are given 𝑓(𝑥) = ln(9 + 𝑥) and we are to construct the first three Taylor polynomials centered at a=4. First, let's find the first three derivatives as follows: 𝑓(𝑥) = ln(9 + 𝑥) ⇒ 𝑓(4) = ln(9 + 4) = ln(13) 1 1 1 𝑓 ′ (𝑥) = ⇒ 𝑓 ′ (4) = = 9+𝑥 −1 9+4 13 −1 𝑓 ′′ (𝑥) = (9+𝑥)2 ⇒ 𝑓 ′′ (4) = (9+4)2 = − 2 2 𝑓 ′′′ (𝑥) = (9+𝑥)3 ⇒ 𝑓 ′′′ (4) = (9+4)3 = 1 169 2 133 We can now construct the first Taylor polynomials of orders 0,1,2, and 3 as follows: 𝑃0 (𝑥) = 𝑓(4) = ln(13) 𝑃1 (𝑥) = 𝑓(4) + 𝑃2 (𝑥) = 𝑓(4) + 𝑃3 (𝑥) = 𝑓(4) + 1 6591 𝑓 ′(4) 1! 𝑓 ′ (4) 1! 𝑓 ′ (4) 1! (𝑥 − 4)1 = ln(13) + (𝑥 − 4)1 + (𝑥 − 4)1 + 𝑓 ′′ (4) 2! 𝑓 ′′ (4) 2! 1 13 (𝑥 − 4) (𝑥 − 4)2 = ln(13) + (𝑥 − 4)2 + 𝑓 ′′′ (4) 3! 1 13 (𝑥 − 4) − 1 338 (𝑥 − 4)2 (𝑥 − 4)3 = ln(13) + 1 13 (𝑥 − 4) − 1 338 (𝑥 − 4)2 + (𝑥 − 4)3 Question #10 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. -1,0,3+i; degree 4 Answer: Given Roots of the polynomial of degree 4 is -1, 0, 3+i To find the polynomial f (x) with real coefficients and leading coefficient 1 has the given zero(s) and degree. Also, to Express f (x) as a product of linear and quadratic polynomials with real coefficients that are irreducible over R. Definition Used We know if a polynomial having a complex root then its complex conjugate will also be the root of the given polynomial. Formation of the polynomial f(x) Since by the definition if 3+i is one root then 3-i will also be the root of the given polynomial In this way, we have a total number of 4 zeroes i.e. -1,0,3+i and 3-i Then the polynomial f(x) is 𝑓(𝑥) = (𝑥 − (−1))(𝑥 − 0)(𝑥 − (3 + 𝑖)(𝑥 − (3 − 𝑖)) = 𝑥(𝑥 + 1)(𝑥 − 3 − 𝑖)(𝑥 − 3 + 𝑖) = 𝑥(𝑥 + 𝑖){(𝑥 − 3)2 − (𝑖)2 } = 𝑥(𝑥 + 1){𝑥 2 − 6𝑥 + 9 + 1} = (𝑥 2 + 𝑥)(𝑥 2 − 6𝑥 + 10) = 𝑥 4 − 6𝑥 3 + 10𝑥 2 + 𝑥 3 − 6𝑥 2 + 10𝑥 = 𝑥 4 − 5𝑥 3 + 4𝑥 2 + 10𝑥